PDE13

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Solve:

\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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