# Exponential method 3 variables

• Dec 31st 2010, 02:13 PM
dwsmith
Exponential method 3 variables
Solve using exponential method.

My book as a good example for separation of variables but neglected one for the exponential method.

$u_{xx}+u_{yy}+u_{zz}-u=0$

$\exp{(rx+sy+tz)}$

$\exp{(rx+sy+tz)}[r^2+s^2+t^2-1]=0$

I usually solve for r. Do I just continue working as I would with 2 variables?

Thanks.

Does the fact that $r^2+s^2+t^2-1=0$ is a sphere of radius help or make solving this easier?

$r=\pm\sqrt{1-s^2-t^2}$

$\exp{(x\pm\sqrt{1-s^2-t^2}+sy+tz)}\Rightarrow$ $\exp{(x\sqrt{1-s^2-t^2}+sy+tz)}+\exp{(-x\sqrt{1-s^2-t^2}+sy+tz)}$

I haven't the slightest on how to proceed now.
• Jan 1st 2011, 02:10 AM
Sudharaka
Quote:

Originally Posted by dwsmith
Solve using exponential method.

My book as a good example for separation of variables but neglected one for the exponential method.

$u_{xx}+u_{yy}+u_{zz}-u=0$

$\exp{(rx+sy+tz)}$

$\exp{(rx+sy+tz)}[r^2+s^2+t^2-1]=0$

I usually solve for r. Do I just continue working as I would with 2 variables?

Thanks.

Does the fact that $r^2+s^2+t^2-1=0$ is a sphere of radius help or make solving this easier?

$r=\pm\sqrt{1-s^2-t^2}$

$\exp{(x\pm\sqrt{1-s^2-t^2}+sy+tz)}\Rightarrow$ $\exp{(x\sqrt{1-s^2-t^2}+sy+tz)}+\exp{(-x\sqrt{1-s^2-t^2}+sy+tz)}$

I haven't the slightest on how to proceed now.

Dear dwsmith,

Yes. You can solve for r, proceed the same way you have done with 2 variable problems. In this case the exponent of the exponentials will contain s and t. Remember when we deal with two variable PDEs' the exponential terms only contained s.

The solution would be, $u(x,y,z)=A\exp{(x\sqrt{1-s^2-t^2}+sy+tz)}+B\exp{(-x\sqrt{1-s^2-t^2}+sy+tz)}$

By the way can you please tell me the book you are refering? I am having trouble finding a good book for PDEs'.
• Jan 1st 2011, 11:39 AM
dwsmith
I am using Elementary Partial Differential Equations by Berg and McGregor. It is an older book but my professor likes it.

Amazon.com: Elementary Partial Differential Equations (9780070048508): P. W. Berg: Books

Since it is out of print, it is really cheap. Mine, surprisingly, is in mint condition for being from the 1966 and only cost 25 dollars.
• Jan 1st 2011, 01:34 PM
dwsmith
The question now becomes what is the correct substitution for the exponential method so that it is equal to the separation of variables method (see below).

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

$\varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi''(y)\o mega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)=0$

$\displaystyle \frac{\varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi' '(y)\omega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)}{\varphi(x)\psi(y)\omeg a(z)}=0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}+\frac{\omega''(z)}{\omega(z)}-1=0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda\Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow m^2-\lambda\Rightarrow m=\pm\sqrt{\lambda}$

$\displaystyle \lambda=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$

$\displaystyle \frac{\psi''(y)}{\psi(y)}=\mu\Rightarrow \psi''(y)-\mu\psi(y)=0\Rightarrow n^2-\mu\Rightarrow n=\pm\sqrt{\mu}$

$\displaystyle \mu=1-\lambda-\frac{\omega''(z)}{\omega(z)}$

$\displaystyle \omega''(z)-\omega(z)+\lambda\omega(z)+\mu\omega(z)=0\Rightarr ow a^2-1+\lambda+\mu\Rightarrow a=\pm\sqrt{1-\lambda-\mu}$

$u(x,y,z;\lambda,\mu)=\exp{(x\pm\sqrt{\lambda})}*\e xp{(y\pm\sqrt{\mu})}*\exp{(z\pm\sqrt{1-\lambda-\mu})}$

I can't figure out the correct substitution for the exponential method to match the separations method.
• Jan 1st 2011, 04:05 PM
Sudharaka
Quote:

Originally Posted by dwsmith
The question now becomes what is the correct substitution for the exponential method so that it is equal to the separation of variables method (see below).

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

$\varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi''(y)\o mega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)=0$

$\displaystyle \frac{\varphi''(x)\psi(y)\omega(z)+\varphi(x)\psi' '(y)\omega(z)+\varphi(x)\psi(y)\omega''(z)-\varphi(x)\psi(y)\omega(z)}{\varphi(x)\psi(y)\omeg a(z)}=0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}+\frac{\psi''(y)}{\ psi(y)}+\frac{\omega''(z)}{\omega(z)}-1=0$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$

$\displaystyle \frac{\varphi''(x)}{\varphi(x)}=\lambda\Rightarrow \varphi''(x)-\lambda\varphi(x)=0\Rightarrow m^2-\lambda\Rightarrow m=\pm\sqrt{\lambda}$

$\displaystyle \lambda=1-\frac{\psi''(y)}{\psi(y)}-\frac{\omega''(z)}{\omega(z)}$

$\displaystyle \frac{\psi''(y)}{\psi(y)}=\mu\Rightarrow \psi''(y)-\mu\psi(y)=0\Rightarrow n^2-\mu\Rightarrow n=\pm\sqrt{\mu}$

$\displaystyle \mu=1-\lambda-\frac{\omega''(z)}{\omega(z)}$

$\displaystyle \omega''(z)-\omega(z)+\lambda\omega(z)+\mu\omega(z)=0\Rightarr ow a^2-1+\lambda+\mu\Rightarrow a=\pm\sqrt{1-\lambda-\mu}$

$u(x,y,z;\lambda,\mu)=\exp{(x\pm\sqrt{\lambda})}*\e xp{(y\pm\sqrt{\mu})}*\exp{(z\pm\sqrt{1-\lambda-\mu})}$

I can't figure out the correct substitution for the exponential method to match the separations method.

Dear dwsmith,

Substitute, $\mu=s^2~and~\lambda=1-s^2-t^2$. Hope you can continue.