Unique weak solution to the biharmonic equation I am attempting to solve some problems from Evans, I need some help with the following question. 

Suppose $u\in H^2_0(\Omega)$, where $\Omega$ is open, bounded subset of $\mathbb{R}^n$. 
  
  
*
  
*How can I solve the biharmonic equation
  $$\begin{cases}
\Delta^2u=f \quad\text{in } \Omega, \\
u  =\frac {\partial u } {\partial n }=0\quad \text{on }\partial\Omega.
\end{cases}
$$ where $n$ is the normal vector such that $\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. 
 A: 
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.
A: 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 :-)
A: 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.
A: 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}\Delta u\Delta\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, ([Edited]be cautious: this step is highly nontrivial as pointed out in the comment)
$$B(u,u)=\int|\Delta u|^2=\color{blue}{\|\Delta u\|^2_{L^2(\Omega)}\geq C\|u\|^2_{H^2_0(\Omega)}}.$$
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 $\varphi\in H^2_0(\Omega)$.
A: 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. 
