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Quasilinear Characteristics and Cauchy Kovalevskaya

1 · Prerequisites

2 · Summary

Starting with a quasilinear Cauchy problem, this page separates ODE solvability from the inverse-projection step that actually constructs a PDE graph. It then makes projection failure precise through Burgers caustics, develops both Charpit invariants for fully nonlinear equations. The closing scope remark links the locally proved Cauchy–Kovalevskaya on a noncharacteristic analytic hypersurface: analytic Cauchy data on a noncharacteristic analytic hypersurface yield a unique local analytic solution germ. For an implicit nonlinear equation the theorem fixes a compatible jet and its selected analytic branch.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-06Open item page →

Semilinear and quasilinear first-order Cauchy problems on a parametrised hypersurface

Definition

Let n≥1 and V⊆Rn−1 be open, let γ:V→Rn and ϕ:V→R be C1, with rank⁡Dγ(y)=n−1 for every y∈V, and let a:O→Rn and b:O→R be smooth on an open set O⊆Rn×R. Assume that (γ(y),ϕ(y))∈O for every y∈V. The Cauchy problem for the quasilinear equation is

a(x,u(x))⋅Du(x)=b(x,u(x)),u(γ(y))=ϕ(y).

It is semilinear when a=a(x) is independent of u. At y0∈V, a classical local solution is a C1 function u on an open neighbourhood Ω of γ(y0) such that (x,u(x))∈O and the PDE holds for every x∈Ω, and such that there is a neighbourhood W⊆V of y0 with γ(W)⊆Ω and u(γ(y))=ϕ(y) for every y∈W. This specializes the first-order classification in Linear, semilinear, quasilinear, and fully nonlinear partial differential equations; the word noncharacteristic will be tested by the rank calculation below, rather than by importing the space-time transport convention of Noncharacteristic Cauchy surfaces for first-order transport.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The augmented characteristic system for a quasilinear first-order PDE

Definition

For the problem in Semilinear and quasilinear first-order Cauchy problems on a parametrised hypersurface, its augmented characteristic strip is a map (X,Z)(s,y) satisfying

X˙=a(X,Z),Z˙=b(X,Z),(X,Z)(0,y)=(γ(y),ϕ(y)).

Here s is the characteristic parameter and y∈V labels the initial point. The system concerns only (X,Z); no differential equation for a putative gradient is included in this definition.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Local solvability and C1 parameter dependence for the augmented characteristic ODE

Statement

At every y0∈V, the system in The augmented characteristic system for a quasilinear first-order PDE has, after shrinking to ∣s∣<ϵ and a neighbourhood W of y0, a unique common solution (X,Z) for y∈W. It is C1 in (s,y).

Facts & Assumptions

Given: The smooth coefficients and C1 initial strip in the definition, and y0∈V.

Proof

technique · direct
1.1givenconstruct

Put Y:=(X,Z) and G(x,z):=(a(x,z),b(x,z)). The coefficients in the defining Cauchy problem are smooth, so G is smooth.

2.1step 1.1given∎

Apply Smooth dependence of ODE solutions on parameters to Y′=G(Y) with initial value Y(0,y)=(γ(y),ϕ(y)). It gives a common local interval, uniqueness, and C1 dependence on y; including the ODE variable gives C1 dependence on (s,y).

LemmaStatement: Literature-sourcedProof: AI-generatedaudited 2026-09-06Open item page →

A quasilinear solution lifts to augmented characteristics

Statement

Let u∈C1(Ω) solve a(x,u)⋅Du=b(x,u). If X solves X˙=a(X,u(X)) while it stays in Ω, then Z:=u(X) satisfies Z˙=b(X,Z). Thus (X,Z) solves the augmented characteristic ODE system.

Facts & Assumptions

Given: A C1 solution u, a differentiable curve X with the stated ODE, and Z=u∘X.

Proof

technique · direct
1.1givenalgebra

The chain rule gives Z˙=Du(X)⋅X˙.

2.1step 1.1givenalgebra∎

Substitute X˙=a(X,u(X)) and the PDE: Z˙=Du(X)⋅a(X,u(X))=b(X,u(X))=b(X,Z).

LemmaStatement: Literature-sourcedProof: AI-generatedaudited 2026-09-06Open item page →

Compatibility of a characteristic strip with Cauchy data

Statement

If a C1 solution has u(γ(y))=ϕ(y) and p0(y):=Du(γ(y)), then

p0(y)Dγ(y)=Dϕ(y),p0(y)⋅a(γ(y),ϕ(y))=b(γ(y),ϕ(y)).

The first condition is tangential compatibility. The rank condition for the projected strip is a separate hypothesis; it is not derived from this identity.

Facts & Assumptions

Given: The C1 Cauchy data and a C1 solution attaining them.

Proof

technique · direct
1.1givenalgebra

Differentiate u∘γ=ϕ with respect to y to obtain Du(γ)Dγ=Dϕ.

2.1givenalgebra∎

Evaluate a(x,u)⋅Du=b(x,u) at x=γ(y) and use u(γ(y))=ϕ(y) to obtain the second identity.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Jacobian of a characteristic strip at its initial surface

Statement

For the local strip (X,Z), the derivative of Ψ(s,y):=X(s,y) at s=0 is

D(s,y)Ψ(0,y)=[a(γ(y),ϕ(y)),Dγ(y)].

Consequently D(s,y)Ψ(0,y) is invertible exactly when this n×n matrix has rank n.

Facts & Assumptions

Given: The C1 local strip supplied by the preceding lemma.

Proof

technique · direct
1.1givenalgebra

Its initial condition gives DyX(0,y)=Dγ(y), and its ODE gives ∂sX(0,y)=a(γ(y),ϕ(y)).

2.1step 1.1algebra∎

These are precisely the first and remaining columns of D(s,y)Ψ(0,y), proving the formula; an n×n derivative is invertible exactly when it has rank n.

TheoremStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Local quasilinear characteristic graph construction

Statement

At y0∈V, suppose [a(γ(y0),ϕ(y0)),Dγ(y0)] has rank n. Then, after shrinking the characteristic strip, Ψ(s,y)=X(s,y) is a C1 diffeomorphism onto an open set U, and

u(x):=Z(Ψ−1(x))

is the unique C1 function obtained by this inverse-projection construction. It attains u(γ(y))=ϕ(y); the next lemma verifies its PDE.

Facts & Assumptions

Given: The smooth coefficients, C1 data, and the stated full-rank condition at y0.

Proof

technique · direct
1.1givenalgebra

The local ODE lemma supplies a C1 strip, and the Jacobian lemma makes DΨ(0,y0) invertible.

2.1step 1.1construct

By The Euclidean inverse function theorem, shrink to a neighbourhood on which Ψ has a C1 inverse. Define u=Z∘Ψ−1 there.

3.1step 2.1given∎

At s=0, Ψ(0,y)=γ(y) and Z(0,y)=ϕ(y), hence u(γ(y))=ϕ(y). Any inverse-projected function from this strip has the same formula and is therefore identical to u.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The inverse-projected characteristic graph satisfies the quasilinear PDE

Statement

The function u=Z∘Ψ−1 constructed in Local quasilinear characteristic graph construction satisfies a(x,u(x))⋅Du(x)=b(x,u(x)) on its local domain.

Facts & Assumptions

Given: The local inverse-projected graph and its characteristic identities Xs=a(X,Z) and Zs=b(X,Z).

Proof

technique · direct
1.1givenalgebra

The identity u(X(s,y))=Z(s,y) differentiates in s to Du(X)⋅Xs=Zs.

2.1step 1.1givenalgebra

Substitute the two characteristic identities to get Du(X)⋅a(X,Z)=b(X,Z).

3.1step 2.1given∎

Since X=Ψ(s,y) ranges over the local domain and Z=u(X), this is a(x,u(x))⋅Du(x)=b(x,u(x)) for every x there.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Characteristic crossing and caustic for a first-order PDE

Definition

For a characteristic strip, a crossing or caustic is a point or time at which the projected map (s,y)↦X(s,y) loses local rank or local one-to-one graphing. It is a failure of the projection needed to define u=Z∘X−1, not a claim that the lifted ODE (X,Z) has reached its maximal lifespan. For a forward-time family, a first crossing time is an infimum in [0,∞], with value ∞ when the relevant set is empty.

LemmaStatement: Literature-sourcedProof: AI-generatedaudited 2026-09-06Open item page →

The Burgers slope obeys a Riccati law along characteristics

Statement

Let u∈C2(Ω) solve ut+uux=0 on an open Ω⊆R2. Let I be an open interval containing zero and X∈C1(I) satisfy (t,X(t))∈Ω and X˙=u(t,X). Then q(t):=ux(t,X(t)) satisfies

q˙=−q2,q(t)=q(0)1+tq(0)

for every t∈I; in particular 1+tq(0) cannot vanish in I.

Facts & Assumptions

Given: A C2 classical Burgers solution and one of its projected characteristics.

Proof

technique · direct
1.1givenalgebra

Put r(t,x)=ux(t,x). Differentiate ut+uux=0 in x to obtain rt+urx+r2=0; equality of mixed partials holds since u is C2.

2.1step 1.1givenalgebra

Along X˙=u, the total-derivative chain rule gives q˙=rt(t,X)+u(t,X)rx(t,X), so q˙=−q2.

3.1step 2.1algebra∎

Set c=q(0) and h(t)=(1+tc)q(t)−c. Then h(0)=0 and h′=−qh, so differentiation shows h(t)exp⁡(∫0tq(s) ds) is constant and hence zero. Thus (1+tc)q(t)=c. If c=0, this gives q=0; if c≠0, the identity rules out a zero denominator in I and yields the displayed formula.

TheoremStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Inviscid Burgers characteristic formula and first crossing time

Statement

For ut+uux=0 with C1 datum u(0,ξ)=u0(ξ), every classical characteristic has

u(t,X(t,ξ))=u0(ξ),X(t,ξ)=ξ+tu0(ξ).

Thus Xξ=1+tu0′(ξ) and its first forward crossing time is

T∗:=inf⁡{t≥0:∃ξ, 1+tu0′(ξ)=0}∈[0,∞],

where the infimum of the empty set is ∞.

Facts & Assumptions

Given: A classical Burgers solution with the stated C1 initial datum.

Proof

technique · direct
1.1givenalgebra

Along X˙=u(t,X), the chain rule and the PDE give d(u(t,X))/dt=ut+uux=0.

2.1step 1.1givenalgebra

The initial condition makes u(t,X)=u0(ξ), hence X˙=u0(ξ) and integration from X(0,ξ)=ξ gives X=ξ+tu0(ξ).

3.1step 2.1given∎

Differentiating in ξ gives Xξ=1+tu0′(ξ), so the definition of caustic gives exactly the displayed set and its stated infimum convention.

CorollaryStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Monotone Burgers data have no forward characteristic crossing

Statement

If u0′(ξ)≥0 for all ξ, then 1+tu0′(ξ)≥1 for t≥0. Hence the Burgers projection has no forward crossing. This says nothing about any unrelated global lifespan condition.

Facts & Assumptions

Given: The characteristic formula and u0′≥0.

Proof

technique · direct
1.1givenalgebra

For t≥0, multiplication gives tu0′(ξ)≥0.

2.1step 1.1given∎

Therefore 1+tu0′(ξ)≥1, so the zero set defining T∗ is empty and the projection has no forward crossing.

TheoremStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Uniqueness of a classical quasilinear solution before characteristic crossing

Statement

Let u,v∈C1 solve the same quasilinear Cauchy problem. On a connected region where their common characteristic projection from the data is a diffeomorphism, u=v.

Facts & Assumptions

Given: Two C1 solutions with identical data, and a connected region on which the common projection is a diffeomorphism.

Proof

technique · direct
1.1givenconstruct

The lifting lemma sends the restrictions of both solutions along each data-labelled characteristic to the same augmented initial-value problem.

2.1step 1.1given

Uniqueness of that ODE gives equal lifted values u(X(s,y))=v(X(s,y)) on the strip.

3.1step 2.1given∎

The diffeomorphic projection represents every point of the region as one X(s,y), so step 2.1 gives u=v there.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Fully nonlinear first-order PDEs and complete integrals

Definition

Let F:O⊆Rn×R×Rn→R. A general first-order equation has the form F(x,u,Du)=0; it is fully nonlinear when its dependence on the highest-order variable p=Du is not affine after (x,u) is fixed. A local complete integral on open sets U,A⊆Rn is a C2 family S:U×A→R such that (x,S(x;α),DxS(x;α))∈O,F(x,S(x;α),DxS(x;α))=0,det⁡Dxα2S(x;α)≠0 for every (x,α)∈U×A. Thus the n parameters enter essentially rather than merely labelling repeated copies of one solution. A stationary envelope is a value u(x)=S(x;α(x)) selected by Sα(x;α(x))=0. This definition asserts neither global representation nor differentiability of an envelope.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

A nondegenerate stationary envelope solves the Hamilton–Jacobi equation

Statement

If Sα(x0;α0)=0 and Sαα(x0;α0) is invertible, then locally Sα(x;α(x))=0 defines a C1 parameter α(x). The envelope u=S(x;α(x)) has Du=Sx(x;α(x)) and satisfies F(x,u,Du)=0.

Facts & Assumptions

Given: A C2 complete integral, a stationary point, and an invertible parameter Hessian there.

Proof

technique · direct
1.1givenconstruct

Apply the implicit function theorem to G(x,α)=Sα(x;α); its derivative in α is the invertible Sαα.

2.1step 1.1givenalgebra

The chain rule gives Du=Sx+SαDα=Sx on the stationary branch.

3.1step 2.1given∎

The complete-integral identity F(x,S,Sx)=0, evaluated at α(x) and using step 2.1, is F(x,u,Du)=0.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The Lagrange–Charpit characteristic system

Definition

For F∈C2(O) on O⊆Rn×R×Rn, the Lagrange–Charpit system is

X˙=Fp,Z˙=P⋅Fp,P˙=−Fx−PFz,

with all derivatives evaluated at (X,Z,P). The alternative contact normalization Z˙=P⋅Fp−F agrees with this one only along F=0; it is not silently substituted off that constraint.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The Charpit flow preserves the PDE constraint

Statement

Along a Charpit curve,

ddsF(X,Z,P)=0.

Consequently an initial point in F=0 remains in F=0 while the curve exists.

Facts & Assumptions

Given: A C1 Charpit curve for a C2 function F.

Proof

technique · direct
1.1givenalgebra

The chain rule gives F˙=Fx⋅X˙+FzZ˙+Fp⋅P˙.

2.1step 1.1givenalgebra

Substitute X˙=Fp, Z˙=P⋅Fp, and P˙=−Fx−PFz; the three terms cancel pairwise.

3.1step 2.1given∎

Hence F(X,Z,P) is constant, so an initial zero stays zero.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Charpit contact compatibility is preserved

Statement

Let (X,Z,P)(s,y) be a C1 Charpit strip lying in F=0. If Zy(0,y)=P(0,y)⋅Xy(0,y), then

Zy(s,y)=P(s,y)⋅Xy(s,y)

while the strip exists.

Facts & Assumptions

Given: A C1 Charpit strip, its initial contact identity, and the constraint F(X,Z,P)=0.

Proof

technique · direct
1.1givenalgebra

Set D:=Zy−P⋅Xy. Differentiate the Charpit equations in y and apply the product and chain rules to obtain Ds=Fy−FzD.

2.1step 1.1givenalgebra

Differentiating F(X,Z,P)=0 in y gives Fy=0, so Ds=−FzD and D(0,y)=0.

3.1step 2.1algebra∎

Multiplication by the integrating factor exp⁡(∫0sFz(r,y) dr) makes D constant; its initial value is zero, hence D=0.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

The Charpit momentum equation from differentiating Hamilton–Jacobi

Statement

If u∈C2 solves F(x,u,Du)=0, put P=Du(X) and let X˙=Fp(X,u(X),Du(X)). Then

P˙=−Fx−PFz,

so the momentum equation is forced by differentiating the PDE.

Facts & Assumptions

Given: A C2 classical solution and the stated projected characteristic.

Proof

technique · direct
1.1givenalgebra

Differentiate F(x,u(x),Du(x))=0 in x to get Fx+FzDu+D2u Fp=0.

1.2givenalgebra

Along X, the chain rule gives P˙=D2u(X)X˙=D2u(X)Fp.

2.1step 1.1step 1.2algebra∎

Combining the two identities and P=Du(X) gives P˙=−Fx−PFz.

TheoremStatement: Literature-sourcedProof: AI-generatedaudited 2026-09-06Open item page →

Local fully nonlinear Charpit graph construction

Statement

Let O⊆Rn×R×Rn be open and F∈C2(O). Let V⊆Rn−1 be open, let y0∈V, and let γ:V→Rn, ϕ:V→R, and p0:V→Rn be C1 maps such that (γ(y),ϕ(y),p0(y))∈O, F(γ(y),ϕ(y),p0(y))=0, and Dϕ(y)=p0(y)Dγ(y) for every y∈V. Suppose also that rank⁡[Fp(γ(y0),ϕ(y0),p0(y0)),Dγ(y0)]=n. Then the Charpit strip through (γ,ϕ,p0) projects locally to a classical graph u satisfying F(x,u,Du)=0. It is unique while that projection is locally invertible among graphs obtained by inverse-projecting this fixed Charpit strip.

Facts & Assumptions

Given: The stated C2 equation, compatible C1 strip data, and full-rank condition at y0.

Proof

technique · direct
1.1givenconstruct

The Charpit vector field is C1 because F∈C2, hence locally Lipschitz. Continuous dependence gives a unique common local strip through the C1 initial data (γ,ϕ,p0).

2.1step 1.1givenalgebra

The C1-dependence theorem makes this strip C1 in (s,y). Its projected derivative at s=0 has columns [Fp,Dγ], hence is invertible by the rank hypothesis.

3.1step 2.1construct

The inverse function theorem supplies a local inverse of (s,y)↦X(s,y); define u(X(s,y)):=Z(s,y).

4.1step 3.1givenalgebra

Constraint preservation gives F(X,Z,P)=0, while contact preservation gives DyZ=P⋅DyX; the s identity is Zs=P⋅Xs. Since D(s,y)X is invertible, these identities imply Du(X)=P.

5.1step 4.1given∎

Substitution in the preserved constraint gives F(x,u(x),Du(x))=0. The fixed strip and its local inverse determine this inverse-projected graph uniquely while the projection remains locally invertible.

RemarkRemark: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Characteristics do not select a post-crossing weak solution

A crossing makes the characteristic projection fail to define a single-valued classical graph. It does not by itself choose a continuation: entropy conditions for conservation laws and viscosity inequalities for Hamilton–Jacobi equations are extra selection principles. No such weak-solution theorem is claimed here.

RemarkRemark: Literature-sourcedProof: Not applicableOpen item page →

The proof boundary of Cauchy–Kovalevskaya

Cauchy–Kovalevskaya on a noncharacteristic analytic hypersurface is proved in this library by analytic flattening, normal-form reduction and coefficient-majorant arguments. Its conclusion is local existence and uniqueness of an analytic solution germ for analytic Cauchy data on a noncharacteristic analytic hypersurface. For an implicit nonlinear equation it fixes a compatible jet and its selected analytic branch. These hypotheses do not assert an extension to arbitrary smooth coefficients or data.

5 · Examples, counterexamples and false statements

None yet.

Sources