Let $\Omega\subset \mathbb{R}^d$ be a bounded domain and $a,\hat{a}\in L^{\infty}(\Omega)$ such that $0<a \leq a(x),\hat{a}(x)\leq A<\infty$ on $\Omega$.

For some $f \in H^{-1}{\Omega}$ let $u \in H^1{(\Omega)}$ be the solution of the PDE $$ −\text{div}(a\nabla{u})=f \text{ on } Ω,\\ u=0 \text{ on } \partial\Omega,$$ and $\hat{u}\in H_0^1(\Omega)$ the solution of

$$-\text{div}(\hat{a}\nabla{u})=f \text{ on } \Omega,\\ u=0 \text{ on } \partial \Omega.$$
Show that it holds $$||u-\hat{u}||_{H^1(\Omega)} \leq c||a-\hat{a}||_{L^{\infty}}$$ with some constant $c > 0$ and write explicitly down this constant $c$ for the given data.

So $u-\hat u$ is a solution of the problem $$\text{div}((\hat{a}-a)\nabla{u})=f \text{ on } \Omega,\\ u=0 \text{ on } \partial \Omega.$$ I think I need derive a weak formulation of the problem (which I struggle with) and by Lax-Milgram I can bound $||u-\hat{u}||$ using $f$. However I don't see how I can bound it using $a$ and $\hat{a}$.


1 Answer 1


Your difference problem is not correct. By subtracting both state equations, we get $$ -\operatorname{div}(a \nabla u) + \operatorname{div}(\hat a \nabla \hat u) = f - f,$$ i.e., $z = u - \hat u$ satisfies $$ -\operatorname{div}(a \nabla z) = -\operatorname{div}((\hat a - a) \nabla \hat u) $$ and $z = 0$ on $\partial\Omega$. For this equation you can use standard elliptic estimates to derive an estimate for $\|z\|_{H^1}$.


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