{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0001","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4306775265fe9691c12b6c8647893ba0319562008e79dacd84db68e5378f3720","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246001,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-12,-2,-3,1,-1,-2],\"operator\":[1,-2],\"source\":-12,\"x\":-2,\"y\":-3}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-12, -2, -3, 1, -1, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[1, -2] means a*u_x+b*u_y; source=-12, x=-2, y=-3. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-12, -2, -3, 1, -1, -2], [1, -2], -2, -3, -3, 11, -12, -13, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-12, -2, -3, 1, -1, -2], [1, -2], -2, -3, -3, 11, -12, -13, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0002","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"698e0129612ae018dcd2fcc0268e18b782804d6d3d211838278a93601bc64043","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246002,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-11,-1,-2,2,0,-1],\"operator\":[2,-1],\"source\":-11,\"x\":-1,\"y\":-2}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-11, -1, -2, 2, 0, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[2, -1] means a*u_x+b*u_y; source=-11, x=-1, y=-2. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-11, -1, -2, 2, 0, -1], [2, -1], -1, -2, -5, 2, -11, -1, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-11, -1, -2, 2, 0, -1], [2, -1], -1, -2, -5, 2, -11, -1, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0003","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5b44110d9a385bd415f2870f765925e6e4171919cfdb2fe330d79f8c1befaf90","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246003,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-10,0,-1,3,1,0],\"operator\":[3,0],\"source\":-10,\"x\":0,\"y\":-1}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-10, 0, -1, 3, 1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[3, 0] means a*u_x+b*u_y; source=-10, x=0, y=-1. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-10, 0, -1, 3, 1, 0], [3, 0], 0, -1, -1, -1, -10, 7, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-10, 0, -1, 3, 1, 0], [3, 0], 0, -1, -1, -1, -10, 7, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0004","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7bdacdb70fd45e5475a5d75397f84c74e56cc5bd4eb1a48188d70fd524971dec","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246004,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-9,1,0,4,-1,1],\"operator\":[1,1],\"source\":-9,\"x\":1,\"y\":0}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-9, 1, 0, 4, -1, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[1, 1] means a*u_x+b*u_y; source=-9, x=1, y=0. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-9, 1, 0, 4, -1, 1], [1, 1], 1, 0, 9, -1, -9, 17, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-9, 1, 0, 4, -1, 1], [1, 1], 1, 0, 9, -1, -9, 17, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0005","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2490851d898d31452f997771ff4990f7c160039b3a7eb8be325e8717a6fa80b3","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246005,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-8,2,1,1,0,2],\"operator\":[2,2],\"source\":-8,\"x\":2,\"y\":1}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-8, 2, 1, 1, 0, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[2, 2] means a*u_x+b*u_y; source=-8, x=2, y=1. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-8, 2, 1, 1, 0, 2], [2, 2], 2, 1, 6, 5, -8, 30, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-8, 2, 1, 1, 0, 2], [2, 2], 2, 1, 6, 5, -8, 30, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0006","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e70d47a78c59d9ce28afb0fb773595f5c3cd5c6e321bd5c2dca26ef794f20930","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246006,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-7,-2,2,2,1,-2],\"operator\":[3,-2],\"source\":-7,\"x\":-2,\"y\":2}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-7, -2, 2, 2, 1, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[3, -2] means a*u_x+b*u_y; source=-7, x=-2, y=2. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-7, -2, 2, 2, 1, -2], [3, -2], -2, 2, -8, -8, -7, -1, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-7, -2, 2, 2, 1, -2], [3, -2], -2, 2, -8, -8, -7, -1, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0007","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81f5c18b65b96cff6102dd6e96196eb860bc9d7996de600a15c4dbb099a77983","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246007,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-6,-1,3,3,-1,-1],\"operator\":[1,-1],\"source\":-6,\"x\":-1,\"y\":3}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-6, -1, 3, 3, -1, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[1, -1] means a*u_x+b*u_y; source=-6, x=-1, y=3. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-6, -1, 3, 3, -1, -1], [1, -1], -1, 3, -10, -2, -6, -2, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-6, -1, 3, 3, -1, -1], [1, -1], -1, 3, -10, -2, -6, -2, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0008","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"993a605a7510bfc7c9e7bb4c53cd83a5d871e027e6d1c8dd26b2f1a3ee182342","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246008,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-5,0,-3,4,0,0],\"operator\":[2,0],\"source\":-5,\"x\":0,\"y\":-3}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-5, 0, -3, 4, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[2, 0] means a*u_x+b*u_y; source=-5, x=0, y=-3. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-5, 0, -3, 4, 0, 0], [2, 0], 0, -3, 0, -3, -5, 5, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-5, 0, -3, 4, 0, 0], [2, 0], 0, -3, 0, -3, -5, 5, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0009","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c3894e80a6c5cb6ae0f2a6ae4130a2c410cc63b6dd13de0316d457e566b1374","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246009,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-4,1,-2,1,1,1],\"operator\":[3,1],\"source\":-4,\"x\":1,\"y\":-2}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-4, 1, -2, 1, 1, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[3, 1] means a*u_x+b*u_y; source=-4, x=1, y=-2. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-4, 1, -2, 1, 1, 1], [3, 1], 1, -2, 1, -5, -4, 2, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-4, 1, -2, 1, 1, 1], [3, 1], 1, -2, 1, -5, -4, 2, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0010","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0558a10972aa00359e5efd1cdea7db285448f08e3cb6c035c01f0580ada887cd","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246010,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-3,2,-1,2,-1,2],\"operator\":[1,2],\"source\":-3,\"x\":2,\"y\":-1}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-3, 2, -1, 2, -1, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[1, 2] means a*u_x+b*u_y; source=-3, x=2, y=-1. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-3, 2, -1, 2, -1, 2], [1, 2], 2, -1, 11, -7, -3, 0, Yes)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-3, 2, -1, 2, -1, 2], [1, 2], 2, -1, 11, -7, -3, 0, Yes)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0011","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c0292b26e774ae0543986301324b8d36c743f558a4822a52b4f8123e0ac1eeed","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246011,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-2,-2,0,3,0,-2],\"operator\":[2,-2],\"source\":-2,\"x\":-2,\"y\":0}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-2, -2, 0, 3, 0, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[2, -2] means a*u_x+b*u_y; source=-2, x=-2, y=0. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-2, -2, 0, 3, 0, -2], [2, -2], -2, 0, -14, 0, -2, -26, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-2, -2, 0, 3, 0, -2], [2, -2], -2, 0, -14, 0, -2, -26, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0012","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee0f9b3afed17339b2b3e22baa0e42c3ff22ea19a9763d1a609510f639e18257","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246012,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[-1,-1,1,4,1,-1],\"operator\":[3,-1],\"source\":-1,\"x\":-1,\"y\":1}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-1, -1, 1, 4, 1, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[3, -1] means a*u_x+b*u_y; source=-1, x=-1, y=1. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([-1, -1, 1, 4, 1, -1], [3, -1], -1, 1, -8, -2, -1, -21, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-1, -1, 1, 4, 1, -1], [3, -1], -1, 1, -8, -2, -1, -21, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0013","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f338619a0a42b4e1760923d80335a1cb7b9aaae33868b6e495c251d9e8dc1ea5","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246013,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[0,0,2,1,-1,0],\"operator\":[1,0],\"source\":0,\"x\":0,\"y\":2}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[0, 0, 2, 1, -1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[1, 0] means a*u_x+b*u_y; source=0, x=0, y=2. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([0, 0, 2, 1, -1, 0], [1, 0], 0, 2, -2, 2, 0, -2, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([0, 0, 2, 1, -1, 0], [1, 0], 0, 2, -2, 2, 0, -2, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0014","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"649187d7e0e8a304ce59d5324a51048febb5db63e735f5b807063a3a59ce4222","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246014,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[1,1,3,2,0,1],\"operator\":[2,1],\"source\":1,\"x\":1,\"y\":3}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[1, 1, 3, 2, 0, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[2, 1] means a*u_x+b*u_y; source=1, x=1, y=3. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([1, 1, 3, 2, 0, 1], [2, 1], 1, 3, 5, 9, 1, 18, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, 1, 3, 2, 0, 1], [2, 1], 1, 3, 5, 9, 1, 18, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0015","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2640e8e965ebc7137b6e4f41426dc3644122486d9c040e280c43956e820fad6b","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246015,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[2,2,-3,3,1,2],\"operator\":[3,2],\"source\":2,\"x\":2,\"y\":-3}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[2, 2, -3, 3, 1, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[3, 2] means a*u_x+b*u_y; source=2, x=2, y=-3. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([2, 2, -3, 3, 1, 2], [3, 2], 2, -3, 11, -13, 2, 5, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, 2, -3, 3, 1, 2], [3, 2], 2, -3, 11, -13, 2, 5, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0016","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4745139afbea5a29d0c139010292759ed88c83bfdec01c77d569cd555f519ac3","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246016,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[3,-2,-2,4,-1,-2],\"operator\":[1,-2],\"source\":3,\"x\":-2,\"y\":-2}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[3, -2, -2, 4, -1, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[1, -2] means a*u_x+b*u_y; source=3, x=-2, y=-2. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([3, -2, -2, 4, -1, -2], [1, -2], -2, -2, -16, 8, 3, -35, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, -2, -2, 4, -1, -2], [1, -2], -2, -2, -16, 8, 3, -35, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0017","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a1cbd101ded6cd53543f3edfc3cb66a69c648b53fba6d42f6fa13f73b20c871","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246017,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[4,-1,-1,1,0,-1],\"operator\":[2,-1],\"source\":4,\"x\":-1,\"y\":-1}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[4, -1, -1, 1, 0, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[2, -1] means a*u_x+b*u_y; source=4, x=-1, y=-1. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([4, -1, -1, 1, 0, -1], [2, -1], -1, -1, -3, 1, 4, -11, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([4, -1, -1, 1, 0, -1], [2, -1], -1, -1, -3, 1, 4, -11, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0018","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c580497b22a113a0614520fe9e3c49c2981b099a67079e3110ba2cc8659ca8b0","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246018,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[5,0,0,2,1,0],\"operator\":[3,0],\"source\":5,\"x\":0,\"y\":0}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[5, 0, 0, 2, 1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[3, 0] means a*u_x+b*u_y; source=5, x=0, y=0. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([5, 0, 0, 2, 1, 0], [3, 0], 0, 0, 0, 0, 5, -5, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([5, 0, 0, 2, 1, 0], [3, 0], 0, 0, 0, 0, 5, -5, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0019","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d4bdbcb6764bb64e18b2fd78bc2720d3f7fe766140d4926fa40036f04950d60","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246019,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[6,1,1,3,-1,1],\"operator\":[1,1],\"source\":6,\"x\":1,\"y\":1}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[6, 1, 1, 3, -1, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[1, 1] means a*u_x+b*u_y; source=6, x=1, y=1. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([6, 1, 1, 3, -1, 1], [1, 1], 1, 1, 6, 2, 6, 2, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([6, 1, 1, 3, -1, 1], [1, 1], 1, 1, 6, 2, 6, 2, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0020","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5de338b2573002be4bf8571b18ab23a6ec4117432a04e9d422f34528c3a22271","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246020,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[7,2,2,4,0,2],\"operator\":[2,2],\"source\":7,\"x\":2,\"y\":2}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[7, 2, 2, 4, 0, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[2, 2] means a*u_x+b*u_y; source=7, x=2, y=2. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([7, 2, 2, 4, 0, 2], [2, 2], 2, 2, 18, 10, 7, 49, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([7, 2, 2, 4, 0, 2], [2, 2], 2, 2, 18, 10, 7, 49, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0021","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5bebd942d9585acd2198bd725efb5d0f41627348f332cedd6b2d565dbd25188b","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246021,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[8,-2,3,1,1,-2],\"operator\":[3,-2],\"source\":8,\"x\":-2,\"y\":3}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[8, -2, 3, 1, 1, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[3, -2] means a*u_x+b*u_y; source=8, x=-2, y=3. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([8, -2, 3, 1, 1, -2], [3, -2], -2, 3, -3, -11, 8, 5, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([8, -2, 3, 1, 1, -2], [3, -2], -2, 3, -3, -11, 8, 5, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0022","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ac791b8cb6554ac07ee06f92ec28aaaa8b51ac2f94630ee835072b877082037","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246022,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[9,-1,-3,2,-1,-1],\"operator\":[1,-1],\"source\":9,\"x\":-1,\"y\":-3}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[9, -1, -3, 2, -1, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[1, -1] means a*u_x+b*u_y; source=9, x=-1, y=-3. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([9, -1, -3, 2, -1, -1], [1, -1], -1, -3, -2, 4, 9, -15, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([9, -1, -3, 2, -1, -1], [1, -1], -1, -3, -2, 4, 9, -15, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0023","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b68859349af6d270ea597b294ab6bbfa07458d5a60b5268c6893b019d60c916e","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246023,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[10,0,-2,3,0,0],\"operator\":[2,0],\"source\":10,\"x\":0,\"y\":-2}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[10, 0, -2, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[2, 0] means a*u_x+b*u_y; source=10, x=0, y=-2. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([10, 0, -2, 3, 0, 0], [2, 0], 0, -2, 0, -2, 10, -10, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([10, 0, -2, 3, 0, 0], [2, 0], 0, -2, 0, -2, 10, -10, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0024","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ea8b9ab20337f88d114eb156f566794d1216bd956ddb1db3d92c1f537430033","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246024,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[11,1,-1,4,1,1],\"operator\":[3,1],\"source\":11,\"x\":1,\"y\":-1}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[11, 1, -1, 4, 1, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[3, 1] means a*u_x+b*u_y; source=11, x=1, y=-1. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([11, 1, -1, 4, 1, 1], [3, 1], 1, -1, 8, -2, 11, 11, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([11, 1, -1, 4, 1, 1], [3, 1], 1, -1, 8, -2, 11, 11, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0025","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45126e8b13e2bae6cdda8dc42063da29873e52c54e5588206c128a63e89dba68","family_id":"qf-mod24-pde-first-order-point-residual","seed":202610246025,"semantic_template":"qf-mod24-pde-first-order-point-residual/{\"coefficients\":[12,2,0,1,-1,2],\"operator\":[1,2],\"source\":12,\"x\":2,\"y\":0}"},"fields":{"title":"One-Point Residual of a Declared First-Order PDE Operator","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[12, 2, 0, 1, -1, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. operator [a,b]=[1, 2] means a*u_x+b*u_y; source=12, x=2, y=0. Give (coefficients,operator,x,y,u_x,u_y,source,operator value−source,satisfied at this point?).","answer":{"kind":"expression","value":"([12, 2, 0, 1, -1, 2], [1, 2], 2, 0, 6, -2, 12, -10, No)","explanation":"Compute both exact partials and the signed residual at the requested point. A zero point residual does not prove a global solution. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([12, 2, 0, 1, -1, 2], [1, 2], 2, 0, 6, -2, 12, -10, No)."},"skills":["One-Point Residual of a Declared First-Order PDE Operator"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0026","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8f922b24dcc0b48b37871caa32a5132b85b5e48d16abf83d425446e24718da6c","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246026,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-12,-2,-3,1,-1,-2],\"principal\":[1,-2,1],\"source\":-12,\"x\":-2,\"y\":-3}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-12, -2, -3, 1, -1, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[1, -2, 1] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-12, x=-2, y=-3. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-12, -2, -3, 1, -1, -2], [1, -2, 1], -2, -3, 2, -1, -4, -12, 12, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-12, -2, -3, 1, -1, -2], [1, -2, 1], -2, -3, 2, -1, -4, -12, 12, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0027","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"abc4a45fb63ccd7fe284d798040ca5d42f7fcff5a42ece1b22605ece9ea6517b","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246027,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-11,-1,-2,2,0,-1],\"principal\":[2,-1,2],\"source\":-11,\"x\":-1,\"y\":-2}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-11, -1, -2, 2, 0, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[2, -1, 2] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-11, x=-1, y=-2. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-11, -1, -2, 2, 0, -1], [2, -1, 2], -1, -2, 4, 0, -2, -11, 15, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-11, -1, -2, 2, 0, -1], [2, -1, 2], -1, -2, 4, 0, -2, -11, 15, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0028","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68ccdf9e076c9e75add7d8a39d70337c22253f7c84205160e7531b88ec8f23c0","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246028,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-10,0,-1,3,1,0],\"principal\":[3,0,3],\"source\":-10,\"x\":0,\"y\":-1}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-10, 0, -1, 3, 1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, 0, 3] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-10, x=0, y=-1. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-10, 0, -1, 3, 1, 0], [3, 0, 3], 0, -1, 6, 1, 0, -10, 28, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-10, 0, -1, 3, 1, 0], [3, 0, 3], 0, -1, 6, 1, 0, -10, 28, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0029","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb29efff0f97fb12e524fc89a3947a0f2276243545f51d446daad367bf1e8003","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246029,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-9,1,0,4,-1,1],\"principal\":[1,1,4],\"source\":-9,\"x\":1,\"y\":0}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-9, 1, 0, 4, -1, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[1, 1, 4] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-9, x=1, y=0. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-9, 1, 0, 4, -1, 1], [1, 1, 4], 1, 0, 8, -1, 2, -9, 24, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-9, 1, 0, 4, -1, 1], [1, 1, 4], 1, 0, 8, -1, 2, -9, 24, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0030","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4e4b506d0b7aee41426f0d4ff6fd6be2cf8be7aa860e8c70094309716012c414","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246030,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-8,2,1,1,0,2],\"principal\":[2,2,1],\"source\":-8,\"x\":2,\"y\":1}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-8, 2, 1, 1, 0, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[2, 2, 1] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-8, x=2, y=1. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-8, 2, 1, 1, 0, 2], [2, 2, 1], 2, 1, 2, 0, 4, -8, 16, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-8, 2, 1, 1, 0, 2], [2, 2, 1], 2, 1, 2, 0, 4, -8, 16, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0031","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"704e316a4354262bc160d5df849a7722087d0dce81d8de061bd45ffd930301cf","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246031,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-7,-2,2,2,1,-2],\"principal\":[3,-2,2],\"source\":-7,\"x\":-2,\"y\":2}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-7, -2, 2, 2, 1, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, -2, 2] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-7, x=-2, y=2. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-7, -2, 2, 2, 1, -2], [3, -2, 2], -2, 2, 4, 1, -4, -7, 9, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-7, -2, 2, 2, 1, -2], [3, -2, 2], -2, 2, 4, 1, -4, -7, 9, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0032","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"668dd1cd2233b125221fb17b79c0025eb577dca344067b512df3c15937eb38c3","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246032,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-6,-1,3,3,-1,-1],\"principal\":[1,-1,3],\"source\":-6,\"x\":-1,\"y\":3}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-6, -1, 3, 3, -1, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[1, -1, 3] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-6, x=-1, y=3. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-6, -1, 3, 3, -1, -1], [1, -1, 3], -1, 3, 6, -1, -2, -6, 7, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-6, -1, 3, 3, -1, -1], [1, -1, 3], -1, 3, 6, -1, -2, -6, 7, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0033","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"311c7e416c424a4cd750a504a10ccb636459373757da08a7b8e72fceb34114fe","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246033,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-5,0,-3,4,0,0],\"principal\":[2,0,4],\"source\":-5,\"x\":0,\"y\":-3}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-5, 0, -3, 4, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[2, 0, 4] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-5, x=0, y=-3. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-5, 0, -3, 4, 0, 0], [2, 0, 4], 0, -3, 8, 0, 0, -5, 21, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-5, 0, -3, 4, 0, 0], [2, 0, 4], 0, -3, 8, 0, 0, -5, 21, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0034","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3973cec3f57d740b3f3319f3711518aa3740aa4b26f07926c792bacc0bf46425","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246034,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-4,1,-2,1,1,1],\"principal\":[3,1,1],\"source\":-4,\"x\":1,\"y\":-2}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-4, 1, -2, 1, 1, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, 1, 1] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-4, x=1, y=-2. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-4, 1, -2, 1, 1, 1], [3, 1, 1], 1, -2, 2, 1, 2, -4, 13, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-4, 1, -2, 1, 1, 1], [3, 1, 1], 1, -2, 2, 1, 2, -4, 13, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0035","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e8ca9e3eb308f9c15cf6045dde07da012a146e5c9cea8d2597e1487ec28549b5","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246035,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-3,2,-1,2,-1,2],\"principal\":[1,2,2],\"source\":-3,\"x\":2,\"y\":-1}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-3, 2, -1, 2, -1, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[1, 2, 2] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-3, x=2, y=-1. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-3, 2, -1, 2, -1, 2], [1, 2, 2], 2, -1, 4, -1, 4, -3, 13, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-3, 2, -1, 2, -1, 2], [1, 2, 2], 2, -1, 4, -1, 4, -3, 13, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0036","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c9944e4ad18371a451291442dc45068c7d130af13ab50feacb9f8f231e4cf8c","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246036,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-2,-2,0,3,0,-2],\"principal\":[2,-2,3],\"source\":-2,\"x\":-2,\"y\":0}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-2, -2, 0, 3, 0, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[2, -2, 3] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-2, x=-2, y=0. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-2, -2, 0, 3, 0, -2], [2, -2, 3], -2, 0, 6, 0, -4, -2, 2, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-2, -2, 0, 3, 0, -2], [2, -2, 3], -2, 0, 6, 0, -4, -2, 2, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0037","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05d8cf416d34f2ca0dda60b9a20e850a5bca18c1f0e4a302c544bead18ba5644","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246037,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[-1,-1,1,4,1,-1],\"principal\":[3,-1,4],\"source\":-1,\"x\":-1,\"y\":1}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-1, -1, 1, 4, 1, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, -1, 4] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-1, x=-1, y=1. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([-1, -1, 1, 4, 1, -1], [3, -1, 4], -1, 1, 8, 1, -2, -1, 16, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-1, -1, 1, 4, 1, -1], [3, -1, 4], -1, 1, 8, 1, -2, -1, 16, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0038","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4951af5a4df5dec28ddf6629b976d84ff8e877cd400005eae8049ea09dd9b59","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246038,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[0,0,2,1,-1,0],\"principal\":[1,0,1],\"source\":0,\"x\":0,\"y\":2}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[0, 0, 2, 1, -1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[1, 0, 1] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=0, x=0, y=2. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([0, 0, 2, 1, -1, 0], [1, 0, 1], 0, 2, 2, -1, 0, 0, 2, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([0, 0, 2, 1, -1, 0], [1, 0, 1], 0, 2, 2, -1, 0, 0, 2, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0039","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2cea78b3c787fce23ad0178aacc1b44bd375d9f21b6e9fa73c94e1330142af8c","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246039,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[1,1,3,2,0,1],\"principal\":[2,1,2],\"source\":1,\"x\":1,\"y\":3}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[1, 1, 3, 2, 0, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[2, 1, 2] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=1, x=1, y=3. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([1, 1, 3, 2, 0, 1], [2, 1, 2], 1, 3, 4, 0, 2, 1, 11, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, 1, 3, 2, 0, 1], [2, 1, 2], 1, 3, 4, 0, 2, 1, 11, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0040","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dd571cbfe0b7e815d675ceab665693571f494668a768d80c31ad8399bc0572a","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246040,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[2,2,-3,3,1,2],\"principal\":[3,2,3],\"source\":2,\"x\":2,\"y\":-3}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[2, 2, -3, 3, 1, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, 2, 3] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=2, x=2, y=-3. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([2, 2, -3, 3, 1, 2], [3, 2, 3], 2, -3, 6, 1, 4, 2, 30, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, 2, -3, 3, 1, 2], [3, 2, 3], 2, -3, 6, 1, 4, 2, 30, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0041","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be3fd2f31919c14e6f88f23596604285a092b2a05e6d6db2145425541cb78d19","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246041,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[3,-2,-2,4,-1,-2],\"principal\":[1,-2,4],\"source\":3,\"x\":-2,\"y\":-2}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[3, -2, -2, 4, -1, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[1, -2, 4] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=3, x=-2, y=-2. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([3, -2, -2, 4, -1, -2], [1, -2, 4], -2, -2, 8, -1, -4, 3, -9, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, -2, -2, 4, -1, -2], [1, -2, 4], -2, -2, 8, -1, -4, 3, -9, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0042","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2232cc4f1feef6613c90709e04e0bd9a550c6700a9005155caed9a4ca2bbcfc7","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246042,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[4,-1,-1,1,0,-1],\"principal\":[2,-1,1],\"source\":4,\"x\":-1,\"y\":-1}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[4, -1, -1, 1, 0, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[2, -1, 1] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=4, x=-1, y=-1. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([4, -1, -1, 1, 0, -1], [2, -1, 1], -1, -1, 2, 0, -2, 4, -2, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([4, -1, -1, 1, 0, -1], [2, -1, 1], -1, -1, 2, 0, -2, 4, -2, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0043","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45d917aced6c831ed89aa3193d2b75355fdd658e1d33930c0981290015d2d4e1","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246043,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[5,0,0,2,1,0],\"principal\":[3,0,2],\"source\":5,\"x\":0,\"y\":0}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[5, 0, 0, 2, 1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, 0, 2] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=5, x=0, y=0. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([5, 0, 0, 2, 1, 0], [3, 0, 2], 0, 0, 4, 1, 0, 5, 7, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([5, 0, 0, 2, 1, 0], [3, 0, 2], 0, 0, 4, 1, 0, 5, 7, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0044","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"324494ffc2304786070bf1f849b3e41a9fc71fb3e2db123e254b414a12e0e5e9","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246044,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[6,1,1,3,-1,1],\"principal\":[1,1,3],\"source\":6,\"x\":1,\"y\":1}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[6, 1, 1, 3, -1, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[1, 1, 3] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=6, x=1, y=1. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([6, 1, 1, 3, -1, 1], [1, 1, 3], 1, 1, 6, -1, 2, 6, 5, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([6, 1, 1, 3, -1, 1], [1, 1, 3], 1, 1, 6, -1, 2, 6, 5, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0045","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3052fde8eec2247f84404b7228d60e98c7482083df7b266cfb29792b4f4066b3","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246045,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[7,2,2,4,0,2],\"principal\":[2,2,4],\"source\":7,\"x\":2,\"y\":2}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[7, 2, 2, 4, 0, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[2, 2, 4] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=7, x=2, y=2. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([7, 2, 2, 4, 0, 2], [2, 2, 4], 2, 2, 8, 0, 4, 7, 25, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([7, 2, 2, 4, 0, 2], [2, 2, 4], 2, 2, 8, 0, 4, 7, 25, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0046","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36c90333c860b6cb9506075f392a17dc887b1521f8033e1e3b6855bb354f12dc","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246046,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[8,-2,3,1,1,-2],\"principal\":[3,-2,1],\"source\":8,\"x\":-2,\"y\":3}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[8, -2, 3, 1, 1, -2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, -2, 1] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=8, x=-2, y=3. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([8, -2, 3, 1, 1, -2], [3, -2, 1], -2, 3, 2, 1, -4, 8, -8, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([8, -2, 3, 1, 1, -2], [3, -2, 1], -2, 3, 2, 1, -4, 8, -8, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0047","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"41c3be0365c6687df7a279649a63823ce351618412639565129c31935a7a4a8e","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246047,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[9,-1,-3,2,-1,-1],\"principal\":[1,-1,2],\"source\":9,\"x\":-1,\"y\":-3}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[9, -1, -3, 2, -1, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[1, -1, 2] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=9, x=-1, y=-3. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([9, -1, -3, 2, -1, -1], [1, -1, 2], -1, -3, 4, -1, -2, 9, -8, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([9, -1, -3, 2, -1, -1], [1, -1, 2], -1, -3, 4, -1, -2, 9, -8, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0048","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd2144bf967ff684ac07a389c3aaabccaccf8a0c8aa09534fd46802eb83c08f5","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246048,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[10,0,-2,3,0,0],\"principal\":[2,0,3],\"source\":10,\"x\":0,\"y\":-2}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[10, 0, -2, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[2, 0, 3] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=10, x=0, y=-2. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([10, 0, -2, 3, 0, 0], [2, 0, 3], 0, -2, 6, 0, 0, 10, 2, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([10, 0, -2, 3, 0, 0], [2, 0, 3], 0, -2, 6, 0, 0, 10, 2, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0049","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f8705e3d838dfedbd71c90c3dc8760a22d7c57c5434d88a4ddac8fe5aa36b77f","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246049,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[11,1,-1,4,1,1],\"principal\":[3,1,4],\"source\":11,\"x\":1,\"y\":-1}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[11, 1, -1, 4, 1, 1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, 1, 4] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=11, x=1, y=-1. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([11, 1, -1, 4, 1, 1], [3, 1, 4], 1, -1, 8, 1, 2, 11, 22, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([11, 1, -1, 4, 1, 1], [3, 1, 4], 1, -1, 8, 1, 2, 11, 22, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0050","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b25cff28d8306283f7b63f3d4d8511097014baeb9972cf3b028d17324e169070","family_id":"qf-mod24-pde-second-order-point-residual","seed":202610246050,"semantic_template":"qf-mod24-pde-second-order-point-residual/{\"coefficients\":[12,2,0,1,-1,2],\"principal\":[1,2,1],\"source\":12,\"x\":2,\"y\":0}"},"fields":{"title":"Second-Order Residual with an Explicit Mixed-Term Convention","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[12, 2, 0, 1, -1, 2] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[1, 2, 1] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=12, x=2, y=0. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).","answer":{"kind":"expression","value":"([12, 2, 0, 1, -1, 2], [1, 2, 1], 2, 0, 2, -1, 4, 12, -8, No)","explanation":"Use u_xx=2*c_xx, u_xy=c_xy and u_yy=2*c_yy in this polynomial model. The requested result is a residual audit, not a general PDE solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([12, 2, 0, 1, -1, 2], [1, 2, 1], 2, 0, 2, -1, 4, 12, -8, No)."},"skills":["Second-Order Residual with an Explicit Mixed-Term Convention"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0051","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf150a65db2227da3a308ea53b283d9c6d45c35075b20bd9816aa69e670e5644","family_id":"qf-mod24-pde-principal-classification","seed":202610246051,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[1,-12,1]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[1, -12, 1] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([1, -12, 1], a*u_xx+b*u_xy+c*u_yy, 140, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, -12, 1], a*u_xx+b*u_xy+c*u_yy, 140, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0052","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17909ce2a1311fe48272c79148f78c7c18e74012ca9354bebd95320b3329c017","family_id":"qf-mod24-pde-principal-classification","seed":202610246052,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[2,-11,2]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[2, -11, 2] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([2, -11, 2], a*u_xx+b*u_xy+c*u_yy, 105, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, -11, 2], a*u_xx+b*u_xy+c*u_yy, 105, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0053","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"323313f6b7a36094b806372e49ca9181e26225c29069359abeb9049b1fb1f98c","family_id":"qf-mod24-pde-principal-classification","seed":202610246053,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[3,-10,3]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[3, -10, 3] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([3, -10, 3], a*u_xx+b*u_xy+c*u_yy, 64, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, -10, 3], a*u_xx+b*u_xy+c*u_yy, 64, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0054","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee63ce1ef081fd50549bdbcd9e3070631055627033a432640d48a609457a7a34","family_id":"qf-mod24-pde-principal-classification","seed":202610246054,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[1,-9,4]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[1, -9, 4] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([1, -9, 4], a*u_xx+b*u_xy+c*u_yy, 65, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, -9, 4], a*u_xx+b*u_xy+c*u_yy, 65, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0055","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c75b06e234ac369480e24ff55d243d1104f2c68b08aa7c83fae317c1107af5a","family_id":"qf-mod24-pde-principal-classification","seed":202610246055,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[2,-8,1]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[2, -8, 1] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([2, -8, 1], a*u_xx+b*u_xy+c*u_yy, 56, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, -8, 1], a*u_xx+b*u_xy+c*u_yy, 56, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0056","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe58d89fa85378054ddb0ef1703b77905169a2c026515b74a61fc733c39899eb","family_id":"qf-mod24-pde-principal-classification","seed":202610246056,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[3,-7,2]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[3, -7, 2] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([3, -7, 2], a*u_xx+b*u_xy+c*u_yy, 25, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, -7, 2], a*u_xx+b*u_xy+c*u_yy, 25, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0057","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"47f967d58c2832ac1f6a687820bd97260cb81af441d8ca04fb6eca21a859741d","family_id":"qf-mod24-pde-principal-classification","seed":202610246057,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[1,-6,3]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[1, -6, 3] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([1, -6, 3], a*u_xx+b*u_xy+c*u_yy, 24, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, -6, 3], a*u_xx+b*u_xy+c*u_yy, 24, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0058","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c177c6e1077cd3fe154cd3e1b4dd8d1916ad91f2210d8f981948c3ecdae2a0aa","family_id":"qf-mod24-pde-principal-classification","seed":202610246058,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[2,-5,4]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[2, -5, 4] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([2, -5, 4], a*u_xx+b*u_xy+c*u_yy, -7, elliptic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, -5, 4], a*u_xx+b*u_xy+c*u_yy, -7, elliptic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0059","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"135ab3430c916d1b69efe3f83af6fda0024df1a8f1e59884dbd688ea3941d406","family_id":"qf-mod24-pde-principal-classification","seed":202610246059,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[3,-4,1]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[3, -4, 1] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([3, -4, 1], a*u_xx+b*u_xy+c*u_yy, 4, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, -4, 1], a*u_xx+b*u_xy+c*u_yy, 4, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0060","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"94589271565fa5cdc77cf1f688d7e291162640ae3ff05db3a19c88391a9e7213","family_id":"qf-mod24-pde-principal-classification","seed":202610246060,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[1,-3,2]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[1, -3, 2] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([1, -3, 2], a*u_xx+b*u_xy+c*u_yy, 1, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, -3, 2], a*u_xx+b*u_xy+c*u_yy, 1, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0061","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0cb8f71b99339808e0e59538a06fb089db2f8ad76765086fe80c50dcf28e0d4a","family_id":"qf-mod24-pde-principal-classification","seed":202610246061,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[2,-2,3]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[2, -2, 3] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([2, -2, 3], a*u_xx+b*u_xy+c*u_yy, -20, elliptic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, -2, 3], a*u_xx+b*u_xy+c*u_yy, -20, elliptic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0062","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4210d9db8990e2cc36f73dc8eb57e577aebe92eb564c9090fc5617f99693ede1","family_id":"qf-mod24-pde-principal-classification","seed":202610246062,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[3,-1,4]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[3, -1, 4] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([3, -1, 4], a*u_xx+b*u_xy+c*u_yy, -47, elliptic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, -1, 4], a*u_xx+b*u_xy+c*u_yy, -47, elliptic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0063","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2afde3a804973c861df98aa81b2e51f6ffaf629e5bd6dcc1cf2495b9d2326228","family_id":"qf-mod24-pde-principal-classification","seed":202610246063,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[1,0,1]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[1, 0, 1] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([1, 0, 1], a*u_xx+b*u_xy+c*u_yy, -4, elliptic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, 0, 1], a*u_xx+b*u_xy+c*u_yy, -4, elliptic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0064","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"22e966dcf8d32ee00f4bebf988267f0e9edaf790db6999a028e5dc148c76c634","family_id":"qf-mod24-pde-principal-classification","seed":202610246064,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[2,1,2]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[2, 1, 2] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([2, 1, 2], a*u_xx+b*u_xy+c*u_yy, -15, elliptic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, 1, 2], a*u_xx+b*u_xy+c*u_yy, -15, elliptic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0065","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"598f06c911a02696c4a4e47a7434a6fe58ded6e43e3f9d86d23fc8f0e052880a","family_id":"qf-mod24-pde-principal-classification","seed":202610246065,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[3,2,3]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[3, 2, 3] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([3, 2, 3], a*u_xx+b*u_xy+c*u_yy, -32, elliptic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, 2, 3], a*u_xx+b*u_xy+c*u_yy, -32, elliptic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0066","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb633e083c65851932aa0251c86e262b92e485c374d53ae3e27ac040217225f7","family_id":"qf-mod24-pde-principal-classification","seed":202610246066,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[1,3,4]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[1, 3, 4] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([1, 3, 4], a*u_xx+b*u_xy+c*u_yy, -7, elliptic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, 3, 4], a*u_xx+b*u_xy+c*u_yy, -7, elliptic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0067","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7fc0537671d32a8a64e1b2e712ee8c931321ea7de7c393d5649c9cdddbf7bab5","family_id":"qf-mod24-pde-principal-classification","seed":202610246067,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[2,4,1]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[2, 4, 1] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([2, 4, 1], a*u_xx+b*u_xy+c*u_yy, 8, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, 4, 1], a*u_xx+b*u_xy+c*u_yy, 8, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0068","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ce10f1b56e988b22a084aae0990cd4889e26bdbe623808dc69483883137b1c3","family_id":"qf-mod24-pde-principal-classification","seed":202610246068,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[3,5,2]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[3, 5, 2] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([3, 5, 2], a*u_xx+b*u_xy+c*u_yy, 1, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, 5, 2], a*u_xx+b*u_xy+c*u_yy, 1, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0069","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8f9da5e22d5b1c98a034f31eb2877ec8b403c74e33ae826c016353d121da3789","family_id":"qf-mod24-pde-principal-classification","seed":202610246069,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[1,6,3]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[1, 6, 3] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([1, 6, 3], a*u_xx+b*u_xy+c*u_yy, 24, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, 6, 3], a*u_xx+b*u_xy+c*u_yy, 24, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0070","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9649dc962c3a592612c846f4c85a81ca0f71eaa083fba03988fcfca94daffef8","family_id":"qf-mod24-pde-principal-classification","seed":202610246070,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[2,7,4]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[2, 7, 4] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([2, 7, 4], a*u_xx+b*u_xy+c*u_yy, 17, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, 7, 4], a*u_xx+b*u_xy+c*u_yy, 17, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0071","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5630effef3ee50bcbd9e49e294c5d12f4b738e9e507c0b914d3504cd50015e79","family_id":"qf-mod24-pde-principal-classification","seed":202610246071,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[3,8,1]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[3, 8, 1] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([3, 8, 1], a*u_xx+b*u_xy+c*u_yy, 52, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, 8, 1], a*u_xx+b*u_xy+c*u_yy, 52, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0072","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d86c79ffd34bf41a466cfe62df224376f36436295dfec9f39d87483b2ffcf18","family_id":"qf-mod24-pde-principal-classification","seed":202610246072,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[1,9,2]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[1, 9, 2] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([1, 9, 2], a*u_xx+b*u_xy+c*u_yy, 73, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, 9, 2], a*u_xx+b*u_xy+c*u_yy, 73, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0073","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d2abc5d74b0b3ade756be5ebfef74afc44a7cdc970b96fcb95ebe1a5cdafc301","family_id":"qf-mod24-pde-principal-classification","seed":202610246073,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[2,10,3]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[2, 10, 3] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([2, 10, 3], a*u_xx+b*u_xy+c*u_yy, 76, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, 10, 3], a*u_xx+b*u_xy+c*u_yy, 76, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0074","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d4079a34987d8fe5950c1264dcf8e347005301e32fc1515b473a433e44b36e2a","family_id":"qf-mod24-pde-principal-classification","seed":202610246074,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[3,11,4]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[3, 11, 4] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([3, 11, 4], a*u_xx+b*u_xy+c*u_yy, 73, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, 11, 4], a*u_xx+b*u_xy+c*u_yy, 73, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0075","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87d1fd0e11d49facf218c030251c19115f051b7dfe5311516b77fca842b582bd","family_id":"qf-mod24-pde-principal-classification","seed":202610246075,"semantic_template":"qf-mod24-pde-principal-classification/{\"principal\":[1,12,1]}"},"fields":{"title":"Classify Only after Fixing b*u_xy rather than 2*b*u_xy","stem":"Use the nonzero second-order principal part [a,b,c]=[1, 12, 1] with convention a*u_xx+b*u_xy+c*u_yy. Give (principal,that exact convention string,discriminant b²−4*a*c,classification). Positive, negative and zero discriminants mean hyperbolic, elliptic and parabolic respectively. The all-zero principal part is excluded.","answer":{"kind":"expression","value":"([1, 12, 1], a*u_xx+b*u_xy+c*u_yy, 140, hyperbolic)","explanation":"Keep the mixed-term convention fixed when computing the discriminant; classification alone proves no existence, uniqueness or well-posedness. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, 12, 1], a*u_xx+b*u_xy+c*u_yy, 140, hyperbolic)."},"skills":["Classify Only after Fixing b*u_xy rather than 2*b*u_xy"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0076","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"115fa1c96a7054e24be13de34803b192c4902e7cbba1976eb287b3aa1e00d8fa","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246076,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-12,-2,2,1,1,0],\"diffusivity\":1}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-12, -2, 2, 1, 1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=1. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-12, -2, 2, 1, 1, 0], 1, [0, 1, 0], No)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-12, -2, 2, 1, 1, 0], 1, [0, 1, 0], No)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0077","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af929de129a661c5eceb0c5a45c98f2f5e44de6274f2152bc6f6645e5d3a2035","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246077,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-11,-1,8,2,0,0],\"diffusivity\":2}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-11, -1, 8, 2, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=2. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-11, -1, 8, 2, 0, 0], 2, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-11, -1, 8, 2, 0, 0], 2, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0078","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ab51c2f09b4b2ef67992f09cc649a3ce426b866bf15f7339fef9fff2c839ef7","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246078,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-10,0,18,3,0,0],\"diffusivity\":3}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-10, 0, 18, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=3. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-10, 0, 18, 3, 0, 0], 3, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-10, 0, 18, 3, 0, 0], 3, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0079","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"12c4eda2c74c8bf330a81014b6f7ab96a622a1a6a817e68f068c734a8d3bedc2","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246079,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-9,1,8,1,0,0],\"diffusivity\":4}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-9, 1, 8, 1, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=4. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-9, 1, 8, 1, 0, 0], 4, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-9, 1, 8, 1, 0, 0], 4, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0080","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cb4450f152946f4dc4445a581625495051b2df7abd63371fe71b4ce7f1f03d9","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246080,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-8,2,4,2,0,0],\"diffusivity\":1}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-8, 2, 4, 2, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=1. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-8, 2, 4, 2, 0, 0], 1, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-8, 2, 4, 2, 0, 0], 1, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0081","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"002170387fc4dc01cd45da38da5c1093adf3f9d17f8a768bceea7e3d1a2d513e","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246081,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-7,-2,12,3,0,0],\"diffusivity\":2}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-7, -2, 12, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=2. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-7, -2, 12, 3, 0, 0], 2, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-7, -2, 12, 3, 0, 0], 2, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0082","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"151395a1a8192094adb519b7bf456d31376cf5b6723e8109c9dc4e5e9b33e155","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246082,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-6,-1,6,1,1,0],\"diffusivity\":3}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-6, -1, 6, 1, 1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=3. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-6, -1, 6, 1, 1, 0], 3, [0, 1, 0], No)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-6, -1, 6, 1, 1, 0], 3, [0, 1, 0], No)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0083","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b64ac825240f7ab6f455f635a53862b12f72969c286dc15612168bbb4b75299d","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246083,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-5,0,16,2,0,0],\"diffusivity\":4}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-5, 0, 16, 2, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=4. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-5, 0, 16, 2, 0, 0], 4, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-5, 0, 16, 2, 0, 0], 4, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0084","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fac119f9d9893d814f447c7135bd19f662beae277c765cd96b0e46b3159bc753","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246084,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-4,1,6,3,0,0],\"diffusivity\":1}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-4, 1, 6, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=1. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-4, 1, 6, 3, 0, 0], 1, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-4, 1, 6, 3, 0, 0], 1, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0085","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0cba6f77d4c82cdd610a3885fe3683e4f382e5e80def4d329b06831fd6ce59a2","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246085,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-3,2,4,1,0,0],\"diffusivity\":2}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-3, 2, 4, 1, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=2. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-3, 2, 4, 1, 0, 0], 2, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-3, 2, 4, 1, 0, 0], 2, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0086","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80bcfbe974a1423dcf19bfd8bd388904314ad5f9a1b76e77e390ba9d18665f75","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246086,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-2,-2,12,2,0,0],\"diffusivity\":3}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-2, -2, 12, 2, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=3. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-2, -2, 12, 2, 0, 0], 3, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-2, -2, 12, 2, 0, 0], 3, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0087","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"130760f82634fb7defd2577ac1173d73014eb78d57a0beec559cb820a877c12f","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246087,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[-1,-1,24,3,0,0],\"diffusivity\":4}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-1, -1, 24, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=4. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([-1, -1, 24, 3, 0, 0], 4, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([-1, -1, 24, 3, 0, 0], 4, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0088","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80b46b7a9444d0c04e0cc0a99aa9a9f7554253d555a0923138225de689a78437","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246088,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[0,0,2,1,1,0],\"diffusivity\":1}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[0, 0, 2, 1, 1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=1. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([0, 0, 2, 1, 1, 0], 1, [0, 1, 0], No)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([0, 0, 2, 1, 1, 0], 1, [0, 1, 0], No)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0089","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"54e5febbcc31b152aa17ee06980891ea6cf1448aea6113343e26bd67e1c8caca","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246089,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[1,1,8,2,0,0],\"diffusivity\":2}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[1, 1, 8, 2, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=2. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([1, 1, 8, 2, 0, 0], 2, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([1, 1, 8, 2, 0, 0], 2, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0090","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e64c808f8d7ae1cf9125bb381568bbb51162dcf18169f3ce580793b90285149","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246090,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[2,2,18,3,0,0],\"diffusivity\":3}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[2, 2, 18, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=3. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([2, 2, 18, 3, 0, 0], 3, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([2, 2, 18, 3, 0, 0], 3, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0091","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae2dc6240607a14e64b5caabd075c92e6af59398cec2834a63253ff31f89dc4b","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246091,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[3,-2,8,1,0,0],\"diffusivity\":4}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[3, -2, 8, 1, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=4. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([3, -2, 8, 1, 0, 0], 4, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([3, -2, 8, 1, 0, 0], 4, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0092","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d2e8bb40110ac19bcef3c8d2c7d955089f3057e7a4f5e67f263782ad5e6fa588","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246092,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[4,-1,4,2,0,0],\"diffusivity\":1}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[4, -1, 4, 2, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=1. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([4, -1, 4, 2, 0, 0], 1, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([4, -1, 4, 2, 0, 0], 1, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0093","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7aed3b5e13583125a9da2d4c95046beda416e2b826eebaaddf365bac972a30a1","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246093,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[5,0,12,3,0,0],\"diffusivity\":2}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[5, 0, 12, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=2. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([5, 0, 12, 3, 0, 0], 2, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([5, 0, 12, 3, 0, 0], 2, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0094","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fde202c01fc6ba68764d0f534731fdadee353e75ee318190c012eba531a8cc04","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246094,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[6,1,6,1,1,0],\"diffusivity\":3}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[6, 1, 6, 1, 1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=3. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([6, 1, 6, 1, 1, 0], 3, [0, 1, 0], No)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([6, 1, 6, 1, 1, 0], 3, [0, 1, 0], No)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0095","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e10e54b7a005513bb2ed703d098beb95b5975e8eb747a6a466aa2a7fd34a750f","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246095,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[7,2,16,2,0,0],\"diffusivity\":4}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[7, 2, 16, 2, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=4. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([7, 2, 16, 2, 0, 0], 4, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([7, 2, 16, 2, 0, 0], 4, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0096","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"961b4c4d279ce7ef7841920e1aed079a3c75ebd1057fde450e549f698ab4949c","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246096,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[8,-2,6,3,0,0],\"diffusivity\":1}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[8, -2, 6, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=1. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([8, -2, 6, 3, 0, 0], 1, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([8, -2, 6, 3, 0, 0], 1, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0097","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f125d00705dd9f8a09d11b40b1d438b269e807da49ad184846586965c9f5c8c1","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246097,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[9,-1,4,1,0,0],\"diffusivity\":2}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[9, -1, 4, 1, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=2. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([9, -1, 4, 1, 0, 0], 2, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([9, -1, 4, 1, 0, 0], 2, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0098","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9fc42c8a5cfebd4721f582608590cde0409c88f7cc0b53f2fc523d4feb71125c","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246098,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[10,0,12,2,0,0],\"diffusivity\":3}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[10, 0, 12, 2, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=3. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([10, 0, 12, 2, 0, 0], 3, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([10, 0, 12, 2, 0, 0], 3, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0099","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"abcfa73268152addb88ebbd6e122a680b39dea98e5a18e4bfbfa43fca9e24949","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246099,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[11,1,24,3,0,0],\"diffusivity\":4}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[11, 1, 24, 3, 0, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=4. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([11, 1, 24, 3, 0, 0], 4, [0, 0, 0], Yes)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([11, 1, 24, 3, 0, 0], 4, [0, 0, 0], Yes)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261013-advance-20261024-r06-partial-differential-equation-0100","batch_id":"batch-20261013-advance-20261024-r06-partial-differential-equation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe519523e3f6f9e3a91133ed266762fb13894d808f451bb6648813130b5ef527","family_id":"qf-mod24-pde-heat-polynomial-identity","seed":202610246100,"semantic_template":"qf-mod24-pde-heat-polynomial-identity/{\"coefficients\":[12,2,2,1,1,0],\"diffusivity\":1}"},"fields":{"title":"A Symbolic Heat Residual Identity, Not a Point Sample","stem":"Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[12, 2, 2, 1, 1, 0] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. Here y is time, x is space and positive integer diffusivity=1. The equation is u_y=diffusivity*u_xx. Test the residual identically as a polynomial, not only at sampled points. Give (coefficients,diffusivity,residual coefficients in basis [1,x,y],all residual coefficients zero?). The residual coefficients are [c_y−2*diffusivity*c_xx,c_xy,2*c_yy].","answer":{"kind":"expression","value":"([12, 2, 2, 1, 1, 0], 1, [0, 1, 0], No)","explanation":"A zero residual coefficient vector proves this specified polynomial satisfies the operator equation everywhere. A nonzero residual that vanishes at some sampled points is not an identity; neither case supplies a general solution theory. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is ([12, 2, 2, 1, 1, 0], 1, [0, 1, 0], No)."},"skills":["A Symbolic Heat Residual Identity, Not a Point Sample"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}