Exam

For $\l,r > 0$ and $\nu\in\C$: $$ K(\l,\nu,r)\colon=\int_0^\infty t^{-\nu-1}e^{-\l^2 t-r^2/4t}\,dt $$ Prove that Bessel's potential operators are given by $$ U_\l^zf(x)=\frac1{(4\pi)^{d/2}\G(z)}\int_{\R^d}K(\l,d/2-z,\norm{y-x})f(y)\,dy~. $$ Moreover we have:

$K(\l,\nu,r)=\l^{2\nu}K(1,\nu,\l r)$.
$\pa_\l K(\l,\nu,r)=-2\l K(\l,\nu-1,r)$.
$\pa_rK(\l,\nu,r)=-\frac12rK(\l,\nu+1,r)$.
$r\pa_r^2K+(2\nu+1)\pa_rK-\l^2 rK=0$.
$K(\l,1/2,r)=(4\pi)^{1/2}e^{-\l r}/r$ und $K(\l,-1/2,r)=(4\pi)^{1/2}e^{-\l r}/\l$.

4. Differentiation of 2. by $\l$ gives: $$ -2\l K(\l,\nu-1,r) =\pa_\l K(\l,\nu,r) =2\nu\l^{2\nu-1}K(1,\nu,\l r) +\l^{2\nu}r\pa_3K(1,\nu,\l r) $$ Differentiation of 2. by $r$ gives: $$ \pa_rK(\l,\nu,r)=\l^{2\nu+1}\pa_3K(1,\nu,\l r)~. $$ Thus we obtain $$ -2\l K(\l,\nu-1,r) =2\nu\l^{-1}K(\l,\nu,r)+\l^{-1}r\pa_rK(\l,\nu,r) $$ and differentiation by $r$: $$ \l rK =-2\l\pa_rK(\l,\nu-1,r) =2\nu\l^{-1}\pa_rK +\l^{-1}\pa_rK +\l^{-1}r\pa_r^2K $$ 6. The function $y(r)\colon=\sqrt{\pi}e^{-\l r}\l^{-1}$ solves the ODE $y^\dprime-\l^2y=0$ and we have: $$ y(0)=\sqrt{\pi}\l^{-1}=\int t^{-1/2}e^{-\l^2t}\,dt=K(\l,-1/2,0) $$ and $y(\infty)=0$.