UMD PDE Qualifying Exams/Jan2007PDE

Problem 1

a) Show that the function $G(x)=1/2e^{-|x|}$ is a solution in the distribution sense of the equation

$-G''+G=\delta (x);\quad -\infty <x<\infty$ .

b) Use part (a) to write a solution of

$-u''+u=f(x);\quad -\infty <x<\infty$

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:

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

A similar calculation gives

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

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 $L^{2}(B)$ .

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