Abstract
The explicit presence of second derivatives in the \(\infty \)-Laplace Equation leads to difficulties, because they do not always exist when they are called for. This lack is the source of many problems. No weak formulation involving only first derivatives seems to be possible. The use of viscosity solutions circumvents this problem. First, let us explain why one does not restrict oneself only to \(C^2\)-solutions. The reason is that in the Dirichlet problem one can prescribe smooth boundary values so that no \(C^2\)-solution can attain them. The critical points (\(\nabla u = 0\)) do not allow smoothness!
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Notes
- 1.
In the proof of everywhere differentiability in [ES] the approach with the vanishing term \(\varepsilon \Delta u_{\varepsilon }\) is essential.
- 2.
Needless to say, the Principle of Superposition is not generally valid for our nonlinear equation!
- 3.
The Comparison Principle is easy to verify. Insert the test function \(\eta = [\psi -v]_+\) into
$$\begin{aligned} \int _{D_{\rho }}\!\langle | \nabla \psi |^{p-2}\nabla \psi ,\nabla \eta \rangle \,dx \le 0, \quad \int _{D_{\rho }}\!\langle | \nabla v|^{p-2}\nabla v,\nabla \eta \rangle \,dx \ge 0 \end{aligned}$$and subtract to arrive at
$$\begin{aligned} \int _{D_{\rho }}\!\langle | \nabla \psi |^{p-2}\nabla \psi - | \nabla v|^{p-2}\nabla v,\nabla (\psi -v)_{+}\rangle \,dx \le 0. \end{aligned}$$By an elementary inequality, the integral is greater than or equal to
$$\begin{aligned} 2^{p-2}\int _{D_{\rho }}\!|\nabla (\psi -v)_+|^p\,dx. \end{aligned}$$Hence \(\nabla (\psi -v)_+ =0.\) The result follows.
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Lindqvist, P. (2016). Viscosity Solutions. In: Notes on the Infinity Laplace Equation. SpringerBriefs in Mathematics. Springer, Cham. https://doi.org/10.1007/978-3-319-31532-4_4
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DOI: https://doi.org/10.1007/978-3-319-31532-4_4
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