# Explicit PDE Conventions: Point Residuals and Symbolic Identities

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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261013-advance-20261024-r06-partial-differential-equation-foundations · 2026-10-13 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.



## 1. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-12, -2, -3, 1, -1, -2], [1, -2], -2, -3, -3, 11, -12, -13, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0001/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 2. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-11, -1, -2, 2, 0, -1], [2, -1], -1, -2, -5, 2, -11, -1, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0002/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 3. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-10, 0, -1, 3, 1, 0], [3, 0], 0, -1, -1, -1, -10, 7, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0003/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 4. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-9, 1, 0, 4, -1, 1], [1, 1], 1, 0, 9, -1, -9, 17, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0004/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 5. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-8, 2, 1, 1, 0, 2], [2, 2], 2, 1, 6, 5, -8, 30, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0005/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 6. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-7, -2, 2, 2, 1, -2], [3, -2], -2, 2, -8, -8, -7, -1, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0006/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 7. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-6, -1, 3, 3, -1, -1], [1, -1], -1, 3, -10, -2, -6, -2, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0007/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 8. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-5, 0, -3, 4, 0, 0], [2, 0], 0, -3, 0, -3, -5, 5, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0008/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 9. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-4, 1, -2, 1, 1, 1], [3, 1], 1, -2, 1, -5, -4, 2, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0009/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 10. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-3, 2, -1, 2, -1, 2], [1, 2], 2, -1, 11, -7, -3, 0, Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0010/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 11. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-2, -2, 0, 3, 0, -2], [2, -2], -2, 0, -14, 0, -2, -26, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0011/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 12. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([-1, -1, 1, 4, 1, -1], [3, -1], -1, 1, -8, -2, -1, -21, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0012/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 13. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([0, 0, 2, 1, -1, 0], [1, 0], 0, 2, -2, 2, 0, -2, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0013/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 14. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([1, 1, 3, 2, 0, 1], [2, 1], 1, 3, 5, 9, 1, 18, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0014/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 15. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([2, 2, -3, 3, 1, 2], [3, 2], 2, -3, 11, -13, 2, 5, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0015/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 16. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([3, -2, -2, 4, -1, -2], [1, -2], -2, -2, -16, 8, 3, -35, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0016/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 17. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([4, -1, -1, 1, 0, -1], [2, -1], -1, -1, -3, 1, 4, -11, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0017/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 18. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([5, 0, 0, 2, 1, 0], [3, 0], 0, 0, 0, 0, 5, -5, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0018/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 19. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([6, 1, 1, 3, -1, 1], [1, 1], 1, 1, 6, 2, 6, 2, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0019/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 20. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([7, 2, 2, 4, 0, 2], [2, 2], 2, 2, 18, 10, 7, 49, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0020/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 21. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([8, -2, 3, 1, 1, -2], [3, -2], -2, 3, -3, -11, 8, 5, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0021/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 22. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([9, -1, -3, 2, -1, -1], [1, -1], -1, -3, -2, 4, 9, -15, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0022/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 23. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([10, 0, -2, 3, 0, 0], [2, 0], 0, -2, 0, -2, 10, -10, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0023/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 24. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([11, 1, -1, 4, 1, 1], [3, 1], 1, -1, 8, -2, 11, 11, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0024/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 25. One-Point Residual of a Declared First-Order PDE Operator

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?).

View answer and explanation([12, 2, 0, 1, -1, 2], [1, 2], 2, 0, 6, -2, 12, -10, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0025/)

Partial differential equation · qf-mod24-pde-first-order-point-residual · candidate





## 26. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-12, -2, -3, 1, -1, -2], [1, -2, 1], -2, -3, 2, -1, -4, -12, 12, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0026/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 27. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-11, -1, -2, 2, 0, -1], [2, -1, 2], -1, -2, 4, 0, -2, -11, 15, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0027/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 28. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-10, 0, -1, 3, 1, 0], [3, 0, 3], 0, -1, 6, 1, 0, -10, 28, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0028/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 29. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-9, 1, 0, 4, -1, 1], [1, 1, 4], 1, 0, 8, -1, 2, -9, 24, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0029/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 30. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-8, 2, 1, 1, 0, 2], [2, 2, 1], 2, 1, 2, 0, 4, -8, 16, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0030/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 31. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-7, -2, 2, 2, 1, -2], [3, -2, 2], -2, 2, 4, 1, -4, -7, 9, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0031/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 32. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-6, -1, 3, 3, -1, -1], [1, -1, 3], -1, 3, 6, -1, -2, -6, 7, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0032/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 33. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-5, 0, -3, 4, 0, 0], [2, 0, 4], 0, -3, 8, 0, 0, -5, 21, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0033/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 34. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-4, 1, -2, 1, 1, 1], [3, 1, 1], 1, -2, 2, 1, 2, -4, 13, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0034/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 35. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-3, 2, -1, 2, -1, 2], [1, 2, 2], 2, -1, 4, -1, 4, -3, 13, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0035/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 36. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-2, -2, 0, 3, 0, -2], [2, -2, 3], -2, 0, 6, 0, -4, -2, 2, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0036/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 37. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([-1, -1, 1, 4, 1, -1], [3, -1, 4], -1, 1, 8, 1, -2, -1, 16, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0037/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 38. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([0, 0, 2, 1, -1, 0], [1, 0, 1], 0, 2, 2, -1, 0, 0, 2, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0038/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 39. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([1, 1, 3, 2, 0, 1], [2, 1, 2], 1, 3, 4, 0, 2, 1, 11, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0039/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 40. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([2, 2, -3, 3, 1, 2], [3, 2, 3], 2, -3, 6, 1, 4, 2, 30, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0040/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 41. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([3, -2, -2, 4, -1, -2], [1, -2, 4], -2, -2, 8, -1, -4, 3, -9, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0041/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 42. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([4, -1, -1, 1, 0, -1], [2, -1, 1], -1, -1, 2, 0, -2, 4, -2, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0042/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 43. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([5, 0, 0, 2, 1, 0], [3, 0, 2], 0, 0, 4, 1, 0, 5, 7, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0043/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 44. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([6, 1, 1, 3, -1, 1], [1, 1, 3], 1, 1, 6, -1, 2, 6, 5, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0044/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 45. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([7, 2, 2, 4, 0, 2], [2, 2, 4], 2, 2, 8, 0, 4, 7, 25, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0045/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 46. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([8, -2, 3, 1, 1, -2], [3, -2, 1], -2, 3, 2, 1, -4, 8, -8, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0046/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 47. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([9, -1, -3, 2, -1, -1], [1, -1, 2], -1, -3, 4, -1, -2, 9, -8, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0047/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 48. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([10, 0, -2, 3, 0, 0], [2, 0, 3], 0, -2, 6, 0, 0, 10, 2, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0048/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 49. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([11, 1, -1, 4, 1, 1], [3, 1, 4], 1, -1, 8, 1, 2, 11, 22, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0049/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 50. Second-Order Residual with an Explicit Mixed-Term Convention

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?).

View answer and explanation([12, 2, 0, 1, -1, 2], [1, 2, 1], 2, 0, 2, -1, 4, 12, -8, No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0050/)

Partial differential equation · qf-mod24-pde-second-order-point-residual · candidate





## 51. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([1, -12, 1], a*u_xx+b*u_xy+c*u_yy, 140, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0051/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 52. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([2, -11, 2], a*u_xx+b*u_xy+c*u_yy, 105, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0052/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 53. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([3, -10, 3], a*u_xx+b*u_xy+c*u_yy, 64, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0053/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 54. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([1, -9, 4], a*u_xx+b*u_xy+c*u_yy, 65, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0054/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 55. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([2, -8, 1], a*u_xx+b*u_xy+c*u_yy, 56, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0055/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 56. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([3, -7, 2], a*u_xx+b*u_xy+c*u_yy, 25, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0056/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 57. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([1, -6, 3], a*u_xx+b*u_xy+c*u_yy, 24, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0057/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 58. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([2, -5, 4], a*u_xx+b*u_xy+c*u_yy, -7, elliptic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0058/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 59. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([3, -4, 1], a*u_xx+b*u_xy+c*u_yy, 4, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0059/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 60. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([1, -3, 2], a*u_xx+b*u_xy+c*u_yy, 1, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0060/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 61. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([2, -2, 3], a*u_xx+b*u_xy+c*u_yy, -20, elliptic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0061/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 62. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([3, -1, 4], a*u_xx+b*u_xy+c*u_yy, -47, elliptic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0062/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 63. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([1, 0, 1], a*u_xx+b*u_xy+c*u_yy, -4, elliptic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0063/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 64. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([2, 1, 2], a*u_xx+b*u_xy+c*u_yy, -15, elliptic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0064/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 65. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([3, 2, 3], a*u_xx+b*u_xy+c*u_yy, -32, elliptic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0065/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 66. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([1, 3, 4], a*u_xx+b*u_xy+c*u_yy, -7, elliptic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0066/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 67. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([2, 4, 1], a*u_xx+b*u_xy+c*u_yy, 8, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0067/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 68. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([3, 5, 2], a*u_xx+b*u_xy+c*u_yy, 1, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0068/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 69. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([1, 6, 3], a*u_xx+b*u_xy+c*u_yy, 24, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0069/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 70. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([2, 7, 4], a*u_xx+b*u_xy+c*u_yy, 17, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0070/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 71. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([3, 8, 1], a*u_xx+b*u_xy+c*u_yy, 52, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0071/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 72. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([1, 9, 2], a*u_xx+b*u_xy+c*u_yy, 73, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0072/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 73. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([2, 10, 3], a*u_xx+b*u_xy+c*u_yy, 76, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0073/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 74. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([3, 11, 4], a*u_xx+b*u_xy+c*u_yy, 73, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0074/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 75. Classify Only after Fixing b*u_xy rather than 2*b*u_xy

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.

View answer and explanation([1, 12, 1], a*u_xx+b*u_xy+c*u_yy, 140, hyperbolic)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0075/)

Partial differential equation · qf-mod24-pde-principal-classification · candidate





## 76. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-12, -2, 2, 1, 1, 0], 1, [0, 1, 0], No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0076/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 77. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-11, -1, 8, 2, 0, 0], 2, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0077/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 78. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-10, 0, 18, 3, 0, 0], 3, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0078/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 79. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-9, 1, 8, 1, 0, 0], 4, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0079/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 80. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-8, 2, 4, 2, 0, 0], 1, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0080/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 81. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-7, -2, 12, 3, 0, 0], 2, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0081/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 82. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-6, -1, 6, 1, 1, 0], 3, [0, 1, 0], No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0082/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 83. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-5, 0, 16, 2, 0, 0], 4, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0083/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 84. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-4, 1, 6, 3, 0, 0], 1, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0084/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 85. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-3, 2, 4, 1, 0, 0], 2, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0085/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 86. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-2, -2, 12, 2, 0, 0], 3, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0086/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 87. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([-1, -1, 24, 3, 0, 0], 4, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0087/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 88. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([0, 0, 2, 1, 1, 0], 1, [0, 1, 0], No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0088/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 89. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([1, 1, 8, 2, 0, 0], 2, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0089/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 90. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([2, 2, 18, 3, 0, 0], 3, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0090/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 91. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([3, -2, 8, 1, 0, 0], 4, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0091/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 92. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([4, -1, 4, 2, 0, 0], 1, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0092/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 93. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([5, 0, 12, 3, 0, 0], 2, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0093/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 94. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([6, 1, 6, 1, 1, 0], 3, [0, 1, 0], No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0094/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 95. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([7, 2, 16, 2, 0, 0], 4, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0095/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 96. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([8, -2, 6, 3, 0, 0], 1, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0096/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 97. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([9, -1, 4, 1, 0, 0], 2, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0097/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 98. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([10, 0, 12, 2, 0, 0], 3, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0098/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 99. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([11, 1, 24, 3, 0, 0], 4, [0, 0, 0], Yes)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0099/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate





## 100. A Symbolic Heat Residual Identity, Not a Point Sample

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].

View answer and explanation([12, 2, 2, 1, 1, 0], 1, [0, 1, 0], No)

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).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261013-advance-20261024-r06-partial-differential-equation-0100/)

Partial differential equation · qf-mod24-pde-heat-polynomial-identity · candidate



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