UMD PDE Qualifying Exams/Jan2007PDE

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Problem 1[edit]

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<x<\infty.

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

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



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:

\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).\\

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.


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[edit]

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,

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.


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).