Partial differential equation: principal-part classification and symbolic residuals of quadratic polynomial candidates

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 u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw.
• 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

L(u)=u_{xx}+2u_{xy}-3u_{yy}+u_x

a=1,\ b=2,\ c=-3;\ \text{lower-order part } u_x;\ g=2x+y-2

D=b^2-4ac

D=2^2-4\cdot 1\cdot(-3)=16

D=16

L\ \text{is hyperbolic}

A u_{xx}+2B u_{xy}+C u_{yy}=u_{xx}+2u_{xy}-3u_{yy}

A=1,\ B=1,\ C=-3,\ B^2-AC=4=\tfrac{16}{4}

u_A=x^2+xy+y^2

(u_A)_x=2x+y,\ (u_A)_y=x+2y

(u_A)_x=2x+y,\ (u_A)_y=x+2y

(u_A)_{xx}=2,\ (u_A)_{xy}=1,\ (u_A)_{yy}=2

L(u_A)=(u_A)_{xx}+2(u_A)_{xy}-3(u_A)_{yy}+(u_A)_x

L(u_A)=2x+y-2

R_{u_A}=L(u_A)-g

R_{u_A}=(2-2)x+(1-1)y+(-2+2)=0

R_{u_A}\equiv 0

\forall (x,y)\in\mathbb{R}^2:\ L(u_A)(x,y)=g(x,y)

u_B=x^2+2xy

(u_B)_x=2x+2y,\ (u_B)_y=2x,\ (u_B)_{xx}=2,\ (u_B)_{xy}=2,\ (u_B)_{yy}=0

R_{u_B}=L(u_B)-g

R_{u_B}=(2+4-0+2x+2y)-(2x+y-2)=y+8

R_{u_B}(x,-8)

R_{u_B}(x,-8)=-8+8=0\ \text{for every } x\in\mathbb{R}

R_{u_B}(0,0)

R_{u_B}(0,0)=8\neq 0,\ \text{so}\ \neg\,\forall (x,y):\ L(u_B)(x,y)=g(x,y)

u_C=x^2+xy+y^2+y

R_{u_C}\equiv 0,\quad u_C-u_A=y\not\equiv 0

\text{results of S03, S09, S13, S14}

L\ \text{hyperbolic};\ u_A,u_C\ \text{solve } L(u)=g\ \text{on}\ \mathbb{R}^2;\ u_B\ \text{does not}

Substeps and lemmas

(2B)^2-4AC

4(B^2-AC)

2+2\cdot 1-3\cdot 2

-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 solution

The common mistake named by the MKO is to claim a global solution because sampled point residuals are zero. The residual of u_B is zero on the whole line y=−8, even at infinitely many points, yet the residual polynomial y+8 is not the zero polynomial and equals 8 at (0,0). Only when every coefficient of the symbolic residual is 0 (S08) is the equation proved on all of ℝ².

Mixing the b and 2B discriminants

Another misconception identified in the MKO is mixing the discriminants for 2b u_xy and b u_xy. Example: the operator u_xx+2u_xy+2u_yy. By G2, a=1, b=2, c=2 and D=4−8=−4<0, elliptic; with A u_xx+2B u_xy+C u_yy, B=1 and B²−AC=1−2=−1<0, also elliptic. If the mixed coefficient 2 is plugged directly into B²−AC as B, one gets 4−2=2>0 and wrongly calls it hyperbolic. So the convention must be stated first and then used with its own formula.

Neither the class nor a passing candidate gives uniqueness

The hyperbolic class of S03 describes only the principal part; S09 says only that u_A is one solution. S14 finds a second solution u_C=u_A+y, showing that without boundary or initial conditions the solution is not unique. Existence, uniqueness and well-posedness need further theorems and conditions, which this guide does not treat.

• 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).

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