UMD PDE Qualifying Exams/Jan2006PDE

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Problem 1[edit]


Problem 2[edit]

Problem 3[edit]

Problem 4[edit]

A weak solution of the biharmonic equation,

\Delta^2 u = f, & x\in U\\
u=\frac{\partial u}{\partial \nu}=0, & x\in \partial U

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.


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

\| 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]

which establishes coercivity.

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

Problem 5[edit]

Problem 6[edit]

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