Let $\Omega$ be a bounded open set in $\mathbb{R}^N$ with smooth boundary conditions. Let $A_n \in C^\infty(\bar{\Omega})$ be a $N$ by $N$ symmetric matrix with $v^TA_nv \geq \alpha|v|^2$ for all $v \in \mathbb{R}^N$. Define $T_n u = -\operatorname{div}(A_n \nabla u)$ for $u \in H^1(\Omega)$ and let $\phi_n \in L^2(\Omega)$.

If $A_n \to A$ in $L^\infty$, $\phi_n \to \phi$ in $L^2$ and $u_n \in H^1_0$ are weak solutions to $T_n u_n = \phi_n$. Then $u_n$ is Cauchy in $H^1$.

I am not sure how to show this: I can not upper bound $\| u_n - u_m \|_{H^1}$ in a meaningful way so I can use $A_n \to A$ and $\phi_n \to \phi$. Any suggestions would be helpful.


1 Answer 1


Testing the weak formulation of $u_n$ with $u_n$ shows that $(u_n)$ is bounded in $H^1$.

Testing the weak formulations for $u_n$ and $u_m$ with $u_n-u_m$ und subtracting them yields $$ \int_\Omega (\phi_n-\phi_m)(u_n-u_m) = \int_\Omega \nabla(u_n-u_m) (A_n\nabla u_n - A_m\nabla u_m)\\ = \int_\Omega \nabla(u_n-u_m) (A_n\nabla (u_n-u_m) - (A_m-A_n)\nabla u_m) $$ which implies $$ \alpha \| \nabla(u_n-u_m)\|_{L^2}^2 \le \int_\Omega (\phi_n-\phi_m)(u_n-u_m) + \int_\Omega \nabla(u_n-u_m)(A_m-A_n)\nabla u_m. $$ Using Cauchy-Schwarz, Poincare inequality, and boundedness of $(u_n)$ proves the claim.

  • $\begingroup$ I realize that I had a typo in the original problem and just fixed it. Could you check how it changes the argument? I am sorry for the inconvenience. $\endgroup$
    – Laund Kez
    Sep 23 at 23:49
  • $\begingroup$ It changes nothing, as I misread this definition and took the new one: this answer is now valid, while it was not for the old version ;) $\endgroup$
    – daw
    Sep 24 at 6:12

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