{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-boundary-value-problem-01",
  "mko_id": "mko-boundary-value-problem",
  "title": {
    "zh-Hant": "邊界值問題：矩形上 Poisson–Dirichlet 問題的內部與四邊分開核對",
    "en": "Boundary-value problem: checking the interior and all four edges separately for a Poisson–Dirichlet problem on a rectangle"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-boundary-value-problem-01",
    "source": {
      "mko_id": "mko-boundary-value-problem",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "4ffaeecac13219cb387fd43a669823ff3c2b402bfadf2e3d04f3938d83bf3f0d",
      "english_sha256": "a666e5292df9be234d0c6cad5ac693748b1c2a78e2b39dd8dc732b0eafe2e60b"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "numeric_evaluation",
      "distributive",
      "eq_replace",
      "poly_partial",
      "set_operations_def",
      "quantifier_negation"
    ],
    "goal": {
      "zh-Hant": "在矩形 Ω=(0,2)×(0,1) 上宣告 BVP：內部 Δu=u_xx+u_yy=4；Dirichlet 資料為左邊 u(0,y)=y²、右邊 u(2,y)=y²+2y+4（0≤y≤1）、下邊 u(x,0)=x²、上邊 u(x,1)=x²+x+1（0≤x≤2）。對二次多項式候選 u=x²+xy+y² 與 v=2x²+xy−2x+y，分別記錄內部 Poisson 義務（符號 Laplacian 恆等式）與四條邊的 Dirichlet 義務（精確限制到整條邊後逐係數比較），判定誰滿足此宣告的 BVP，並說明 v 為何在內部方程成立、四個角點也都吻合的情況下仍不滿足。",
      "en": "On the rectangle Ω=(0,2)×(0,1) declare the BVP: in the interior Δu=u_xx+u_yy=4; Dirichlet data u(0,y)=y² on the left, u(2,y)=y²+2y+4 on the right (0≤y≤1), u(x,0)=x² on the bottom and u(x,1)=x²+x+1 on the top (0≤x≤2). For the quadratic polynomial candidates u=x²+xy+y² and v=2x²+xy−2x+y, record separately the interior Poisson obligation (a symbolic Laplacian identity) and the Dirichlet obligation on all four edges (exact restriction to each whole edge, then coefficient comparison); decide which candidate satisfies this declared BVP, and explain why v fails although its interior equation holds and it matches the data at all four corners."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "Ω=(0,2)×(0,1)，其閉包為 [0,2]×[0,1]；邊界恰由四條閉邊組成：Γ_L={0}×[0,1]、Γ_R={2}×[0,1]、Γ_B=[0,2]×{0}、Γ_T=[0,2]×{1}。",
          "en": "Ω=(0,2)×(0,1), with closure [0,2]×[0,1]; the boundary consists exactly of four closed edges: Γ_L={0}×[0,1], Γ_R={2}×[0,1], Γ_B=[0,2]×{0}, Γ_T=[0,2]×{1}."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "宣告的 BVP：在 Ω 內 Δu=u_xx+u_yy=4；Dirichlet 資料 g_L(y)=y²、g_R(y)=y²+2y+4（y∈[0,1]），g_B(x)=x²、g_T(x)=x²+x+1（x∈[0,2]），要求 u 在每條邊上等於對應資料。",
          "en": "The declared BVP: Δu=u_xx+u_yy=4 in Ω; Dirichlet data g_L(y)=y² and g_R(y)=y²+2y+4 (y∈[0,1]), g_B(x)=x² and g_T(x)=x²+x+1 (x∈[0,2]), requiring u to equal the corresponding data on each edge."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "兩個二次多項式候選：u=x²+xy+y²，v=2x²+xy−2x+y。",
          "en": "Two quadratic polynomial candidates: u=x²+xy+y² and v=2x²+xy−2x+y."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2",
          "mko_bvp"
        ],
        "before_tex": "\\text{declared BVP on } \\Omega=(0,2)\\times(0,1)",
        "after_tex": "(I):\\ \\Delta u-4\\equiv 0\\ \\text{in}\\ \\Omega;\\quad (B):\\ u|_{\\Gamma_L}=y^2,\\ u|_{\\Gamma_R}=y^2+2y+4,\\ u|_{\\Gamma_B}=x^2,\\ u|_{\\Gamma_T}=x^2+x+1",
        "relation": "definition",
        "rule": "mko_bvp",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 MKO，「候選滿足宣告的 BVP」由兩組不同的義務組成：(I) 內部 Poisson 方程 Δu=4 在 Ω 的每一點成立；(B) u 在四條邊的每一點等於 Dirichlet 資料。先把兩組義務分開寫下，之後各自取得證據，最後才合併。",
          "en": "By the MKO, 'the candidate satisfies the declared BVP' consists of two distinct groups of obligations: (I) the interior Poisson equation Δu=4 holds at every point of Ω; (B) u equals the Dirichlet data at every point of the four edges. The two groups are written down separately first; each gets its own evidence, and only then are they combined."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_bvp"
          ],
          "explanation": {
            "zh-Hant": "此步只整理宣告，不推出新事實。四條邊必須完整列出（MKO a1），否則 (B) 會漏掉一部分邊界。",
            "en": "This step only organizes the declaration and derives no new fact. All four edges must be listed (MKO assumption a1); otherwise (B) would miss part of the boundary."
          }
        },
        "leaf_basis": "mko_bvp + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G3",
          "poly_partial"
        ],
        "before_tex": "u=x^2+xy+y^2",
        "after_tex": "u_x=2x+y,\\ u_y=x+2y",
        "relation": "implies",
        "rule": "poly_partial",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐項求偏導：∂x(x²)=2x、∂x(xy)=y、∂x(y²)=0，得 u_x=2x+y；∂y(x²)=0、∂y(xy)=x、∂y(y²)=2y，得 u_y=x+2y。",
          "en": "Differentiate term by term: ∂x(x²)=2x, ∂x(xy)=y, ∂x(y²)=0, giving u_x=2x+y; ∂y(x²)=0, ∂y(xy)=x, ∂y(y²)=2y, giving u_y=x+2y."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "poly_partial"
          ],
          "explanation": {
            "zh-Hant": "只用多項式求導規則；不宣稱可由偏導唯一回推 u。",
            "en": "Only the differentiation rule for polynomials is used; it is not claimed that u can be recovered uniquely from its partials."
          }
        },
        "leaf_basis": "poly_partial"
      },
      {
        "id": "S03",
        "uses": [
          "S02",
          "poly_partial"
        ],
        "before_tex": "u_x=2x+y,\\ u_y=x+2y",
        "after_tex": "u_{xx}=2,\\ u_{yy}=2",
        "relation": "implies",
        "rule": "poly_partial",
        "conditions": [],
        "explanation": {
          "zh-Hant": "u_xx=∂x(2x+y)=2，u_yy=∂y(x+2y)=2。Laplacian 只需要這兩個純二階偏導。",
          "en": "u_xx=∂x(2x+y)=2 and u_yy=∂y(x+2y)=2. The Laplacian needs only these two pure second partials."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S02",
            "poly_partial"
          ],
          "explanation": {
            "zh-Hant": "需要 S02 的一階偏導；不宣稱反向。",
            "en": "It needs the first partials of S02; no converse is claimed."
          }
        },
        "leaf_basis": "poly_partial"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "S01",
          "poly_coeff_zero"
        ],
        "before_tex": "\\Delta u-4=u_{xx}+u_{yy}-4",
        "after_tex": "\\Delta u-4=2+2-4=0\\ \\text{(zero polynomial)};\\ (I)\\ \\text{holds}",
        "relation": "implies",
        "rule": "poly_coeff_zero",
        "conditions": [],
        "explanation": {
          "zh-Hant": "Δu−4=2+2−4=0。內部殘差是零多項式（常數 0），所以在 Ω 的每一點、事實上在 ℝ² 的每一點都等於 0。這是符號恆等，不是在內部取幾個點核對。記錄：u 滿足義務 (I)。",
          "en": "Δu−4=2+2−4=0. The interior residual is the zero polynomial (the constant 0), so it equals 0 at every point of Ω, indeed of ℝ². This is a symbolic identity, not a check at a few interior points. Record: u satisfies obligation (I)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S03",
            "poly_coeff_zero"
          ],
          "explanation": {
            "zh-Hant": "需要兩個二階偏導與零多項式處處為零。本步只處理 (I)，對邊界沒有任何結論。",
            "en": "It needs the two second partials and that the zero polynomial vanishes everywhere. This step settles only (I) and says nothing about the boundary."
          }
        },
        "leaf_basis": "poly_coeff_zero + numeric_evaluation + forall_intro"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "S01",
          "set_operations_def"
        ],
        "before_tex": "u=g\\ \\text{on}\\ \\partial\\Omega",
        "after_tex": "u=g_L\\ \\text{on}\\ \\Gamma_L\\ \\land\\ u=g_R\\ \\text{on}\\ \\Gamma_R\\ \\land\\ u=g_B\\ \\text{on}\\ \\Gamma_B\\ \\land\\ u=g_T\\ \\text{on}\\ \\Gamma_T",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "∂Ω=Γ_L∪Γ_R∪Γ_B∪Γ_T（G1）。一個點在聯集中，當且僅當它在某一條邊上；所以「在 ∂Ω 的每一點 u 等於資料」等價於「在四條邊的每一條的每一點都相等」。四個合取項缺一不可。",
          "en": "∂Ω=Γ_L∪Γ_R∪Γ_B∪Γ_T (G1). A point lies in the union exactly when it lies on some edge, so 'u equals the data at every point of ∂Ω' is equivalent to 'u equals the data at every point of each of the four edges'. None of the four conjuncts may be dropped."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "G1"
          ],
          "explanation": {
            "zh-Hant": "聯集的定義是雙向的；四條邊的條件合起來恰好是整個邊界的條件。",
            "en": "The definition of union is two-way; the four edge conditions together are exactly the condition on the whole boundary."
          }
        },
        "leaf_basis": "set_operations_def + and_intro_elim"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "G2",
          "G3",
          "edge_restrict",
          "poly_coeff_zero"
        ],
        "before_tex": "u|_{\\Gamma_L}=u(0,y)",
        "after_tex": "u(0,y)=y^2=g_L(y)\\ \\ (\\text{coefficients } (1,0,0)=(1,0,0))",
        "relation": "implies",
        "rule": "edge_restrict",
        "conditions": [
          "0\\le y\\le 1"
        ],
        "explanation": {
          "zh-Hant": "左邊 x=0：u(0,y)=0²+0·y+y²=y²。按 (y², y, 1) 的係數，限制結果為 (1,0,0)，資料 g_L 也是 (1,0,0)，逐係數相同，所以在整條左邊（每個 y∈[0,1]）u=g_L。這是全邊恆等，不是抽查幾個 y。",
          "en": "Left edge x=0: u(0,y)=0²+0·y+y²=y². In terms of the coefficients of (y², y, 1), the restriction is (1,0,0) and the data g_L are also (1,0,0); the coefficients agree, so u=g_L on the whole left edge (every y∈[0,1]). This is an identity on the whole edge, not a check at a few values of y."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "edge_restrict",
            "poly_coeff_zero",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "需要精確代入 x=0 與係數相同推出處處相等。只宣稱左邊這一條義務。",
            "en": "It needs the exact substitution x=0 and that equal coefficients give equality everywhere. Only the left-edge obligation is claimed."
          }
        },
        "leaf_basis": "edge_restrict + poly_coeff_zero + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S05",
          "G2",
          "G3",
          "edge_restrict",
          "poly_coeff_zero"
        ],
        "before_tex": "u|_{\\Gamma_R}=u(2,y)",
        "after_tex": "u(2,y)=y^2+2y+4=g_R(y)\\ \\ (\\text{coefficients } (1,2,4)=(1,2,4))",
        "relation": "implies",
        "rule": "edge_restrict",
        "conditions": [
          "0\\le y\\le 1"
        ],
        "explanation": {
          "zh-Hant": "右邊 x=2：u(2,y)=2²+2y+y²=y²+2y+4。係數 (1,2,4) 與 g_R 的 (1,2,4) 相同，所以在整條右邊 u=g_R。",
          "en": "Right edge x=2: u(2,y)=2²+2y+y²=y²+2y+4. The coefficients (1,2,4) equal those of g_R, (1,2,4), so u=g_R on the whole right edge."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "edge_restrict",
            "poly_coeff_zero",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "同 S06，代入 x=2 後逐係數比較；只宣稱右邊義務。",
            "en": "As in S06: substitute x=2 and compare coefficients; only the right-edge obligation is claimed."
          }
        },
        "leaf_basis": "edge_restrict + poly_coeff_zero + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S05",
          "G2",
          "G3",
          "edge_restrict",
          "poly_coeff_zero"
        ],
        "before_tex": "u|_{\\Gamma_B}=u(x,0)",
        "after_tex": "u(x,0)=x^2=g_B(x)\\ \\ (\\text{coefficients } (1,0,0)=(1,0,0))",
        "relation": "implies",
        "rule": "edge_restrict",
        "conditions": [
          "0\\le x\\le 2"
        ],
        "explanation": {
          "zh-Hant": "下邊 y=0：u(x,0)=x²+x·0+0²=x²。按 (x², x, 1) 的係數 (1,0,0) 與 g_B 相同，所以在整條下邊 u=g_B。",
          "en": "Bottom edge y=0: u(x,0)=x²+x·0+0²=x². The coefficients of (x², x, 1), namely (1,0,0), equal those of g_B, so u=g_B on the whole bottom edge."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "edge_restrict",
            "poly_coeff_zero",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "代入 y=0 後逐係數比較；只宣稱下邊義務。",
            "en": "Substitute y=0 and compare coefficients; only the bottom-edge obligation is claimed."
          }
        },
        "leaf_basis": "edge_restrict + poly_coeff_zero + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S05",
          "G2",
          "G3",
          "edge_restrict",
          "poly_coeff_zero"
        ],
        "before_tex": "u|_{\\Gamma_T}=u(x,1)",
        "after_tex": "u(x,1)=x^2+x+1=g_T(x)\\ \\ (\\text{coefficients } (1,1,1)=(1,1,1))",
        "relation": "implies",
        "rule": "edge_restrict",
        "conditions": [
          "0\\le x\\le 2"
        ],
        "explanation": {
          "zh-Hant": "上邊 y=1：u(x,1)=x²+x·1+1²=x²+x+1。係數 (1,1,1) 與 g_T 相同，所以在整條上邊 u=g_T。",
          "en": "Top edge y=1: u(x,1)=x²+x·1+1²=x²+x+1. The coefficients (1,1,1) equal those of g_T, so u=g_T on the whole top edge."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "edge_restrict",
            "poly_coeff_zero",
            "G2"
          ],
          "explanation": {
            "zh-Hant": "代入 y=1 後逐係數比較；只宣稱上邊義務。",
            "en": "Substitute y=1 and compare coefficients; only the top-edge obligation is claimed."
          }
        },
        "leaf_basis": "edge_restrict + poly_coeff_zero + numeric_evaluation"
      },
      {
        "id": "S10",
        "uses": [
          "S05",
          "S06",
          "S07",
          "S08",
          "S09"
        ],
        "before_tex": "u=g_L,\\ u=g_R,\\ u=g_B,\\ u=g_T\\ \\text{on their edges}",
        "after_tex": "(B)\\ \\text{holds for } u",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四條邊的恆等都已證明（S06 至 S09）；依合取引入與 S05 的等價，u 在整個 ∂Ω 上等於 Dirichlet 資料。記錄：u 滿足義務 (B)。",
          "en": "The identities on all four edges are proved (S06 to S09); by conjunction introduction and the equivalence of S05, u equals the Dirichlet data on all of ∂Ω. Record: u satisfies obligation (B)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S06",
            "S07",
            "S08",
            "S09",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "需要四條邊全部；少核任一條且沒有其他涵蓋它的證明時，(B) 就尚未建立；不是據此斷言邊界條件為假。",
            "en": "All four edges are needed; if an edge is unverified and no other proof covers it, (B) is not established; this does not assert that the boundary condition is false."
          }
        },
        "leaf_basis": "and_intro_elim + set_operations_def"
      },
      {
        "id": "S11",
        "uses": [
          "S04",
          "S10",
          "S01"
        ],
        "before_tex": "(I)\\ \\text{for } u,\\ (B)\\ \\text{for } u",
        "after_tex": "u\\ \\text{satisfies the declared BVP}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "兩組義務都已分開記錄（S04 與 S10），依 MKO 此時才可說 u 滿足宣告的 BVP。附帶觀察：每個角點的資料來自兩條邊，而且一致，例如 (2,1) 處 g_R(1)=1+2+4=7=g_T(2)=4+2+1，與它們都是同一個函數的限制相符。",
          "en": "Both groups of obligations have been recorded separately (S04 and S10), so by the MKO one may now say that u satisfies the declared BVP. A side observation: the data at each corner come from two edges and agree, for example at (2,1) g_R(1)=1+2+4=7=g_T(2)=4+2+1, consistent with both being restrictions of one function."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S10",
            "mko_bvp"
          ],
          "explanation": {
            "zh-Hant": "只是合取兩組已證的義務。不宣稱這個 BVP 的解唯一、穩定或適定（MKO a2）。",
            "en": "This only conjoins the two proved obligations. It does not claim that the BVP has a unique, stable or well-posed solution (MKO assumption a2)."
          }
        },
        "leaf_basis": "and_intro_elim + mko_bvp"
      },
      {
        "id": "S12",
        "uses": [
          "G3",
          "poly_partial"
        ],
        "before_tex": "v=2x^2+xy-2x+y",
        "after_tex": "\\Delta v-4=4+0-4=0;\\ (I)\\ \\text{holds for } v",
        "relation": "implies",
        "rule": "poly_partial",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依 S12.1、S12.2，v_xx=4、v_yy=0，所以 Δv−4=0 是零多項式，v 也滿足內部義務 (I)。注意 v=u+h，其中 h=x²−y²−2x+y 是調和的（h_xx+h_yy=2−2=0），所以內部義務不能區分 u 與 v。",
          "en": "By S12.1 and S12.2, v_xx=4 and v_yy=0, so Δv−4=0 is the zero polynomial and v also satisfies the interior obligation (I). Note that v=u+h with h=x²−y²−2x+y harmonic (h_xx+h_yy=2−2=0), so the interior obligation cannot tell u and v apart."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G3",
            "poly_partial",
            "poly_coeff_zero"
          ],
          "explanation": {
            "zh-Hant": "需要 v 的兩個純二階偏導（在子步 S12.1、S12.2 中算出）與零多項式處處為零。本步只處理 v 的 (I)。",
            "en": "It needs the two pure second partials of v (computed in sub-steps S12.1 and S12.2) and that the zero polynomial vanishes everywhere. This step settles only (I) for v."
          }
        },
        "leaf_basis": "poly_partial + poly_coeff_zero"
      },
      {
        "id": "S12.1",
        "uses": [
          "G3",
          "poly_partial"
        ],
        "before_tex": "v=2x^2+xy-2x+y",
        "after_tex": "v_x=4x+y-2,\\ v_y=x+1",
        "relation": "implies",
        "rule": "poly_partial",
        "conditions": [],
        "explanation": {
          "zh-Hant": "∂x(2x²)=4x、∂x(xy)=y、∂x(−2x)=−2、∂x(y)=0，得 v_x=4x+y−2；∂y(2x²)=0、∂y(xy)=x、∂y(−2x)=0、∂y(y)=1，得 v_y=x+1。",
          "en": "∂x(2x²)=4x, ∂x(xy)=y, ∂x(−2x)=−2, ∂x(y)=0, giving v_x=4x+y−2; ∂y(2x²)=0, ∂y(xy)=x, ∂y(−2x)=0, ∂y(y)=1, giving v_y=x+1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "poly_partial"
          ],
          "explanation": {
            "zh-Hant": "只用多項式求導規則。",
            "en": "Only the differentiation rule for polynomials is used."
          }
        },
        "leaf_basis": "poly_partial"
      },
      {
        "id": "S12.2",
        "uses": [
          "S12.1",
          "poly_partial"
        ],
        "before_tex": "v_x=4x+y-2,\\ v_y=x+1",
        "after_tex": "v_{xx}=4,\\ v_{yy}=0,\\ \\Delta v=4",
        "relation": "implies",
        "rule": "poly_partial",
        "conditions": [],
        "explanation": {
          "zh-Hant": "v_xx=∂x(4x+y−2)=4，v_yy=∂y(x+1)=0，所以 Δv=4+0=4。",
          "en": "v_xx=∂x(4x+y−2)=4 and v_yy=∂y(x+1)=0, so Δv=4+0=4."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S12.1",
            "poly_partial"
          ],
          "explanation": {
            "zh-Hant": "需要 S12.1 的一階偏導。",
            "en": "It needs the first partials of S12.1."
          }
        },
        "leaf_basis": "poly_partial + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "G2",
          "G3"
        ],
        "before_tex": "v(0,0),\\ v(2,0),\\ v(0,1),\\ v(2,1)",
        "after_tex": "0,\\ 4,\\ 1,\\ 7\\ =\\ g\\ \\text{at the four corners}",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "四個角點：v(0,0)=0，資料 g_B(0)=0；v(2,0)=8+0−4+0=4，資料 g_B(2)=4；v(0,1)=0+0−0+1=1，資料 g_L(1)=1；v(2,1)=8+2−4+1=7，資料 g_R(1)=7。四個角點全部吻合，因為 h 在四個角點都為 0。只抽查角點會誤以為 v 符合邊界資料。",
          "en": "The four corners: v(0,0)=0 with data g_B(0)=0; v(2,0)=8+0−4+0=4 with data g_B(2)=4; v(0,1)=0+0−0+1=1 with data g_L(1)=1; v(2,1)=8+2−4+1=7 with data g_R(1)=7. All four corners match, because h vanishes at all four corners. Checking only the corners would wrongly suggest that v meets the boundary data."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G2",
            "G3",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只是四個點的精確求值；不推出 v 在任何整條邊上等於資料。",
            "en": "This is exact evaluation at four points only; it does not imply that v equals the data on any whole edge."
          }
        },
        "leaf_basis": "numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "S05",
          "G2",
          "G3",
          "edge_restrict"
        ],
        "before_tex": "v|_{\\Gamma_B}=v(x,0)",
        "after_tex": "v(x,0)=2x^2-2x\\neq x^2;\\ v(1,0)=0\\neq 1=g_B(1)",
        "relation": "implies",
        "rule": "edge_restrict",
        "conditions": [
          "0\\le x\\le 2"
        ],
        "explanation": {
          "zh-Hant": "下邊 y=0：v(x,0)=2x²+0−2x+0=2x²−2x。係數 (2,−2,0) 與 g_B 的 (1,0,0) 不同。取下邊上的點 x=1∈[0,2]：v(1,0)=2−2=0，而 g_B(1)=1，所以 v 在下邊不等於資料。兩者之差 v(x,0)−x²=x(x−2) 只在 x=0 與 x=2（兩個角點）為零。",
          "en": "Bottom edge y=0: v(x,0)=2x²+0−2x+0=2x²−2x. The coefficients (2,−2,0) differ from those of g_B, (1,0,0). Take the point x=1∈[0,2] on the bottom edge: v(1,0)=2−2=0 while g_B(1)=1, so v does not equal the data on the bottom edge. Their difference v(x,0)−x²=x(x−2) vanishes only at x=0 and x=2, the two corners."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "edge_restrict",
            "G2",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "否證只需邊上一個精確的點；x=1 確實在 [0,2] 內。係數不同提示失敗，具體的點證實它。",
            "en": "A refutation needs only one exact point on the edge; x=1 does lie in [0,2]. The differing coefficients signal the failure and the specific point confirms it."
          }
        },
        "leaf_basis": "edge_restrict + numeric_evaluation"
      },
      {
        "id": "S15",
        "uses": [
          "S05",
          "G2",
          "G3",
          "edge_restrict"
        ],
        "before_tex": "v|_{\\Gamma_L}=v(0,y)",
        "after_tex": "v(0,y)=y\\neq y^2;\\ v(0,\\tfrac12)=\\tfrac12\\neq\\tfrac14=g_L(\\tfrac12)",
        "relation": "implies",
        "rule": "edge_restrict",
        "conditions": [
          "0\\le y\\le 1"
        ],
        "explanation": {
          "zh-Hant": "左邊 x=0：v(0,y)=0+0−0+y=y，係數 (0,1,0) 與 g_L 的 (1,0,0) 不同；在 y=1/2∈[0,1]，v=1/2 而 g_L=1/4。左邊同樣失敗。（右邊 v(2,y)=3y+4 與上邊 v(x,1)=2x²−x+1 也與資料不同，本稿不再逐一展開。）",
          "en": "Left edge x=0: v(0,y)=0+0−0+y=y, with coefficients (0,1,0) differing from those of g_L, (1,0,0); at y=1/2∈[0,1], v=1/2 while g_L=1/4. The left edge fails as well. (On the right, v(2,y)=3y+4, and on the top, v(x,1)=2x²−x+1, also differ from the data; this guide does not expand them further.)"
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "edge_restrict",
            "G2",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "一個邊上的點就足以否證該邊義務。括號中的另外兩條邊只是補充，S16 不依賴它們。",
            "en": "One point on the edge suffices to refute that edge's obligation. The two further edges in parentheses are a remark only; S16 does not depend on them."
          }
        },
        "leaf_basis": "edge_restrict + numeric_evaluation"
      },
      {
        "id": "S16",
        "uses": [
          "S05",
          "S12",
          "S13",
          "S14",
          "quantifier_negation"
        ],
        "before_tex": "(I)\\ \\text{for } v,\\ v(1,0)\\neq g_B(1)",
        "after_tex": "(B)\\ \\text{fails for } v;\\ v\\ \\text{does not satisfy the declared BVP}",
        "relation": "implies",
        "rule": "quantifier_negation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "(B) 是「邊界上每一點都相等」的全稱命題；S14 給出一個不相等的邊界點，依量詞否定 (B) 對 v 不成立。BVP 要求 (I) 與 (B) 同時成立，所以 v 不滿足宣告的 BVP，儘管 (I) 成立（S12）而且四個角點都吻合（S13）。",
          "en": "(B) is the universal statement 'equal at every boundary point'; S14 gives one boundary point where they differ, so by quantifier negation (B) fails for v. The BVP requires (I) and (B) together, so v does not satisfy the declared BVP, even though (I) holds (S12) and all four corners match (S13)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S05",
            "S14",
            "quantifier_negation",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "需要一個精確的反例點與「兩組義務都必須成立」。不宣稱 v 在邊界的每一點都不等於資料（角點就相等）。",
            "en": "It needs one exact counterexample point and the requirement that both groups hold. It does not claim that v differs from the data at every boundary point (it agrees at the corners)."
          }
        },
        "leaf_basis": "quantifier_negation + and_intro_elim + exists_intro"
      }
    ],
    "result": {
      "zh-Hant": "u=x²+xy+y² 滿足宣告的 BVP：內部 Δu−4 是零多項式（義務 (I)），且限制到四條邊後逐係數等於 y²、y²+2y+4、x²、x²+x+1（義務 (B)）。v=2x²+xy−2x+y 雖然 Δv=4 且在四個角點都吻合，但 v(x,0)=2x²−2x 在 x=1 處為 0≠1，所以不滿足此 BVP。已核對的 u 是此特定 BVP 的解存在見證；本結果不給一般存在定理，也不證唯一性或適定性。",
      "en": "u=x²+xy+y² satisfies the declared BVP: in the interior Δu−4 is the zero polynomial (obligation (I)), and its restrictions to the four edges equal y², y²+2y+4, x² and x²+x+1 coefficient by coefficient (obligation (B)). v=2x²+xy−2x+y has Δv=4 and matches the data at all four corners, but v(x,0)=2x²−2x equals 0≠1 at x=1, so v does not satisfy this BVP. The checked u witnesses existence for this particular BVP. No general existence theorem, uniqueness or well-posedness result is proved."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "v 滿足 Δv=4，所以 v 解出這個 BVP",
          "en": "v satisfies Δv=4, so v solves this BVP"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "邊界義務失敗：v(1,0)=0≠1=g_B(1)。",
          "en": "The boundary obligation fails: v(1,0)=0≠1=g_B(1)."
        }
      },
      {
        "claim": {
          "zh-Hant": "v 在四個角點都等於邊界資料，所以 v 滿足 Dirichlet 條件",
          "en": "v equals the boundary data at all four corners, so v satisfies the Dirichlet condition"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "下邊在 x=1 不符（0 對 1），左邊在 y=1/2 不符（1/2 對 1/4）。",
          "en": "The bottom edge fails at x=1 (0 vs 1) and the left edge at y=1/2 (1/2 vs 1/4)."
        }
      },
      {
        "claim": {
          "zh-Hant": "w=(x+y)² 滿足 Δw=4 且在左邊、下邊都等於資料，所以 w 滿足此 BVP",
          "en": "w=(x+y)² satisfies Δw=4 and equals the data on the left and bottom edges, so w satisfies this BVP"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "四條邊都要核：右邊 w(2,1)=9≠7=g_R(1)。",
          "en": "All four edges must be checked: on the right w(2,1)=9≠7=g_R(1)."
        }
      },
      {
        "claim": {
          "zh-Hant": "u 滿足此 BVP，所以這個 BVP 的解唯一且對資料穩定",
          "en": "u satisfies this BVP, so the BVP has a unique solution that is stable under changes of the data"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "符號恆等只核指定候選；唯一性與穩定性需要另外的定理，MKO a2 明確排除這類推論。",
          "en": "Symbolic identities check only the specified candidate; uniqueness and stability need further theorems, and MKO assumption a2 explicitly excludes such inferences."
        }
      }
    ],
    "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": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "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": "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": "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": "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": "mko_bvp",
        "kind": "cited",
        "source": {
          "mko_id": "mko-boundary-value-problem",
          "sha256": "4ffaeecac13219cb387fd43a669823ff3c2b402bfadf2e3d04f3938d83bf3f0d",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "本輪固定矩形域上的二次多項式候選 u。內部 Poisson 責任以符號 Laplacian 恆等核對，Dirichlet 資料則把 u 精確限制到四條邊並逐係數比較；兩類責任分開記錄後，才可說候選滿足該宣告 BVP。符號恆等只核指定候選與指定方程／邊界，不證一般解存在、唯一、穩定、正則或適定。",
          "en": "This slice fixes a quadratic polynomial candidate u on a rectangle. A symbolic Laplacian identity checks the interior Poisson obligation; Dirichlet data are checked by restricting u exactly to all four edges and comparing polynomial coefficients. Only after separately recording both groups may the candidate be said to satisfy the declared BVP. Symbolic identities check only the specified candidate, equation and boundary data; they prove no general existence, uniqueness, stability, regularity or well-posedness theorem."
        }
      },
      {
        "id": "poly_partial",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "多項式的偏導數：∂/∂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。",
          "en": "Partial derivatives of polynomials: ∂/∂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."
        }
      },
      {
        "id": "poly_coeff_zero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "係數全為 0 的多項式（一變數或二變數）在每一點取值都是 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 (in one or two variables) whose coefficients are all 0 takes the value 0 at every point; hence two polynomials with identical coefficients take equal values at every point."
        }
      },
      {
        "id": "edge_restrict",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "把二變數多項式限制到一條邊，就是把該邊固定的座標（x=0、x=2、y=0 或 y=1）代入；結果是剩下變數的多項式，在該邊的每一點都給出 u 的值。",
          "en": "Restricting a bivariate polynomial to an edge means substituting the coordinate fixed on that edge (x=0, x=2, y=0 or y=1); the result is a polynomial in the remaining variable that gives the value of u at every point of the edge."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S04",
        "lhs": "2+2-4",
        "rhs": "0"
      },
      {
        "id": "N2",
        "step": "S06",
        "vars": [
          "y"
        ],
        "degree": 2,
        "lhs": "0**2+0*y+y**2",
        "rhs": "y**2"
      },
      {
        "id": "N3",
        "step": "S07",
        "vars": [
          "y"
        ],
        "degree": 2,
        "lhs": "2**2+2*y+y**2",
        "rhs": "y**2+2*y+4"
      },
      {
        "id": "N4",
        "step": "S08",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "x**2+x*0+0**2",
        "rhs": "x**2"
      },
      {
        "id": "N5",
        "step": "S09",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "x**2+x*1+1**2",
        "rhs": "x**2+x+1"
      },
      {
        "id": "N6",
        "step": "S11",
        "lhs": "1**2+2*1+4",
        "rhs": "2**2+2+1"
      },
      {
        "id": "N7",
        "step": "S12",
        "vars": [
          "x",
          "y"
        ],
        "degree": 2,
        "lhs": "(x**2+x*y+y**2)+(x**2-y**2-2*x+y)",
        "rhs": "2*x**2+x*y-2*x+y"
      },
      {
        "id": "N8",
        "step": "S12.2",
        "lhs": "4+0-4",
        "rhs": "0"
      },
      {
        "id": "N9",
        "step": "S13",
        "lhs": "2*0**2+0*0-2*0+0",
        "rhs": "0**2"
      },
      {
        "id": "N10",
        "step": "S13",
        "lhs": "2*2**2+2*0-2*2+0",
        "rhs": "2**2"
      },
      {
        "id": "N11",
        "step": "S13",
        "lhs": "2*0**2+0*1-2*0+1",
        "rhs": "1**2"
      },
      {
        "id": "N12",
        "step": "S13",
        "lhs": "2*2**2+2*1-2*2+1",
        "rhs": "1**2+2*1+4"
      },
      {
        "id": "N13",
        "step": "S14",
        "vars": [
          "x"
        ],
        "degree": 2,
        "lhs": "2*x**2+x*0-2*x+0",
        "rhs": "2*x**2-2*x"
      },
      {
        "id": "N14",
        "step": "S14",
        "lhs": "2*1**2-2*1",
        "relation": "!=",
        "rhs": "1**2"
      },
      {
        "id": "N15",
        "step": "S15",
        "lhs": "2*0**2+0*(1/2)-2*0+1/2",
        "rhs": "1/2"
      },
      {
        "id": "N16",
        "step": "S15",
        "lhs": "1/2",
        "relation": "!=",
        "rhs": "(1/2)**2"
      },
      {
        "id": "N17",
        "lhs": "(2+1)**2",
        "relation": "!=",
        "rhs": "1**2+2*1+4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "停在 MKO 的切片內：矩形域、二次多項式候選、符號 Laplacian 恆等式、四條邊的精確限制與係數比較。符號約定明示為 Δu=u_xx+u_yy，方程寫成 Δu=4。核對成功的候選給出此特定 BVP 的存在見證，但不證一般存在定理或唯一、穩定、正則與適定性；不處理弱解或函數空間；不改動 canonical MKO 或英文 edition。",
      "en": "Stays inside the MKO's slice: a rectangular domain, quadratic polynomial candidates, a symbolic Laplacian identity, exact restriction to the four edges and coefficient comparison. The sign convention is explicit: Δu=u_xx+u_yy, and the equation is Δu=4. A successfully checked candidate witnesses existence for this particular BVP, but no general existence theorem, uniqueness, stability, regularity or well-posedness result is proved; weak solutions and function spaces are not treated; the canonical MKO and its English edition are not changed."
    },
    "md_symbols": [
      [
        "\\Omega=(0,2)\\times(0,1)",
        "開矩形（內部）",
        "the open rectangle (interior)"
      ],
      [
        "\\Gamma_L,\\Gamma_R,\\Gamma_B,\\Gamma_T",
        "左、右、下、上四條閉邊",
        "the closed left, right, bottom and top edges"
      ],
      [
        "\\Delta u=u_{xx}+u_{yy}",
        "Laplacian",
        "the Laplacian"
      ],
      [
        "u|_{\\Gamma}",
        "u 限制到邊 Γ 上所得的單變數多項式",
        "the one-variable polynomial obtained by restricting u to the edge Γ"
      ],
      [
        "(I),\\ (B)",
        "內部義務與邊界義務",
        "the interior obligation and the boundary obligation"
      ],
      [
        "h=x^2-y^2-2x+y",
        "調和多項式，v=u+h",
        "a harmonic polynomial, with v=u+h"
      ]
    ],
    "md_error_controls": [
      {
        "title": "只核內部方程不夠 / Checking only the interior equation is not enough",
        "zh": "MKO 指出的誤解是只核內部 PDE 就宣稱 BVP 已解。v 的 Δv=4 恆成立，與 u 一樣通過 (I)，但違反邊界資料。內部義務與邊界義務必須分開記錄、分別證明。",
        "en": "The mistake named by the MKO is to declare the BVP solved after checking only the interior PDE. v has Δv=4 identically and passes (I) just as u does, yet it violates the boundary data. The interior and boundary obligations must be recorded separately and proved separately."
      },
      {
        "title": "角點不足；全邊可用係數或次數界證書 / Corners are insufficient; whole-edge certificates need coefficients or degree bounds",
        "zh": "v 在四個角點都吻合（S13），但下邊差 v(x,0)−x²=x(x−2) 只在兩個角點為零。此稿採整條邊的多項式係數比較作證書；沒有額外前提的幾個點不能取代全邊證明。但若已證差多項式次數至多 2，三個相異點為零就可由多項式根數界推出恆零；這是另一種帶次數界的完整證書，不是角點抽樣本身足夠。",
        "en": "v matches all four corners (S13), but its bottom-edge difference v(x,0)−x²=x(x−2) vanishes only at the two corners. This guide uses whole-edge polynomial coefficient comparison as its certificate. A few points without additional premises do not replace a whole-edge proof. If the difference is proved to have degree at most 2, however, three distinct zeros force it to vanish identically by the polynomial root bound. That is a different complete certificate with a degree bound, not a claim that corners alone suffice."
      },
      {
        "title": "滿足宣告的 BVP 不是適定性 / Satisfying the declared BVP is not well-posedness",
        "zh": "S11 只說 u 是這一組資料下的一個候選解。是否唯一、對資料擾動是否穩定，需要另外的定理（例如最大值原理類結果），本稿沒有證明也沒有引用。",
        "en": "S11 says only that u is a candidate solution for this data. Whether it is unique, and whether it is stable under perturbations of the data, needs further theorems (for example results of maximum-principle type), which this guide neither proves nor cites."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不證明也不引用 Dirichlet–Poisson 問題的唯一性或穩定性結果；這些屬於 MKO 的 formal next。",
        "en": "This guide neither proves nor cites uniqueness or stability results for the Dirichlet–Poisson problem; these belong to the MKO's formal next."
      },
      {
        "zh-Hant": "S15 括號中右邊與上邊的失敗只是補充說明，N 系列只核了下邊與左邊的見證點。",
        "en": "The failures on the right and top edges mentioned in parentheses in S15 are a remark only; the numeric checks cover only the witness points on the bottom and left edges."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "bbbc47a9eaa572e972cb4678073ca74403dc1f89cbba8026999be6bdd3be0414",
    "original_source_sha256": "95cd648e4ae4e97fe1b121a19464dc1e55260bc5fc03d901cdf37eea033e9390",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [
      "BR-M066",
      "BR-M067",
      "BR-M068"
    ],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}