UMD PDE Qualifying Exams/Jan2006PDE

Problem 4

A weak solution of the biharmonic equation, $\{\begin{array}{cc}{\mathrm{\Delta }}^{2}u=f,& x\in U\\ u=\frac{\partial u}{\partial \nu }=0,& x\in \partial U\end{array}$ is a function $u\in H_0^1(U)$ such that $\underset{U}{\int }\mathrm{\Delta }u\mathrm{\Delta }vdx=\underset{U}{\int }fvdx$ for all $v\in H_0^1(U)$. Assume that $U$ is a bounded subset of ${ℝ}^n$ with smooth boundary and use the weak formulation of the problem to prove the existence of a unique weak solution.

Solution

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

$|B[u,v]|=\left|\underset{U}{\int }\mathrm{\Delta }u\mathrm{\Delta }v\right|\le \Vert \mathrm{\Delta }u{\Vert }_{L^2(U)}\Vert \mathrm{\Delta }v{\Vert }_{L^2(U)}\le \Vert \mathrm{\Delta }u{\Vert }_{H_0^1(U)}\Vert \mathrm{\Delta }v{\Vert }_{H_0^1(U)}$ 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, $\underset{U}{\int }{u}_{ij}{u}_{ij}=-\underset{U}{\int }{u}_i{u}_{ijj}=\underset{U}{\int }{u}_{ii}{u}_{jj}$ which gives

$\begin{array}{cc}\Vert u{\Vert }_{H_0^1(U)}^2=& \underset{U}{\int }{u}^2+\sum _{j=1}^n(|\frac{\partial }{\partial j}u{|}^2)+\sum _{i,j=1}^n(|\frac{{\partial }^2}{\partial i\partial j}u{|}^2)dx\\ =& \underset{U}{\int }{u}^2+|\mathrm{\nabla }u{|}^2+\sum _{i,j=1}^{\mathrm{\infty }}(|u_{ii}{u}_{jj}{|}^2)dx\\ \le & \underset{U}{\int }0+0+|\mathrm{\Delta }u{|}^2\le B[u,u]\end{array}$

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.