08.3 – Kreissektor in Polarkoordinaten

 

Sei \Omega ein Kreissektor mit dem Innenwinkel \beta, also

\Omega  = \left\{ {\left( {r,\varphi } \right):0 < r < 1,\quad 0 < \varphi  < \beta } \right\}

wobei r und \varphi die ebenen Polarkoordinaten sind. Sei weiter \lambda  = \frac{\pi } {\beta } und

{u_s}\left( {r,\varphi } \right) = {r^\lambda }\sin \left( {\lambda \varphi } \right),\quad \quad u\left( {r,\varphi } \right) = \left( {1-{r^2}} \right){u_s}

  1. Zeigen Sie, dass u_s harmonisch ist, d.h. -\Delta {u_s} = 0 bzw in Polarkoordinaten

    -\frac{1} {r}\frac{\partial } {{\partial r}}\left( {r\frac{{\partial {u_s}}} {{\partial r}}} \right)-\frac{1} {{{r^2}}}\frac{{{\partial ^2}{u_s}}} {{\partial {\varphi ^2}}} = 0

  2. Zeigen Sie, dass u=0 auf \Gamma  = \partial \Omega
  3. Zeigen Sie, dass u die Gleichung -\Delta u = 4\left( {1+\lambda } \right){u_s} löst. Also ist u \in {H^1}\left( \Omega  \right)
  4. Berechnen Sie \frac{{{\partial ^2}u}} {{\partial {r^2}}} und {\left\| {\frac{{{\partial ^2}u}} {{\partial {r^2}}}} \right\|_{{L^2}\left( \Omega  \right)}}.

Lösung

a )

Wir wollen -\Delta {u_s} = 0 zeigen. Dazu benötigen wir:

\frac{{\partial {u_s}}} {{\partial \varphi }} = \lambda {r^\lambda }\cos \left( {\lambda \varphi } \right)

\frac{{{\partial ^2}{u_s}}} {{\partial {\varphi ^2}}} = -{\lambda ^2}{r^\lambda }\sin \left( {\lambda \varphi } \right)

\frac{{\partial {u_s}}} {{\partial r}} = \lambda {r^{\lambda -1}}\sin \left( {\lambda \varphi } \right)

Es ist also:

r\frac{{\partial {u_s}}} {{\partial r}} = \lambda {r^\lambda }\sin \left( {\lambda \varphi } \right)

und somit

\frac{\partial } {{\partial r}}\left( {r\frac{{\partial {u_s}}} {{\partial r}}} \right) = {\lambda ^2}{r^{\lambda -1}}\sin \left( {\lambda \varphi } \right)

Eingesetzt:

-\frac{1} {r}\frac{\partial } {{\partial r}}\left( {r\frac{{\partial {u_s}}} {{\partial r}}} \right)-\frac{1} {{{r^2}}}\frac{{{\partial ^2}{u_s}}} {{\partial {\varphi ^2}}} = -{\lambda ^2}{r^{\lambda -2}}\sin \left( {\lambda \varphi } \right)-\left( {-{\lambda ^2}{r^{\lambda -2}}\sin \left( {\lambda \varphi } \right)} \right) = 0

womit u_s harmonisch ist.

b )

\sin 0 = 0,\quad \sin \lambda \beta  = \sin \pi  = 0\quad \quad u\left( {1,\varphi } \right) = \left({1-1}\right){u_S} = 0

c )

-\Delta u = -\Delta \left( {1-{r^2}} \right){u_s} = -\Delta {u_s}+\Delta {r^2}{u_s}

Wir haben bereits berechnet:

-\Delta {u_s} = 0

Es bleibt übrig:

\Delta {r^2}{u_s} = \frac{1} {r}\frac{\partial } {{\partial r}}\left( {r\frac{{\partial {r^2}{u_s}}} {{\partial r}}} \right)+\underbrace {\frac{1} {{{r^2}}}\frac{{{\partial ^2}{r^2}{u_s}}} {{\partial {\varphi ^2}}}}_{ = -{\lambda ^2}{u_s}}

Also:

\frac{1} {r}\frac{\partial } {{\partial r}}\left( {r\frac{{\partial {r^2}{r^\lambda }\sin \left( {\lambda \varphi } \right)}} {{\partial r}}} \right) = \frac{1} {r}{\left( {\lambda +2} \right)^2}{r^{\lambda +1}}\sin \left( {\lambda \varphi } \right) = {\left( {\lambda +2} \right)^2}\underbrace {{r^\lambda }\sin \left( {\lambda \varphi } \right)}_{{u_s}}

Damit ist:

-\Delta u = {u_s}\left( {{\lambda ^2}+{{\left( {\lambda +2} \right)}^2}} \right) = {u_s}\left( {-{\lambda ^2}+{\lambda ^2}+4\lambda +4} \right) = {u_s}\left( {4\lambda +4} \right)

Offensichtlich ist u \in {H^1}\left( \Omega  \right), da schwache Lösung einer PDGL mit rechter Seite in L^2.

d )

\frac{{{\partial ^2}u}} {{\partial {r^2}}} = \frac{{{\partial ^2}{r^\lambda }\sin \left( {\lambda \varphi } \right)-{r^{2+\lambda }}\sin \left( {\lambda \varphi } \right)}} {{\partial {r^2}}}

= \frac{\partial } {{\partial r}}\left( {\lambda {r^{\lambda -1}}\sin \left( {\lambda \varphi } \right)-\left( {2+\lambda } \right){r^{\lambda +1}}\sin \left( {\lambda \varphi } \right)} \right)

= \lambda \left( {\lambda -1} \right){r^{\lambda -2}}\sin \left( {\lambda \varphi } \right)-\left( {2+\lambda } \right)\left( {\lambda -1} \right){r^\lambda }\sin \left( {\lambda \varphi } \right)

Es ist

\int_0^\beta  {\int_0^1 {{\lambda ^2}{{\left( {\lambda -1} \right)}^2}{r^{2\lambda -4}}{{\sin }^2}\left( {\lambda \varphi } \right)rdrd\varphi } }

= \underbrace {\int_0^\beta  {{{\sin }^2}\left( {\lambda \varphi } \right)d\varphi } }_{ = \frac{\beta } {2}}\int_0^1 {{\lambda ^2}{{\left( {\lambda -1} \right)}^2}{r^{2\lambda -3}}dr}

= \frac{\beta } {2}{\lambda ^2}{\left( {\lambda -1} \right)^2}\left[ {\frac{1} {{2\lambda -2}}{r^{2\lambda -2}}} \right]_0^1 < \infty falls 2\lambda -2 > 0

\lambda  > 1

\frac{\pi } {\beta } > 1\quad  \Rightarrow \quad \pi  > \beta

Lösung in H^2, falls \beta < \pi, d.h. der Kreissektor muss konvex sein, damit u\in H^2. Sonst ist nur u\in H^1.