# Use separation of variables

• Jan 4th 2011, 03:52 PM
dwsmith
Use separation of variables
$\displaystyle u_x+u_y+u_z+u=0$

$\displaystyle u(x,y,z)=\varphi(x)\psi(y)\omega(z)$

$\displaystyle \varphi'(x)\psi(y)\omega(z)+\varphi(x)\psi'(y)\ome ga(z)+\varphi(x)\psi(y)\omega'(z)+\varphi(x)\psi(y )\omega(z)=0$

$\displaystyle \varphi'(x)\psi(y)\omega(z)+\varphi(x)[\psi'(y)\omega(z)+\psi(y)\omega'(z)+\psi(y)\omega( z)]=0$

$\displaystyle \displaystyle\frac{\varphi'(x)}{\varphi(x)}=-\left[\frac{\psi'(y)\omega(z)+\psi(y)\omega'(z)+\psi(y)\ omega(z)}{\psi(y)\omega(z)}\right]$

$\displaystyle \varphi'(x)-\lambda\varphi(x)=0\Rightarrow m=\lambda$

$\displaystyle \varphi(x)=\exp{(x\lambda)}$

$\displaystyle \psi'(y)\omega(z)+\psi(y)\omega'(z)+\psi(y)\omega( z)+\lambda\psi(y)\omega(z)=0$

$\displaystyle \psi'(y)\omega(z)+\psi(y)[\omega'(z)+\omega(z)+\lambda\omega(z)]=0$

$\displaystyle \displaystyle\frac{\psi'(y)}{\psi(y)}=-\left[\frac{\omega'(z)+\omega(z)+\lambda\omega(z)}{\omeg a(z)}\right]$

$\displaystyle \psi'(y)-\mu\psi(y)=0\Rightarrow n=\mu$

$\displaystyle \psi(y)=\exp{(y\mu)}$

$\displaystyle \omega'(z)+\omega(z)+\lambda\omega(z)+\mu\omega(z) =0$

$\displaystyle t+1+\lambda+\mu=0\Rightarrow t=-(1+\lambda+\mu)$

$\displaystyle \omega(z)=\exp{-z(1+\lambda+\mu)}$

$\displaystyle u(x,y,z)=C\exp{[x\lambda+y\mu-z(1+\lambda+\mu)]}$

I was original going to ask why this didn't work when I checked it but now it worked. However, since I typed it all out, I am going to post it for others to look at.