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

Solution[edit]

(a)[edit]

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)[edit]

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,
\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[edit]

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).