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\Delta u={\frac  {1}{r}}(ru_{r})_{r}+{\frac  {1}{r^{2}}}u_{{\theta \theta }}=0\,

u(a,\theta )=0\,
u(b,\theta )=0\,
u(r,\alpha )=0\,

a<r<b,\,\,\,\,\,0<\theta <\alpha \,

Separate variables:

u(r,\theta )=R(r)\Theta (\theta )\,

{\frac  {r(rR'(r))'}{R(r)}}=-{\frac  {\Theta ''(\theta )}{\Theta (\theta )}}=\lambda \, (constant)

r^{2}R''+rR'-\lambda R=0\,

Try R(r)=r^{m}\,. If \lambda \neq 0\,, m^{2}=\lambda \, so m=\pm {\sqrt  {\lambda }}\,.

R(r)=Ar^{{\sqrt  {\lambda }}}+Br^{{-{\sqrt  {\lambda }}}}\,

Apply the boundary conditions:

  • Aa^{{\sqrt  {\lambda }}}+Ba^{{-{\sqrt  {\lambda }}}}=0\,

  • Ab^{{\sqrt  {\lambda }}}+Bb^{{-{\sqrt  {\lambda }}}}=0\,

Since the system is homogeneous we must have \det {{\begin{bmatrix}a^{{\sqrt  {\lambda }}}&a^{{-{\sqrt  {\lambda }}}}\\b^{{\sqrt  {\lambda }}}&b^{{-{\sqrt  {\lambda }}}}\end{bmatrix}}}=0.\,

This can be written as

\left({\frac  {a}{b}}\right)^{{\sqrt  {\lambda }}}-\left({\frac  {b}{a}}\right)^{{\sqrt  {\lambda }}}=0\,

Let {\frac  {b}{a}}=e^{c}\, so that c=\ln(b)-\ln(a)>0\,

e^{{c{\sqrt  {\lambda }}}}-e^{{-c{\sqrt  {\lambda }}}}=0\, so \lambda \, must be negative. Let \lambda =-\delta ^{2}\,

e^{{c\,i\,\delta }}-e^{{-c\,i\,\delta }}=(\cos(c\delta )+i\sin(c\delta ))-(\cos(c\delta )-i\sin(c\delta ))=2i\sin(c\delta )=0\Rightarrow \sin(c\delta )=0\,

\delta ={\frac  {n\pi }{c}},n=1,2,...\,

So the DE for Theta can be written:

\Theta ''(\theta )+\left({\frac  {n\pi }{c}}\right)^{2}\Theta (\theta )\,=0

\Theta (\theta )=C\sinh \left({\frac  {n\pi }{c}}\theta \right)+D\cosh \left({\frac  {n\pi }{c}}\theta \right)=E\sinh \left({\frac  {n\pi }{c}}(\theta -\beta )\right)\,

Using the initial conditon,

\Theta (\alpha )=E\sinh \left({\frac  {n\pi }{c}}(\alpha -\beta )\right)=0\, so let \beta =\alpha \,

\Theta (\theta )=E\sinh \left({\frac  {n\pi }{c}}(\theta -\alpha )\right)\,

If \lambda =0,R(r)=A\ln r+B\, then the boundary conditions cannot be satisfied.

From the starred equations we have B=-Aa^{{2{\sqrt  {\lambda }}}}\,

R(r)=Ar^{{i\delta }}+Br^{{-i\delta }}\,

=A(r^{{i\delta }}-r^{{-i\delta }}a^{{2i\delta }}\,

Rewrite: r^{{i\delta }}=e^{{i\delta \ln r}}\,

R(r)=Ae^{{i\delta \ln r}}[e^{{i\delta (\ln r-\ln a)}}-e^{{i\delta (\ln a-\ln r)}}]\,

=2iAe^{{i\delta \ln A}}\sin(\delta (\ln r-\ln a))\,

The solution before boundary conditions are applied is:

u(r,\theta )=\sum _{{n=1}}^{\infty }d_{n}\sin({\frac  {n\pi }{c}}(\ln r-\ln a))\sinh({\frac  {n\pi }{c}}(\theta -\alpha ))\,

When \theta =0\,,

f(r)=-\sum _{{n=1}}^{\infty }d_{n}\sin({\frac  {n\pi }{c}}(\ln r-\ln a)\sinh({\frac  {n\pi \alpha }{c}})\,

Let c_{n}=-d_{n}\sinh({\frac  {n\pi \alpha }{c}})\,

u(r,\theta )=\sum _{{n=1}}^{\infty }c_{n}{\frac  {\sin({\frac  {n\pi }{c}}(\ln r-\ln a))\sinh({\frac  {n\pi }{c}}(\alpha -\theta ))}{\sinh({\frac  {n\pi \alpha }{c}})}}\,

f(r)=\sum _{{n=1}}^{\infty }c_{n}\sin {\frac  {n\pi }{c}}(\ln r-\ln a)\,

Let \phi =\ln r-\ln a\,. When r=b,\phi -\ln b-\ln a=c\,, so o\leq \phi \leq c\,

c_{n}={\frac  {2}{c}}\int _{0}^{c}g(\phi )\sin({\frac  {n\pi }{c}}\phi )\,d\phi \, where g(\phi )=f(r)\,

={\frac  {2}{c}}\int _{a}^{b}f(r)\sin({\frac  {n\pi }{c}}(\ln r-\ln a)){\frac  {dr}{r}}\,

Partial Differential Equations

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