# Proving unique weak solution.

I am attempting to solve some problems from Evans , I need some help with the following question. If $u\in H^2_0\Omega$, where $\Omega$ is open, bounded subset of $\mathbb{R}^n$. How can I solve the biharmonic equation

$\Delta^2u=f$ in $\Omega$, $u =\frac {\partial u } {\partial n }=0$ on $\partial \Omega$, $n$ is the normal vector .

provided $\int _\Omega \Delta u \Delta v \, \,dx =\int _\Omega fv$ for all $v\in H^2_0\Omega$. Given $f \in L^2(\Omega)$ , and prove that the weak solution is unique. Any kind of help would be great .

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$u=\frac{\partial u}{\partial n} = 0$ on $\partial \Omega$? Instead of $u \frac{\partial u}{\partial n}$? – user20266 Jun 10 '12 at 7:44
@Thomas : Thank you for pointing out the mistake . – Theorem Jun 10 '12 at 7:58
Do you still need an answer for this question? – Shuhao Cao Jun 18 '13 at 21:35

Suppose that $u \in C_0^\infty(\Omega)$.

Then $$\int_\Omega |D^2 u|^2 \, dx = \int_\Omega \sum_{j,k=1}^n (u_{x_jx_k})^2 \, dx = \sum_{j,k=1}^n \int_\Omega u_{x_jx_k} u_{x_j x_k} \, dx.$$ You can integrate by parts twice to get $$\int_\Omega u_{x_jx_k} u_{x_jx_k} \, dx = - \int_\Omega u_{x_jx_kx_j}u_{x_k} \, dx = \int_\Omega u_{x_j x_j}u_{x_kx_k}\, dx$$ taking into account that $u$ is smooth and vanishes near the boundary of $\Omega$. Thus $$\int_\Omega |D^2 u|^2 \, dx = \sum_{j,k=1}^n \int_\Omega u_{x_j x_j}u_{x_kx_k}\, dx = \int_\Omega \left( \sum_{j=1}^n u_{x_jx_j} \right) \left( \sum_{k=1}^n u_{x_k x_k} \right) \, dx = \int_\Omega |\Delta u|^2 \, dx.$$ Thus $\|D^2 u\|_2^2 = \|\Delta u\|_2^2$. You can use the Poincare inequality to find a constant $C = C(n,\Omega)$ with the property that $\|u\|_2^2 \le C \|Du\|_2^2.$ On the other hand, for any $\epsilon > 0$ you have $$\|Du\|_2^2 = \int_\Omega |Du|^2 \, dx = \int_\Omega Du \cdot Du \, dx = - \int_\Omega u (\Delta u) \, dx \le \frac \epsilon 2 \|u\|_2^2 + \frac 1{2\epsilon} \|\Delta u\|_2^2$$ by Young's inequality. Thus $$\|Du\|_2^2 \le \frac{\epsilon C}{2} \|D u\|_2^2 + \frac{1}{2\epsilon} \|\Delta u\|_2^2.$$ With e.g. $\epsilon = \dfrac 1 C$ it follows that $\|Du\|_2^2 \le \dfrac{C}{2} \|\Delta u\|_2^2$ and consequently $\|u\|_2^2 \le \dfrac{C^2}{2} \|\Delta u\|_2^2$.

Finally we obtain $$\|u\|_2^2 + \|Du\|_2^2 + \|D^2u\|_2^2 \le \left(1 + \frac C2 + \frac{C^2}{2} \right) \|\Delta u\|_2^2.$$ This can be extended to $u \in H_0^2(\Omega)$ using the density of $C_0^\infty(\Omega)$ in that space.

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It seems this answer is trying to prove that $u \in H_0^2(\Omega)$, rather than asserting the existence of a weak solution, which was what the question asked. – Cookie Feb 1 '15 at 20:20
IIRC this answer was in response to a bounty asking why the operator $$B[u,v] = \int_\Omega \Delta u \Delta v \, dx$$ is coercive. – Umberto P. Feb 1 '15 at 20:27
@UmbertoP. Could you help me with the density argument? Let $u\in H^2_0$. Then there is $(u_n)$ in $C_c^\infty$ such that $u_n\to u$ in $H^2$. By the proved part, $\|u_n\|_{H^2}\leq \|\Delta u_n\|_{L^2}$ for all $n\in\mathbb{N}$. So, we have to pass to the limit in the last inequality. I see that $\lim\|u_n\|_{H^2}=\|u\|_{H^2}$. But, since $\|\Delta u_n\|$ isn't exactly a term of the $H^2$-norm, how can we conclude that $\lim\|\Delta u_n\|=\|\Delta u\|$? – Pedro Oct 26 '15 at 23:20
You have $$|\|\Delta u_n\|_2 - \|\Delta u\|_2| \le \|\Delta u_n - \Delta u\|_2 \le \|u - u_n\|_{H^2}.$$ – Umberto P. Oct 27 '15 at 10:36

We will use the Lax-Milgram Theorem. A weak solution of your problem is a $u\in H^2(\Omega)$ such that

\begin{eqnarray} \Delta^2u=f & \Rightarrow & [\Delta^2u]\varphi=f\varphi\\ & \Rightarrow & \int_{\Omega}\Delta(\Delta u)\varphi=\int_{\Omega} f\varphi\\ & \Rightarrow & -\int_{\Omega}\nabla(\Delta u)\nabla\varphi=\int_{\Omega} f\varphi\\ & \Rightarrow & \int_{\Omega}\nabla u\nabla\varphi=\int_{\Omega} f\varphi, \end{eqnarray} for all $\varphi\in H^2_0(\Omega)$. Define the bilinear operator $B:H^2_0(\Omega)\times H^2_0(\Omega)\rightarrow\mathbb{R}$, $$B(u,\varphi)=\int_{\Omega}\Delta u\Delta\varphi.$$ Statement 1 This bilinear operator is continuos.

In fact,

\begin{eqnarray} |B(u,\varphi)| & \leq & \int_{\Omega}|\Delta u||\Delta\varphi|\\ & \leq & \|\Delta u\|^2_{L^2(\Omega)}\|\Delta \varphi\|^2_{L^2(\Omega)}\\ & \leq & C\|u\|^2_{H^2_0(\Omega)}\|\varphi\|^2_{H^2_0(\Omega)} \end{eqnarray} You can prove easily this last inequality.

Statemant 2 The bilinear operator is coercive.

In fact, $$B(u,u)=\int|\Delta u|^2=\|\Delta u\|^2_{L^2(\Omega)}\geq C\|u\|^2_{H^2_0(\Omega)}.$$ We used that $\|\Delta u\|_{L^2(\Omega)}$ defines a norm on $H^2_0(\Omega)$ equivalent to the usual norm.

Then, by the Lax-Milgram Theorem, for each $f\in H^2_0(\Omega)$, exists an unique function $u\in H^2_0(\Omega)$ such that $$B(u,\varphi)=\int_{\Omega}\Delta u\Delta\varphi=\int_{\Omega}f\varphi,$$ for all $\varphi\in H^2_0(\Omega)$.

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Proof of $\|\Delta u\|_{L^2} \geq \|u\|_{H^2}$ is totally non-trivial! You might wanna address that problem. For you used $\|\Delta u\|_{L^2} \geq \|D^2 u\|_{L^2}$, $\|\Delta u\|_{L^2} \geq \|\nabla u\|_{L^2} \leq \|u\|_{L^2}$. And a suggestion on notation is: there is no such term as "$H^2_0(\Omega)$-norm" in your subscript for norms, only $H^2$-norm. – Shuhao Cao Jun 16 '13 at 7:40
Thank you for this observations. You are right. I knew that the proof of this inequality is non-trivial, but as I dont know to prove this, I wrote just what I did in my exercise. – José Carlos Jun 17 '13 at 4:12

Four hints:

i) what kind of functional is $v\mapsto \int fv$? It's obviously linear, but is it continuous?

ii) Assume $u,\bar{u}$ solve the problem. Then $\int\Delta(u-\bar{u})\Delta v dx =0$ for all admissible $v$. What is the image of $\Delta$ when applied to admissible $v$? I.e. for which $\phi$ can you solve $\Delta v = \phi$? All these $\phi$ are admissible test functions. What does that tell you about $u-\bar{u}$?

iii) What does ii) tell you about $(u,v)\mapsto \langle u, v\rangle := \int\Delta u \Delta v dx$ ? Could this possibly be a scalar product? If yes, on which space?

iv) Now try to combine i) and iii). Does any representation theorem for linear functionals apply?

This comes without any kind of warranty, I do explicitly state that I did not check the details, it's just the roadmap I'd try first (with quite some confidence, I'd like to add, though). But you asked for assistance not for a solution :-)

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Thank you very much :) . First I will try with the hints. – Theorem Jun 10 '12 at 8:53
I have been trying to get some idea but i am not able at all . If you could help me with detailed solution it would be great . – Theorem Jun 10 '12 at 10:20

The Lax-Milgram theorem

Given a Hilbert space $V$ with scalar product $(.,.)_V$ and corresponding norm $\|\cdot\|_V$, a continuous and coercive bilinear form $a(.,.)$ on $V \times V$ and a continuous linear functional $L$ on $V$, there exists unique $u \in V$ s.t.

$a(u,v) = L(v)\ \ \ \ \forall v \in V$.

Some of the proofs required if you want to use this:

• $a(.,.)$ is symmetric, ie, it holds that $a(u,v) = a(v,u) \ \ \ \forall u,v \in V$
• $a(.,.)$ is continuous, ie, there exists a constant $C>0$ s.t. $|a(u,v)|\leq C\|u\|_V\|v\|_V \ \ \forall u,v \in V$
• $a(.,.)$ is coercive or V-elliptic, ie, there exists a constant $\alpha>0$ s.t. $a(u,u) \geq \alpha \|u\|^2_V \ \ \ \forall u \in V$
• $L$ is continuous, ie, there exists a constant $\Lambda>0$ s.t. $|L(v)| \leq \lambda \|v\|_V \ \ \ \forall v \in V$

I hope this gets you started.

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I have some ideas about the first question.You can use the method that we find the solution of the Possion's Equation.

First, you can find a spherical symmetry solution of the biharmonic equation. Thus, you will get a fundamental solution. Then you can get solution of $\Delta^2u=f$ by Green's Identity.

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For the second question,you can use the representation theorem in Hilbert Space.For more details,you can read Page 118 in the PDE by Fritz John. – Juntao Huang Jun 10 '12 at 8:34