UMD PDE Qualifying Exams/Jan2007PDE

(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 $$G'(x)=1/2 (sgn(-x)) e^{-|x|}$$. 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),

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

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 $$L^2(B)$$.