UMD PDE Qualifying Exams/Jan2006PDE

Problem 1

Solution

Problem 2

Problem 3

Problem 4

A weak solution of the biharmonic equation,

$\left\{{\begin{aligned}\Delta ^{2}u=f,&x\in U\\u={\frac {\partial u}{\partial \nu }}=0,&x\in \partial U\end{aligned}}\right.$

is a function  $u\in H_{0}^{1}(U)$  such that

$\int _{U}\Delta u\Delta v\,dx=\int _{U}fv\,dx$ for all $v\in H_{0}^{1}(U)$ .

Assume that $U$ is a bounded subset of $\mathbb {R} ^{n}$ with smooth boundary and use the weak formulation of the problem to prove the existence of a unique weak solution.

Solution

Consider the functional $B[u,v]=\int _{U}\Delta u\Delta v=\int _{U}fv$ . $B$ 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

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

which establishes coercivity.

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

Problem 5

Problem 6

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

This article is issued from Wikibooks. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.