# Partial differential equation: principal-part classification and symbolic residuals of quadratic polynomial candidates [← Guide index](https://ocme.evemisslab.com/en/derivations/) · [Original MKO](https://ocme.evemisslab.com/en/mko/mko-partial-differential-equation/) This is an owner-adopted natural-language teaching guide. Finite/arithmetic checks are not universal proofs; no Lean was run, and review scope is recorded per revision. Original source status labels are retained as historical data. Formulas retain exact TeX; no new semantic AST is invented. derivation-mko-partial-differential-equation-01 · production-20261019-adopted-96 ## Goal and scope On ℝ², take the linear constant-coefficient operator L(u)=u_xx+2u_xy−3u_yy+u_x and the right-hand side g(x,y)=2x+y−2. First classify L using the explicit principal-part convention a u_xx+b u_xy+c u_yy with discriminant b²−4ac; then, for the three quadratic polynomial candidates u_A=x²+xy+y², u_B=x²+2xy and u_C=x²+xy+y²+y, compute the symbolic residual R=L(u)−g exactly, decide which candidates satisfy L(u)=g on all of ℝ², and explain why u_B is still not a solution although its point residual vanishes on the whole line y=−8. Stays inside the MKO's slice: bivariate quadratic polynomial candidates, a linear constant-coefficient operator, symbolic partial derivatives of polynomials, point residuals, principal-part classification and polynomial identities. The principal-part convention is fixed as a u_xx+b u_xy+c u_yy (the mixed coefficient is called b, not 2b). General solutions, boundary or initial conditions, existence and uniqueness, weak solutions and well-posedness are not treated; the canonical MKO and its English edition are not changed. ## Givens and symbols - numeric_evaluation - distributive - eq_replace - poly_partial - quantifier_negation - G1 — Operator and equation: for twice differentiable u on ℝ², L(u)=u_xx+2u_xy−3u_yy+u_x, where u_xy means differentiate first in x and then in y; the right-hand side is g(x,y)=2x+y−2; the equation asked about is L(u)=g on all of ℝ². - G2 — Principal-part convention (the MKO requires it to be explicit): the second-order principal part is written a u_xx+b u_xy+c u_yy, with b the full coefficient of u_xy; the discriminant is D=b²−4ac. - G3 — Three quadratic polynomial candidates: u_A=x²+xy+y², u_B=x²+2xy, u_C=x²+xy+y²+y. - u_x,\ u_y — first partial derivatives of u in x and in y - u_{xx},\ u_{xy},\ u_{yy} — second partial derivatives; u_{xy} means differentiate in x, then in y - L — the linear constant-coefficient operator of this guide - a,\ b,\ c — coefficients of the principal part a u_{xx}+b u_{xy}+c u_{yy} - D=b^2-4ac — the discriminant under this convention - A,\ B,\ C — coefficients of the other convention A u_{xx}+2B u_{xy}+C u_{yy} - R_u=L(u)-g — the symbolic residual polynomial of candidate u ## Rules and sources - numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable. - distributive · prerequisite — Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu. - eq_replace · prerequisite — Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true. - quantifier_negation · prerequisite — Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x). - def_unfold · prerequisite — Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true. - neg_mul · prerequisite — Signs and multiplication: (−u)v=−(uv) and (−u)(−v)=uv. - order_basics · prerequisite — Order of the reals: exactly one of uv holds; u0 give uwvw. - power_laws · prerequisite — Integer exponent laws (base nonzero, or exponents positive integers): a^m·a^n=a^{m+n}, (a^m)^n=a^{mn}, (ab)^n=a^n b^n, a^0=1. - mul_comm · prerequisite — Commutativity of multiplication: uv=vu (in commutative number systems such as ℤ, ℚ, ℝ, ℂ). - add_inverse · prerequisite — Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v). - add_comm · prerequisite — Commutativity of addition: u+v=v+u. - forall_intro · prerequisite — ∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain. - exists_intro · prerequisite — ∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x). - and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q. - mko_pde · cited — This slice fixes bivariate quadratic polynomial candidates u and linear constant-coefficient operators and computes first- or second-order residuals exactly. The second-order principal part is a u_xx+b u_xy+c u_yy, with discriminant b²−4ac; this convention must be explicit. A zero point residual proves only that point check; only an identically zero symbolic residual polynomial proves that the specified polynomial satisfies the operator equation globally. · mko-partial-differential-equation - poly_partial · prerequisite — Partial derivatives of polynomials: a bivariate polynomial has partial derivatives of every order; ∂/∂x acts termwise and linearly, treating y as a constant, with ∂/∂x(x^m y^k)=m x^(m−1) y^k for m≥1 and 0 for m=0; ∂/∂y acts symmetrically. For polynomials the mixed partials do not depend on the order: u_xy=u_yx. - pde_class · prerequisite — Classification of constant-coefficient second-order linear operators (standard definition): if the coefficients of the principal part a u_xx+b u_xy+c u_yy are not all 0, the operator is called hyperbolic when b²−4ac>0, parabolic when b²−4ac=0 and elliptic when b²−4ac<0. The class depends only on the principal part, not on lower-order terms or the right-hand side. - poly_coeff_zero · prerequisite — A polynomial whose coefficients are all 0 takes the value 0 at every point of ℝ²; hence two polynomials with identical coefficients take equal values at every point. - L_conv · lemma — If the same principal part is also written A u_xx+2B u_xy+C u_yy (that is, a=A, b=2B, c=C), then b²−4ac=4(B²−AC); since 4>0, the two discriminants have the same sign. ## Main derivation S01 · definitionBefore L(u)=u_{xx}+2u_{xy}-3u_{yy}+u_x After a=1,\ b=2,\ c=-3;\ \text{lower-order part } u_x;\ g=2x+y-2 Conditions: none Rule: mko_pde · Uses: G1, G2 Leaf basis: mko_pde + def_unfold Under the convention of G2, the principal part is the second-order part a u_xx+b u_xy+c u_yy of L. Matching term by term: the coefficient of u_xx is a=1, the full coefficient of u_xy is b=2 (not halved), and the coefficient of u_yy is c=−3. The term u_x is first order (lower order) and does not enter the classification; g is the right-hand side. Stating the convention first is an explicit requirement of the MKO, because under another way of writing the same operator the mixed coefficient would be read as 1. S02 · equivalentBefore D=b^2-4ac After D=2^2-4\cdot 1\cdot(-3)=16 Conditions: none Rule: numeric_evaluation · Uses: S01 Leaf basis: numeric_evaluation + neg_mul Substitute a=1, b=2, c=−3: b²=4; 4ac=4·1·(−3)=−12 (a negative times a positive is negative); hence D=4−(−12)=4+12=16. Subtracting a negative number adds its opposite. S03 · impliesBefore D=16 After L\ \text{is hyperbolic} Conditions: (a,b,c)\neq(0,0,0); D>0 Rule: pde_class · Uses: S01, S02, pde_class Leaf basis: pde_class + order_basics The classification definition requires the principal coefficients not to be all 0: a=1≠0, so this holds. D=16>0, so by definition L is hyperbolic. The lower-order term u_x and the right-hand side g do not affect the class. S04 · impliesBefore A u_{xx}+2B u_{xy}+C u_{yy}=u_{xx}+2u_{xy}-3u_{yy} After A=1,\ B=1,\ C=-3,\ B^2-AC=4=\tfrac{16}{4} Conditions: b=2B Rule: L_conv · Uses: S01, S02, L_conv Leaf basis: L_conv + numeric_evaluation Under the other convention the mixed coefficient is written 2B, so 2B=2 and B=1; A=1 and C=−3. Then B²−AC=1−(1)(−3)=4. By Lemma L_conv, b²−4ac=4(B²−AC), that is 16=4·4, so the two conventions give the same sign and the same class. This requires each formula to be used with its own coefficients: putting b=2 into B²−AC gives 4+3=7, and putting B=1 into b²−4ac gives 1+12=13; neither is the discriminant under either convention. S05 · impliesBefore u_A=x^2+xy+y^2 After (u_A)_x=2x+y,\ (u_A)_y=x+2y Conditions: none Rule: poly_partial · Uses: G3, poly_partial Leaf basis: poly_partial Differentiate term by term. In x: ∂x(x²)=2x, ∂x(xy)=y (y treated as a constant), ∂x(y²)=0, so (u_A)_x=2x+y. In y: ∂y(x²)=0, ∂y(xy)=x, ∂y(y²)=2y, so (u_A)_y=x+2y. S06 · impliesBefore (u_A)_x=2x+y,\ (u_A)_y=x+2y After (u_A)_{xx}=2,\ (u_A)_{xy}=1,\ (u_A)_{yy}=2 Conditions: none Rule: poly_partial · Uses: S05, poly_partial Leaf basis: poly_partial Differentiate once more: (u_A)_xx=∂x(2x+y)=2; (u_A)_xy=∂y(2x+y)=1 (by G1, first x and then y); (u_A)_yy=∂y(x+2y)=2. Cross-check: (u_A)_yx=∂x(x+2y)=1, the same as (u_A)_xy, as expected because mixed partials of polynomials do not depend on the order. S07 · equivalentBefore L(u_A)=(u_A)_{xx}+2(u_A)_{xy}-3(u_A)_{yy}+(u_A)_x After L(u_A)=2x+y-2 Conditions: none Rule: eq_replace · Uses: G1, S05, S06 Leaf basis: eq_replace + distributive + numeric_evaluation Substitute the results of S06 and S05 into the operator of G1: L(u_A)=2+2·1−3·2+(2x+y). The constant part is 2+2−6=−2 (S07.1), so L(u_A)=2x+y−2. This is a symbolic computation, valid at every point (x,y), not a sample at chosen points. S08 · equivalentBefore R_{u_A}=L(u_A)-g After R_{u_A}=(2-2)x+(1-1)y+(-2+2)=0 Conditions: none Rule: distributive · Uses: S07, G1 Leaf basis: distributive + add_inverse + add_comm R_{u_A}=(2x+y−2)−(2x+y−2). Collecting like terms by distributivity: the coefficient of x is 2−2=0, of y is 1−1=0, and the constant is −2−(−2)=0. All three coefficients are 0, so the symbolic residual is the zero polynomial; this is the kind of evidence that, by MKO assumption a2, proves a global statement. S09 · impliesBefore R_{u_A}\equiv 0 After \forall (x,y)\in\mathbb{R}^2:\ L(u_A)(x,y)=g(x,y) Conditions: none Rule: poly_coeff_zero · Uses: S08, poly_coeff_zero, forall_intro Leaf basis: poly_coeff_zero + forall_intro A polynomial with all coefficients 0 takes the value 0 at every point. Take any (x,y)∈ℝ²: R_{u_A}(x,y)=0, i.e. L(u_A)(x,y)−g(x,y)=0, so L(u_A)(x,y)=g(x,y). The argument puts no extra assumption on (x,y), so by ∀-introduction u_A satisfies L(u)=g on all of ℝ². S10 · impliesBefore u_B=x^2+2xy After (u_B)_x=2x+2y,\ (u_B)_y=2x,\ (u_B)_{xx}=2,\ (u_B)_{xy}=2,\ (u_B)_{yy}=0 Conditions: none Rule: poly_partial · Uses: G3, poly_partial Leaf basis: poly_partial First order: ∂x(x²)=2x and ∂x(2xy)=2y, so (u_B)_x=2x+2y; ∂y(x²)=0 and ∂y(2xy)=2x, so (u_B)_y=2x. Second order: (u_B)_xx=∂x(2x+2y)=2, (u_B)_xy=∂y(2x+2y)=2, (u_B)_yy=∂y(2x)=0. Cross-check: (u_B)_yx=∂x(2x)=2. S11 · equivalentBefore R_{u_B}=L(u_B)-g After R_{u_B}=(2+4-0+2x+2y)-(2x+y-2)=y+8 Conditions: none Rule: eq_replace · Uses: G1, S10 Leaf basis: eq_replace + distributive + numeric_evaluation Substitute: L(u_B)=2+2·2−3·0+(2x+2y)=2x+2y+6. Subtract g: the coefficient of x is 2−2=0, of y is 2−1=1, and the constant is 6−(−2)=8. So R_{u_B}=y+8. The coefficient of y is 1≠0, so this is not the zero polynomial. S12 · impliesBefore R_{u_B}(x,-8) After R_{u_B}(x,-8)=-8+8=0\ \text{for every } x\in\mathbb{R} Conditions: none Rule: eq_replace · Uses: S11 Leaf basis: eq_replace + numeric_evaluation Substituting y=−8 into R_{u_B}=y+8 gives 0, independently of x. So the point residual of u_B vanishes at infinitely many points, the whole line y=−8, for example (0,−8) and (5,−8). This is exactly the situation that is easily misread as 'u_B is a solution'. S13 · impliesBefore R_{u_B}(0,0) After R_{u_B}(0,0)=8\neq 0,\ \text{so}\ \neg\,\forall (x,y):\ L(u_B)(x,y)=g(x,y) Conditions: none Rule: quantifier_negation · Uses: S11, quantifier_negation Leaf basis: quantifier_negation + exists_intro + numeric_evaluation At (0,0): R_{u_B}(0,0)=0+8=8≠0; that is, L(u_B)(0,0)=6 while g(0,0)=−2. The negation of the universal statement 'for all (x,y), L(u_B)=g' is equivalent to 'there is a point where it fails', and (0,0) is such a point. So u_B is not a solution of L(u)=g on ℝ², even though by S12 its point residual vanishes on a whole line. S14 · impliesBefore u_C=x^2+xy+y^2+y After R_{u_C}\equiv 0,\quad u_C-u_A=y\not\equiv 0 Conditions: none Rule: poly_partial · Uses: G1, G3, poly_partial, S09 Leaf basis: poly_partial + eq_replace + poly_coeff_zero (u_C)_x=2x+y and (u_C)_y=x+2y+1; the second partials (u_C)_xx=2, (u_C)_xy=1, (u_C)_yy=2 are the same as for u_A. So L(u_C)=2+2−6+(2x+y)=2x+y−2, R_{u_C} is the zero polynomial, and u_C also satisfies L(u)=g on all of ℝ² (for the same reason as S09). But u_C−u_A=y is not the zero polynomial, so u_C≠u_A. The equation L(u)=g by itself, with no boundary or initial conditions, has at least two distinct polynomial solutions: one passing candidate gives neither the general solution nor uniqueness. S15 · impliesBefore \text{results of S03, S09, S13, S14} After L\ \text{hyperbolic};\ u_A,u_C\ \text{solve } L(u)=g\ \text{on}\ \mathbb{R}^2;\ u_B\ \text{does not} Conditions: none Rule: and_intro_elim · Uses: S03, S09, S13, S14 Leaf basis: and_intro_elim Combine the separate conclusions: the classification (S03), the global symbolic certificates for u_A and u_C (S09, S14), and the one-point refutation for u_B (S13). Each conclusion has its own evidence and none substitutes for another: the class says nothing about solutions, point residuals say nothing global, and a passing candidate says nothing about uniqueness. ## Substeps and lemmas S04.1 · equivalentBefore (2B)^2-4AC After 4(B^2-AC) Conditions: none Rule: distributive · Uses: power_laws, distributive Leaf basis: power_laws + distributive + mul_comm + order_basics By the exponent laws (2B)²=2²B²=4B²; by distributivity 4B²−4AC=4(B²−AC). Substituting a=A, b=2B, c=C gives b²−4ac=4(B²−AC). Since 4>0, multiplying by 4 preserves whether a value is positive, zero or negative, so the two discriminants have the same sign. This proves Lemma L_conv (N3 checks the same polynomial identity on a grid). S07.1 · equivalentBefore 2+2\cdot 1-3\cdot 2 After -2 Conditions: none Rule: numeric_evaluation · Uses: numeric_evaluation Leaf basis: numeric_evaluation 2+2=4, 3·2=6, 4−6=−2. ## Reverse audits - S15 · premise_audit · S03, S09, S13, S14, and_intro_elim — This only conjoins proved conclusions. It does not claim a general solution, existence and uniqueness, or well-posedness, which lie outside this slice (MKO a2, para2). - S14 · premise_audit · G3, poly_partial, poly_coeff_zero, S09 — It needs the zero symbolic residual of u_C and that u_C−u_A is not the zero polynomial. It claims only non-uniqueness without side conditions, and nothing about what happens once boundary or initial conditions are added. - S13 · premise_audit · S11, quantifier_negation — A refutation needs only one exact point; what is needed is the residual formula from S11 and quantifier negation. This does not say u_B fails the equation at every point (S12 shows it holds on y=−8). - S12 · premise_audit · S11, eq_replace — This only evaluates on the line y=−8. The conclusion concerns these points only and does not give that R_{u_B} is the zero polynomial; S13 shows it is not. - S11 · inverse_equivalence · S10, eq_replace, distributive — Substitution and collecting like terms are rewrites into equal expressions and can be undone; both are the same polynomial. - S10 · premise_audit · G3, poly_partial — As in S05 and S06, only the differentiation rule for polynomials is used; no converse is claimed. - S09 · premise_audit · S08, poly_coeff_zero, forall_intro — It needs the residual to be the zero polynomial (S08), not merely zero at some points. It claims only that u_A is one solution of this equation on ℝ², not that it is the unique or the general solution. - S08 · inverse_equivalence · S07, distributive, add_inverse — Collecting like terms rewrites a polynomial into an equal one and can be undone; both expressions are the same polynomial. - S07.1 · inverse_equivalence · numeric_evaluation — Backwards: −2+6=4=2+2·1; both sides are the same integer. - S07 · inverse_equivalence · S05, S06, eq_replace — Replacement of equals works both ways: 2x+y−2 splits back into 2+2·1−3·2+(2x+y), and each term equals the corresponding partial derivative. - S06 · premise_audit · S05, poly_partial — It needs the first partials from S05 and the differentiation rule; all second partials are constants because u_A is quadratic. No converse is claimed. - S05 · premise_audit · G3, poly_partial — Only the termwise differentiation rule for polynomials is used. No converse is claimed: recovering a function from one partial derivative fixes it only up to a function of the other variable, so the step is not an equivalence. - S04.1 · inverse_equivalence · power_laws, distributive — Expanding backwards, 4(B²−AC)=4B²−4AC=(2B)²−4AC; both are the same polynomial, valid for all A, B, C with no side condition. - S04 · premise_audit · S01, S02, L_conv — The conclusion depends on the consistent reading b=2B. Only the agreement in sign of the two correct formulas is claimed, not that any mixed formula that happens to have the same sign is acceptable; an error control gives an example where mixing changes the sign. - S03 · premise_audit · S01, S02, pde_class — It needs a nonzero principal part and the sign of D. The class is a property of the principal part only; this step does not claim that L(u)=g has a solution, or a unique one, under any boundary or initial conditions. - S02 · inverse_equivalence · S01, numeric_evaluation, neg_mul — Recomputing backwards: 16−4=12=−4ac, consistent with 4ac=−12; both sides are the same number, so the equality holds both ways. - S01 · premise_audit · G1, G2, mko_pde — This step only reads off coefficients under the convention and derives no new fact. The reading depends on G2: under A u_xx+2B u_xy+C u_yy the mixed coefficient would be B=1, and the discriminant formula must change with it (see S04). ## Error controls and negative controls A zero point residual is not a solutionThe common mistake named by the MKO is to claim a global solution because sampled point residuals are zero. The residual of u_B is zero on the whole line y=−8, even at infinitely many points, yet the residual polynomial y+8 is not the zero polynomial and equals 8 at (0,0). Only when every coefficient of the symbolic residual is 0 (S08) is the equation proved on all of ℝ². Mixing the b and 2B discriminantsAnother misconception identified in the MKO is mixing the discriminants for 2b u_xy and b u_xy. Example: the operator u_xx+2u_xy+2u_yy. By G2, a=1, b=2, c=2 and D=4−8=−4<0, elliptic; with A u_xx+2B u_xy+C u_yy, B=1 and B²−AC=1−2=−1<0, also elliptic. If the mixed coefficient 2 is plugged directly into B²−AC as B, one gets 4−2=2>0 and wrongly calls it hyperbolic. So the convention must be stated first and then used with its own formula. Neither the class nor a passing candidate gives uniquenessThe hyperbolic class of S03 describes only the principal part; S09 says only that u_A is one solution. S14 finds a second solution u_C=u_A+y, showing that without boundary or initial conditions the solution is not unique. Existence, uniqueness and well-posedness need further theorems and conditions, which this guide does not treat. - The residual of u_B is 0 at (0,−8) and at (5,−8), so u_B satisfies L(u)=g on ℝ² → reject — R_{u_B}=y+8 is not the zero polynomial; R_{u_B}(0,0)=8≠0. - The mixed coefficient of u_xx+2u_xy+2u_yy is 2; plugging it into B²−AC gives 2>0, so it is hyperbolic → reject — Under the 2B convention B=1 and B²−AC=−1; under the b convention D=4−8=−4; both say elliptic. - u_A satisfies L(u)=g, so u_A is the unique solution of this equation → reject — u_C=u_A+y also satisfies it, and u_C≠u_A. - L is hyperbolic, so L(u)=g has a unique solution for any boundary data → reject — The class looks only at the principal part; nothing here proves any existence, uniqueness or well-posedness result, and MKO assumption a2 rules such claims out. ## Result and limitations Under the convention a u_xx+b u_xy+c u_yy with D=b²−4ac, L has a=1, b=2, c=−3 and D=16>0, so L is hyperbolic (with A u_xx+2B u_xy+C u_yy one gets B=1 and B²−AC=4, the same sign). The symbolic residuals of u_A=x²+xy+y² and u_C=x²+xy+y²+y are the zero polynomial, so both satisfy L(u)=g on all of ℝ²; the residual of u_B=x²+2xy is y+8, which vanishes on the whole line y=−8 but equals 8 at (0,0), so u_B is not a solution. - poly_partial, pde_class and poly_coeff_zero are premises of this guide (standard definitions and polynomial facts) and are not derived here from more basic definitions. - This guide does not treat general solutions, boundary or initial conditions, existence and uniqueness, or well-posedness (the MKO's formal next). ## Verification and source binding 14 declared checks for this guide; only the listed expressions are covered. - source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f - adopted source SHA-256: d7f1dca1ded3622b8bc0d17369236ab7e35c0df919f547af5b7b66b3c462a9b0 - original source SHA-256: d7f1dca1ded3622b8bc0d17369236ab7e35c0df919f547af5b7b66b3c462a9b0 - Lean: NOT_RUN · independent AI: NOT_MEASURED [Same-source full JSON (formulas, ASTs, controls and revision log)](https://ocme.evemisslab.com/data/derivations/production-20261019/guides/derivation-mko-partial-differential-equation-01.json)