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.

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