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

$\frac{1}{r}(r u_r)_r + \frac{1}{r^2}u_{\theta\theta} = 0,\,\,\, a<r<b\,$

The BC's are

$u(a,\theta) = g(\theta),\,\,\,\theta\isin\Re\,$

$u(b,\theta) = f(\theta),\,\,\,\theta\isin\Re\,$

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.

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

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

$\frac{r}{R(r)}(r R'(r))_r + \frac{H''(\theta)}{H(\theta)} = 0\,$

$\frac{H''(\theta)}{H(\theta)} = \frac{-r}{R(r)}(r R'(r))_r = -\lambda\,$

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

$\frac{-r}{R(r)}(r R'(r))_r = -\lambda_n\,$

If $n = 0\,$ so that $\lambda_n = 0\,$ , the solution is:

$r(r R'(r))' = 0\,$

$r R'(r) = c_1\,$

$R'(r) = \frac{c_1}{r}\,$

$R(r) - R(0) = c_1 \log r + c_2\,$

$R(r) = c_1 \log r + c_3,\,\,\, c_3 = c_2 + R(0)\,$

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

$\frac{r}{R(r)}(R'(r) + r R''(r)) = \lambda_n\,$

$r R'(r) + r^2 R''(r) - \lambda_n R(r) = 0\,$

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

$r a r^{a-1} + r^2 a (a-1) r^{a-2} - \lambda_n r^a = 0\,$

$a + a^2 - a = \lambda_n\,$

$a = \pm \sqrt{\lambda_n},\,\,\,n = 1,2,...\,$

$R(r) = \begin{cases}c_1 \log r + c_3 & n=0 \\c_{4,n}r^n + c_{5,n}r^{-n} & n=1,2,...\, \end{cases}$

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:

$A_0+B_0\log a = \frac{1}{\pi}\int_{-\pi}^\pi g(\theta) \,d\theta\,$

$A_0+B_0\log b = \frac{1}{\pi}\int_{-\pi}^\pi f(\theta) \,d\theta\,$

$A_na^n + B_na^{-n} = \frac{1}{\pi}\int_{-\pi}^\pi g(\theta) \cos n \theta \, d\theta\,$

$A_nb^n + B_nb^{-n} = \frac{1}{\pi}\int_{-\pi}^\pi f(\theta) \cos n \theta \, d\theta\,$

$C_na^n + D_na^{-n} = \frac{1}{\pi}\int_{-\pi}^\pi g(\theta) \sin n \theta \, d\theta\,$

$C_nb^n + D_nb^{-n} = \frac{1}{\pi}\int_{-\pi}^\pi f(\theta) \sin n \theta \, d\theta\,$

Partial Differential Equations

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PDE12

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