PDE26

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\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 sx\,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 sxF_{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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