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.

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.

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 $\varphi\in H^2_0(\Omega)$.