PDE26

From Exampleproblems

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

$u_y(x,a)=0\,$
$u_x(0,y)=0\,$
$u(x,0) = f(x)\,$

$0<x<\infty,\,\,\,\,\, 0<y<a\,$

Since $x$ tends towards infinity, transform it into the variable $s$ . Since the first partials are zero, use the cosine transform.

$U(s,y) = \int_{-\infty}^\infty u(x,y) \cos sx\,dx\,$

$\frac{\partial^2}{\partial y^2}U(s,y) = \int_{-\infty}^\infty u_{yy}(x,y)\cos s x\,dx = -\int_{-\infty}^\infty u_{xx}(x,y)\cos sx\,dx\, = s^2 U(s,y)$

This gives the ODE:

$U_{yy} = s^2 U\,$

$U(s,y) = c_1\cosh sy + c_2\sinh sy\,$

Transforming the boundary conditions,

$u(x,0) = f(x)\,$

$U(s,0) = c_1 = F(s)\,$

$u_y(x,a) = 0\,$

$U_y(s,a) = sF(s) \sinh sa + c_2 s \cosh sa = 0\,$

$c_2 = \frac{-sF(s)\sinh sa}{s \cosh sa}\,$

So the transformed solution is

$U(s,y) = F(s)\cosh sy + \frac{-F(s)\sinh sa}{\cosh sa} \sinh sy\,$

$= \frac{F(s)\cosh(a-y)s}{\cosh sa} = F(s)G(s,y),\,\,G(s,y) = \frac{\cosh(a-y)s}{\cosh sa}$

Using the convolution theorem

$\int_{-\infty}^\infty \cos sx F_c(s)G_c(s,y)\,ds = \frac{1}{2}\int_{-\infty}^\infty f(\xi)\left[ g(|t-\xi|,y)+g(|t+\xi|,y)\right]\,d\xi\,$

where $g(x,y) = F^{-1}\{G(s,y),s\rightarrow x\}\,$

The final solution is

$u(x,y) = \frac{1}{\sqrt{2\pi}}\int_0^\infty f(\xi)\left[g(|x+\xi|,y)+g(|x-\xi|,y)\right]\,d\xi\,$

Partial Differential Equations

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