PDE13

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$\Delta u = \frac{1}{r}(ru_r)_r + \frac{1}{r^2}u_{\theta\theta}=0\,$

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

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

Separate variables:

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

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

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

Try $R(r) = r^m\,$ . If $\lambda\ne 0\,$ , $m^2=\lambda\,$ 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 $B=-A a^{2\sqrt{\lambda}}\,$

$R(r) = A r^{i\delta} + B r^{-i \delta}\,$

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

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

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

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

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 $o\le\phi\le c\,$

$c_n = \frac{2}{c}\int_0^c g(\phi)\sin(\frac{n\pi}{c}\phi)\,d\phi\,$ 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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PDE13

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