# PDEFT5

 $u_{xx}+u_{yy}+u_{zz}=0\,$ $u(x,y,0) = f(x,y)\,$Auxiliary condition: u is bounded. $t>0,\,\,x,y\isin\Re,\,\,\,z>0\,$

Since $x\,$ and $y\,$ both range over the whole line, they should both be transformed.

Assume that a solution exists in the form

$u(x,y,z) = \int\!\!\!\int_\Re e^{i\lambda x + i \mu y} U(\lambda,\mu,z)\,d\lambda d\mu\,$

$U(\lambda,\mu,z) = \left(\frac{1}{2\pi}\right)^2\int\!\!\!\int_\Re e^{-i\lambda x - i \mu y} u(x,y,z)\,dx dy\,$

Take the second partial derivative with respect to z of both sides.

$\frac{\partial^2}{\partial z}U(\lambda,\mu,z) = -\left(\frac{1}{2\pi}\right)^2\int\!\!\!\int_\Re e^{-i\lambda x - i \mu y} u_{zz}(x,y,z)\,dx dy = -\left(\frac{1}{2\pi}\right)^2\int\!\!\!\int_\Re e^{-i\lambda x - i \mu y} (u_{xx}+u_{yy})\,dx dy\,$

$= (\lambda^2+\mu^2)\frac{1}{4\pi^2}\int\!\!\!\int_\Re e^{-i\lambda x - i \mu y}u(x,y,z)\,dx dy\,$

$= (\lambda^2+\mu^2) U\,$

The solution to this ODE is:

$U(\lambda,\mu,z) = A(\lambda,\mu) e^{\sqrt{\lambda^2+\mu^2}\,z} + B(\lambda,\mu) e^{-\sqrt{\lambda^2+\mu^2}\,z}\,$

To satisfy the auxiliary boundedness condition, we must set $A(\lambda,\mu) = 0\,$.

$U(\lambda,\mu,z) = B(\lambda,\mu) e^{-\sqrt{\lambda^2+\mu^2}\,z}\,$

So the solution is the inverse transform.

• $u(x,y,z) = \int\!\!\!\int_\Re e^{i \lambda x + i \mu y} B(\lambda,\mu) e^{-\sqrt{\lambda^2+\mu^2}\,z}\,d\lambda d\mu\,$

Use the boundary condition now.

$u(x,y,0) = \int\!\!\!\int_\Re e^{i \lambda x + i \mu y} B(\lambda,\mu) \,d\lambda d\mu = f(x,y)\,$

$B(\lambda,\mu) = \frac{1}{4\pi^2}\int\!\!\!\int_\Re e^{-i \lambda x - i \mu y} f(x,y)\,dx dy\,$

##### Toolbox

 Get A Wifi Network Switcher Widget for Android