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Let $\Omega$ an open and bounded set of $\mathbb{R}^N$. Consider $f\in L^2(\Omega)$, and $\varepsilon >0$. Let us call $u_\varepsilon$ the unique weak solution in $H^1_0(\Omega)$ of the problem

$-\varepsilon \Delta u + u = f \; \mbox{ in } \Omega,\;\; u=0 \;\; \mbox{ on }\partial \Omega.$

Prove that $u_\varepsilon$ converges to $f$ in $L^2(\Omega)$, as $\varepsilon$ goes to $0$.

Can you help me with to prove this result?

So far, I have this estimates:

$\|\nabla u_\varepsilon\|_{L^2(\Omega)}\leq \|f\|_{L^2(\Omega)}\varepsilon^{-1}$ and

$\|u_\varepsilon\|_{L^2(\Omega)}\leq \|f\|_{L^2(\Omega)}$.

Charles
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1 Answers1

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This PDE is the Euler-Lagrange equation for the convex functional $$\Phi_\epsilon(u)=\int_{\Omega} \left(\epsilon |\nabla u|^2+\frac12(u-f)^2\right) \tag{1}$$ If you can find some function $v\in H_0^1(\Omega)$ for which $\Phi_\epsilon(v)$ is small, then the solution of PDE will satisfy $$\frac12\int_\Omega (u_\epsilon-f)^2 \le \Phi_\epsilon (u)\le \Phi_\epsilon (v)\tag{2}$$ because the solution minimizes $\Phi_\epsilon$ in $H_0^1(\Omega)$.

The rest of solution is hidden below.

Given $\delta>0$, let $v$ be an element of of $C^\infty_c(\Omega)$ such that $ \int_\Omega (v-f)^2<\delta$ (smooth compactly supported functions are dense in $L^2$). Since $\int_{\Omega} \epsilon |\nabla v|^2 \to 0$ as $\epsilon\to 0$, for all sufficiently small $\epsilon$ we have $\Phi_\epsilon(v)<\delta /2$. By (2), the latter implies $\int_\Omega (u_\epsilon-f)^2<\delta$. $\quad\Box$

Community
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user111146
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