# Math Help - Exponential method 3 variables

1. ## 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.

2. 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'.

3. 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.

4. 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.

5. 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.