Limit of minimizers of a class of functionals

Question

Assume that $ \Omega $ is a smooth bounded domain in $ \mathbb{R}^n $. Consider a functional $$ \mathcal{F}(u)=\int_\Omega(|\nabla u|^2+h^{-1}|u-u_0|^2) \, dx $$ where $ h>0 $ is a parameter and $ u_0\in H_0^1(\Omega) $. For the minimizing problem $$ \min_{u\in H_0^1(\Omega)}\mathcal{F}(u), $$ it is easy to get by standard arguments that there is a unique minimizer $ u_h\in H_0^1(\Omega) $, i.e. $ u_h=\operatorname{argmin}_{u\in H_0^1(\Omega)} \mathcal{F}(u) $. A natural observation is that $ \|u_h-u_0\|_{L^2(\Omega)}\to 0 $. I want to ask the following question.

Let $ h_0>0 $. Assume that $ h\to h_0 $. If there is the result $ \|u_{h}-u_{h_0}\|_{L^2(\Omega)}\to 0 $?

Such question is natural and intuitively true. However I cannot solve it, can you give me some hints or references?

Answer 1

By simple calculations, we can obtain that $ u_{h} $ and $ u_{h_0} $ satisfy the following Euler-Lagrange equation \begin{align} \frac{u_h-u_0}{h}-\Delta u_{h}=0,\\ \frac{u_{h_0}-u_0}{h}-\Delta u_{h_0}=0. \end{align} Then it can be got that $$ \begin{aligned} 0&=\frac{u_h-u_0}{h}-\frac{u_{h_0}-u_0}{h_0}-\Delta(u_{h}-u_{h_0})\\ &=\frac{u_h-u_0}{h}-\frac{u_h-u_0}{h_0}+\frac{u_h-u_0}{h_0}-\frac{u_{h_0}-u_0}{h_0}-\Delta(u_{h}-u_{h_0})\\ &=(u_h-u_0)\frac{h_0-h}{hh_0}+\frac{u_h-u_{h_0}}{h_0}-\Delta(u_h-u_{h_0}). \end{aligned} $$ Testing the above equation by the function $ u_h-u_{h_0} $, we obtain that $$ \int_{\Omega}|\nabla(u_h-u_{h_0})|^2dx+\int_{\Omega}\frac{(u_{h}-u_{h_0})^2}{h_0}dx\leq\int_{\Omega}\frac{h-h_0}{hh_0}(u_h-u_0)(u_h-u_{h_0})dx. $$ By Cauchy inequality, we have $$ \frac{1}{2}\int_{\Omega}(u_h-u_{h_0})^2dx\leq\frac{(h-h_0)^2}{h^2}\int_{\Omega}(u_h-u_0)^2dx. $$ From this, we can complete the proof.