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Uniqueness and sup-norm contraction for the Cauchy problem

Statement

Let n≥1, T>0, and let H:Rn×[0,T]×Rn→R satisfy the Lipschitz conditions of part (a) of Comparison for first-order Hamilton--Jacobi equations. (1) If u,v are bounded viscosity solutions of the Cauchy problem in Z=Rn×(0,T) with the same bounded continuous initial datum u0, then u=v on Z; in particular the classical and the Hopf--Lax solutions of later sections are the unique ones in the bounded class whenever H satisfies those conditions. (2) More generally, if u,v are bounded viscosity solutions with initial data u0,v0, then for every T>0 sup⁡z∈Z(u−v)≤max⁡(0, sup⁡x∈Rn(u0(x)−v0(x))), and applying the same bound to v−u gives sup⁡Z∣u−v∣≤sup⁡Rn∣u0−v0∣ when the initial difference is bounded. In particular the solution operator is a contraction in the supremum norm on the bounded initial data. No choice principle is used.

Facts & Assumptions

Given: A Hamiltonian H satisfying the Lipschitz conditions of comparison case (a), T>0, bounded viscosity solutions u,v of the Cauchy problem in Z=Rn×(0,T) with bounded continuous data u0,v0.

[F1]

Comparison, case (a): if U is a bounded upper semicontinuous subsolution and V a bounded lower semicontinuous supersolution on the closed slab with U(x,0)≤V(x,0) pointwise, then U≤V on Z (Comparison for first-order Hamilton--Jacobi equations).

[F2]

For a bounded viscosity solution u, its upper envelope u∗ is a bounded upper semicontinuous subsolution and its lower envelope u∗ is a bounded lower semicontinuous supersolution, each satisfying the corresponding relaxed initial inequality (Discontinuous viscosity solutions through the two envelopes). Adding a constant K shifts both envelopes by K and preserves their one-sided viscosity inequalities because H is independent of the unknown (Discontinuous viscosity solutions through the two envelopes, Comparison for first-order Hamilton--Jacobi equations for the equation class).

[F3]

Boundedness of u,v and of their continuous initial data is assumed in the statement and Given, so the displayed suprema are finite. The relaxed joint initial limsup/liminf conditions are part of Discontinuous viscosity solutions through the two envelopes, as recorded in [F2]. No uniform-continuity hypothesis is needed for this comparison consequence.

Proof

technique · comparison applied twice, plus the constant-shift invariance of the equation
1.1F1F2F3

Uniqueness. Let u,v be bounded viscosity solutions with the same datum u0. By [F2], u∗ is a bounded upper semicontinuous subsolution and v∗ is a bounded lower semicontinuous supersolution. Extend them to the initial face by U(x,0)=u0(x) and V(x,0)=u0(x). Their relaxed initial inequalities and continuity of u0 make U upper semicontinuous and V lower semicontinuous on the closed slab, with ordered pointwise initial values. Comparison [F1] gives u∗≤v∗ on Z. Since u≤u∗ and v∗≤v, this yields u≤v. Applying the same argument to (v,u) gives v≤u, hence u=v on Z; in fact all four envelopes and functions coincide. In particular, whenever a classical or Hopf--Lax solution is known to be a bounded viscosity solution of the same Cauchy problem, it is the unique bounded solution.

1.2F1F2F3algebra

The one-sided bound for general data. Put K:=max⁡{0,sup⁡Rn(u0−v0)}<∞. By [F2], u∗ is a bounded upper semicontinuous subsolution and (v+K)∗=v∗+K is a bounded lower semicontinuous supersolution. Extend these envelopes to t=0 by u0 and v0+K, respectively; their relaxed initial inequalities and continuity of the data make the extensions semicontinuous on the closed slab with ordered pointwise initial values. Comparison [F1] gives u∗≤v∗+K on Z. Since u≤u∗ and v∗≤v, this implies u−v≤K; taking the supremum over Z gives sup⁡Z(u−v)≤K.

2.1step 1.1step 1.2∎

Conclusion. Applying step 1.2 to the pair (u,v) and to the exchanged pair (v,u) gives sup⁡Z(u−v)≤max⁡(0,sup⁡(u0−v0)) and sup⁡Z(v−u)≤max⁡(0,sup⁡(v0−u0)); when the initial difference is bounded, both right-hand sides are at most sup⁡Rn∣u0−v0∣, hence sup⁡Z∣u−v∣≤sup⁡Rn∣u0−v0∣ and the solution operator is a contraction in the supremum norm.

Remarks

  • Domain. The corollary is stated on O=Rn because comparison case (a) is; on a bounded domain without lateral data uniqueness fails, as the companion counterexample shows.
  • Choice. Only comparison and the constant shift are used, both choice-free.

Depends on

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