# UMD PDE Qualifying Exams/Jan2007PDE

## Problem 1

 a) Show that the function ${\displaystyle G(x)=1/2e^{-|x|}}$ is a solution in the distribution sense of the equation ${\displaystyle -G''+G=\delta (x);\quad -\infty . b) Use part (a) to write a solution of ${\displaystyle -u''+u=f(x);\quad -\infty

### Solution

#### (a)

We want to show ${\displaystyle \int _{\mathbb {R} }G(-\varphi ''+\varphi )\,dx=\varphi (0)}$ for every test function ${\displaystyle \varphi \in C_{c}^{\infty }(\mathbb {R} )}$.

One can compute ${\displaystyle G(x)=G''(x)}$ and ${\displaystyle G'(x)=1/2(sgn(-x))e^{(sgn(-x))}}$. Therefore, away from 0, we have ${\displaystyle -G''+G=0}$, that is, ${\displaystyle -G''+G=0}$ a.e. and ${\displaystyle \int -G''+G=0}$.

We now compute by an integration by parts:

{\displaystyle {\begin{aligned}\int _{-\infty }^{0}(-G''+G)\varphi =&\int _{-\infty }^{0}G'\varphi '+G\varphi \,dx-\left.\varphi G\right|_{-\infty }^{0}\\=&\int _{-\infty }^{0}G'\varphi '+G\varphi \,dx-\varphi (0)\lim _{x\to 0^{-}}G(x)\\=&\int _{-\infty }^{0}G(-\varphi ''+\varphi )\,dx-1/2\varphi (0)+\left.\varphi 'G\right|_{-\infty }^{0}\\=&\int _{-\infty }^{0}G(-\varphi ''+\varphi )\,dx-1/2\varphi (0)+1/2\varphi '(0).\\\end{aligned}}}

A similar calculation gives

${\displaystyle \int _{0}^{\infty }(-G''+G)\varphi =\int _{0}^{\infty }G(-\varphi ''+\varphi )\,dx-1/2\varphi (0)-1/2\varphi '(0).}$

So we have shown that for all ${\displaystyle \varphi \in C_{c}^{\infty }(\mathbb {R} )}$

${\displaystyle 0=\int \varphi (-G''+G)=\int G(-\varphi ''+\varphi )\,dx-\varphi (0)}$ which gives the desired result.

#### (b)

We guess ${\displaystyle u=G\ast f}$. Then by part (a),

${\displaystyle -u''(x)+u(x)=\int _{\mathbb {R} }[-G''(x-y)+G(x-y)]f(y)\,dy=\int _{\mathbb {R} }\delta (x-y)f(y)\,dy=f(x)}$.

## Problem 6

 Let ${\displaystyle B}$ be the unit ball in ${\displaystyle \mathbb {R} ^{3}}$. Consider the eigenvalue problem, ${\displaystyle \left\{{\begin{array}{r l}-\Delta u=\lambda u,&x\in B\\\partial _{\nu }u+u=0,&x\in \partial B,\end{array}}\right.}$ where ${\displaystyle \partial _{\nu }}$ denotes the normal derivative on the boundary ${\displaystyle \partial B}$. Show that all eigenvalues are positive and the eigenfunctions corresponding to different eigenvalues are orthogonal to each other.

### Solution

Multiply the PDE by ${\displaystyle u}$ and integrate:

${\displaystyle \lambda \int _{B}u^{2}=-\int _{B}u\Delta u=\int _{B}|Du|^{2}-\int _{\partial B}u\partial _{\nu }u=\int _{B}|Du|^{2}+\int _{\partial B}u^{2}\geq 0}$.

Of course we know that ${\displaystyle \lambda =0}$ is an eigenvalue of ${\displaystyle -\Delta }$ corresponding to a constant eigenfunction. But a constant function has ${\displaystyle \partial _{v}u\equiv 0}$ which implies ${\displaystyle u\equiv 0}$ by the boundary condition. Hence ${\displaystyle \lambda =0}$ is no longer an eigenvalue. This forces ${\displaystyle \lambda >0}$.

To see orthogonality of the eigenfunctions, let ${\displaystyle \varphi _{n},\varphi _{m}}$ be two eigenfunctions corresponding to distinct eigenvalues ${\displaystyle \lambda _{n},\lambda _{m}}$, respectively. Then by an integration of parts,

${\displaystyle \int _{B}-\varphi _{n}\Delta \varphi _{m}+\varphi _{m}\Delta \varphi _{n}=\int _{B}D\varphi _{n}D\varphi _{m}-D\varphi _{n}D\varphi _{m}\,dx+\int _{\partial B}-\varphi _{n}\partial _{\nu }\varphi _{m}+\partial _{\nu }\varphi _{n}\varphi _{m}\,dS=0}$

So by the PDE,

${\displaystyle 0=\int _{B}-\varphi _{n}\Delta \varphi _{m}+\varphi _{m}\Delta \varphi _{n}=\int _{B}(\lambda _{n}-\lambda _{m})\varphi _{n}\varphi _{m}}$.

Since ${\displaystyle \lambda _{n}-\lambda _{m}\neq 0}$ this implies that ${\displaystyle \{\varphi _{n}\}}$ are pairwise orthogonal in ${\displaystyle L^{2}(B)}$.