Question 11687: Second-Order Residual with an Explicit Mixed-Term Convention

Original question and givens

Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-1, -1, 1, 4, 1, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, -1, 4] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-1, x=-1, y=1. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).

• G001 — Use the explicit linear constant-coefficient PDE operator and polynomial candidate coefficients=[-1, -1, 1, 4, 1, -1] in order [1,x,y,x²,x*y,y²]. Fix the operator convention before evaluating residuals. A point residual proves only a point check; general solution, existence, uniqueness and well-posedness remain separate obligations. principal [a,b,c]=[3, -1, 4] means a*u_xx+b*u_xy+c*u_yy, not 2*b*u_xy. source=-1, x=-1, y=1. Give (coefficients,principal,x,y,u_xx,u_xy,u_yy,source,signed residual,satisfied at the point?).

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

• mko-partial-differential-equation

Step-by-step derivation

original coefficients=[-1, -1, 1, 4, 1, -1];basis=[1,x,y,x^2,xy,y^2]

u(x,y)=(-1)+(-1)*x+(1)*y+(4)*x^2+(1)*x*y+(-1)*y^2

u_xx constant := mul(2, 4)

u_xx constant = 8

u_yy constant := mul(2, -1)

u_yy constant = -2

u coefficients=[-1, -1, 1, 4, 1, -1];u_x=c1+2*c3*x+c4*y;u_y=c2+c4*x+2*c5*y

u_xx=8;u_xy=1;u_yy=-2

principal=[3, -1, 4];point=[-1, 1]

operator=a*u_xx+b*u_xy+c*u_yy;not 2*b*u_xy

second order term0 := mul(3, 8)

second order term0 = 24

second order term1 := mul(-1, 1)

second order term1 = -1

second order term2 := mul(4, -2)

second order term2 = -8

second order operator sum := sum(24, -1, -8)

second order operator sum = 15

second order signed residual := sub(15, -1)

second order signed residual = 16

query point=[-1, 1];operator value=15;source=-1;signed residual=16

satisfied at declared point=No;no general solution claim

Reverse checks

• S011 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S010 · premise_and_result_audit — Recompute second order signed residual with exact rational arithmetic, not rounded decimals.
• S009 · premise_and_result_audit — Recompute second order operator sum with exact rational arithmetic, not rounded decimals.
• S008 · premise_and_result_audit — Recompute second order term2 with exact rational arithmetic, not rounded decimals.
• S007 · premise_and_result_audit — Recompute second order term1 with exact rational arithmetic, not rounded decimals.
• S006 · premise_and_result_audit — Recompute second order term0 with exact rational arithmetic, not rounded decimals.
• S005 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S004 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S003 · premise_and_result_audit — Recompute u_yy constant with exact rational arithmetic, not rounded decimals.
• S002 · premise_and_result_audit — Recompute u_xx constant with exact rational arithmetic, not rounded decimals.
• S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

([-1, -1, 1, 4, 1, -1], [3, -1, 4], -1, 1, 8, 1, -2, -1, 16, No)

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.