{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-partial-differential-equation-01",
  "mko_id": "mko-partial-differential-equation",
  "title": {
    "zh-Hant": "偏微分方程：常係數二階算子的主部分類與二次多項式候選的符號殘差",
    "en": "Partial differential equation: principal-part classification and symbolic residuals of quadratic polynomial candidates"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-partial-differential-equation-01",
    "source": {
      "mko_id": "mko-partial-differential-equation",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "b78ef1b1a613523b784d17e0507361cf263e495850e2b7aa880558f1294758b8",
      "english_sha256": "62e82ace9246a9a6a884ee7e0a1a332257e56eba985caac8dbc5c8d5efa7df75"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "numeric_evaluation",
      "distributive",
      "eq_replace",
      "poly_partial",
      "quantifier_negation"
    ],
    "goal": {
      "zh-Hant": "在 ℝ² 上給定線性常係數算子 L(u)=u_xx+2u_xy−3u_yy+u_x 與右端 g(x,y)=2x+y−2。先依明示的主部約定 a u_xx+b u_xy+c u_yy 與判別式 b²−4ac 對 L 分類；再對三個二次多項式候選 u_A=x²+xy+y²、u_B=x²+2xy、u_C=x²+xy+y²+y 精確計算符號殘差 R=L(u)−g，判定哪些候選在全 ℝ² 上滿足 L(u)=g，並說明 u_B 的點殘差在整條直線 y=−8 上為零，為何仍不構成解。",
      "en": "On ℝ², take the linear constant-coefficient operator L(u)=u_xx+2u_xy−3u_yy+u_x and the right-hand side g(x,y)=2x+y−2. First classify L using the explicit principal-part convention a u_xx+b u_xy+c u_yy with discriminant b²−4ac; then, for the three quadratic polynomial candidates u_A=x²+xy+y², u_B=x²+2xy and u_C=x²+xy+y²+y, compute the symbolic residual R=L(u)−g exactly, decide which candidates satisfy L(u)=g on all of ℝ², and explain why u_B is still not a solution although its point residual vanishes on the whole line y=−8."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "算子與方程：對 ℝ² 上的二次可微函數 u，L(u)=u_xx+2u_xy−3u_yy+u_x，其中 u_xy 表示先對 x、再對 y 求偏導；右端 g(x,y)=2x+y−2；所問的方程是在全 ℝ² 上 L(u)=g。",
          "en": "Operator and equation: for twice differentiable u on ℝ², L(u)=u_xx+2u_xy−3u_yy+u_x, where u_xy means differentiate first in x and then in y; the right-hand side is g(x,y)=2x+y−2; the equation asked about is L(u)=g on all of ℝ²."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "主部約定（MKO 要求明示）：二階主部寫成 a u_xx+b u_xy+c u_yy，b 是 u_xy 的完整係數；判別式 D=b²−4ac。",
          "en": "Principal-part convention (the MKO requires it to be explicit): the second-order principal part is written a u_xx+b u_xy+c u_yy, with b the full coefficient of u_xy; the discriminant is D=b²−4ac."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "三個二次多項式候選：u_A=x²+xy+y²、u_B=x²+2xy、u_C=x²+xy+y²+y。",
          "en": "Three quadratic polynomial candidates: u_A=x²+xy+y², u_B=x²+2xy, u_C=x²+xy+y²+y."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "L(u)=u_{xx}+2u_{xy}-3u_{yy}+u_x",
        "after_tex": "a=1,\\ b=2,\\ c=-3;\\ \\text{lower-order part } u_x;\\ g=2x+y-2",
        "relation": "definition",
        "rule": "mko_pde",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 G2 的約定，主部是 L 的二階部分 a u_xx+b u_xy+c u_yy。逐項對照：u_xx 的係數 a=1，u_xy 的完整係數 b=2（不除以 2），u_yy 的係數 c=−3。u_x 是一階（低階）項，不進入分類；g 是右端。先把約定寫明，是 MKO 的明確要求，因為同一個算子在另一種寫法下混合項係數會被讀成 1。",
          "en": "Under the convention of G2, the principal part is the second-order part a u_xx+b u_xy+c u_yy of L. Matching term by term: the coefficient of u_xx is a=1, the full coefficient of u_xy is b=2 (not halved), and the coefficient of u_yy is c=−3. The term u_x is first order (lower order) and does not enter the classification; g is the right-hand side. Stating the convention first is an explicit requirement of the MKO, because under another way of writing the same operator the mixed coefficient would be read as 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_pde"
          ],
          "explanation": {
            "zh-Hant": "此步只依約定讀出係數，不推出新事實。讀法依賴 G2：若改用 A u_xx+2B u_xy+C u_yy，混合項係數是 B=1，判別式公式也要跟著換（見 S04）。",
            "en": "This step only reads off coefficients under the convention and derives no new fact. The reading depends on G2: under A u_xx+2B u_xy+C u_yy the mixed coefficient would be B=1, and the discriminant formula must change with it (see S04)."
          }
        },
        "leaf_basis": "mko_pde + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "S01"
        ],
        "before_tex": "D=b^2-4ac",
        "after_tex": "D=2^2-4\\cdot 1\\cdot(-3)=16",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入 a=1、b=2、c=−3：b²=4；4ac=4·1·(−3)=−12（負數乘正數為負）；所以 D=4−(−12)=4+12=16。減去一個負數等於加上它的相反數。",
          "en": "Substitute a=1, b=2, c=−3: b²=4; 4ac=4·1·(−3)=−12 (a negative times a positive is negative); hence D=4−(−12)=4+12=16. Subtracting a negative number adds its opposite."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S01",
            "numeric_evaluation",
            "neg_mul"
          ],
          "explanation": {
            "zh-Hant": "反向重算：16−4=12=−4ac，與 4ac=−12 一致；兩邊是同一個數，等式雙向成立。",
            "en": "Recomputing backwards: 16−4=12=−4ac, consistent with 4ac=−12; both sides are the same number, so the equality holds both ways."
          }
        },
        "leaf_basis": "numeric_evaluation + neg_mul"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02",
          "pde_class"
        ],
        "before_tex": "D=16",
        "after_tex": "L\\ \\text{is hyperbolic}",
        "relation": "implies",
        "rule": "pde_class",
        "conditions": [
          "(a,b,c)\\neq(0,0,0)",
          "D>0"
        ],
        "explanation": {
          "zh-Hant": "分類定義要求主部係數不全為 0：a=1≠0，成立。D=16>0，所以依定義 L 是雙曲型。低階項 u_x 與右端 g 不影響分類。",
          "en": "The classification definition requires the principal coefficients not to be all 0: a=1≠0, so this holds. D=16>0, so by definition L is hyperbolic. The lower-order term u_x and the right-hand side g do not affect the class."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S02",
            "pde_class"
          ],
          "explanation": {
            "zh-Hant": "需要主部非零與 D 的符號。分類只是主部的性質；本步不宣稱 L(u)=g 在任何邊界或初始條件下有解或解唯一。",
            "en": "It needs a nonzero principal part and the sign of D. The class is a property of the principal part only; this step does not claim that L(u)=g has a solution, or a unique one, under any boundary or initial conditions."
          }
        },
        "leaf_basis": "pde_class + order_basics"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "S02",
          "L_conv"
        ],
        "before_tex": "A u_{xx}+2B u_{xy}+C u_{yy}=u_{xx}+2u_{xy}-3u_{yy}",
        "after_tex": "A=1,\\ B=1,\\ C=-3,\\ B^2-AC=4=\\tfrac{16}{4}",
        "relation": "implies",
        "rule": "L_conv",
        "conditions": [
          "b=2B"
        ],
        "explanation": {
          "zh-Hant": "在另一種約定下，混合項係數寫成 2B，所以 2B=2，B=1；A=1、C=−3。此時 B²−AC=1−(1)(−3)=4。依引理 L_conv，b²−4ac=4(B²−AC)，即 16=4·4，兩約定給出同號、同類型。前提是每個公式配它自己的係數；若把 b=2 放進 B²−AC 得 4+3=7，或把 B=1 放進 b²−4ac 得 1+12=13，兩者都不是任何一種約定下的判別式。",
          "en": "Under the other convention the mixed coefficient is written 2B, so 2B=2 and B=1; A=1 and C=−3. Then B²−AC=1−(1)(−3)=4. By Lemma L_conv, b²−4ac=4(B²−AC), that is 16=4·4, so the two conventions give the same sign and the same class. This requires each formula to be used with its own coefficients: putting b=2 into B²−AC gives 4+3=7, and putting B=1 into b²−4ac gives 1+12=13; neither is the discriminant under either convention."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S01",
            "S02",
            "L_conv"
          ],
          "explanation": {
            "zh-Hant": "結論依賴 b=2B 這個一致的讀法。只宣稱兩個正確公式同號，不宣稱任何恰好同號的混用公式都可接受；錯誤控制給出混用導致符號改變的例子。",
            "en": "The conclusion depends on the consistent reading b=2B. Only the agreement in sign of the two correct formulas is claimed, not that any mixed formula that happens to have the same sign is acceptable; an error control gives an example where mixing changes the sign."
          }
        },
        "leaf_basis": "L_conv + numeric_evaluation"
      },
      {
        "id": "S04.1",
        "uses": [
          "power_laws",
          "distributive"
        ],
        "before_tex": "(2B)^2-4AC",
        "after_tex": "4(B^2-AC)",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "由指數律 (2B)²=2²B²=4B²；再由分配律 4B²−4AC=4(B²−AC)。以 a=A、b=2B、c=C 代入，得 b²−4ac=4(B²−AC)。因 4>0，乘以 4 不改變正、零、負，所以兩個判別式同號。這證明了引理 L_conv（N3 是同一恆等式在網格上的多項式核對）。",
          "en": "By the exponent laws (2B)²=2²B²=4B²; by distributivity 4B²−4AC=4(B²−AC). Substituting a=A, b=2B, c=C gives b²−4ac=4(B²−AC). Since 4>0, multiplying by 4 preserves whether a value is positive, zero or negative, so the two discriminants have the same sign. This proves Lemma L_conv (N3 checks the same polynomial identity on a grid)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "power_laws",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "反向展開 4(B²−AC)=4B²−4AC=(2B)²−4AC；兩式是同一個多項式，對所有 A、B、C 成立，沒有附加條件。",
            "en": "Expanding backwards, 4(B²−AC)=4B²−4AC=(2B)²−4AC; both are the same polynomial, valid for all A, B, C with no side condition."
          }
        },
        "leaf_basis": "power_laws + distributive + mul_comm + order_basics"
      },
      {
        "id": "S05",
        "uses": [
          "G3",
          "poly_partial"
        ],
        "before_tex": "u_A=x^2+xy+y^2",
        "after_tex": "(u_A)_x=2x+y,\\ (u_A)_y=x+2y",
        "relation": "implies",
        "rule": "poly_partial",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐項求偏導。對 x：∂x(x²)=2x，∂x(xy)=y（y 視為常數），∂x(y²)=0，所以 (u_A)_x=2x+y。對 y：∂y(x²)=0，∂y(xy)=x，∂y(y²)=2y，所以 (u_A)_y=x+2y。",
          "en": "Differentiate term by term. In x: ∂x(x²)=2x, ∂x(xy)=y (y treated as a constant), ∂x(y²)=0, so (u_A)_x=2x+y. In y: ∂y(x²)=0, ∂y(xy)=x, ∂y(y²)=2y, so (u_A)_y=x+2y."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "poly_partial"
          ],
          "explanation": {
            "zh-Hant": "只用多項式的逐項求導規則。不反向宣稱：由偏導數回推原函數只能定到另一變數的函數，所以此步不是等價。",
            "en": "Only the termwise differentiation rule for polynomials is used. No converse is claimed: recovering a function from one partial derivative fixes it only up to a function of the other variable, so the step is not an equivalence."
          }
        },
        "leaf_basis": "poly_partial"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "poly_partial"
        ],
        "before_tex": "(u_A)_x=2x+y,\\ (u_A)_y=x+2y",
        "after_tex": "(u_A)_{xx}=2,\\ (u_A)_{xy}=1,\\ (u_A)_{yy}=2",
        "relation": "implies",
        "rule": "poly_partial",
        "conditions": [],
        "explanation": {
          "zh-Hant": "再求一次：(u_A)_xx=∂x(2x+y)=2；(u_A)_xy=∂y(2x+y)=1（依 G1，先對 x 再對 y）；(u_A)_yy=∂y(x+2y)=2。交叉核對：(u_A)_yx=∂x(x+2y)=1，與 (u_A)_xy 相同，符合多項式混合偏導與次序無關。",
          "en": "Differentiate once more: (u_A)_xx=∂x(2x+y)=2; (u_A)_xy=∂y(2x+y)=1 (by G1, first x and then y); (u_A)_yy=∂y(x+2y)=2. Cross-check: (u_A)_yx=∂x(x+2y)=1, the same as (u_A)_xy, as expected because mixed partials of polynomials do not depend on the order."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "poly_partial"
          ],
          "explanation": {
            "zh-Hant": "需要 S05 的一階偏導與求導規則；二階偏導都是常數，因為 u_A 是二次多項式。不宣稱反向。",
            "en": "It needs the first partials from S05 and the differentiation rule; all second partials are constants because u_A is quadratic. No converse is claimed."
          }
        },
        "leaf_basis": "poly_partial"
      },
      {
        "id": "S07",
        "uses": [
          "G1",
          "S05",
          "S06"
        ],
        "before_tex": "L(u_A)=(u_A)_{xx}+2(u_A)_{xy}-3(u_A)_{yy}+(u_A)_x",
        "after_tex": "L(u_A)=2x+y-2",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 S06 與 S05 的結果代入 G1 的算子：L(u_A)=2+2·1−3·2+(2x+y)。常數部分 2+2−6=−2（S07.1），所以 L(u_A)=2x+y−2。這是符號計算，結果對每一點 (x,y) 都成立，不是在個別點取樣。",
          "en": "Substitute the results of S06 and S05 into the operator of G1: L(u_A)=2+2·1−3·2+(2x+y). The constant part is 2+2−6=−2 (S07.1), so L(u_A)=2x+y−2. This is a symbolic computation, valid at every point (x,y), not a sample at chosen points."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S05",
            "S06",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "等值代換可雙向進行：把 2x+y−2 拆回 2+2·1−3·2+(2x+y)，每一項又等於對應的偏導數。",
            "en": "Replacement of equals works both ways: 2x+y−2 splits back into 2+2·1−3·2+(2x+y), and each term equals the corresponding partial derivative."
          }
        },
        "leaf_basis": "eq_replace + distributive + numeric_evaluation"
      },
      {
        "id": "S07.1",
        "uses": [
          "numeric_evaluation"
        ],
        "before_tex": "2+2\\cdot 1-3\\cdot 2",
        "after_tex": "-2",
        "relation": "equivalent",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "2+2=4，3·2=6，4−6=−2。",
          "en": "2+2=4, 3·2=6, 4−6=−2."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：−2+6=4=2+2·1，兩邊是同一個整數。",
            "en": "Backwards: −2+6=4=2+2·1; both sides are the same integer."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S07",
          "G1"
        ],
        "before_tex": "R_{u_A}=L(u_A)-g",
        "after_tex": "R_{u_A}=(2-2)x+(1-1)y+(-2+2)=0",
        "relation": "equivalent",
        "rule": "distributive",
        "conditions": [],
        "explanation": {
          "zh-Hant": "R_{u_A}=(2x+y−2)−(2x+y−2)。依分配律合併同類項：x 的係數 2−2=0，y 的係數 1−1=0，常數 −2−(−2)=0。三個係數全為 0，所以符號殘差是零多項式；這是 MKO a2 所說能證全域的那種證據。",
          "en": "R_{u_A}=(2x+y−2)−(2x+y−2). Collecting like terms by distributivity: the coefficient of x is 2−2=0, of y is 1−1=0, and the constant is −2−(−2)=0. All three coefficients are 0, so the symbolic residual is the zero polynomial; this is the kind of evidence that, by MKO assumption a2, proves a global statement."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S07",
            "distributive",
            "add_inverse"
          ],
          "explanation": {
            "zh-Hant": "合併同類項是多項式的等值改寫，可以逆向拆開；兩式是同一個多項式。",
            "en": "Collecting like terms rewrites a polynomial into an equal one and can be undone; both expressions are the same polynomial."
          }
        },
        "leaf_basis": "distributive + add_inverse + add_comm"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "poly_coeff_zero",
          "forall_intro"
        ],
        "before_tex": "R_{u_A}\\equiv 0",
        "after_tex": "\\forall (x,y)\\in\\mathbb{R}^2:\\ L(u_A)(x,y)=g(x,y)",
        "relation": "implies",
        "rule": "poly_coeff_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "係數全為 0 的多項式在每一點取值 0。取任意 (x,y)∈ℝ²：R_{u_A}(x,y)=0，即 L(u_A)(x,y)−g(x,y)=0，L(u_A)(x,y)=g(x,y)。推導對 (x,y) 沒有任何額外假設，依 ∀ 引入，u_A 在全 ℝ² 上滿足 L(u)=g。",
          "en": "A polynomial with all coefficients 0 takes the value 0 at every point. Take any (x,y)∈ℝ²: R_{u_A}(x,y)=0, i.e. L(u_A)(x,y)−g(x,y)=0, so L(u_A)(x,y)=g(x,y). The argument puts no extra assumption on (x,y), so by ∀-introduction u_A satisfies L(u)=g on all of ℝ²."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "poly_coeff_zero",
            "forall_intro"
          ],
          "explanation": {
            "zh-Hant": "需要殘差是零多項式（S08），而不是在某些點為零。只宣稱 u_A 是這個方程在 ℝ² 上的一個解；不宣稱它是唯一解或一般解。",
            "en": "It needs the residual to be the zero polynomial (S08), not merely zero at some points. It claims only that u_A is one solution of this equation on ℝ², not that it is the unique or the general solution."
          }
        },
        "leaf_basis": "poly_coeff_zero + forall_intro"
      },
      {
        "id": "S10",
        "uses": [
          "G3",
          "poly_partial"
        ],
        "before_tex": "u_B=x^2+2xy",
        "after_tex": "(u_B)_x=2x+2y,\\ (u_B)_y=2x,\\ (u_B)_{xx}=2,\\ (u_B)_{xy}=2,\\ (u_B)_{yy}=0",
        "relation": "implies",
        "rule": "poly_partial",
        "conditions": [],
        "explanation": {
          "zh-Hant": "一階：∂x(x²)=2x、∂x(2xy)=2y，所以 (u_B)_x=2x+2y；∂y(x²)=0、∂y(2xy)=2x，所以 (u_B)_y=2x。二階：(u_B)_xx=∂x(2x+2y)=2，(u_B)_xy=∂y(2x+2y)=2，(u_B)_yy=∂y(2x)=0。交叉核對 (u_B)_yx=∂x(2x)=2。",
          "en": "First order: ∂x(x²)=2x and ∂x(2xy)=2y, so (u_B)_x=2x+2y; ∂y(x²)=0 and ∂y(2xy)=2x, so (u_B)_y=2x. Second order: (u_B)_xx=∂x(2x+2y)=2, (u_B)_xy=∂y(2x+2y)=2, (u_B)_yy=∂y(2x)=0. Cross-check: (u_B)_yx=∂x(2x)=2."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "poly_partial"
          ],
          "explanation": {
            "zh-Hant": "與 S05、S06 相同，只用多項式求導規則；不宣稱反向。",
            "en": "As in S05 and S06, only the differentiation rule for polynomials is used; no converse is claimed."
          }
        },
        "leaf_basis": "poly_partial"
      },
      {
        "id": "S11",
        "uses": [
          "G1",
          "S10"
        ],
        "before_tex": "R_{u_B}=L(u_B)-g",
        "after_tex": "R_{u_B}=(2+4-0+2x+2y)-(2x+y-2)=y+8",
        "relation": "equivalent",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "代入：L(u_B)=2+2·2−3·0+(2x+2y)=2x+2y+6。減去 g：x 的係數 2−2=0，y 的係數 2−1=1，常數 6−(−2)=8。所以 R_{u_B}=y+8。y 的係數是 1≠0，這不是零多項式。",
          "en": "Substitute: L(u_B)=2+2·2−3·0+(2x+2y)=2x+2y+6. Subtract g: the coefficient of x is 2−2=0, of y is 2−1=1, and the constant is 6−(−2)=8. So R_{u_B}=y+8. The coefficient of y is 1≠0, so this is not the zero polynomial."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "S10",
            "eq_replace",
            "distributive"
          ],
          "explanation": {
            "zh-Hant": "代換與合併同類項都是等值改寫，可逆向還原；兩式是同一個多項式。",
            "en": "Substitution and collecting like terms are rewrites into equal expressions and can be undone; both are the same polynomial."
          }
        },
        "leaf_basis": "eq_replace + distributive + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S11"
        ],
        "before_tex": "R_{u_B}(x,-8)",
        "after_tex": "R_{u_B}(x,-8)=-8+8=0\\ \\text{for every } x\\in\\mathbb{R}",
        "relation": "implies",
        "rule": "eq_replace",
        "conditions": [],
        "explanation": {
          "zh-Hant": "以 y=−8 代入 R_{u_B}=y+8，得 0，與 x 無關。所以 u_B 的點殘差在整條直線 y=−8 的無限多個點上都為零，例如 (0,−8)、(5,−8)。這正是容易被誤讀成「u_B 是解」的情形。",
          "en": "Substituting y=−8 into R_{u_B}=y+8 gives 0, independently of x. So the point residual of u_B vanishes at infinitely many points, the whole line y=−8, for example (0,−8) and (5,−8). This is exactly the situation that is easily misread as 'u_B is a solution'."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "eq_replace"
          ],
          "explanation": {
            "zh-Hant": "只是在直線 y=−8 上求值。結論只關於這些點，不推出 R_{u_B} 是零多項式；S13 顯示它不是。",
            "en": "This only evaluates on the line y=−8. The conclusion concerns these points only and does not give that R_{u_B} is the zero polynomial; S13 shows it is not."
          }
        },
        "leaf_basis": "eq_replace + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S11",
          "quantifier_negation"
        ],
        "before_tex": "R_{u_B}(0,0)",
        "after_tex": "R_{u_B}(0,0)=8\\neq 0,\\ \\text{so}\\ \\neg\\,\\forall (x,y):\\ L(u_B)(x,y)=g(x,y)",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "在 (0,0)：R_{u_B}(0,0)=0+8=8≠0；也就是 L(u_B)(0,0)=6，而 g(0,0)=−2。全稱命題「對所有 (x,y)，L(u_B)=g」的否定等價於「存在一點使其不成立」，(0,0) 就是這樣的點。所以 u_B 不是 L(u)=g 在 ℝ² 上的解，儘管 S12 的點殘差在整條直線上為零。",
          "en": "At (0,0): R_{u_B}(0,0)=0+8=8≠0; that is, L(u_B)(0,0)=6 while g(0,0)=−2. The negation of the universal statement 'for all (x,y), L(u_B)=g' is equivalent to 'there is a point where it fails', and (0,0) is such a point. So u_B is not a solution of L(u)=g on ℝ², even though by S12 its point residual vanishes on a whole line."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S11",
            "quantifier_negation"
          ],
          "explanation": {
            "zh-Hant": "否證只需一個精確的點；需要的是 S11 的殘差公式與量詞否定。這不說 u_B 在任何點都不滿足方程（S12 顯示它在 y=−8 上滿足）。",
            "en": "A refutation needs only one exact point; what is needed is the residual formula from S11 and quantifier negation. This does not say u_B fails the equation at every point (S12 shows it holds on y=−8)."
          }
        },
        "leaf_basis": "quantifier_negation + exists_intro + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "G1",
          "G3",
          "poly_partial",
          "S09"
        ],
        "before_tex": "u_C=x^2+xy+y^2+y",
        "after_tex": "R_{u_C}\\equiv 0,\\quad u_C-u_A=y\\not\\equiv 0",
        "relation": "implies",
        "rule": "poly_partial",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(u_C)_x=2x+y，(u_C)_y=x+2y+1；二階偏導 (u_C)_xx=2、(u_C)_xy=1、(u_C)_yy=2，與 u_A 相同。所以 L(u_C)=2+2−6+(2x+y)=2x+y−2，R_{u_C} 是零多項式，u_C 也在全 ℝ² 上滿足 L(u)=g（理由同 S09）。但 u_C−u_A=y 不是零多項式，所以 u_C≠u_A。方程 L(u)=g 本身（不加任何邊界或初始條件）至少有兩個相異的多項式解：一個候選通過，並不給出一般解或唯一性。",
          "en": "(u_C)_x=2x+y and (u_C)_y=x+2y+1; the second partials (u_C)_xx=2, (u_C)_xy=1, (u_C)_yy=2 are the same as for u_A. So L(u_C)=2+2−6+(2x+y)=2x+y−2, R_{u_C} is the zero polynomial, and u_C also satisfies L(u)=g on all of ℝ² (for the same reason as S09). But u_C−u_A=y is not the zero polynomial, so u_C≠u_A. The equation L(u)=g by itself, with no boundary or initial conditions, has at least two distinct polynomial solutions: one passing candidate gives neither the general solution nor uniqueness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "poly_partial",
            "poly_coeff_zero",
            "S09"
          ],
          "explanation": {
            "zh-Hant": "需要 u_C 的符號殘差為零與 u_C−u_A 不為零多項式。只宣稱沒有附加條件時解不唯一；不宣稱加上邊界或初始條件後的任何情況。",
            "en": "It needs the zero symbolic residual of u_C and that u_C−u_A is not the zero polynomial. It claims only non-uniqueness without side conditions, and nothing about what happens once boundary or initial conditions are added."
          }
        },
        "leaf_basis": "poly_partial + eq_replace + poly_coeff_zero"
      },
      {
        "id": "S15",
        "uses": [
          "S03",
          "S09",
          "S13",
          "S14"
        ],
        "before_tex": "\\text{results of S03, S09, S13, S14}",
        "after_tex": "L\\ \\text{hyperbolic};\\ u_A,u_C\\ \\text{solve } L(u)=g\\ \\text{on}\\ \\mathbb{R}^2;\\ u_B\\ \\text{does not}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把各部分結論合在一起：分類（S03）、u_A 與 u_C 的全域符號證書（S09、S14）、u_B 的單點否證（S13）。每一項結論各有自己的證據，彼此不互相替代：分類不說解，點殘差不說全域，候選通過不說唯一。",
          "en": "Combine the separate conclusions: the classification (S03), the global symbolic certificates for u_A and u_C (S09, S14), and the one-point refutation for u_B (S13). Each conclusion has its own evidence and none substitutes for another: the class says nothing about solutions, point residuals say nothing global, and a passing candidate says nothing about uniqueness."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "S09",
            "S13",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "只是合取已證的結論。不宣稱一般解、存在唯一性或適定性，這些不在本切片內（MKO a2、para2）。",
            "en": "This only conjoins proved conclusions. It does not claim a general solution, existence and uniqueness, or well-posedness, which lie outside this slice (MKO a2, para2)."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "在約定 a u_xx+b u_xy+c u_yy、D=b²−4ac 下，L 的係數為 a=1、b=2、c=−3，D=16>0，L 是雙曲型（改用 A u_xx+2B u_xy+C u_yy 時 B=1、B²−AC=4，同號）。u_A=x²+xy+y² 與 u_C=x²+xy+y²+y 的符號殘差都是零多項式，所以都在全 ℝ² 上滿足 L(u)=g；u_B=x²+2xy 的殘差是 y+8，雖然在整條直線 y=−8 上為零，但在 (0,0) 等於 8，所以 u_B 不是解。",
      "en": "Under the convention a u_xx+b u_xy+c u_yy with D=b²−4ac, L has a=1, b=2, c=−3 and D=16>0, so L is hyperbolic (with A u_xx+2B u_xy+C u_yy one gets B=1 and B²−AC=4, the same sign). The symbolic residuals of u_A=x²+xy+y² and u_C=x²+xy+y²+y are the zero polynomial, so both satisfy L(u)=g on all of ℝ²; the residual of u_B=x²+2xy is y+8, which vanishes on the whole line y=−8 but equals 8 at (0,0), so u_B is not a solution."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "u_B 在 (0,−8) 與 (5,−8) 的殘差都是 0，所以 u_B 在 ℝ² 上滿足 L(u)=g",
          "en": "The residual of u_B is 0 at (0,−8) and at (5,−8), so u_B satisfies L(u)=g on ℝ²"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "R_{u_B}=y+8 不是零多項式；R_{u_B}(0,0)=8≠0。",
          "en": "R_{u_B}=y+8 is not the zero polynomial; R_{u_B}(0,0)=8≠0."
        }
      },
      {
        "claim": {
          "zh-Hant": "u_xx+2u_xy+2u_yy 的混合項係數是 2，代入 B²−AC 得 2>0，所以它是雙曲型",
          "en": "The mixed coefficient of u_xx+2u_xy+2u_yy is 2; plugging it into B²−AC gives 2>0, so it is hyperbolic"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "在 2B 約定下 B=1，B²−AC=−1；在 b 約定下 D=4−8=−4；兩者都是橢圓型。",
          "en": "Under the 2B convention B=1 and B²−AC=−1; under the b convention D=4−8=−4; both say elliptic."
        }
      },
      {
        "claim": {
          "zh-Hant": "u_A 滿足 L(u)=g，所以 u_A 是此方程的唯一解",
          "en": "u_A satisfies L(u)=g, so u_A is the unique solution of this equation"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "u_C=u_A+y 也滿足，且 u_C≠u_A。",
          "en": "u_C=u_A+y also satisfies it, and u_C≠u_A."
        }
      },
      {
        "claim": {
          "zh-Hant": "L 是雙曲型，所以 L(u)=g 對任何邊界資料都有唯一解",
          "en": "L is hyperbolic, so L(u)=g has a unique solution for any boundary data"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "分類只看主部；本稿沒有證明任何存在唯一或適定性結果，MKO a2 也排除這類宣稱。",
          "en": "The class looks only at the principal part; nothing here proves any existence, uniqueness or well-posedness result, and MKO assumption a2 rules such claims out."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "distributive",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分配律：u(v+w)=uv+uw；配合乘法交換律也有 (v+w)u=vu+wu。",
          "en": "Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu."
        }
      },
      {
        "id": "eq_replace",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。",
          "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true."
        }
      },
      {
        "id": "quantifier_negation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "量詞否定：¬(∀x, P(x)) 等價於 ∃x, ¬P(x)；¬(∃x, P(x)) 等價於 ∀x, ¬P(x)。",
          "en": "Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x)."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "neg_mul",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "負號與乘法：(−u)v=−(uv)，(−u)(−v)=uv。",
          "en": "Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "power_laws",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整數指數律（底數非零或指數為正整數）：a^m·a^n=a^{m+n}，(a^m)^n=a^{mn}，(ab)^n=a^n b^n，a^0=1。",
          "en": "Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1."
        }
      },
      {
        "id": "mul_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "乘法交換律：uv=vu（在交換的數系中，例如 ℤ、ℚ、ℝ、ℂ）。",
          "en": "Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ)."
        }
      },
      {
        "id": "add_inverse",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。",
          "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)."
        }
      },
      {
        "id": "add_comm",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "加法交換律：u+v=v+u。",
          "en": "Commutativity of addition: u+v=v+u."
        }
      },
      {
        "id": "forall_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。",
          "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain."
        }
      },
      {
        "id": "exists_intro",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "∃引入：若找到 a∈D 使 P(a) 成立，則 ∃x∈D, P(x)。",
          "en": "∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x)."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_pde",
        "kind": "cited",
        "source": {
          "mko_id": "mko-partial-differential-equation",
          "sha256": "b78ef1b1a613523b784d17e0507361cf263e495850e2b7aa880558f1294758b8",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "本輪固定二變數二次多項式候選 u 與線性常係數算子，精確計算一階或二階殘差。二階主部採 a u_xx+b u_xy+c u_yy，判別式為 b²−4ac；這個約定必須明示。點殘差為零只證該點；只有符號殘差多項式全零，才證指定多項式在全域滿足該算子方程。",
          "en": "This slice fixes bivariate quadratic polynomial candidates u and linear constant-coefficient operators and computes first- or second-order residuals exactly. The second-order principal part is a u_xx+b u_xy+c u_yy, with discriminant b²−4ac; this convention must be explicit. A zero point residual proves only that point check; only an identically zero symbolic residual polynomial proves that the specified polynomial satisfies the operator equation globally."
        }
      },
      {
        "id": "poly_partial",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "多項式的偏導數：二變數多項式有各階偏導數；∂/∂x 逐項且線性地作用，把 y 視為常數，∂/∂x(x^m y^k)=m x^(m−1) y^k（m≥1），m=0 時為 0；∂/∂y 對稱地作用。多項式的混合偏導與求導次序無關：u_xy=u_yx。",
          "en": "Partial derivatives of polynomials: a bivariate polynomial has partial derivatives of every order; ∂/∂x acts termwise and linearly, treating y as a constant, with ∂/∂x(x^m y^k)=m x^(m−1) y^k for m≥1 and 0 for m=0; ∂/∂y acts symmetrically. For polynomials the mixed partials do not depend on the order: u_xy=u_yx."
        }
      },
      {
        "id": "pde_class",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "常係數二階線性算子的分類（標準定義）：若主部 a u_xx+b u_xy+c u_yy 的係數不全為 0，則 b²−4ac>0 稱雙曲型，b²−4ac=0 稱拋物型，b²−4ac<0 稱橢圓型。分類只看主部，與低階項及右端無關。",
          "en": "Classification of constant-coefficient second-order linear operators (standard definition): if the coefficients of the principal part a u_xx+b u_xy+c u_yy are not all 0, the operator is called hyperbolic when b²−4ac>0, parabolic when b²−4ac=0 and elliptic when b²−4ac<0. The class depends only on the principal part, not on lower-order terms or the right-hand side."
        }
      },
      {
        "id": "poly_coeff_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "係數全為 0 的多項式在 ℝ² 的每一點取值都是 0；因此兩個係數逐一相同的多項式在每一點取值相同。",
          "en": "A polynomial whose coefficients are all 0 takes the value 0 at every point of ℝ²; hence two polynomials with identical coefficients take equal values at every point."
        }
      },
      {
        "id": "L_conv",
        "kind": "lemma",
        "proved_in": [
          "S04.1"
        ],
        "statement": {
          "zh-Hant": "若同一個主部也寫成 A u_xx+2B u_xy+C u_yy（即 a=A、b=2B、c=C），則 b²−4ac=4(B²−AC)；因 4>0，兩個判別式同號。",
          "en": "If the same principal part is also written A u_xx+2B u_xy+C u_yy (that is, a=A, b=2B, c=C), then b²−4ac=4(B²−AC); since 4>0, the two discriminants have the same sign."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S02",
        "lhs": "2**2-4*1*(-3)",
        "rhs": "16"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "1**2-1*(-3)",
        "rhs": "4"
      },
      {
        "id": "N3",
        "step": "S04.1",
        "vars": [
          "A",
          "B",
          "C"
        ],
        "degree": 2,
        "lhs": "(2*B)**2-4*A*C",
        "rhs": "4*(B**2-A*C)"
      },
      {
        "id": "N4",
        "step": "S07",
        "vars": [
          "x",
          "y"
        ],
        "degree": 1,
        "lhs": "2+2*1-3*2+(2*x+y)",
        "rhs": "2*x+y-2"
      },
      {
        "id": "N5",
        "step": "S07.1",
        "lhs": "2+2*1-3*2",
        "rhs": "-2"
      },
      {
        "id": "N6",
        "step": "S08",
        "vars": [
          "x",
          "y"
        ],
        "degree": 1,
        "lhs": "(2*x+y-2)-(2*x+y-2)",
        "rhs": "0"
      },
      {
        "id": "N7",
        "step": "S11",
        "vars": [
          "x",
          "y"
        ],
        "degree": 1,
        "lhs": "(2+2*2-3*0+(2*x+2*y))-(2*x+y-2)",
        "rhs": "y+8"
      },
      {
        "id": "N8",
        "step": "S12",
        "lhs": "-8+8",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S13",
        "lhs": "0+8",
        "relation": "!=",
        "rhs": "0"
      },
      {
        "id": "N10",
        "step": "S13",
        "lhs": "(2*0+2*0+6)-(2*0+0-2)",
        "rhs": "8"
      },
      {
        "id": "N11",
        "step": "S14",
        "vars": [
          "x",
          "y"
        ],
        "degree": 1,
        "lhs": "(2+2*1-3*2+(2*x+y))-(2*x+y-2)",
        "rhs": "0"
      },
      {
        "id": "N12",
        "lhs": "2**2-4*1*2",
        "rhs": "-4"
      },
      {
        "id": "N13",
        "lhs": "1**2-1*2",
        "rhs": "-1"
      },
      {
        "id": "N14",
        "lhs": "2**2-1*2",
        "relation": ">",
        "rhs": "0"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "停在 MKO 的切片內：二變數二次多項式候選、線性常係數算子、多項式的符號偏導數、點殘差、主部分類與多項式恆等式。主部約定固定為 a u_xx+b u_xy+c u_yy（混合項係數記為 b，而非 2b）。不討論一般解、邊界或初始條件、存在唯一性、弱解或適定性；不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO's slice: bivariate quadratic polynomial candidates, a linear constant-coefficient operator, symbolic partial derivatives of polynomials, point residuals, principal-part classification and polynomial identities. The principal-part convention is fixed as a u_xx+b u_xy+c u_yy (the mixed coefficient is called b, not 2b). General solutions, boundary or initial conditions, existence and uniqueness, weak solutions and well-posedness are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "u_x,\\ u_y",
        "u 對 x、對 y 的一階偏導數",
        "first partial derivatives of u in x and in y"
      ],
      [
        "u_{xx},\\ u_{xy},\\ u_{yy}",
        "二階偏導數；u_{xy} 表示先對 x 再對 y 微分",
        "second partial derivatives; u_{xy} means differentiate in x, then in y"
      ],
      [
        "L",
        "本稿的線性常係數算子",
        "the linear constant-coefficient operator of this guide"
      ],
      [
        "a,\\ b,\\ c",
        "主部 a u_{xx}+b u_{xy}+c u_{yy} 的係數",
        "coefficients of the principal part a u_{xx}+b u_{xy}+c u_{yy}"
      ],
      [
        "D=b^2-4ac",
        "本約定下的判別式",
        "the discriminant under this convention"
      ],
      [
        "A,\\ B,\\ C",
        "另一約定 A u_{xx}+2B u_{xy}+C u_{yy} 的係數",
        "coefficients of the other convention A u_{xx}+2B u_{xy}+C u_{yy}"
      ],
      [
        "R_u=L(u)-g",
        "候選 u 的符號殘差多項式",
        "the symbolic residual polynomial of candidate u"
      ]
    ],
    "md_error_controls": [
      {
        "title": "點殘差為零不等於解 / A zero point residual is not a solution",
        "zh": "MKO 指出的常見誤解是由取樣點殘差為零宣稱全域解。u_B 的殘差在整條直線 y=−8 上為零，甚至是無限多個點，但殘差多項式 y+8 不是零多項式，(0,0) 處等於 8。只有符號殘差的每個係數都為 0（S08），才證明在全 ℝ² 上成立。",
        "en": "The common mistake named by the MKO is to claim a global solution because sampled point residuals are zero. The residual of u_B is zero on the whole line y=−8, even at infinitely many points, yet the residual polynomial y+8 is not the zero polynomial and equals 8 at (0,0). Only when every coefficient of the symbolic residual is 0 (S08) is the equation proved on all of ℝ²."
      },
      {
        "title": "混用 b 與 2B 的判別式 / Mixing the b and 2B discriminants",
        "zh": "MKO 的另一個誤解是混用 2b u_xy 與 b u_xy 的判別式。例：算子 u_xx+2u_xy+2u_yy。依 G2，a=1、b=2、c=2，D=4−8=−4<0，橢圓型；依 A u_xx+2B u_xy+C u_yy，B=1，B²−AC=1−2=−1<0，同樣橢圓型。若把混合項係數 2 直接當成 B 代入 B²−AC，得 4−2=2>0，會錯判為雙曲型。所以必須先寫明約定，再用該約定自己的公式。",
        "en": "Another misconception identified in the MKO is mixing the discriminants for 2b u_xy and b u_xy. Example: the operator u_xx+2u_xy+2u_yy. By G2, a=1, b=2, c=2 and D=4−8=−4<0, elliptic; with A u_xx+2B u_xy+C u_yy, B=1 and B²−AC=1−2=−1<0, also elliptic. If the mixed coefficient 2 is plugged directly into B²−AC as B, one gets 4−2=2>0 and wrongly calls it hyperbolic. So the convention must be stated first and then used with its own formula."
      },
      {
        "title": "分類與候選通過都不給唯一性 / Neither the class nor a passing candidate gives uniqueness",
        "zh": "S03 的雙曲型只描述主部；S09 只說 u_A 是一個解。S14 找到第二個解 u_C=u_A+y，說明沒有邊界或初始條件時解不唯一。存在唯一性與適定性需要另外的定理與條件，本稿不處理。",
        "en": "The hyperbolic class of S03 describes only the principal part; S09 says only that u_A is one solution. S14 finds a second solution u_C=u_A+y, showing that without boundary or initial conditions the solution is not unique. Existence, uniqueness and well-posedness need further theorems and conditions, which this guide does not treat."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "poly_partial、pde_class 與 poly_coeff_zero 是本稿的前提（標準定義與多項式事實），不在本稿中從更基本的定義推出。",
        "en": "poly_partial, pde_class and poly_coeff_zero are premises of this guide (standard definitions and polynomial facts) and are not derived here from more basic definitions."
      },
      {
        "zh-Hant": "本稿不處理一般解、邊界或初始條件、存在唯一性與適定性（MKO 的 formal next）。",
        "en": "This guide does not treat general solutions, boundary or initial conditions, existence and uniqueness, or well-posedness (the MKO's formal next)."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "d7f1dca1ded3622b8bc0d17369236ab7e35c0df919f547af5b7b66b3c462a9b0",
    "original_source_sha256": "d7f1dca1ded3622b8bc0d17369236ab7e35c0df919f547af5b7b66b3c462a9b0",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}