PDE12

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Solve Dirichlet's problem for a circular annulus. The domain D\, is the space between two concentric circles, C_{1} being the innermost circle with radius a, and C_{2} being the outermost circle with radius b.
 \Delta u=\nabla ^{2}u=0\, in D\,

u=g\, on C_{1}\,
u=f\, on C_{2}\,


First transform the Laplacian operator to polar coordinates, as explained in this problem.

Now the DE is

{\frac  {1}{r}}(ru_{r})_{r}+{\frac  {1}{r^{2}}}u_{{\theta \theta }}=0,\,\,\,a<r<b\,

The BC's are

u(a,\theta )=g(\theta ),\,\,\,\theta \in \Re \,

u(b,\theta )=f(\theta ),\,\,\,\theta \in \Re \,

An auxiliary condition required from the nature of the problem and coordinate system is

u(r,\theta +2\pi )=u(r,\theta )\,

Separate variables and plug back into the original equation, separate variables and set each side equal to a negative (for convenience) constant lambda.

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

{\frac  {1}{r}}(rR'(r))_{r}\,H(\theta )+{\frac  {R(r)}{r^{2}}}H''(\theta )=0\,

{\frac  {r}{R(r)}}(rR'(r))_{r}+{\frac  {H''(\theta )}{H(\theta )}}=0\,

{\frac  {H''(\theta )}{H(\theta )}}={\frac  {-r}{R(r)}}(rR'(r))_{r}=-\lambda \,

The first ODE is

H''+\lambda H=0\, with the auxiliary condition H(\theta +2\pi )=H(\theta )\,

  • H(\theta )=A\cos({\sqrt  {\lambda }}\theta )+B\sin({\sqrt  {\lambda }}\theta )\,

In order for H(\theta )=H(\theta +2\pi )\,, the square root of lambda must be an integer value.

{\sqrt  {\lambda _{n}}}=n,\,\,\,n=0,1,2,...\,

The second ODE is

{\frac  {-r}{R(r)}}(rR'(r))_{r}=-\lambda _{n}\,

If n=0\, so that \lambda _{n}=0\,, the solution is:

r(rR'(r))'=0\,

rR'(r)=c_{1}\,

R'(r)={\frac  {c_{1}}{r}}\,

R(r)-R(0)=c_{1}\log r+c_{2}\,

R(r)=c_{1}\log r+c_{3},\,\,\,c_{3}=c_{2}+R(0)\,

If \lambda >0\,, the solution is:

{\frac  {r}{R(r)}}(R'(r)+rR''(r))=\lambda _{n}\,

rR'(r)+r^{2}R''(r)-\lambda _{n}R(r)=0\,

This ODE is in Cauchy-Euler form, so guess the solution R(r)=r^{a}\, and plug it back in.

rar^{{a-1}}+r^{2}a(a-1)r^{{a-2}}-\lambda _{n}r^{a}=0\,

a+a^{2}-a=\lambda _{n}\,

a=\pm {\sqrt  {\lambda _{n}}},\,\,\,n=1,2,...\,

  • R(r)={\begin{cases}c_{1}\log r+c_{3}&n=0\\c_{{4,n}}r^{n}+c_{{5,n}}r^{{-n}}&n=1,2,...\,\end{cases}}

The solution is

u(r,\theta )={\frac  {1}{2}}(A_{0}+B_{0}\log r)+\sum _{{n=1}}^{\infty }\left[(A_{n}r^{n}+B_{n}r^{{-n}})\cos n\theta +(C_{n}r^{n}+D_{n}r^{{-n}})\sin n\theta \right]\,

The Fourier coefficients are given with the help of the boundary conditions.

u(a,\theta )={\frac  {1}{2}}(A_{0}+B_{0}\log a)+\sum _{{n=1}}^{\infty }\left[(A_{n}a^{n}+B_{n}a^{{-n}})\cos n\theta +(C_{n}a^{n}+D_{n}a^{{-n}})\sin n\theta \right]=g(\theta )\,

u(b,\theta )={\frac  {1}{2}}(A_{0}+B_{0}\log b)+\sum _{{n=1}}^{\infty }\left[(A_{n}b^{n}+B_{n}b^{{-n}})\cos n\theta +(C_{n}b^{n}+D_{n}b^{{-n}})\sin n\theta \right]=f(\theta )\,

Now you have this system of equations to solve to determine the coefficients:

A_{0}+B_{0}\log a={\frac  {1}{\pi }}\int _{{-\pi }}^{\pi }g(\theta )\,d\theta \,

A_{0}+B_{0}\log b={\frac  {1}{\pi }}\int _{{-\pi }}^{\pi }f(\theta )\,d\theta \,

A_{n}a^{n}+B_{n}a^{{-n}}={\frac  {1}{\pi }}\int _{{-\pi }}^{\pi }g(\theta )\cos n\theta \,d\theta \,

A_{n}b^{n}+B_{n}b^{{-n}}={\frac  {1}{\pi }}\int _{{-\pi }}^{\pi }f(\theta )\cos n\theta \,d\theta \,

C_{n}a^{n}+D_{n}a^{{-n}}={\frac  {1}{\pi }}\int _{{-\pi }}^{\pi }g(\theta )\sin n\theta \,d\theta \,

C_{n}b^{n}+D_{n}b^{{-n}}={\frac  {1}{\pi }}\int _{{-\pi }}^{\pi }f(\theta )\sin n\theta \,d\theta \,


Partial Differential Equations

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