# UMD PDE Qualifying Exams/Jan2006PDE

## Problem 4

 A weak solution of the biharmonic equation, {\displaystyle \left\{{\begin{aligned}\Delta ^{2}u=f,&x\in U\\u={\frac {\partial u}{\partial \nu }}=0,&x\in \partial U\end{aligned}}\right.} is a function ${\displaystyle u\in H_{0}^{1}(U)}$ such that ${\displaystyle \int _{U}\Delta u\Delta v\,dx=\int _{U}fv\,dx}$ for all ${\displaystyle v\in H_{0}^{1}(U)}$. Assume that ${\displaystyle U}$ is a bounded subset of ${\displaystyle \mathbb {R} ^{n}}$ with smooth boundary and use the weak formulation of the problem to prove the existence of a unique weak solution.

### Solution

Consider the functional ${\displaystyle B[u,v]=\int _{U}\Delta u\Delta v=\int _{U}fv}$. ${\displaystyle B}$ is bilinear by linearity of the Laplacian. Now, we claim that ${\displaystyle B}$ is also continuous and coercive.

${\displaystyle |B[u,v]|=\left|\int _{U}\Delta u\Delta v\right|\leq \|\Delta u\|_{L^{2}(U)}\|\Delta v\|_{L^{2}(U)}\leq \|\Delta u\|_{H_{0}^{1}(U)}\|\Delta v\|_{H_{0}^{1}(U)}}$ where the first inequality is due to Holder and the second is by the definition of the Sobolev norm. And so ${\displaystyle B[u,v]}$ is a continuous functional.

To show coercivity, we use the fact that by two uses of integration by parts, ${\displaystyle \int _{U}u_{ij}u_{ij}=-\int _{U}u_{i}u_{ijj}=\int _{U}u_{ii}u_{jj}}$ which gives

{\displaystyle {\begin{aligned}\|u\|_{H_{0}^{1}(U)}^{2}=&\int _{U}u^{2}+\sum _{j=1}^{n}\left(|{\frac {\partial }{\partial j}}u|^{2}\right)+\sum _{i,j=1}^{n}\left(|{\frac {\partial ^{2}}{\partial i\partial j}}u|^{2}\right)\,dx\\=&\int _{U}u^{2}+|\nabla u|^{2}+\sum _{i,j=1}^{\infty }\left(|u_{ii}u_{jj}|^{2}\right)\,dx\\\leq &\int _{U}0+0+|\Delta u|^{2}\leq B[u,u]\end{aligned}}}

which establishes coercivity.

Thus, by the Lax-Milgram Theorem, the weak solution exists and is unique.

## Problem 6

 This problem has a typo and can't be solved by characteristics as it is written.