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Proof of Uniqueness for the Dirichlet Problem for Second-Order Equations

corollarycor:uniqueness-dirichlet-second-order-2026a
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· 1,422 chars · 5 deps · depth 23 Reason: Proof of uniqueness: continuity makes each solution both upper and lower semicontinuous, hence both a subsolution and a supersolution up to the boundary, and the comparison principle applies in both directions.

Continuity makes each function both upper and lower semicontinuous, so each is a viscosity subsolution and a viscosity supersolution up to the boundary; the comparison principle applied in both directions gives the two inequalities.

Proof

By claim 2 of Semicontinuity Under Negation and Characterization of Continuity, applied at each point of Ω\overline{\Omega}, the function uu is both upper semicontinuous and lower semicontinuous on Ω\overline{\Omega}, and likewise vv.

By Viscosity Solution of a Second-Order Equation, uΩu|_{\Omega} is both a viscosity subsolution and a viscosity supersolution of FF on Ω\Omega, and likewise vΩv|_{\Omega}. Hence, by Viscosity Subsolution and Supersolution up to the Boundary, each of uu and vv is both a viscosity subsolution and a viscosity supersolution of FF up to the boundary of Ω\Omega.

The operator FF, the constant γ\gamma and the modulus ω\omega satisfy conditions 1, 2 and 3 of Comparison Principle for the Dirichlet Problem for Second-Order Equations by hypothesis. Applying that theorem to the viscosity subsolution uu and the viscosity supersolution vv, whose boundary hypothesis u(x)v(x)u(x)\le v(x) for xΩx\in\partial\Omega follows from u(x)=v(x)u(x)=v(x) there, gives

u(x)v(x)for every xΩ.u(x)\le v(x)\qquad\text{for every }x\in\overline{\Omega}.

Applying it to the viscosity subsolution vv and the viscosity supersolution uu, whose boundary hypothesis v(x)u(x)v(x)\le u(x) for xΩx\in\partial\Omega follows in the same way, gives

v(x)u(x)for every xΩ.v(x)\le u(x)\qquad\text{for every }x\in\overline{\Omega}.

By the antisymmetry of \le, part of the total order structure of R\mathbb{R}, we conclude that u(x)=v(x)u(x)=v(x) for every xΩx\in\overline{\Omega}.

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