After some computations I delivered the following variational equality

$$\int_{\Omega}(\mbox{div}\mu\nabla u+f_{0})v\,dx=0, \qquad \forall v\in V,$$ where $\mu\nabla u\in H^{1}(\Omega, \mathbb{R}^2), f_{0}\in L^{2}(\Omega)$ and $V=\{v\in H^{1}(\Omega): v=0 \quad \mbox{on}\quad \Gamma_{1}\}$, $\partial \Omega=\Gamma_{1}\cup \Gamma_{2}$, $\Omega$ is an open and connected subset of $\mathbb{R}^{2}$. I want to prove that u verifies the partial differential equation $$\mbox{div}\mu\nabla u+f_{0}=0\qquad (1)$$ in $L^{2}(\Omega)$. So the solution is regular in Lebesque space sense instead of the classical one. How to prove that?

My idea is very simple: since we have quantification i.e., $\forall v\in V$ we may pick $v=\varphi\in C^{\infty}_{0}(\Omega)$ and have $$\int_{\Omega}(\mbox{div}\mu\nabla u+f_{0})\varphi\,dx=0, \qquad \forall \varphi\in C^{\infty}_{0}(\Omega).$$ I recall that $C^{\infty}_{0}(\Omega)$ is dense in $L^{2}(\Omega)$. Can anybody show me how to use that fact in details in here?

  • $\begingroup$ You have $\mbox{div}\mu\nabla u+f_{0}=0$ in $L^2$, and you want to know it is true pointwise ? $\endgroup$ – reuns Apr 19 '17 at 11:54
  • $\begingroup$ Now I started to wonder what it means to verify (1) in $L^{2}$? Can it be proven that $\mbox{div}\mu\nabla u+f_{0}=0$ pointwise? $\endgroup$ – zorro47 Apr 19 '17 at 13:14
  • $\begingroup$ Since both the divergence of your $H^1$ function and $f_0$ are in $L^2$, you cannot make any pointwise statements about $u$ without any further regularity assumptions. $\endgroup$ – Miguel Apr 19 '17 at 13:53

You want to prove $$ (w, \varphi)_{L^2} = 0 \text{ for all } \varphi \in D \Rightarrow w = 0. $$ where $D$ is a dense subset of $L^2$ and we write $w = L u$, where $L$ is your differential operator, for convenience. Because $D$ is dense, you can find a sequence $(\varphi_n) \subset D$ converging to $w$ in $L^2$. Then, by Cauchy-Schwarz: $$ (w, w) = (w, w - \varphi_n) + (w, \varphi_n) \leqslant | w | | w - \varphi_n | + (w, \varphi_n) \rightarrow 0. $$ The last term is zero by assumption and the first consists of a bounded factor $|w|$ and one that goes to zero by construction: $|w-\varphi_n|$.

Note that if you had your integral equation for all $v \in L^2$, then the statement would be trivial, because you could just take $v$ to be $L u$ and then the integral means $||Lu||=0$, which means $Lu=0$ in $L^2$. But you only have it for test functions $v \in H^1_0$, a smaller space, hence the need of going through a dense subset.

  • $\begingroup$ Thank you for your answer and remark. There is one thing that bothers me. You proved that $(w,\varphi)_{L^{2}}=0\quad\mbox{for all}\quad \varphi \in D$ then $w=0$. In what sense is the equality understood? It cannot be pointwise, because $w$ is from $L^{2}$. Hence, it must be understood with respect to the space where it takes place - so $w=0$ a.e. ? $\endgroup$ – zorro47 Apr 19 '17 at 16:09
  • $\begingroup$ Exactly. Or rather: $w=0$ in $L^2$, and consequently a.e. $\endgroup$ – Miguel Apr 19 '17 at 16:11
  • $\begingroup$ Let's make our problem more complicated and add boundary condition: $$\mu\partial_{\nu}u=f_{2}$$ on $\Gamma_{2}$. After some computations I obtained $$\int_{\Gamma_{2}}(\mu\partial_{\nu}u-f_{2})vd\Gamma=0$$ for all $v\in V$ (trace operator included inside the integral). Once again I'd like to show that $\mu\partial_{\nu}u=f_{2}$ in $L^{2}(\Gamma_{2})$. As far as I know, the set of traces of functions in $V$ is dense in $L^{2}(\Gamma_{2})$. But what does that really mean? $\endgroup$ – zorro47 Apr 21 '17 at 12:34
  • $\begingroup$ Does that mean that I can choose a sequance $(v_{n})\subset V$ such that $\gamma(v_{n})\rightarrow (\gamma(\mu\partial_{\nu}u)-f_{2})$ with respect to $L^{2}(\Gamma_{2})$ norm, where $\gamma$ denotes the trace operator? If so, I can use the same argumentation as above and prove my statement. $\endgroup$ – zorro47 Apr 21 '17 at 13:07
  • $\begingroup$ Yes, it's the same argument. $\endgroup$ – Miguel Apr 21 '17 at 14:34

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