PDE16

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$\Delta u = u_{xx}+u_{yy}=0\,$

$u_x(0,y) = 0\,$
$u_x(\pi,y) = 0\,$
$u(x,0) = K \cos x\,$
$u(x,1) = K \cos^2 x\,$
$t>0,\,\,0<x<\pi\,$

$u(x,y) = X(x)Y(y)\,$

$\frac{X''}{X} = -\frac{Y''}{Y} = -\lambda^2\,$

New BC's:

$X'(0) = 0, X'(\pi) = 0\,$

$Y(0) = K \cos x, Y(1) = K \cos^2 x\,$

$\frac{X''}{X} = -\frac{Y''}{Y} = -\lambda^2\,$

$X(x) = c_1 \cos\lambda x + c_2 \sin\lambda x\,$

$X'(x) = -c_1\lambda\sin\lambda x + c_2 \lambda \cos\lambda x\,$

$X'(0) = c_2 \lambda = 0 \rightarrow c_2=0\,$

$X'(\pi) = -c_2\lambda\sin\lambda\pi = 0 \rightarrow \lambda = n, n=1,2,...\,$

If $n=0,\,\, X''=0,\,\, X_0(x) = c_3 x + c_4\,$ ,

$X_0'(0) = c_3 = 0\,$

$X_0(x) = c_4\,$ . For mathematical ease, let $c_4=1\,$

So the solution for $X(x)\,$ is

$X_n(x) = d_n \cos n x,\,\,n=1,2,...\,$

$Y''-n^2Y=0\,$

$Y(y) = c_5\cosh n y + c_6 \sinh n y\,$

If $n=0,\,\,Y''=0,\,\,Y_0(y) = c_7 y + c_8\,$

Using superposition to add all these solutions together,

$u(x,t) = c_7 y + c_8 + \sum_{n=1}^\infty (A_n \cosh n y + B_n \sinh n y)\cos n x\,$

$u(x,0) = c_8 + \sum_{n=1}^\infty A_n \cos n x = K \cos x\,$

We are able to determine the coefficients just by comparing like terms.

$c_8=0,\,\,\,A_1 = K,\,\,\,A_n=0,\,\,\,n>1\,$

$u(x,y) = c_7 y + K \cosh y \cos x + \sum_{n=1}^\infty B_n \sinh n y \cos n x\,$

$u(x,1) = c_7 + K\cosh 1 \cos x + \sum_{n=1}^\infty B_n\sinh n\cos n x = K\cos^2 x\,$

Using $\cos^2x = \frac{1+\cos2x}{2}\,$

$\frac{K}{2}(1+\cos2x)-K\cosh 1 \cos x = c_7 + \sum_{n=1}^\infty B_n \sinh n \cos n x\,$

$c_7 = \frac{K}{2}\,$

$B_1 = \frac{-K\cosh 1}{\sinh 1}\,$

$B_2 = \frac{K}{2\sinh 2}\,$

$B_n = 0,\,\,n>2\,$

Now the solution is

$u(x,y) = \frac{K y}{2} + K\cosh y\cos x + \frac{-K\cosh 1}{\sinh 1}\sinh y \cos x + \frac{K}{2\sinh 2}\sinh2y\cos2x\,$

Using the trig identity $\sinh(A-B) = \sinh A\cosh B - \cosh A \sinh B\,$ the solution can be simplified to:

$u(x,y) = K\left[\frac{1}{2}y + \frac{\sinh(1-y)}{\sinh 1}\cos x + \frac{\sinh 2y}{2\sinh 2}\cos2x\right]\,$

Partial Differential Equations

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