# u_{xx}+u_{yy}=0

• Dec 29th 2010, 07:03 PM
dwsmith
u_{xx}+u_{yy}=0
Separation of Variables

$\displaystyle u_{xx}+u_{yy}=0$

$\displaystyle \displaystyle e^{rx+sy}=e^{rx}e^{sy}$

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

$\displaystyle \varphi''(x)\psi(y)+\varphi(x)\psi''(y)=0$

$\displaystyle \displaystyle\frac{\varphi''(x)\psi(y)+\varphi(x)\ psi''(y)}{\varphi(x)\psi(y)}=0$

$\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}=0$

$\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=-\frac{\psi''(y)}{\psi(y)}$

$\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda \ \mbox{and} \ \frac{\psi''(y)}{\psi(y)}=-\lambda$

$\displaystyle \Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow \mbox{sol.s} \ \cos{(x\sqrt{\lambda})} \ \mbox{and} \ \sin{(x\sqrt{\lambda})}$

$\displaystyle \Rightarrow \psi''(y)+\lambda\psi(y)=0\Rightarrow \mbox{sol.s} \ e^{\pm y\sqrt{\lambda}}$

How are those solutions obtained?

Also, I can't tell if the books has $\displaystyle e^{\pm y\sqrt{\lambda}}$ or $\displaystyle e^{\pm y^{\sqrt{\lambda}}}$.
• Dec 29th 2010, 07:08 PM
Prove It
Quote:

Originally Posted by dwsmith
Separation of Variables

$\displaystyle u_{xx}+u_{yy}=0$

$\displaystyle \displaystyle e^{rx+sy}=e^{rx}e^{sy}$

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

$\displaystyle \varphi''(x)\psi(y)+\varphi(x)\psi''(y)=0$

$\displaystyle \displaystyle\frac{\varphi''(x)\psi(y)+\varphi(x)\ psi''(y)}{\varphi(x)\psi(y)}=0$

$\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}=0$

$\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=-\frac{\psi''(y)}{\psi(y)}$

$\displaystyle \displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda \ \mbox{and} \ \frac{\psi''(y)}{\psi(y)}=-\lambda$

$\displaystyle \Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow \mbox{sol.s} \ \cos{(x\sqrt{\lambda})} \ \mbox{and} \ \sin{(x\sqrt{\lambda})}$

$\displaystyle \Rightarrow \psi''(y)+\lambda\psi(y)=0\Rightarrow \mbox{sol.s} \ e^{\pm y\sqrt{\lambda}}$

How are those solutions obtained?

Also, I can't tell if the books has $\displaystyle e^{\pm y\sqrt{\lambda}}$ or $\displaystyle e^{\pm y^{\sqrt{\lambda}}}$.

The two final DEs are second-order linear constant coefficient homogeneous. I suggest you read Differential Equations Tutorial to refresh your memory how to solve DEs of this type.
• Dec 29th 2010, 07:34 PM
dwsmith
Quote:

Originally Posted by Prove It
The two final DEs are second-order linear constant coefficient homogeneous.

Thanks