# Show $||u-\hat{u}||_{H^1(\Omega)} \leq c||a-\hat{a}||_{L^{\infty}}$ for $−\text{div}(a\nabla{u})=f, \text{div}(\hat{a}\nabla{u})=f$

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

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 $$\|z\|_{H^1}$$.