## Lax Milgram application

Let ${I=(0,2)}$ and ${V=H^1(I)}$. Consider the bilinear form

$\displaystyle a(u,v)= \int_0^2 u'(t)v'(t)dt +\left( \int_0^1 u(t)dt\right)\left( \int_0^1 v(t)dt\right).$

1. Check that ${a(u,v)}$ is a continuous symmetric bilinear form and that ${a(u,u)=0}$ implies ${u=0}$.

2. Prove that ${a}$ is coercive.

3. Deduce that for every ${f \in L^2(I)}$ there exists a unique ${u \in H^1(I)}$ satisfying

$\displaystyle (1) \ \ \ \ a(u,v)=\int_0^2 fv,\ \forall v \in H^1(I).$

What is the corresponding minimization problem?

4. Show that the solution of ${(1)}$ belongs to ${H^2(I)}$ (and in particular ${u \in C^1(\overline{I})}$). Determine the equation and the boundary conditions satisfied by ${u}$.

5. Assume that ${f \in C(\overline{I})}$, and let ${u}$ be the solution of ${(1)}$. Prove that ${u}$ belongs to ${W^{2,p}(I)}$ for every ${p<\infty}$. Show that ${u \in C^2(\overline{I})}$ if and only if ${\int_If=0}$.

6. Determine explicitly the solution ${u}$ of ${(1)}$ when ${f}$ is a constant.

7. Set ${u=Tf}$, where ${u}$ is the solution of ${(1)}$ and ${f \in L^2(I)}$. Check that ${T}$ is a self-adjoint compact operator from ${L^2(I)}$ into itself.

8. Study the eigenvalues of ${T}$.

H. Brezis, Functional Analysis

Solution: 1. It is easy to see that ${a}$ is symmetric and bilinear. To prove the continuity part apply Holder’s inequality to see that

$\displaystyle |a(u,v)|\leq \|u\|_{H^1}\|v\|_{H^1}.$

If ${u \in H^1(I)}$ such that ${a(u,u)=0}$ then we first have ${u'=0}$, which implies that ${u}$ is constant and secondly ${\int_0^1 u(t)=0}$ which implies that ${u=0}$.

2. Suppose there exists a sequence ${(u_n)\subset H^1(I)}$ such that ${\|u_n\|_{H^1}=1}$ and ${a(u_n,u_n)=0}$. Using the compacity of the unit ball in the weak convergence it is not restrictive to assume that there exists ${u \in H^1(I)}$ with ${u_n \rightharpoonup u}$ in ${H^1(I)}$ and ${u_n \rightarrow u}$ in ${L^2(I)}$.

Since we have ${\|u'\|_{L^2(I)}^2\leq \liminf \|u_n'\|_{L^2(I)}^2\leq a(u_n,u_n)}$ we get that ${u'=0}$, and therefore ${u}$ is a constant.

Since ${u_n \rightarrow u}$ in ${L^2(I)}$ implies that ${\int_0^1 u_n \rightarrow \int_0^1 u}$ it follows that ${u=0}$. Note that we have

$\displaystyle a(u_n,u_n)=1-\int_0^2 u_n^2+\left(\int_0^1 u_n\right)^2\rightarrow 1,$

which is a contradiction to our assumption. Therefore ${a}$ is coercive.

3. To get the desired result we apply Lax Milgram Theorem. Since ${a}$ is symmetric, the solution is also a minimizer for the problem

$\displaystyle \min_{v \in H^1(I)} \left\{ \frac{1}{2}a(v,v)-f(v)\right\}$

4. Let ${u}$ be the solution of ${(1)}$. Then for every ${\varphi \in C_0^\infty(I)}$ we have

$\displaystyle \int_I u'\varphi'=\int_I (f-g)\varphi,$

where ${g=\int_0^1 u(t)dt \chi_{(0,1)}}$. This implies that ${u' \in H^1(I)}$ and therefore ${u \in H^2(I)}$. The equation satisfied by ${u}$ is

$\displaystyle -u''=f-g,$

and the necessary boundary conditions can be obtained by multiplying the equation with ${v \in H^1(I)}$ and integrating on ${I}$. The boundary conditions are

$\displaystyle u'(0)=u'(2)=0.$

5. The differential equation satisfied by ${u}$ can be written as

$\displaystyle -u''=f-\int_0^1 u \cdot \chi_{(0,1)}.$

First note that since ${u \in C^1(\overline{I})}$ we know that ${u,u' \in L^p(I)}$ for all ${1\leq p<\infty}$. Therefore ${u \in W^{1,p}(I)}$. Moreover, since ${f}$ is continuous, using the differential equation above we can see that ${u'' \in L^p}$ for every ${p<\infty}$. Therefore ${u \in W^{2,p}(I)}$ for ${p<\infty}$.

We have the following equivalences:

$\displaystyle u \in C^2(\overline{I}) \Leftrightarrow \int_0^1 u=0 \text{ and } -u''=f \Leftrightarrow \int_If=0.$

7. Let ${f,g \in L^2(I)}$ and denote ${u=Tf, v=Tg}$. Then we have

$\displaystyle \int_I Tf g=\int_I gu=a(v,u)=a(u,v)=\int_I fv=\int_I f Tg,$

which proves that ${T}$ is self-adjoint.

Suppose that ${(f_n)\subset L^2(I)}$ and ${\|f_n\|_{L^2(I)}\leq 1}$. Denote by ${u_n=Tf_n}$ and note that by the coercivity of ${a(\cdot,\cdot)}$ we have

$\displaystyle c\|u_n\|_{H^1(I)}^2 \leq a(u_n,u_n)=\int_I fnu_n \leq \|fn\|_{L^2}\|u_n\|_{L^2} \leq \|u_n\|_{L^2}\leq \|u_n\|_{H^1(I)}.$

This means that ${(u_n)}$ is bounded in ${H_1(I)}$ and therefore it contains a subsequence which converges weakly to ${u \in H^1(I)}$ which implies that ${u_n \rightarrow u}$ in ${L^2(I)}$. Therefore ${(Tf_n)}$ contains a subsequence converging strongly in ${L^2(I)}$. This proves that ${T}$ is compact.