# Characterization of weak solution

## 5 Nonlinear elliptic variational inequalities

### Preliminaries

In order to explain the importance of elliptic variational inequalities, first consider the weak solution of the linear elliptic function $(1.1)$ with homogeneous Dirichlet boundary condition, i.e. a function $u\in H_0^1(\Omega)$ satisfying for all $v\in H^1_0(\Omega)$

$$\langle Au,v\rangle=\sum^n_{j,k=1}\int_{\Omega}a_{jk}(D_ku)(D_jv)dx+\int_\Omega cuvdx=\int_\Omega fvdx=\langle F,v \rangle\tag{5.1}$$

It is well-known (see, e.g., $[67]$) that if $c\ge0,\;a_{kj}=a_{kj}\in L^\infty(\Omega)$ satisfy the uniform ellipticity condition then the unique solution is $u\in H^1_0(\Omega)$ of $(5.1)$ is the unique function $u=u^\star\in H^1_0(\Omega)$ which minimizes the quadratic functional

$$E(u)=\langle Au,u\rangle-2\langle F,u\rangle=\tag{5.2}$$ $$\sum^n_{j,k=1}\int_{\Omega}a_{jk}(D_ku)(D_ju)dx+\int_\Omega cu^2dx-2\int_\Omega fudx.$$ (Here $A:V\to V^\star$ is a linear operator, $V=H^1_0(\Omega)$.)

Similarly, the weak solution of the Neumann problem with homogeneous boundary condition, i.e. the solution $u\in H^1(\Omega)$ of $(5.1)$ for all $v\in H^1(\Omega)$ , is the unique $u=u^\star\in H^1(\Omega)$ where $E$ attains its minimum in $H^1(\Omega)$. By using similar arguments as in $[67]$, one can show the following generalization of the above statements.

In the weak formulation of a linear elliptic PDE we have the following characterization of the unique solution $u \in H_{0}^{1}(\Omega)$. In the attached image it is the part that reads "...the unique solution $u \in H_{0}^{1}(\Omega)$ of (5.1) is the unique function $u=u^{*} \in H_{0}^{1}(\Omega)$ which minimizes the quadratic functional..." Does anyone know how this characterization follows or a good book as a reference? The book that is recommended is in Hungarian and not easily available. Thanks.

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From the note it looks like for unique $u \in H_{0}^{1}(\Omega)$ solving $Au = f$ iff unique $u$ minimizes $E(u) := \frac{1}{2}\langle Au,u \rangle - f(u)$...so it is a sufficient and necessary condition...? The answer below just says necessary condition. –  Moses Jun 26 '14 at 17:27
It is necessary and sufficient, see the edited answer. –  daw Jun 27 '14 at 5:26
@daw Okay I see. Do you know why the $u$ is taken as being unique? –  Moses Jul 5 '14 at 16:57
It follows from strict convexity of $E$, see also my answer below. –  daw Jul 6 '14 at 13:34

Edit: Rewrote the answer for more clarity.

Assume that $A$ is symmetric and coercive: $$\langle Au,v\rangle = \langle Av,u\rangle\quad \forall u,v\in H^1_0(\Omega),$$ and $$\langle Au,u\rangle \ge \delta \|u\|^2 \quad \forall u\in H^1_0(\Omega)$$ with some $\delta>0$.

We want to prove the following: Let $u^*\in H_0^1(\Omega)$. Then these two points are equivalent:

1. $Au^*=f$.
2. $u^*$ minimizes $E(u):=\frac12\langle Au,u\rangle - f(u)$.

First observe, that the following identity holds: $$E(v) - E(u)= \langle Au-f, v-u\rangle + \frac12\langle A(v-u), v-u\rangle$$ for all $v,u\in H^1_0(\Omega)$. This can be derived by equivalent algebraic operations. If you know about differentiability, this is just the Taylor expansion of $E$ at $u$.

Let now (1.) be satisfied. Hence, it holds $$\begin{split} E(v) - E(u^*)& =\langle Au^*-f, v-u^*\rangle + \frac12\langle A(v-u^*), v-u^*\rangle\\ &\ge \delta \|v-u^*\|^2, \end{split}$$ which implies $E(v)>E(u)$ for all $v\ne u$, hence (2.) holds.

Let (2.) be satisfied. Then $$E(v) - E(u^*) =\langle Au^*-f, v-u^*\rangle + \frac12\langle A(v-u^*) \ge 0$$ for all $v$. Take $h\in H^1_0(\Omega)$. Set $v:=u^*+th$ with $t>0$, divide the resulting inequality by $t$, let $t$ go to zero, to conclude $$Au^*=f,$$ which is (1.).

Hope this helps.

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Coercivuty is used to ensure the existence of a minimum for $E$. –  Siminore Jun 25 '14 at 12:56
@daw So the idea is to show that $E(v) - E(u) \geq 0$ for all $v \in H_{0}^{1}$ –  John Doe Jun 25 '14 at 13:55
@daw thanks for your response. Is the first line some kind of taylor expansion? How did you evaluate $E'(u)$? –  John Doe Jun 25 '14 at 15:44
It is Taylor expansion. –  daw Jun 26 '14 at 11:30
@daw Okay but why did you only expand to three terms? Also could you lastly tell me how you expanded $E^{'}(u)$? Thanks for all the assistance. –  John Doe Jun 26 '14 at 14:39