# Prove there exist a unique $u\in H^1$ such that $\int_{\Omega}(\kappa\nabla u\cdot\nabla v + \frac{1}{\kappa}uv) =\ \int_{\Omega}fv$ for $f\in L^2$

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Let $\Omega$ be a bounded open subset of $\mathbb{R}^n$ and let $\kappa : \Omega\rightarrow\mathbb{R}$ be a continuous function, such that there's constants $M, \beta > 0$, such that $\beta \leq \kappa(x) \leq M$ for each $x\in\Omega$. Show, using some duality result, that for each $f\in L^2(\Omega)$, there exists a unique $u\in H^1(\Omega)$ such that $$\int_{\Omega}\left\{\kappa\nabla u\cdot\nabla v + \frac{1}{\kappa}uv\right\}\ =\ \int_{\Omega}fv,\;\;\; \forall\ v\in H^1(\Omega),$$ and $$\|u\|_{H^1(\Omega)}\ \leq\ \left\{\frac{M}{\min\left\{\beta, \frac{1}{M}\right\}}\right\}^{1/2}\|f\|_{L^2(\Omega)}.$$

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Question: Show, using some duality result, that for each $f\in L^2(\Omega)$, there exists a unique $u\in H^1(\Omega)$ such that $u$ is the weak solution to the $\kappa$-weighted elliptic equation, and the solution estimate in $H^1$-norm.

Like you said, this can be proved by Riesz.

First, we have $f\in L^2(\Omega)\subset (H^{1}(\Omega))'$. This is true because (with slight abuse of notation), we can induce a bounded linear functional on $H^1(\Omega)$ by defining $f(v): = \int_{\Omega} fv$ for any $v\in H^1(\Omega)$.

Second we can show that $$(u,v)_{\kappa}:= \int_{\Omega}\left\{\kappa\nabla u\cdot\nabla v + \frac{1}{\kappa}uv\right\}$$ is an inner product on $H^1(\Omega)$. This is shown in your other question: How to prove $\int_{\Omega}\frac{1}{\kappa}uv\ +\ \int_{\Omega}\kappa\nabla u\cdot\nabla v$ is an inner product in $H^1$ .

Now applying Riesz representation theorem, there exists a unique $u$ such that $$(u,v)_{\kappa} = f(v) \quad \text{for any }v\in H^1(\Omega). \tag{1}$$ This gives us the existence of a unique weak solution, which answers your first question.

The estimate is obtained by letting $u=v$ in (1): $$M^{-1}\|u\|_{L^2(\Omega)}^2\leq \int_{\Omega}\frac{1}{\kappa} u^2 \leq (u,u)_{\kappa} = \int_{\Omega}fu\leq \|f\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)} \\ \implies \|u\|_{L^2(\Omega)}\leq M\|f\|_{L^2(\Omega)}.$$ Yet we have $$\min\{\beta,M^{-1}\} \|u\|_{H^1(\Omega)}^2 \leq (u,u)_{\kappa} = \int_{\Omega}fu\leq \|f\|_{L^2(\Omega)}\|u\|_{L^2(\Omega)} \leq M \|f\|_{L^2(\Omega)}^2.$$ Therefore: $$\|u\|_{H^1(\Omega)}\leq \left\{\frac{M}{\min\left\{\beta, M^{-1}\right\}}\right\}^{1/2}\|f\|_{L^2(\Omega)}.$$

To do the second part, let $v=u$ and just bound below by the bounds on k for each term on the left hand side. Then you take the min of them and the $H^1$ norm basically pops out. Use Cauchy-Schwarz on the right.

The first part will follow similarly. Use the above strategy to get a coercive bound on the left hand side and use Lax-Milgram.

• Thank you for the answer, but I can not use, 'coercive' and Lax-Milgram, because we didn't see that yet. I suppose, I need to use the fact of $H^1$ is a Hilbet Space and then use the Riesz Representation Theorem. – FASCH Apr 24 '13 at 2:56
• Indeed, if I can prove that $H^1$ is a Hilbert Space with the inner product involve the function $\kappa$, then I can use Riesz with the functional $F(v) := \int fv$, for each $v$ and each $f$, in order to guarantee the existence of $u$. Am I right? – FASCH Apr 24 '13 at 10:09
• Yes if you could prove that. It's not very hard to show an applicable bilinear form is an inner product. – toypajme Apr 24 '13 at 18:16
• In order to conclude that, it suffices to show that the corresponding norms are equivalent, right? But How do I can do that? Thanks in advance. – FASCH Apr 24 '13 at 19:32
• I have wrote this, as a new question: math.stackexchange.com/questions/372923/… – FASCH Apr 25 '13 at 21:48