u_{xy}+u_{yz}+u_{zx}-u=0

• Jan 1st 2011, 05:29 PM
dwsmith
u_{xy}+u_{yz}+u_{zx}-u=0
Using separation of variables

$\displaystyle u_{xy}+u_{yz}+u_{zx}-u=0$

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

$\displaystyle \varphi'(x)\psi'(y)\omega(z)+\varphi(x)\psi'(y)\om ega'(z)+\varphi'(x)\psi(y)\omega'(z)-\varphi(x)\psi(y)\omega(z)=0$

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

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

Now what?
• Jan 1st 2011, 06:04 PM
Ackbeet
Well, pick a constant of separation:

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

The first equality gives you a DE for $\displaystyle \varphi.$ The second equality gives you another equation that, I believe separates out thus:

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

You can do the same sort of trick you just did. That is, solve for $\displaystyle \psi'(y)/\psi(y),$ and choose another separation constant.

You follow?
• Jan 1st 2011, 08:25 PM
chiph588@
Quote:

Originally Posted by dwsmith
$\displaystyle u(x,y,z)=\varphi(x)\psi(y)\omega(z)$

How do you know this?
• Jan 1st 2011, 08:36 PM
dwsmith
$\displaystyle \varphi'(x)-\lambda\varphi(x)=0\Rightarrow m-\lambda=0\Rightarrow m=\lambda$

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

$\displaystyle \displaystyle\Rightarrow\lambda\psi'(y)\omega'(z)-\lamba\psi(y)\omega(z)}\right)+\psi'(y)\omega(z)+\ psi(y)\omega'(z)=0$

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

$\displaystyle \displaystyle\Rightarrow\mu=\frac{\psi'(y)}{\psi(y )}=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig ht)$

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

$\displaystyle \displaystyle\Rightarrow\mu=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig ht)$

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

$\displaystyle \displaystyle\omega'(z)[1+\mu\lambda]+\omega(z)[\mu-\lambda]=0\Rightarrow a(1+\mu\lambda)+\mu-\lambda=0\Rightarrow a=\frac{\lambda-\mu}{1+\mu\lambda}$

$\displaystyle \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$

I tried the substitution of $\displaystyle \displaystyle\mu=s \ \mbox{and} \ \lambda=\frac{1-\mu t}{\mu+t}$ in order to manipulate the exponential method to the separations method but it fell short.

Exponential solution:

$\displaystyle \displaystyle u(x,y,z;s,t)=\exp{\left(x\frac{1-st}{s+t}\right)}*\exp{(ys)}*\exp{(zt)}$

What substitution do I need to make?

Thanks.
• Jan 1st 2011, 08:40 PM
dwsmith
Quote:

Originally Posted by chiph588@
How do you know this?

This is from Elementary PDE by Berg and McGregor 1966.

...different technique for finding a particular solutions of homogeneous linear PDE is called the method of separation of variables. The exponential solution $\displaystyle \exp{(rx+sy)}=\exp{(rx)}*\exp{(sy)}$, are products of functions of the separable variables.... We seek a solution of the form $\displaystyle u(x,y)=\varphi(x)\psi(y)$ (pg. 14).
• Jan 1st 2011, 10:19 PM
Sudharaka
Quote:

Originally Posted by dwsmith
$\displaystyle \varphi'(x)-\lambda\varphi(x)=0\Rightarrow m-\lambda=0\Rightarrow m=\lambda$

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

$\displaystyle \displaystyle\Rightarrow\lambda\psi'(y)\omega'(z)-\lamba\psi(y)\omega(z)}\right)+\psi'(y)\omega(z)+\ psi(y)\omega'(z)=0$

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

$\displaystyle \displaystyle\Rightarrow\mu=\frac{\psi'(y)}{\psi(y )}=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig ht)$

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

$\displaystyle \displaystyle\Rightarrow\mu=-\left(\frac{\omega'(z)-\lambda\omega(z)}{\lambda\omega'(z)+\omega(z)}\rig ht)$

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

$\displaystyle \displaystyle\omega'(z)[1+\mu\lambda]+\omega(z)[\mu-\lambda]=0\Rightarrow a(1+\mu\lambda)+\mu-\lambda=0\Rightarrow a=\frac{\lambda-\mu}{1+\mu\lambda}$

$\displaystyle \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$

I tried the substitution of $\displaystyle \displaystyle\mu=s \ \mbox{and} \ \lambda=\frac{1-\mu t}{\mu+t}$ in order to manipulate the exponential method to the separations method but it fell short.

Exponential solution:

$\displaystyle \displaystyle u(x,y,z;s,t)=\exp{x\left(\frac{1-st}{s+t}\right)}*\exp{(ys)}*\exp{(zt)}$

What substitution do I need to make?

Thanks.

Dear dwsmith,

Think about it like this. You have the two solutions,

$\displaystyle \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$--------(Seperation method)

$\displaystyle \displaystyle u(x,y,z;s,t)=\exp{x\left(\frac{1-st}{s+t}\right)}*\exp{(ys)}*\exp{(zt)}$---------(Exponential method)

Consider the exponential part with y as the variable. Since the coefficient of y must be equal in both of these equations,

$\displaystyle \mu=s---------(1)$
Now move on to the exponential term with z as the variable. Using the same reasoning,

$\displaystyle t=\dfrac{\lambda-\mu}{1+\mu\lambda}\Rightarrow t=\dfrac{\lambda-s}{1+s\lambda}\Rightarrow \lambda=\dfrac{t+s}{1-ts}--------(2)$

(1) and (2) gives the substitutions necessary to convert the solution obtained from the seperation method to the solution obtained from the exponential method. Hope you understood.
• Jan 2nd 2011, 12:40 PM
dwsmith
By making that substitution, we don't obtain the same solutions.

$\displaystyle \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$

$\displaystyle \displaystyle s=\mu, \ t=\frac{\lambda-\mu}{1+\mu\lambda}, \ \mbox{and} \ \lambda=\frac{t+s}{1-st}$

$\displaystyle \displaystyle u(x,y,z;s,t)\rightarrow u(x,y,z;\lambda,\mu)=\exp{\left(x\frac{1}{\lambda} \right)}*\exp{(\mu s)}*\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$

Any thoughts?

I could say $\displaystyle \displaystyle \frac{1}{\lambda}=\lambda_2$ but would this be ok?
• Jan 2nd 2011, 04:38 PM
Sudharaka
Quote:

Originally Posted by dwsmith
By making that substitution, we don't obtain the same solutions.

$\displaystyle \displaystyle u(x,y,z;\lambda,\mu)=\exp{(x\lambda)}*\exp{(y\mu)} *\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$

$\displaystyle \displaystyle s=\mu, \ t=\frac{\lambda-\mu}{1+\mu\lambda}, \ \mbox{and} \ \lambda=\frac{t+s}{1-st}$

$\displaystyle \displaystyle u(x,y,z;s,t)\rightarrow u(x,y,z;\lambda,\mu)=\exp{\left(x\frac{1}{\lambda} \right)}*\exp{(\mu s)}*\exp{\left(z\frac{\lambda-\mu}{1+\mu\lambda}\right)}$

Any thoughts?

I could say $\displaystyle \displaystyle \frac{1}{\lambda}=\lambda_2$ but would this be ok?

Dear dwsmith,

You should get the same results. The problem is that there is a mistake in your calculations. Its in your first post. The fifth line should be,

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

You will have to do it all over again.....(Rofl)
• Jan 2nd 2011, 04:42 PM
dwsmith
Quote:

Originally Posted by Sudharaka
Dear dwsmith,

You should get the same results. The problem is that there is a mistake in your calculations. Its in your first post. The fifth line should be,

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

You will have to do it all over again.....(Rofl)

Thanks. I am not redoing the problem though haha.