# UMD PDE Qualifying Exams/Jan2007PDE

## §Problem 1

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

### §Solution

#### §(a)

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

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

We now compute by an integration by parts:

\begin{align} \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{align}

A similar calculation gives

$\int_0^{\infty}(-G'' +G)\varphi =\int_0^{\infty} G(-\varphi'' + \varphi) \,dx - 1/2\varphi(0) -1/2 \varphi'(0).$

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

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

#### §(b)

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

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

## §Problem 6

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

### §Solution

Multiply the PDE by $u$ and integrate:

$\lambda \int_B u^2 = - \int_B u \Delta u = \int_B |Du|^2 - \int_{\partial B} u \partial_\nu u =\int_B |Du|^2 + \int_{\partial B} u^2\geq 0$.

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

To see orthogonality of the eigenfunctions, let $\varphi_n,\varphi_m$ be two eigenfunctions corresponding to distinct eigenvalues $\lambda_n,\lambda_m$, respectively. Then by an integration of parts,

$\int_B -\varphi_n\Delta \varphi_m +\varphi_m \Delta\varphi_n = \int_B D\varphi_n D\varphi_m -D\varphi_n D\varphi_m\,dx +\int_{\partial B} -\varphi_n\partial_\nu \varphi_m +\partial_\nu\varphi_n\varphi_m\,dS =0$

So by the PDE,

$0=\int_B -\varphi_n\Delta \varphi_m +\varphi_m \Delta\varphi_n =\int_B (\lambda_n -\lambda_m) \varphi_n\varphi_m$.

Since $\lambda_n-\lambda_m\neq 0$ this implies that $\{\varphi_n\}$ are pairwise orthogonal in $L^2(B)$.