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In using lax-milgram to find a weak solution in an intersection of sobolev spaces the weak solution for $$ -\Delta^2 u = f \in L^2(U)\\ \\ u|_{\partial U}=\Delta u|_{\partial U} = 0 $$ was discussed, I have a question about the week solution of $$ \Delta^2 u + u = f \in L^2(U)\\ \\ u|_{\partial U}=\Delta u|_{\partial U} = 0 $$ I think I should use coupled elliptic PDE theory. Any hint or suggestion is helpful for me.

In advanced thanks from anyone who tries to help me.

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All
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1 Answers1

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I will assume that $U$ is bounded. Let $H=H(U)=H_0^1(U)\cap H^2(U)$. As you can see in the post you have cited, $H$ is a Hibert space with the scalar product $(u,v)=\int_U \Delta u\Delta v$. We also have $$\int |\Delta u|^2\geq\lambda_1^2\int|u|^2,\ \forall\ u\in H\tag{1}$$

Now consider, for example, the bilinear form $a:H\times H\to\mathbb{R}$ given by $$a(u,v)=\int_U\Delta u\Delta v+\int_Uuv$$

You can verify (by using (1)) that $a$ is continuous, coercive and symmetric. You can also verify that $f$ defines a bounded linear functional in $H$ by the expression $$v\mapsto\int_U fv,\ \forall\ v\in H$$

This implies by Lax-Milgram Theorem that for every $f\in L^2$, there exist $u\in H$ such that $$a(u,v)=\int_U fv,\ \forall v\in H\tag{2}$$

i.e. $u$ is a weak solution of your problem.

Remark 1: Lax-Milgram also gives you that $u$ is the minimizer of the functional $$I(u)=\int_U |\Delta u|^2+\int_U u^2-\int_U fu,\ \forall\ u\in H$$

Remark 2: $u\in H_0^1(U)\cap H^4(U)$

Remark 3: Can you see why $\Delta u=0$ in the boundary of $U$?

Edit: To prove Remark 3, first choose $w\in H$ such that $\Delta w=f$ in $U$. This can be done because the problem $$ \left\{\begin{array}{ccc} \Delta w=f &\mbox{ in $U$} \\ w=0 &\mbox{ in $\partial U$} \end{array} \right. $$

has a unique solution $w$ in $H^2$. Moreover, because $v\mapsto \int_U uv$ defines a bounded linear function in $H_0^1(U)$, we have by Riesz Theorem that there exist $\phi\in H_0^1(U)$ such that $$\int_U uv = \int_U \nabla \phi\nabla v,\ \forall\ v\in H_0^1\tag{3}$$

We conclude from $(2)$ and $(3)$ that $$\int_U \Delta u\Delta v+\int_U\nabla \phi\nabla v=\int_U \Delta w v,\ \forall\ v\in H\tag{4}$$

We apply Green Theoren in $(4)$ to conclude that

$$\int_U \Delta u\Delta v-\int_U\phi\Delta v=\int_U w \Delta v,\ \forall\ v\in H\tag{5}$$

"Again", for every $\psi\in L^2(U)$, there exist $v\in H$ such that $\Delta v=\psi$, therefore, from $(5)$ we get

$$\int_U (\Delta u-\phi-w)\psi=0,\ \forall\ \psi\in L^2(U)\tag{6}$$

We apply du Bois-Reymond to conclude that $$\Delta u(x)-\phi(x)-w(x)=0,\ a.e.\ x\in U$$

Because $\phi,w\in H_0^1(U)$, we conclude that $\Delta u=0$ in $\partial U$.

Tomás
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  • Thanks. When I want to verify continuity I confused with the spaces that should be considered, I mean: $$\int_{U} |\Delta u \Delta v| +| uv | \leq \int_{U}|\Delta u \Delta v| + c_1|\Delta u \Delta v| \leq (1+ c_1) |u|{H^2}|v|{H^2},$$ Now can we infer continuity for $H^1_0(U)\cap H^4(U)$? – All Oct 30 '13 at 12:16
  • To prove continuity, we have to prove that $$|a(u,v)|\leq C |u||v|$$ where $|u|=|\Delta u|_2$. Now, note that $$\int_U |\Delta u\Delta v|+\int_U |uv|\leq |u||v|+|u|_2|v|_2$$ To conclude apply inequality $(1)$ and regroup the terms. – Tomás Oct 30 '13 at 12:24
  • You are welcome @schliffen – Tomás Oct 30 '13 at 12:41
  • @schliffen, what do you study? – Tomás Oct 30 '13 at 13:05
  • Applied mathematics, Control and Optimization. – All Oct 30 '13 at 13:08
  • Dear @Tomas, I can't figure out why Δu=0 in the boundary of U? can you give me a clue, – All Oct 30 '13 at 15:40
  • I have edited my question with the answer for your last comment, please take a look on it. – Tomás Oct 30 '13 at 15:51
  • Thank you so much Dear @Tomas. You helped me to solve most of my problems. – All Oct 31 '13 at 06:35
  • You are welcome @schliffen. If you like the answer, you can accept it and vote up in the arrow buttom, next to the beggining of the answer. – Tomás Oct 31 '13 at 11:08