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 $-G^{″}+G=\delta (x);-\mathrm{\infty }<x<\mathrm{\infty }$. b) Use part (a) to write a solution of $-u^{″}+u=f(x);-\mathrm{\infty }<x<\mathrm{\infty }$

Solution

(a)

We want to show $\underset{ℝ}{\int }G(-{\phi }^{″}+\phi )dx=\phi (0)$ for every test function $\phi \in C_c^{\mathrm{\infty }}(ℝ)$.

One can compute $G(x)=G^{″}(x)$ and $G^{\prime }(x)=1/2(sgn(-x))e^{-|x|}$. 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:

$\begin{array}{cc}\underset{-\mathrm{\infty }}{\overset{0}{\int }}(-G^{″}+G)\phi =& \underset{-\mathrm{\infty }}{\overset{0}{\int }}{G}^{\prime }{\phi }^{\prime }+G\phi dx-{\phi G|}_{-\mathrm{\infty }}^0\\ =& \underset{-\mathrm{\infty }}{\overset{0}{\int }}{G}^{\prime }{\phi }^{\prime }+G\phi dx-\phi (0)\underset{x\to 0^{-}}{\lim }G(x)\\ =& \underset{-\mathrm{\infty }}{\overset{0}{\int }}G(-{\phi }^{″}+\phi )dx-1/2\phi (0)+{{\phi }^{\prime }G|}_{-\mathrm{\infty }}^0\\ =& \underset{-\mathrm{\infty }}{\overset{0}{\int }}G(-{\phi }^{″}+\phi )dx-1/2\phi (0)+1/2{\phi }^{\prime }(0).\\ \end{array}$

A similar calculation gives

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(-G^{″}+G)\phi =\underset{0}{\overset{\mathrm{\infty }}{\int }}G(-{\phi }^{″}+\phi )dx-1/2\phi (0)-1/2{\phi }^{\prime }(0).$

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

$0=\int \phi (-G^{″}+G)=\int G(-{\phi }^{″}+\phi )dx-\phi (0)$ which gives the desired result.

(b)

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

$-u^{″}(x)+u(x)=\underset{ℝ}{\int }[-G^{″}(x-y)+G(x-y)]f(y)dy=\underset{ℝ}{\int }\delta (x-y)f(y)dy=f(x)$.

Problem 6

Let $B$ be the unit ball in ${ℝ}^3$. Consider the eigenvalue problem, $\{\begin{array}{cc}-\mathrm{\Delta }u=\lambda u,& x\in B\\ {\partial }_{\nu }u+u=0,& x\in \partial B,\end{array}$ 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:

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

Of course we know that $\lambda =0$ is an eigenvalue of $-\mathrm{\Delta }$ 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 ${\phi }_n,{\phi }_m$ be two eigenfunctions corresponding to distinct eigenvalues ${\lambda }_n,{\lambda }_m$, respectively. Then by an integration of parts,

$\underset{B}{\int }-{\phi }_n\mathrm{\Delta }{\phi }_m+{\phi }_m\mathrm{\Delta }{\phi }_n=\underset{B}{\int }D{\phi }_{n}D{\phi }_m-D{\phi }_{n}D{\phi }_{m}dx+\underset{\partial B}{\int }-{\phi }_n{\partial }_{\nu }{\phi }_m+{\partial }_{\nu }{\phi }_n{\phi }_{m}dS=0$

So by the PDE,

$0=\underset{B}{\int }-{\phi }_n\mathrm{\Delta }{\phi }_m+{\phi }_m\mathrm{\Delta }{\phi }_n=\underset{B}{\int }({\lambda }_n-{\lambda }_m){\phi }_n{\phi }_m$.

Since ${\lambda }_n-{\lambda }_m\ne 0$ this implies that $\{{\phi }_n\}$ are pairwise orthogonal in $L^2(B)$.