Alphabeta Math
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

8 results · all verified · 6 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 2 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Partial Differential Equations and Characteristics — Examples

1 · Prerequisites

2 · Summary

These examples compute characteristic formulas in explicit transport models, show how characteristic data can fail on a tangent surface, and illustrate the second-order classification on the Laplace, heat, wave, and Tricomi operators. The closing counterexample records both failure modes of the naive global trichotomy: mixed-type equations and higher-order equations.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Constant-velocity transport translates the initial profile

Example

For a constant vector bRn and a datum u0C1(Rn), the equation

ut+bxu=0,u(x,0)=u0(x),

has solution

u(x,t)=u0(xbt).

Facts & Assumptions

Given: A constant velocity b, a datum u0C1(Rn), and the homogeneous transport equation above.

[L1]

Homogeneous transport is represented by the inverse characteristic flow (Homogeneous linear transport is solved by the inverse characteristic flow).

Verification

technique · direct
1.1

The characteristic ODE is X(s)=b, so the characteristic through x at time t is X(s;t,x)=x+(st)b, and in particular X(0;t,x)=xbt.

given
2.1

Apply [L1] with the characteristic from step 1.1; the transport solution is therefore u(x,t)=u0(X(0;t,x))=u0(xbt).

L1step 1.1
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Transport with growth and source along straight characteristics

Example

Let bRn be constant and let q:RnR be C1, and let u0C1(Rn). The equation

ut+bxu+u=q(xbt),u(x,0)=u0(x),

has the explicit solution

u(x,t)=etu0(xbt)+(1et)q(xbt).

Facts & Assumptions

Given: The transport equation with constant velocity b, unit zeroth-order coefficient, source q(xbt), and datum u0C1(Rn).

[L1]

The inhomogeneous transport equation is solved by the characteristic integrating-factor formula (The inhomogeneous linear transport equation has the characteristic integrating-factor formula).

Verification

technique · direct
1.1

The characteristics are the straight lines X(s;t,x)=x+(st)b, so X(0;t,x)=xbt, and along such a curve the source becomes q(X(s;t,x)bs)=q(xbt), which is independent of s.

given
2.1

Applying [L1] with c1 gives u(x,t)=etu0(xbt)+0te(ts)q(xbt)ds, and the integral equals (1et)q(xbt), so the displayed closed form follows.

L1step 1.1
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-09-05Open item page →

The stationary equation x dot Du = u is solved by radial characteristics

Example

On Rn{0}, the stationary first-order equation

xu(x)=u(x)

has precisely the C1 solutions of the form

u(rω)=rg(ω),r>0, ωSn1,

with arbitrary C1 data g on the unit sphere.

Facts & Assumptions

Given: The stationary equation xu=u on Rn{0}.

[L1]

Along a characteristic, a transport equation reduces to the scalar ODE from the transport lemma (A transport equation restricts to a linear ODE along each characteristic).

Verification

technique · direct
1.1

Regard the equation as transport with characteristic ODE X(s)=X(s); its solutions are the rays X(s)=esx0, and by [L1] the restricted function v(s)=u(X(s)) satisfies v(s)=v(s), hence v(s)=esv(0).

L1
2.1

Writing x=rω with r=es and ωSn1 gives u(rω)=ru(ω), so every solution has the claimed form with g(ω)=u(ω); conversely, for u(rω)=rg(ω) the radial derivative is u/r, hence xu=u.

step 1.1
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Characteristic Cauchy data may be nonunique or incompatible

Statement refuted

Cauchy data for a first-order linear transport equation always determine a unique local classical solution, even when the data surface is characteristic.

Facts & Assumptions

Given: The transport equation ut+ux=0 and the line Σ={(x,t):x=t}.

[L1]

The local uniqueness theorem for linear transport assumes the data surface is noncharacteristic (Local linear transport has a unique solution from noncharacteristic Cauchy data).

[L2]

A noncharacteristic first-order Cauchy surface is one transverse to the space-time transport vector (Noncharacteristic Cauchy surfaces for first-order transport).

Counterexample

technique · direct
1.1

For ut+ux=0, the space-time transport vector is (1,1), which is tangent to Σ, so Σ is characteristic rather than noncharacteristic by [L2], and [L1] does not apply.

L1L2
2.1

Every C1 function of the form u(x,t)=F(xt) solves ut+ux=0, and on Σ one has xt=0, so the restriction is the constant F(0). Taking F0(s)=0 and F1(s)=s gives two different local classical solutions with the same constant data on Σ, proving nonuniqueness. On the other hand, every classical solution restricts to a constant on Σ, so nonconstant prescribed data such as g(t)=t are incompatible.

step 1.1
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Laplace, heat, and wave equations have elliptic, parabolic, and hyperbolic principal symbols

Example

The model operators

Δ=uxx+uyy,tΔx,ttxx

represent the elliptic, parabolic, and hyperbolic cases respectively.

Facts & Assumptions

Given: The principal parts of Laplace, heat, and one-space-dimensional wave operators.

[L1]

Elliptic, hyperbolic, and parabolic type are read from the principal symbol definitions (Elliptic, hyperbolic, and parabolic principal symbols).

[L2]

In two variables, the discriminant is B2AC (The discriminant for a second-order equation in two variables).

Verification

technique · direct
1.1

For uxx+uyy one has A=C=1 and B=0, so [L2] gives Δ=1<0 and the Laplace operator is elliptic; for uttuxx, the principal polynomial in (τ,ξ) is τ2ξ2, which has two distinct real roots τ=±ξ for ξ0, so [L1] makes it hyperbolic.

L1L2
2.1

For the heat operator tΔx, the spatial quadratic form is ξ2, which is positive definite, while the time derivative is first order, so [L1] identifies it as parabolic.

L1
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Characteristic coordinates reduce a constant-coefficient hyperbolic equation to mixed form

Example

For the wave-type equation

uxxuyy=0,

the characteristic coordinates

ξ=x+y,η=xy

turn the principal part into

uξη=0

up to the nonzero factor 4.

Facts & Assumptions

Given: The constant-coefficient hyperbolic operator uxxuyy.

[L1]

Constant-coefficient hyperbolic principal parts admit canonical linear coordinates (Constant-coefficient second-order equations in two variables have canonical principal forms).

[L2]

Characteristic directions are coordinate invariant and are determined by the characteristic families (In two variables, type and characteristic directions are coordinate invariant).

Verification

technique · direct
1.1

The principal quadratic form is ξx2ξy2, so the characteristic covectors are proportional to d(x+y) and d(xy), and [L2] shows that using ξ=x+y and η=xy follows the two characteristic families.

L2
2.1

In these coordinates, x=ξ+η and y=ξη, so uxxuyy=(ξ+η)2u(ξη)2u=4uξη; this is the mixed canonical form, equivalent to the hyperbolic normal form from [L1] after a further linear recombination of ξ and η.

L1step 1.1
ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-09-05Open item page →

The Tricomi equation changes type across y = 0

Example

The Tricomi equation

yuxx+uyy=0

is elliptic for y>0, hyperbolic for y<0, and parabolic on the line y=0.

Facts & Assumptions

Given: The principal part yuxx+uyy.

[L1]

For a two-variable second-order principal part, the discriminant is B2AC (The discriminant for a second-order equation in two variables).

[L2]

The elliptic/parabolic/hyperbolic trichotomy is only a pointwise second-order classification (Limits of the elliptic-parabolic-hyperbolic trichotomy).

Verification

technique · direct
1.1

Here A=y, B=0, and C=1, so [L1] gives Δ(y)=0y1=y.

L1
2.1

If y>0, then Δ<0 and the equation is elliptic; if y<0, then Δ>0 and it is hyperbolic; and if y=0, then Δ=0 and the principal part has rank one, so this pointwise change of sign is exactly the mixed-type behavior singled out in [L2].

L2step 1.1
CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

The elliptic-parabolic-hyperbolic trichotomy is not a global taxonomy

Statement refuted

Every partial differential equation belongs globally to exactly one of the three classes elliptic, parabolic, or hyperbolic.

Facts & Assumptions

Given: The Tricomi operator and the fourth-order biharmonic operator Δ2.

[L1]

The Tricomi equation changes type across y=0 (The Tricomi equation changes type across y = 0).

[L2]

The threefold classification on this page is only for scalar real second-order principal symbols (Limits of the elliptic-parabolic-hyperbolic trichotomy).

Counterexample

technique · direct
1.1

By [L1], the single operator yuxx+uyy is elliptic on one open region, hyperbolic on another, and degenerate on the separating line, so even within second order one operator need not belong globally to just one of the three names.

L1
2.1

The biharmonic equation Δ2u=0 has order 4, so [L2] says the second-order trichotomy does not classify it at all; the claimed global taxonomy therefore fails both because some operators change type from point to point and because others lie outside the stated second-order scope.

L2step 1.1

Sources