Separate Interior PDE and Complete Rectangle-Boundary 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-20261025-r01-boundary-value-problem-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. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-12, -2, -3, 1, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=-1. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-12, -2, -3, 1, -1, -2], u_xx+u_yy=source, -2, -1, -1, No, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-12, -2, -3, 1, -1, -2], u_xx+u_yy=source, -2, -1, -1, No, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
2. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-11, -1, -2, 2, 0, -1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=2. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-11, -1, -2, 2, 0, -1], u_xx+u_yy=source, 2, 2, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-11, -1, -2, 2, 0, -1], u_xx+u_yy=source, 2, 2, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
3. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-10, 0, -1, 3, 1, 0] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=6. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-10, 0, -1, 3, 1, 0], u_xx+u_yy=source, 6, 6, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-10, 0, -1, 3, 1, 0], u_xx+u_yy=source, 6, 6, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
4. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-9, 1, 0, 4, -1, 1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=10. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-9, 1, 0, 4, -1, 1], u_xx+u_yy=source, 10, 10, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-9, 1, 0, 4, -1, 1], u_xx+u_yy=source, 10, 10, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
5. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-8, 2, 1, 1, 0, 2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=7. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-8, 2, 1, 1, 0, 2], u_xx+u_yy=source, 6, 7, -1, No, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-8, 2, 1, 1, 0, 2], u_xx+u_yy=source, 6, 7, -1, No, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
6. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-7, -2, 2, 2, 1, -2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=0. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-7, -2, 2, 2, 1, -2], u_xx+u_yy=source, 0, 0, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-7, -2, 2, 2, 1, -2], u_xx+u_yy=source, 0, 0, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
7. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-6, -1, 3, 3, -1, -1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=4. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-6, -1, 3, 3, -1, -1], u_xx+u_yy=source, 4, 4, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-6, -1, 3, 3, -1, -1], u_xx+u_yy=source, 4, 4, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
8. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-5, 0, -3, 4, 0, 0] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=8. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-5, 0, -3, 4, 0, 0], u_xx+u_yy=source, 8, 8, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-5, 0, -3, 4, 0, 0], u_xx+u_yy=source, 8, 8, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
9. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-4, 1, -2, 1, 1, 1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=5. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-4, 1, -2, 1, 1, 1], u_xx+u_yy=source, 4, 5, -1, No, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-4, 1, -2, 1, 1, 1], u_xx+u_yy=source, 4, 5, -1, No, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
10. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-3, 2, -1, 2, -1, 2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=8. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-3, 2, -1, 2, -1, 2], u_xx+u_yy=source, 8, 8, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-3, 2, -1, 2, -1, 2], u_xx+u_yy=source, 8, 8, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
11. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-2, -2, 0, 3, 0, -2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=2. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-2, -2, 0, 3, 0, -2], u_xx+u_yy=source, 2, 2, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-2, -2, 0, 3, 0, -2], u_xx+u_yy=source, 2, 2, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
12. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[-1, -1, 1, 4, 1, -1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=6. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-1, -1, 1, 4, 1, -1], u_xx+u_yy=source, 6, 6, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-1, -1, 1, 4, 1, -1], u_xx+u_yy=source, 6, 6, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
13. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[0, 0, 2, 1, -1, 0] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=3. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([0, 0, 2, 1, -1, 0], u_xx+u_yy=source, 2, 3, -1, No, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 0, 2, 1, -1, 0], u_xx+u_yy=source, 2, 3, -1, No, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
14. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[1, 1, 3, 2, 0, 1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=6. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([1, 1, 3, 2, 0, 1], u_xx+u_yy=source, 6, 6, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 1, 3, 2, 0, 1], u_xx+u_yy=source, 6, 6, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
15. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[2, 2, -3, 3, 1, 2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=10. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([2, 2, -3, 3, 1, 2], u_xx+u_yy=source, 10, 10, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 2, -3, 3, 1, 2], u_xx+u_yy=source, 10, 10, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
16. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[3, -2, -2, 4, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=4. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([3, -2, -2, 4, -1, -2], u_xx+u_yy=source, 4, 4, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -2, -2, 4, -1, -2], u_xx+u_yy=source, 4, 4, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
17. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[4, -1, -1, 1, 0, -1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=1. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([4, -1, -1, 1, 0, -1], u_xx+u_yy=source, 0, 1, -1, No, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([4, -1, -1, 1, 0, -1], u_xx+u_yy=source, 0, 1, -1, No, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
18. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[5, 0, 0, 2, 1, 0] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=4. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([5, 0, 0, 2, 1, 0], u_xx+u_yy=source, 4, 4, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([5, 0, 0, 2, 1, 0], u_xx+u_yy=source, 4, 4, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
19. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[6, 1, 1, 3, -1, 1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=8. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([6, 1, 1, 3, -1, 1], u_xx+u_yy=source, 8, 8, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([6, 1, 1, 3, -1, 1], u_xx+u_yy=source, 8, 8, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
20. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[7, 2, 2, 4, 0, 2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=12. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([7, 2, 2, 4, 0, 2], u_xx+u_yy=source, 12, 12, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([7, 2, 2, 4, 0, 2], u_xx+u_yy=source, 12, 12, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
21. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[8, -2, 3, 1, 1, -2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=-1. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([8, -2, 3, 1, 1, -2], u_xx+u_yy=source, -2, -1, -1, No, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([8, -2, 3, 1, 1, -2], u_xx+u_yy=source, -2, -1, -1, No, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
22. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[9, -1, -3, 2, -1, -1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=2. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([9, -1, -3, 2, -1, -1], u_xx+u_yy=source, 2, 2, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([9, -1, -3, 2, -1, -1], u_xx+u_yy=source, 2, 2, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
23. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[10, 0, -2, 3, 0, 0] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=6. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([10, 0, -2, 3, 0, 0], u_xx+u_yy=source, 6, 6, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([10, 0, -2, 3, 0, 0], u_xx+u_yy=source, 6, 6, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
24. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[11, 1, -1, 4, 1, 1] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=10. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([11, 1, -1, 4, 1, 1], u_xx+u_yy=source, 10, 10, 0, Yes, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([11, 1, -1, 4, 1, 1], u_xx+u_yy=source, 10, 10, 0, Yes, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
25. An Interior Poisson Identity without a Boundary Claim
Use the quadratic candidate coefficients=[12, 2, 0, 1, -1, 2] in fixed order [1,x,y,x²,x*y,y²]. Interior equation u_xx+u_yy=source has source=7. Give (coefficients,u_xx+u_yy=source,constant Laplacian,source,signed residual,interior identity?,interior_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([12, 2, 0, 1, -1, 2], u_xx+u_yy=source, 6, 7, -1, No, interior_only)
The Laplacian is 2*c_xx+2*c_yy for this candidate, so a zero constant residual is an interior identity only. No boundary data have been checked. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([12, 2, 0, 1, -1, 2], u_xx+u_yy=source, 6, 7, -1, No, interior_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-interior-pde-identity · candidate
26. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-12, -2, -3, 1, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-2, 2, -1, 3]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=left; supplied coefficients=[-3, -1, -2]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-2, 2, -1, 3], left, y, [-4, -1, -2], [-3, -1, -2], No, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-2, 2, -1, 3], left, y, [-4, -1, -2], [-3, -1, -2], No, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
27. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-11, -1, -2, 2, 0, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-3, 3, -2, 4]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=right; supplied coefficients=[4, -2, -1]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-3, 3, -2, 4], right, y, [4, -2, -1], [4, -2, -1], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-3, 3, -2, 4], right, y, [4, -2, -1], [4, -2, -1], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
28. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-10, 0, -1, 3, 1, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-4, 4, -3, 5]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=bottom; supplied coefficients=[-7, -3, 3]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-4, 4, -3, 5], bottom, x, [-7, -3, 3], [-7, -3, 3], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-4, 4, -3, 5], bottom, x, [-7, -3, 3], [-7, -3, 3], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
29. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-9, 1, 0, 4, -1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-5, 5, -4, 6]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=top; supplied coefficients=[27, -5, 4]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-5, 5, -4, 6], top, x, [27, -5, 4], [27, -5, 4], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-5, 5, -4, 6], top, x, [27, -5, 4], [27, -5, 4], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
30. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-8, 2, 1, 1, 0, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-6, 6, -5, 7]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=left; supplied coefficients=[16, 1, 2]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-6, 6, -5, 7], left, y, [16, 1, 2], [16, 1, 2], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-6, 6, -5, 7], left, y, [16, 1, 2], [16, 1, 2], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
31. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-7, -2, 2, 2, 1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-7, 7, -6, 8]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=right; supplied coefficients=[78, 9, -2]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-7, 7, -6, 8], right, y, [77, 9, -2], [78, 9, -2], No, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-7, 7, -6, 8], right, y, [77, 9, -2], [78, 9, -2], No, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
32. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-6, -1, 3, 3, -1, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-8, 8, -7, 9]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=bottom; supplied coefficients=[-76, 6, 3]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-8, 8, -7, 9], bottom, x, [-76, 6, 3], [-76, 6, 3], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-8, 8, -7, 9], bottom, x, [-76, 6, 3], [-76, 6, 3], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
33. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-5, 0, -3, 4, 0, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-9, 9, -8, 10]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=top; supplied coefficients=[-35, 0, 4]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-9, 9, -8, 10], top, x, [-35, 0, 4], [-35, 0, 4], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-9, 9, -8, 10], top, x, [-35, 0, 4], [-35, 0, 4], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
34. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-4, 1, -2, 1, 1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-10, 10, -9, 11]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=left; supplied coefficients=[86, -12, 1]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-10, 10, -9, 11], left, y, [86, -12, 1], [86, -12, 1], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-10, 10, -9, 11], left, y, [86, -12, 1], [86, -12, 1], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
35. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-3, 2, -1, 2, -1, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-11, 11, -10, 12]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=right; supplied coefficients=[261, -12, 2]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-11, 11, -10, 12], right, y, [261, -12, 2], [261, -12, 2], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-11, 11, -10, 12], right, y, [261, -12, 2], [261, -12, 2], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
36. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-2, -2, 0, 3, 0, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-12, 12, -11, 13]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=bottom; supplied coefficients=[-243, -2, 3]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-12, 12, -11, 13], bottom, x, [-244, -2, 3], [-243, -2, 3], No, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-12, 12, -11, 13], bottom, x, [-244, -2, 3], [-243, -2, 3], No, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
37. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[-1, -1, 1, 4, 1, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-13, 13, -12, 14]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=top; supplied coefficients=[-183, 13, 4]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-13, 13, -12, 14], top, x, [-183, 13, 4], [-183, 13, 4], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-13, 13, -12, 14], top, x, [-183, 13, 4], [-183, 13, 4], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
38. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[0, 0, 2, 1, -1, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-14, 14, -13, 15]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=left; supplied coefficients=[196, 16, 0]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-14, 14, -13, 15], left, y, [196, 16, 0], [196, 16, 0], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-14, 14, -13, 15], left, y, [196, 16, 0], [196, 16, 0], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
39. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[1, 1, 3, 2, 0, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-15, 15, -14, 16]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=right; supplied coefficients=[466, 3, 1]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-15, 15, -14, 16], right, y, [466, 3, 1], [466, 3, 1], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-15, 15, -14, 16], right, y, [466, 3, 1], [466, 3, 1], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
40. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[2, 2, -3, 3, 1, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-16, 16, -15, 17]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=bottom; supplied coefficients=[497, -13, 3]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-16, 16, -15, 17], bottom, x, [497, -13, 3], [497, -13, 3], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-16, 16, -15, 17], bottom, x, [497, -13, 3], [497, -13, 3], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
41. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[3, -2, -2, 4, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-17, 17, -16, 18]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=top; supplied coefficients=[-680, -20, 4]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-17, 17, -16, 18], top, x, [-681, -20, 4], [-680, -20, 4], No, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-17, 17, -16, 18], top, x, [-681, -20, 4], [-680, -20, 4], No, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
42. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[4, -1, -1, 1, 0, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-18, 18, -17, 19]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=left; supplied coefficients=[346, -1, -1]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-18, 18, -17, 19], left, y, [346, -1, -1], [346, -1, -1], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-18, 18, -17, 19], left, y, [346, -1, -1], [346, -1, -1], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
43. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[5, 0, 0, 2, 1, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-19, 19, -18, 20]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=right; supplied coefficients=[727, 19, 0]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-19, 19, -18, 20], right, y, [727, 19, 0], [727, 19, 0], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-19, 19, -18, 20], right, y, [727, 19, 0], [727, 19, 0], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
44. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[6, 1, 1, 3, -1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-20, 20, -19, 21]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=bottom; supplied coefficients=[348, 20, 3]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-20, 20, -19, 21], bottom, x, [348, 20, 3], [348, 20, 3], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-20, 20, -19, 21], bottom, x, [348, 20, 3], [348, 20, 3], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
45. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[7, 2, 2, 4, 0, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-21, 21, -20, 22]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=top; supplied coefficients=[1019, 2, 4]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-21, 21, -20, 22], top, x, [1019, 2, 4], [1019, 2, 4], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-21, 21, -20, 22], top, x, [1019, 2, 4], [1019, 2, 4], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
46. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[8, -2, 3, 1, 1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-22, 22, -21, 23]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=left; supplied coefficients=[537, -19, -2]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-22, 22, -21, 23], left, y, [536, -19, -2], [537, -19, -2], No, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-22, 22, -21, 23], left, y, [536, -19, -2], [537, -19, -2], No, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
47. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[9, -1, -3, 2, -1, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-23, 23, -22, 24]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=right; supplied coefficients=[1044, -26, -1]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-23, 23, -22, 24], right, y, [1044, -26, -1], [1044, -26, -1], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-23, 23, -22, 24], right, y, [1044, -26, -1], [1044, -26, -1], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
48. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[10, 0, -2, 3, 0, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-24, 24, -23, 25]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=bottom; supplied coefficients=[56, 0, 3]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-24, 24, -23, 25], bottom, x, [56, 0, 3], [56, 0, 3], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-24, 24, -23, 25], bottom, x, [56, 0, 3], [56, 0, 3], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
49. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[11, 1, -1, 4, 1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-25, 25, -24, 26]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=top; supplied coefficients=[661, 27, 4]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-25, 25, -24, 26], top, x, [661, 27, 4], [661, 27, 4], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-25, 25, -24, 26], top, x, [661, 27, 4], [661, 27, 4], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
50. A Whole-Edge Dirichlet Coefficient Identity
Use the quadratic candidate coefficients=[12, 2, 0, 1, -1, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-26, 26, -25, 27]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Selected edge=left; supplied coefficients=[636, 26, 2]. Give (bounds,edge,remaining variable,restriction coefficients,supplied coefficients,coefficientwise identity?,declared_edge_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-26, 26, -25, 27], left, y, [636, 26, 2], [636, 26, 2], Yes, declared_edge_only)
Substitute the fixed edge coordinate and compare all three coefficients in the remaining variable. This proves only the declared edge identity, not all four boundaries or the interior equation. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-26, 26, -25, 27], left, y, [636, 26, 2], [636, 26, 2], Yes, declared_edge_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-dirichlet-edge-identity · candidate
51. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-12, -2, -3, 1, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-2, 2, -1, 3]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=-1; supplied [edge,coefficients] records=[[left, [-3, -1, -2]], [right, [-12, -5, -2]], [bottom, [-11, -1, 1]], [top, [-39, -5, 1]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-12, -2, -3, 1, -1, -2], -2, -1, No, [[left, y, [-4, -1, -2], [-3, -1, -2], No], [right, y, [-12, -5, -2], [-12, -5, -2], Yes], [bottom, x, [-11, -1, 1], [-11, -1, 1], Yes], [top, x, [-39, -5, 1], [-39, -5, 1], Yes]], No, No, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-12, -2, -3, 1, -1, -2], -2, -1, No, [[left, y, [-4, -1, -2], [-3, -1, -2], No], [right, y, [-12, -5, -2], [-12, -5, -2], Yes], [bottom, x, [-11, -1, 1], [-11, -1, 1], Yes], [top, x, [-39, -5, 1], [-39, -5, 1], Yes]], No, No, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
52. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-11, -1, -2, 2, 0, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-3, 3, -2, 4]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=2; supplied [edge,coefficients] records=[[left, [10, -2, -1]], [right, [4, -2, -1]], [bottom, [-11, -1, 2]], [top, [-35, -1, 2]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-11, -1, -2, 2, 0, -1], 2, 2, Yes, [[left, y, [10, -2, -1], [10, -2, -1], Yes], [right, y, [4, -2, -1], [4, -2, -1], Yes], [bottom, x, [-11, -1, 2], [-11, -1, 2], Yes], [top, x, [-35, -1, 2], [-35, -1, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-11, -1, -2, 2, 0, -1], 2, 2, Yes, [[left, y, [10, -2, -1], [10, -2, -1], Yes], [right, y, [4, -2, -1], [4, -2, -1], Yes], [bottom, x, [-11, -1, 2], [-11, -1, 2], Yes], [top, x, [-35, -1, 2], [-35, -1, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
53. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-10, 0, -1, 3, 1, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-4, 4, -3, 5]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=6; supplied [edge,coefficients] records=[[left, [38, -5, 0]], [right, [38, 3, 0]], [bottom, [-7, -3, 3]], [top, [-15, 5, 3]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-10, 0, -1, 3, 1, 0], 6, 6, Yes, [[left, y, [38, -5, 0], [38, -5, 0], Yes], [right, y, [38, 3, 0], [38, 3, 0], Yes], [bottom, x, [-7, -3, 3], [-7, -3, 3], Yes], [top, x, [-15, 5, 3], [-15, 5, 3], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-10, 0, -1, 3, 1, 0], 6, 6, Yes, [[left, y, [38, -5, 0], [38, -5, 0], Yes], [right, y, [38, 3, 0], [38, 3, 0], Yes], [bottom, x, [-7, -3, 3], [-7, -3, 3], Yes], [top, x, [-15, 5, 3], [-15, 5, 3], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
54. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-9, 1, 0, 4, -1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-5, 5, -4, 6]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=10; supplied [edge,coefficients] records=[[left, [86, 5, 1]], [right, [96, -5, 1]], [bottom, [7, 5, 4]], [top, [27, -5, 4]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-9, 1, 0, 4, -1, 1], 10, 10, Yes, [[left, y, [86, 5, 1], [86, 5, 1], Yes], [right, y, [96, -5, 1], [96, -5, 1], Yes], [bottom, x, [7, 5, 4], [7, 5, 4], Yes], [top, x, [27, -5, 4], [27, -5, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-9, 1, 0, 4, -1, 1], 10, 10, Yes, [[left, y, [86, 5, 1], [86, 5, 1], Yes], [right, y, [96, -5, 1], [96, -5, 1], Yes], [bottom, x, [7, 5, 4], [7, 5, 4], Yes], [top, x, [27, -5, 4], [27, -5, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
55. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-8, 2, 1, 1, 0, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-6, 6, -5, 7]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=6; supplied [edge,coefficients] records=[[left, [16, 1, 2]], [right, [40, 1, 2]], [bottom, [37, 2, 1]], [top, [97, 2, 1]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-8, 2, 1, 1, 0, 2], 6, 6, Yes, [[left, y, [16, 1, 2], [16, 1, 2], Yes], [right, y, [40, 1, 2], [40, 1, 2], Yes], [bottom, x, [37, 2, 1], [37, 2, 1], Yes], [top, x, [97, 2, 1], [97, 2, 1], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-8, 2, 1, 1, 0, 2], 6, 6, Yes, [[left, y, [16, 1, 2], [16, 1, 2], Yes], [right, y, [40, 1, 2], [40, 1, 2], Yes], [bottom, x, [37, 2, 1], [37, 2, 1], Yes], [top, x, [97, 2, 1], [97, 2, 1], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
56. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-7, -2, 2, 2, 1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-7, 7, -6, 8]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=0; supplied [edge,coefficients] records=[[left, [105, -5, -2]], [right, [77, 9, -2]], [bottom, [-91, -8, 2]], [top, [-119, 6, 2]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-7, -2, 2, 2, 1, -2], 0, 0, Yes, [[left, y, [105, -5, -2], [105, -5, -2], Yes], [right, y, [77, 9, -2], [77, 9, -2], Yes], [bottom, x, [-91, -8, 2], [-91, -8, 2], Yes], [top, x, [-119, 6, 2], [-119, 6, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-7, -2, 2, 2, 1, -2], 0, 0, Yes, [[left, y, [105, -5, -2], [105, -5, -2], Yes], [right, y, [77, 9, -2], [77, 9, -2], Yes], [bottom, x, [-91, -8, 2], [-91, -8, 2], Yes], [top, x, [-119, 6, 2], [-119, 6, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
57. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-6, -1, 3, 3, -1, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-8, 8, -7, 9]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=4; supplied [edge,coefficients] records=[[left, [195, 11, -1]], [right, [178, -5, -1]], [bottom, [-76, 6, 3]], [top, [-60, -10, 3]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-6, -1, 3, 3, -1, -1], 4, 4, Yes, [[left, y, [194, 11, -1], [195, 11, -1], No], [right, y, [178, -5, -1], [178, -5, -1], Yes], [bottom, x, [-76, 6, 3], [-76, 6, 3], Yes], [top, x, [-60, -10, 3], [-60, -10, 3], Yes]], No, No, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-6, -1, 3, 3, -1, -1], 4, 4, Yes, [[left, y, [194, 11, -1], [195, 11, -1], No], [right, y, [178, -5, -1], [178, -5, -1], Yes], [bottom, x, [-76, 6, 3], [-76, 6, 3], Yes], [top, x, [-60, -10, 3], [-60, -10, 3], Yes]], No, No, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
58. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-5, 0, -3, 4, 0, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-9, 9, -8, 10]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=9; supplied [edge,coefficients] records=[[left, [319, -3, 0]], [right, [319, -3, 0]], [bottom, [19, 0, 4]], [top, [-35, 0, 4]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-5, 0, -3, 4, 0, 0], 8, 9, No, [[left, y, [319, -3, 0], [319, -3, 0], Yes], [right, y, [319, -3, 0], [319, -3, 0], Yes], [bottom, x, [19, 0, 4], [19, 0, 4], Yes], [top, x, [-35, 0, 4], [-35, 0, 4], Yes]], Yes, No, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-5, 0, -3, 4, 0, 0], 8, 9, No, [[left, y, [319, -3, 0], [319, -3, 0], Yes], [right, y, [319, -3, 0], [319, -3, 0], Yes], [bottom, x, [19, 0, 4], [19, 0, 4], Yes], [top, x, [-35, 0, 4], [-35, 0, 4], Yes]], Yes, No, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
59. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-4, 1, -2, 1, 1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-10, 10, -9, 11]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=4; supplied [edge,coefficients] records=[[left, [86, -12, 1]], [right, [106, 8, 1]], [bottom, [95, -8, 1]], [top, [95, 12, 1]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-4, 1, -2, 1, 1, 1], 4, 4, Yes, [[left, y, [86, -12, 1], [86, -12, 1], Yes], [right, y, [106, 8, 1], [106, 8, 1], Yes], [bottom, x, [95, -8, 1], [95, -8, 1], Yes], [top, x, [95, 12, 1], [95, 12, 1], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-4, 1, -2, 1, 1, 1], 4, 4, Yes, [[left, y, [86, -12, 1], [86, -12, 1], Yes], [right, y, [106, 8, 1], [106, 8, 1], Yes], [bottom, x, [95, -8, 1], [95, -8, 1], Yes], [top, x, [95, 12, 1], [95, 12, 1], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
60. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-3, 2, -1, 2, -1, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-11, 11, -10, 12]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=8; supplied [edge,coefficients] records=[[left, [217, 10, 2]], [right, [261, -12, 2]], [bottom, [207, 12, 2]], [top, [273, -10, 2]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-3, 2, -1, 2, -1, 2], 8, 8, Yes, [[left, y, [217, 10, 2], [217, 10, 2], Yes], [right, y, [261, -12, 2], [261, -12, 2], Yes], [bottom, x, [207, 12, 2], [207, 12, 2], Yes], [top, x, [273, -10, 2], [273, -10, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-3, 2, -1, 2, -1, 2], 8, 8, Yes, [[left, y, [217, 10, 2], [217, 10, 2], Yes], [right, y, [261, -12, 2], [261, -12, 2], Yes], [bottom, x, [207, 12, 2], [207, 12, 2], Yes], [top, x, [273, -10, 2], [273, -10, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
61. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-2, -2, 0, 3, 0, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-12, 12, -11, 13]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=2; supplied [edge,coefficients] records=[[left, [454, 0, -2]], [right, [406, 0, -2]], [bottom, [-244, -2, 3]], [top, [-340, -2, 3]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-2, -2, 0, 3, 0, -2], 2, 2, Yes, [[left, y, [454, 0, -2], [454, 0, -2], Yes], [right, y, [406, 0, -2], [406, 0, -2], Yes], [bottom, x, [-244, -2, 3], [-244, -2, 3], Yes], [top, x, [-340, -2, 3], [-340, -2, 3], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-2, -2, 0, 3, 0, -2], 2, 2, Yes, [[left, y, [454, 0, -2], [454, 0, -2], Yes], [right, y, [406, 0, -2], [406, 0, -2], Yes], [bottom, x, [-244, -2, 3], [-244, -2, 3], Yes], [top, x, [-340, -2, 3], [-340, -2, 3], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
62. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[-1, -1, 1, 4, 1, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-13, 13, -12, 14]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=6; supplied [edge,coefficients] records=[[left, [688, -12, -1]], [right, [662, 14, -1]], [bottom, [-157, -13, 4]], [top, [-183, 13, 4]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([-1, -1, 1, 4, 1, -1], 6, 6, Yes, [[left, y, [688, -12, -1], [688, -12, -1], Yes], [right, y, [662, 14, -1], [662, 14, -1], Yes], [bottom, x, [-157, -13, 4], [-157, -13, 4], Yes], [top, x, [-183, 13, 4], [-183, 13, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([-1, -1, 1, 4, 1, -1], 6, 6, Yes, [[left, y, [688, -12, -1], [688, -12, -1], Yes], [right, y, [662, 14, -1], [662, 14, -1], Yes], [bottom, x, [-157, -13, 4], [-157, -13, 4], Yes], [top, x, [-183, 13, 4], [-183, 13, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
63. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[0, 0, 2, 1, -1, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-14, 14, -13, 15]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=2; supplied [edge,coefficients] records=[[left, [197, 16, 0]], [right, [196, -12, 0]], [bottom, [-26, 13, 1]], [top, [30, -15, 1]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([0, 0, 2, 1, -1, 0], 2, 2, Yes, [[left, y, [196, 16, 0], [197, 16, 0], No], [right, y, [196, -12, 0], [196, -12, 0], Yes], [bottom, x, [-26, 13, 1], [-26, 13, 1], Yes], [top, x, [30, -15, 1], [30, -15, 1], Yes]], No, No, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([0, 0, 2, 1, -1, 0], 2, 2, Yes, [[left, y, [196, 16, 0], [197, 16, 0], No], [right, y, [196, -12, 0], [196, -12, 0], Yes], [bottom, x, [-26, 13, 1], [-26, 13, 1], Yes], [top, x, [30, -15, 1], [30, -15, 1], Yes]], No, No, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
64. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[1, 1, 3, 2, 0, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-15, 15, -14, 16]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=6; supplied [edge,coefficients] records=[[left, [436, 3, 1]], [right, [466, 3, 1]], [bottom, [155, 1, 2]], [top, [305, 1, 2]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([1, 1, 3, 2, 0, 1], 6, 6, Yes, [[left, y, [436, 3, 1], [436, 3, 1], Yes], [right, y, [466, 3, 1], [466, 3, 1], Yes], [bottom, x, [155, 1, 2], [155, 1, 2], Yes], [top, x, [305, 1, 2], [305, 1, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([1, 1, 3, 2, 0, 1], 6, 6, Yes, [[left, y, [436, 3, 1], [436, 3, 1], Yes], [right, y, [466, 3, 1], [466, 3, 1], Yes], [bottom, x, [155, 1, 2], [155, 1, 2], Yes], [top, x, [305, 1, 2], [305, 1, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
65. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[2, 2, -3, 3, 1, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-16, 16, -15, 17]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=11; supplied [edge,coefficients] records=[[left, [738, -19, 2]], [right, [802, 13, 2]], [bottom, [497, -13, 3]], [top, [529, 19, 3]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([2, 2, -3, 3, 1, 2], 10, 11, No, [[left, y, [738, -19, 2], [738, -19, 2], Yes], [right, y, [802, 13, 2], [802, 13, 2], Yes], [bottom, x, [497, -13, 3], [497, -13, 3], Yes], [top, x, [529, 19, 3], [529, 19, 3], Yes]], Yes, No, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([2, 2, -3, 3, 1, 2], 10, 11, No, [[left, y, [738, -19, 2], [738, -19, 2], Yes], [right, y, [802, 13, 2], [802, 13, 2], Yes], [bottom, x, [497, -13, 3], [497, -13, 3], Yes], [top, x, [529, 19, 3], [529, 19, 3], Yes]], Yes, No, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
66. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[3, -2, -2, 4, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-17, 17, -16, 18]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=4; supplied [edge,coefficients] records=[[left, [1193, 15, -2]], [right, [1125, -19, -2]], [bottom, [-477, 14, 4]], [top, [-681, -20, 4]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([3, -2, -2, 4, -1, -2], 4, 4, Yes, [[left, y, [1193, 15, -2], [1193, 15, -2], Yes], [right, y, [1125, -19, -2], [1125, -19, -2], Yes], [bottom, x, [-477, 14, 4], [-477, 14, 4], Yes], [top, x, [-681, -20, 4], [-681, -20, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([3, -2, -2, 4, -1, -2], 4, 4, Yes, [[left, y, [1193, 15, -2], [1193, 15, -2], Yes], [right, y, [1125, -19, -2], [1125, -19, -2], Yes], [bottom, x, [-477, 14, 4], [-477, 14, 4], Yes], [top, x, [-681, -20, 4], [-681, -20, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
67. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[4, -1, -1, 1, 0, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-18, 18, -17, 19]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=0; supplied [edge,coefficients] records=[[left, [346, -1, -1]], [right, [310, -1, -1]], [bottom, [-268, -1, 1]], [top, [-376, -1, 1]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([4, -1, -1, 1, 0, -1], 0, 0, Yes, [[left, y, [346, -1, -1], [346, -1, -1], Yes], [right, y, [310, -1, -1], [310, -1, -1], Yes], [bottom, x, [-268, -1, 1], [-268, -1, 1], Yes], [top, x, [-376, -1, 1], [-376, -1, 1], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([4, -1, -1, 1, 0, -1], 0, 0, Yes, [[left, y, [346, -1, -1], [346, -1, -1], Yes], [right, y, [310, -1, -1], [310, -1, -1], Yes], [bottom, x, [-268, -1, 1], [-268, -1, 1], Yes], [top, x, [-376, -1, 1], [-376, -1, 1], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
68. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[5, 0, 0, 2, 1, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-19, 19, -18, 20]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=4; supplied [edge,coefficients] records=[[left, [727, -19, 0]], [right, [727, 19, 0]], [bottom, [5, -18, 2]], [top, [5, 20, 2]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([5, 0, 0, 2, 1, 0], 4, 4, Yes, [[left, y, [727, -19, 0], [727, -19, 0], Yes], [right, y, [727, 19, 0], [727, 19, 0], Yes], [bottom, x, [5, -18, 2], [5, -18, 2], Yes], [top, x, [5, 20, 2], [5, 20, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([5, 0, 0, 2, 1, 0], 4, 4, Yes, [[left, y, [727, -19, 0], [727, -19, 0], Yes], [right, y, [727, 19, 0], [727, 19, 0], Yes], [bottom, x, [5, -18, 2], [5, -18, 2], Yes], [top, x, [5, 20, 2], [5, 20, 2], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
69. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[6, 1, 1, 3, -1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-20, 20, -19, 21]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=8; supplied [edge,coefficients] records=[[left, [1187, 21, 1]], [right, [1226, -19, 1]], [bottom, [348, 20, 3]], [top, [468, -20, 3]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([6, 1, 1, 3, -1, 1], 8, 8, Yes, [[left, y, [1186, 21, 1], [1187, 21, 1], No], [right, y, [1226, -19, 1], [1226, -19, 1], Yes], [bottom, x, [348, 20, 3], [348, 20, 3], Yes], [top, x, [468, -20, 3], [468, -20, 3], Yes]], No, No, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([6, 1, 1, 3, -1, 1], 8, 8, Yes, [[left, y, [1186, 21, 1], [1187, 21, 1], No], [right, y, [1226, -19, 1], [1226, -19, 1], Yes], [bottom, x, [348, 20, 3], [348, 20, 3], Yes], [top, x, [468, -20, 3], [468, -20, 3], Yes]], No, No, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
70. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[7, 2, 2, 4, 0, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-21, 21, -20, 22]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=12; supplied [edge,coefficients] records=[[left, [1729, 2, 2]], [right, [1813, 2, 2]], [bottom, [767, 2, 4]], [top, [1019, 2, 4]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([7, 2, 2, 4, 0, 2], 12, 12, Yes, [[left, y, [1729, 2, 2], [1729, 2, 2], Yes], [right, y, [1813, 2, 2], [1813, 2, 2], Yes], [bottom, x, [767, 2, 4], [767, 2, 4], Yes], [top, x, [1019, 2, 4], [1019, 2, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([7, 2, 2, 4, 0, 2], 12, 12, Yes, [[left, y, [1729, 2, 2], [1729, 2, 2], Yes], [right, y, [1813, 2, 2], [1813, 2, 2], Yes], [bottom, x, [767, 2, 4], [767, 2, 4], Yes], [top, x, [1019, 2, 4], [1019, 2, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
71. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[8, -2, 3, 1, 1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-22, 22, -21, 23]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=-2; supplied [edge,coefficients] records=[[left, [536, -19, -2]], [right, [448, 25, -2]], [bottom, [-937, -23, 1]], [top, [-981, 21, 1]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([8, -2, 3, 1, 1, -2], -2, -2, Yes, [[left, y, [536, -19, -2], [536, -19, -2], Yes], [right, y, [448, 25, -2], [448, 25, -2], Yes], [bottom, x, [-937, -23, 1], [-937, -23, 1], Yes], [top, x, [-981, 21, 1], [-981, 21, 1], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([8, -2, 3, 1, 1, -2], -2, -2, Yes, [[left, y, [536, -19, -2], [536, -19, -2], Yes], [right, y, [448, 25, -2], [448, 25, -2], Yes], [bottom, x, [-937, -23, 1], [-937, -23, 1], Yes], [top, x, [-981, 21, 1], [-981, 21, 1], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
72. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[9, -1, -3, 2, -1, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-23, 23, -22, 24]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=3; supplied [edge,coefficients] records=[[left, [1090, 20, -1]], [right, [1044, -26, -1]], [bottom, [-409, 21, 2]], [top, [-639, -25, 2]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([9, -1, -3, 2, -1, -1], 2, 3, No, [[left, y, [1090, 20, -1], [1090, 20, -1], Yes], [right, y, [1044, -26, -1], [1044, -26, -1], Yes], [bottom, x, [-409, 21, 2], [-409, 21, 2], Yes], [top, x, [-639, -25, 2], [-639, -25, 2], Yes]], Yes, No, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([9, -1, -3, 2, -1, -1], 2, 3, No, [[left, y, [1090, 20, -1], [1090, 20, -1], Yes], [right, y, [1044, -26, -1], [1044, -26, -1], Yes], [bottom, x, [-409, 21, 2], [-409, 21, 2], Yes], [top, x, [-639, -25, 2], [-639, -25, 2], Yes]], Yes, No, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
73. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[10, 0, -2, 3, 0, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-24, 24, -23, 25]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=6; supplied [edge,coefficients] records=[[left, [1738, -2, 0]], [right, [1738, -2, 0]], [bottom, [56, 0, 3]], [top, [-40, 0, 3]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([10, 0, -2, 3, 0, 0], 6, 6, Yes, [[left, y, [1738, -2, 0], [1738, -2, 0], Yes], [right, y, [1738, -2, 0], [1738, -2, 0], Yes], [bottom, x, [56, 0, 3], [56, 0, 3], Yes], [top, x, [-40, 0, 3], [-40, 0, 3], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([10, 0, -2, 3, 0, 0], 6, 6, Yes, [[left, y, [1738, -2, 0], [1738, -2, 0], Yes], [right, y, [1738, -2, 0], [1738, -2, 0], Yes], [bottom, x, [56, 0, 3], [56, 0, 3], Yes], [top, x, [-40, 0, 3], [-40, 0, 3], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
74. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[11, 1, -1, 4, 1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-25, 25, -24, 26]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=10; supplied [edge,coefficients] records=[[left, [2486, -26, 1]], [right, [2536, 24, 1]], [bottom, [611, -23, 4]], [top, [661, 27, 4]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([11, 1, -1, 4, 1, 1], 10, 10, Yes, [[left, y, [2486, -26, 1], [2486, -26, 1], Yes], [right, y, [2536, 24, 1], [2536, 24, 1], Yes], [bottom, x, [611, -23, 4], [611, -23, 4], Yes], [top, x, [661, 27, 4], [661, 27, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([11, 1, -1, 4, 1, 1], 10, 10, Yes, [[left, y, [2486, -26, 1], [2486, -26, 1], Yes], [right, y, [2536, 24, 1], [2536, 24, 1], Yes], [bottom, x, [611, -23, 4], [611, -23, 4], Yes], [top, x, [661, 27, 4], [661, 27, 4], Yes]], Yes, Yes, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
75. Close Interior and All Four Boundary Obligations Separately
Use the quadratic candidate coefficients=[12, 2, 0, 1, -1, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-26, 26, -25, 27]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=6; supplied [edge,coefficients] records=[[left, [637, 26, 2]], [right, [740, -26, 2]], [bottom, [1262, 27, 1]], [top, [1470, -25, 1]]] must include all four distinct edges. Preserve their supplied order. Give (coefficients,Laplacian,source,interior identity?,every [edge,remaining variable,restriction,supplied,identity?] row,all boundary identities?,both obligation groups?,candidate_satisfies_declared_BVP_only). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
([12, 2, 0, 1, -1, 2], 6, 6, Yes, [[left, y, [636, 26, 2], [637, 26, 2], No], [right, y, [740, -26, 2], [740, -26, 2], Yes], [bottom, x, [1262, 27, 1], [1262, 27, 1], Yes], [top, x, [1470, -25, 1], [1470, -25, 1], Yes]], No, No, candidate_satisfies_declared_BVP_only)
Recompute the interior residual and every boundary restriction independently before taking their conjunction. This verifies a candidate for the declared problem without claiming general well-posedness. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is ([12, 2, 0, 1, -1, 2], 6, 6, Yes, [[left, y, [636, 26, 2], [637, 26, 2], No], [right, y, [740, -26, 2], [740, -26, 2], Yes], [bottom, x, [1262, 27, 1], [1262, 27, 1], Yes], [top, x, [1470, -25, 1], [1470, -25, 1], Yes]], No, No, candidate_satisfies_declared_BVP_only).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-joint-candidate-audit · candidate
76. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-12, -2, -3, 1, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-2, 2, -1, 3]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=-1, boundary data=[[left, [-4, -1, -2]], [right, [-12, -5, -2]], [bottom, [-11, -1, 1]], [top, [-39, -5, 1]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(-2, -1, No, [[left, y, [-4, -1, -2], [-4, -1, -2], Yes], [right, y, [-12, -5, -2], [-12, -5, -2], Yes], [bottom, x, [-11, -1, 1], [-11, -1, 1], Yes], [top, x, [-39, -5, 1], [-39, -5, 1], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (-2, -1, No, [[left, y, [-4, -1, -2], [-4, -1, -2], Yes], [right, y, [-12, -5, -2], [-12, -5, -2], Yes], [bottom, x, [-11, -1, 1], [-11, -1, 1], Yes], [top, x, [-39, -5, 1], [-39, -5, 1], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
77. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-11, -1, -2, 2, 0, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-3, 3, -2, 4]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=2, boundary data=[[left, [11, -2, -1]], [right, [4, -2, -1]], [bottom, [-11, -1, 2]], [top, [-35, -1, 2]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(2, 2, Yes, [[left, y, [10, -2, -1], [11, -2, -1], No], [right, y, [4, -2, -1], [4, -2, -1], Yes], [bottom, x, [-11, -1, 2], [-11, -1, 2], Yes], [top, x, [-35, -1, 2], [-35, -1, 2], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (2, 2, Yes, [[left, y, [10, -2, -1], [11, -2, -1], No], [right, y, [4, -2, -1], [4, -2, -1], Yes], [bottom, x, [-11, -1, 2], [-11, -1, 2], Yes], [top, x, [-35, -1, 2], [-35, -1, 2], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
78. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-10, 0, -1, 3, 1, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-4, 4, -3, 5]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=7, boundary data=[[left, [38, -5, 0]], [right, [38, 3, 0]], [bottom, [-7, -3, 3]], [top, [-15, 5, 3]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(6, 7, No, [[left, y, [38, -5, 0], [38, -5, 0], Yes], [right, y, [38, 3, 0], [38, 3, 0], Yes], [bottom, x, [-7, -3, 3], [-7, -3, 3], Yes], [top, x, [-15, 5, 3], [-15, 5, 3], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (6, 7, No, [[left, y, [38, -5, 0], [38, -5, 0], Yes], [right, y, [38, 3, 0], [38, 3, 0], Yes], [bottom, x, [-7, -3, 3], [-7, -3, 3], Yes], [top, x, [-15, 5, 3], [-15, 5, 3], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
79. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-9, 1, 0, 4, -1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-5, 5, -4, 6]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=10, boundary data=[[left, [87, 5, 1]], [right, [96, -5, 1]], [bottom, [7, 5, 4]], [top, [27, -5, 4]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(10, 10, Yes, [[left, y, [86, 5, 1], [87, 5, 1], No], [right, y, [96, -5, 1], [96, -5, 1], Yes], [bottom, x, [7, 5, 4], [7, 5, 4], Yes], [top, x, [27, -5, 4], [27, -5, 4], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (10, 10, Yes, [[left, y, [86, 5, 1], [87, 5, 1], No], [right, y, [96, -5, 1], [96, -5, 1], Yes], [bottom, x, [7, 5, 4], [7, 5, 4], Yes], [top, x, [27, -5, 4], [27, -5, 4], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
80. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-8, 2, 1, 1, 0, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-6, 6, -5, 7]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=7, boundary data=[[left, [16, 1, 2]], [right, [40, 1, 2]], [bottom, [37, 2, 1]], [top, [97, 2, 1]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(6, 7, No, [[left, y, [16, 1, 2], [16, 1, 2], Yes], [right, y, [40, 1, 2], [40, 1, 2], Yes], [bottom, x, [37, 2, 1], [37, 2, 1], Yes], [top, x, [97, 2, 1], [97, 2, 1], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (6, 7, No, [[left, y, [16, 1, 2], [16, 1, 2], Yes], [right, y, [40, 1, 2], [40, 1, 2], Yes], [bottom, x, [37, 2, 1], [37, 2, 1], Yes], [top, x, [97, 2, 1], [97, 2, 1], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
81. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-7, -2, 2, 2, 1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-7, 7, -6, 8]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=0, boundary data=[[left, [106, -5, -2]], [right, [77, 9, -2]], [bottom, [-91, -8, 2]], [top, [-119, 6, 2]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(0, 0, Yes, [[left, y, [105, -5, -2], [106, -5, -2], No], [right, y, [77, 9, -2], [77, 9, -2], Yes], [bottom, x, [-91, -8, 2], [-91, -8, 2], Yes], [top, x, [-119, 6, 2], [-119, 6, 2], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 0, Yes, [[left, y, [105, -5, -2], [106, -5, -2], No], [right, y, [77, 9, -2], [77, 9, -2], Yes], [bottom, x, [-91, -8, 2], [-91, -8, 2], Yes], [top, x, [-119, 6, 2], [-119, 6, 2], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
82. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-6, -1, 3, 3, -1, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-8, 8, -7, 9]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=5, boundary data=[[left, [194, 11, -1]], [right, [178, -5, -1]], [bottom, [-76, 6, 3]], [top, [-60, -10, 3]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(4, 5, No, [[left, y, [194, 11, -1], [194, 11, -1], Yes], [right, y, [178, -5, -1], [178, -5, -1], Yes], [bottom, x, [-76, 6, 3], [-76, 6, 3], Yes], [top, x, [-60, -10, 3], [-60, -10, 3], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (4, 5, No, [[left, y, [194, 11, -1], [194, 11, -1], Yes], [right, y, [178, -5, -1], [178, -5, -1], Yes], [bottom, x, [-76, 6, 3], [-76, 6, 3], Yes], [top, x, [-60, -10, 3], [-60, -10, 3], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
83. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-5, 0, -3, 4, 0, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-9, 9, -8, 10]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=8, boundary data=[[left, [320, -3, 0]], [right, [319, -3, 0]], [bottom, [19, 0, 4]], [top, [-35, 0, 4]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(8, 8, Yes, [[left, y, [319, -3, 0], [320, -3, 0], No], [right, y, [319, -3, 0], [319, -3, 0], Yes], [bottom, x, [19, 0, 4], [19, 0, 4], Yes], [top, x, [-35, 0, 4], [-35, 0, 4], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (8, 8, Yes, [[left, y, [319, -3, 0], [320, -3, 0], No], [right, y, [319, -3, 0], [319, -3, 0], Yes], [bottom, x, [19, 0, 4], [19, 0, 4], Yes], [top, x, [-35, 0, 4], [-35, 0, 4], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
84. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-4, 1, -2, 1, 1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-10, 10, -9, 11]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=5, boundary data=[[left, [86, -12, 1]], [right, [106, 8, 1]], [bottom, [95, -8, 1]], [top, [95, 12, 1]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(4, 5, No, [[left, y, [86, -12, 1], [86, -12, 1], Yes], [right, y, [106, 8, 1], [106, 8, 1], Yes], [bottom, x, [95, -8, 1], [95, -8, 1], Yes], [top, x, [95, 12, 1], [95, 12, 1], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (4, 5, No, [[left, y, [86, -12, 1], [86, -12, 1], Yes], [right, y, [106, 8, 1], [106, 8, 1], Yes], [bottom, x, [95, -8, 1], [95, -8, 1], Yes], [top, x, [95, 12, 1], [95, 12, 1], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
85. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-3, 2, -1, 2, -1, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-11, 11, -10, 12]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=8, boundary data=[[left, [218, 10, 2]], [right, [261, -12, 2]], [bottom, [207, 12, 2]], [top, [273, -10, 2]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(8, 8, Yes, [[left, y, [217, 10, 2], [218, 10, 2], No], [right, y, [261, -12, 2], [261, -12, 2], Yes], [bottom, x, [207, 12, 2], [207, 12, 2], Yes], [top, x, [273, -10, 2], [273, -10, 2], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (8, 8, Yes, [[left, y, [217, 10, 2], [218, 10, 2], No], [right, y, [261, -12, 2], [261, -12, 2], Yes], [bottom, x, [207, 12, 2], [207, 12, 2], Yes], [top, x, [273, -10, 2], [273, -10, 2], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
86. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-2, -2, 0, 3, 0, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-12, 12, -11, 13]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=3, boundary data=[[left, [454, 0, -2]], [right, [406, 0, -2]], [bottom, [-244, -2, 3]], [top, [-340, -2, 3]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(2, 3, No, [[left, y, [454, 0, -2], [454, 0, -2], Yes], [right, y, [406, 0, -2], [406, 0, -2], Yes], [bottom, x, [-244, -2, 3], [-244, -2, 3], Yes], [top, x, [-340, -2, 3], [-340, -2, 3], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (2, 3, No, [[left, y, [454, 0, -2], [454, 0, -2], Yes], [right, y, [406, 0, -2], [406, 0, -2], Yes], [bottom, x, [-244, -2, 3], [-244, -2, 3], Yes], [top, x, [-340, -2, 3], [-340, -2, 3], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
87. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[-1, -1, 1, 4, 1, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-13, 13, -12, 14]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=6, boundary data=[[left, [689, -12, -1]], [right, [662, 14, -1]], [bottom, [-157, -13, 4]], [top, [-183, 13, 4]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(6, 6, Yes, [[left, y, [688, -12, -1], [689, -12, -1], No], [right, y, [662, 14, -1], [662, 14, -1], Yes], [bottom, x, [-157, -13, 4], [-157, -13, 4], Yes], [top, x, [-183, 13, 4], [-183, 13, 4], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (6, 6, Yes, [[left, y, [688, -12, -1], [689, -12, -1], No], [right, y, [662, 14, -1], [662, 14, -1], Yes], [bottom, x, [-157, -13, 4], [-157, -13, 4], Yes], [top, x, [-183, 13, 4], [-183, 13, 4], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
88. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[0, 0, 2, 1, -1, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-14, 14, -13, 15]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=3, boundary data=[[left, [196, 16, 0]], [right, [196, -12, 0]], [bottom, [-26, 13, 1]], [top, [30, -15, 1]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(2, 3, No, [[left, y, [196, 16, 0], [196, 16, 0], Yes], [right, y, [196, -12, 0], [196, -12, 0], Yes], [bottom, x, [-26, 13, 1], [-26, 13, 1], Yes], [top, x, [30, -15, 1], [30, -15, 1], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (2, 3, No, [[left, y, [196, 16, 0], [196, 16, 0], Yes], [right, y, [196, -12, 0], [196, -12, 0], Yes], [bottom, x, [-26, 13, 1], [-26, 13, 1], Yes], [top, x, [30, -15, 1], [30, -15, 1], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
89. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[1, 1, 3, 2, 0, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-15, 15, -14, 16]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=6, boundary data=[[left, [437, 3, 1]], [right, [466, 3, 1]], [bottom, [155, 1, 2]], [top, [305, 1, 2]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(6, 6, Yes, [[left, y, [436, 3, 1], [437, 3, 1], No], [right, y, [466, 3, 1], [466, 3, 1], Yes], [bottom, x, [155, 1, 2], [155, 1, 2], Yes], [top, x, [305, 1, 2], [305, 1, 2], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (6, 6, Yes, [[left, y, [436, 3, 1], [437, 3, 1], No], [right, y, [466, 3, 1], [466, 3, 1], Yes], [bottom, x, [155, 1, 2], [155, 1, 2], Yes], [top, x, [305, 1, 2], [305, 1, 2], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
90. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[2, 2, -3, 3, 1, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-16, 16, -15, 17]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=11, boundary data=[[left, [738, -19, 2]], [right, [802, 13, 2]], [bottom, [497, -13, 3]], [top, [529, 19, 3]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(10, 11, No, [[left, y, [738, -19, 2], [738, -19, 2], Yes], [right, y, [802, 13, 2], [802, 13, 2], Yes], [bottom, x, [497, -13, 3], [497, -13, 3], Yes], [top, x, [529, 19, 3], [529, 19, 3], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (10, 11, No, [[left, y, [738, -19, 2], [738, -19, 2], Yes], [right, y, [802, 13, 2], [802, 13, 2], Yes], [bottom, x, [497, -13, 3], [497, -13, 3], Yes], [top, x, [529, 19, 3], [529, 19, 3], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
91. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[3, -2, -2, 4, -1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-17, 17, -16, 18]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=4, boundary data=[[left, [1194, 15, -2]], [right, [1125, -19, -2]], [bottom, [-477, 14, 4]], [top, [-681, -20, 4]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(4, 4, Yes, [[left, y, [1193, 15, -2], [1194, 15, -2], No], [right, y, [1125, -19, -2], [1125, -19, -2], Yes], [bottom, x, [-477, 14, 4], [-477, 14, 4], Yes], [top, x, [-681, -20, 4], [-681, -20, 4], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (4, 4, Yes, [[left, y, [1193, 15, -2], [1194, 15, -2], No], [right, y, [1125, -19, -2], [1125, -19, -2], Yes], [bottom, x, [-477, 14, 4], [-477, 14, 4], Yes], [top, x, [-681, -20, 4], [-681, -20, 4], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
92. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[4, -1, -1, 1, 0, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-18, 18, -17, 19]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=1, boundary data=[[left, [346, -1, -1]], [right, [310, -1, -1]], [bottom, [-268, -1, 1]], [top, [-376, -1, 1]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(0, 1, No, [[left, y, [346, -1, -1], [346, -1, -1], Yes], [right, y, [310, -1, -1], [310, -1, -1], Yes], [bottom, x, [-268, -1, 1], [-268, -1, 1], Yes], [top, x, [-376, -1, 1], [-376, -1, 1], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, No, [[left, y, [346, -1, -1], [346, -1, -1], Yes], [right, y, [310, -1, -1], [310, -1, -1], Yes], [bottom, x, [-268, -1, 1], [-268, -1, 1], Yes], [top, x, [-376, -1, 1], [-376, -1, 1], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
93. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[5, 0, 0, 2, 1, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-19, 19, -18, 20]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=4, boundary data=[[left, [728, -19, 0]], [right, [727, 19, 0]], [bottom, [5, -18, 2]], [top, [5, 20, 2]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(4, 4, Yes, [[left, y, [727, -19, 0], [728, -19, 0], No], [right, y, [727, 19, 0], [727, 19, 0], Yes], [bottom, x, [5, -18, 2], [5, -18, 2], Yes], [top, x, [5, 20, 2], [5, 20, 2], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (4, 4, Yes, [[left, y, [727, -19, 0], [728, -19, 0], No], [right, y, [727, 19, 0], [727, 19, 0], Yes], [bottom, x, [5, -18, 2], [5, -18, 2], Yes], [top, x, [5, 20, 2], [5, 20, 2], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
94. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[6, 1, 1, 3, -1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-20, 20, -19, 21]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=9, boundary data=[[left, [1186, 21, 1]], [right, [1226, -19, 1]], [bottom, [348, 20, 3]], [top, [468, -20, 3]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(8, 9, No, [[left, y, [1186, 21, 1], [1186, 21, 1], Yes], [right, y, [1226, -19, 1], [1226, -19, 1], Yes], [bottom, x, [348, 20, 3], [348, 20, 3], Yes], [top, x, [468, -20, 3], [468, -20, 3], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (8, 9, No, [[left, y, [1186, 21, 1], [1186, 21, 1], Yes], [right, y, [1226, -19, 1], [1226, -19, 1], Yes], [bottom, x, [348, 20, 3], [348, 20, 3], Yes], [top, x, [468, -20, 3], [468, -20, 3], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
95. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[7, 2, 2, 4, 0, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-21, 21, -20, 22]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=12, boundary data=[[left, [1730, 2, 2]], [right, [1813, 2, 2]], [bottom, [767, 2, 4]], [top, [1019, 2, 4]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(12, 12, Yes, [[left, y, [1729, 2, 2], [1730, 2, 2], No], [right, y, [1813, 2, 2], [1813, 2, 2], Yes], [bottom, x, [767, 2, 4], [767, 2, 4], Yes], [top, x, [1019, 2, 4], [1019, 2, 4], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (12, 12, Yes, [[left, y, [1729, 2, 2], [1730, 2, 2], No], [right, y, [1813, 2, 2], [1813, 2, 2], Yes], [bottom, x, [767, 2, 4], [767, 2, 4], Yes], [top, x, [1019, 2, 4], [1019, 2, 4], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
96. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[8, -2, 3, 1, 1, -2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-22, 22, -21, 23]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=-1, boundary data=[[left, [536, -19, -2]], [right, [448, 25, -2]], [bottom, [-937, -23, 1]], [top, [-981, 21, 1]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(-2, -1, No, [[left, y, [536, -19, -2], [536, -19, -2], Yes], [right, y, [448, 25, -2], [448, 25, -2], Yes], [bottom, x, [-937, -23, 1], [-937, -23, 1], Yes], [top, x, [-981, 21, 1], [-981, 21, 1], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (-2, -1, No, [[left, y, [536, -19, -2], [536, -19, -2], Yes], [right, y, [448, 25, -2], [448, 25, -2], Yes], [bottom, x, [-937, -23, 1], [-937, -23, 1], Yes], [top, x, [-981, 21, 1], [-981, 21, 1], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
97. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[9, -1, -3, 2, -1, -1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-23, 23, -22, 24]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=2, boundary data=[[left, [1091, 20, -1]], [right, [1044, -26, -1]], [bottom, [-409, 21, 2]], [top, [-639, -25, 2]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(2, 2, Yes, [[left, y, [1090, 20, -1], [1091, 20, -1], No], [right, y, [1044, -26, -1], [1044, -26, -1], Yes], [bottom, x, [-409, 21, 2], [-409, 21, 2], Yes], [top, x, [-639, -25, 2], [-639, -25, 2], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (2, 2, Yes, [[left, y, [1090, 20, -1], [1091, 20, -1], No], [right, y, [1044, -26, -1], [1044, -26, -1], Yes], [bottom, x, [-409, 21, 2], [-409, 21, 2], Yes], [top, x, [-639, -25, 2], [-639, -25, 2], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
98. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[10, 0, -2, 3, 0, 0] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-24, 24, -23, 25]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=7, boundary data=[[left, [1738, -2, 0]], [right, [1738, -2, 0]], [bottom, [56, 0, 3]], [top, [-40, 0, 3]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(6, 7, No, [[left, y, [1738, -2, 0], [1738, -2, 0], Yes], [right, y, [1738, -2, 0], [1738, -2, 0], Yes], [bottom, x, [56, 0, 3], [56, 0, 3], Yes], [top, x, [-40, 0, 3], [-40, 0, 3], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (6, 7, No, [[left, y, [1738, -2, 0], [1738, -2, 0], Yes], [right, y, [1738, -2, 0], [1738, -2, 0], Yes], [bottom, x, [56, 0, 3], [56, 0, 3], Yes], [top, x, [-40, 0, 3], [-40, 0, 3], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
99. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[11, 1, -1, 4, 1, 1] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-25, 25, -24, 26]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=10, boundary data=[[left, [2487, -26, 1]], [right, [2536, 24, 1]], [bottom, [611, -23, 4]], [top, [661, 27, 4]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(10, 10, Yes, [[left, y, [2486, -26, 1], [2487, -26, 1], No], [right, y, [2536, 24, 1], [2536, 24, 1], Yes], [bottom, x, [611, -23, 4], [611, -23, 4], Yes], [top, x, [661, 27, 4], [661, 27, 4], Yes]], No, interior_pass_boundary_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (10, 10, Yes, [[left, y, [2486, -26, 1], [2487, -26, 1], No], [right, y, [2536, 24, 1], [2536, 24, 1], Yes], [bottom, x, [611, -23, 4], [611, -23, 4], Yes], [top, x, [661, 27, 4], [661, 27, 4], Yes]], No, interior_pass_boundary_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
100. Interior Pass and Boundary Pass Must Not Be Conflated
Use the quadratic candidate coefficients=[12, 2, 0, 1, -1, 2] in fixed order [1,x,y,x²,x*y,y²]. Rectangle bounds [x_min,x_max,y_min,y_max]=[-26, 26, -25, 27]. Left/right fix x, leaving y; bottom/top fix y, leaving x. Boundary polynomials use ascending coefficients [constant,linear,quadratic]. Compare every coefficient for an edge identity, not a few sampled boundary values. Interior source=7, boundary data=[[left, [636, 26, 2]], [right, [740, -26, 2]], [bottom, [1262, 27, 1]], [top, [1470, -25, 1]]] with all four distinct edges. Give (Laplacian,source,interior identity?,boundary rows in supplied order,boundary identities?,one of both_pass / interior_pass_boundary_fail / boundary_pass_interior_fail / both_fail). Interior PDE and boundary data are separate obligations. Symbolic success covers only this candidate and declared BVP, not a general existence, uniqueness, stability, regularity or well-posedness theorem.
View answer and explanation
(6, 7, No, [[left, y, [636, 26, 2], [636, 26, 2], Yes], [right, y, [740, -26, 2], [740, -26, 2], Yes], [bottom, x, [1262, 27, 1], [1262, 27, 1], Yes], [top, x, [1470, -25, 1], [1470, -25, 1], Yes]], Yes, boundary_pass_interior_fail)
Report both groups even when only one succeeds. A boundary pass does not repair a failed interior equation, or conversely. This exact finite slice does not automatically establish a general infinite conclusion. The exact result, retaining the requested order, is (6, 7, No, [[left, y, [636, 26, 2], [636, 26, 2], Yes], [right, y, [740, -26, 2], [740, -26, 2], Yes], [bottom, x, [1262, 27, 1], [1262, 27, 1], Yes], [top, x, [1470, -25, 1], [1470, -25, 1], Yes]], Yes, boundary_pass_interior_fail).
Reviewed step-by-step derivation
Boundary-value problem · qf-mod25-bvp-obligation-separation · candidate
Canonical question JSONL · English JSONL · Validation report