# UMD PDE Qualifying Exams/Jan2006PDE

## Problem 4

 A weak solution of the biharmonic equation, \left\{\begin{align} \Delta^2 u = f, & x\in U\\ u=\frac{\partial u}{\partial \nu}=0, & x\in \partial U \end{align}\right. is a function $u\in H_0^1(U)$ such that  $\int_U \Delta u \Delta v\,dx=\int_U f v\, dx$ for all $v\in H_0^1(U)$. Assume that $U$ is a bounded subset of $\mathbb R^n$ withs mooth boundary and use the weak formulation of the problem to prove the existence of a unique weak solution.

### Solution

Consider the functional $B[u,v]=\int_U \Delta u \Delta v = \int_U f v$. $B$ is bilinear by linearity of the Laplacian. Now, we claim that $B$ is also continuous and coercive.

$|B[u,v]| = \left| \int_U\Delta u\Delta v \right| \leq \|\Delta u \|_{L^2(U)}\|\Delta v \|_{L^2(U)}\leq \|\Delta u \|_{H_0^1(U)}\|\Delta v \|_{H_0^1(U)}$ where the first inequality is due to Holder and the second is by the definition of the Sobolev norm. And so $B[u,v]$ is a continuous functional.

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

\begin{align} \| 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{align}

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.