# Thread: u_{xy}+u_{yz}+u_{zx}-u=0

1. ## u_{xy}+u_{yz}+u_{zx}-u=0

Using separation of variables

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

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

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

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

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

Now what?

2. Well, pick a constant of separation:

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

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

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

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

You follow?

3. Originally Posted by dwsmith
$u(x,y,z)=\varphi(x)\psi(y)\omega(z)$
How do you know this?

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

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

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

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

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

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

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

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

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

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

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

Exponential solution:

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

What substitution do I need to make?

Thanks.

5. 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 $\exp{(rx+sy)}=\exp{(rx)}*\exp{(sy)}$, are products of functions of the separable variables.... We seek a solution of the form $u(x,y)=\varphi(x)\psi(y)$ (pg. 14).

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

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

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

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

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

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

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

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

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

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

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

Exponential solution:

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

What substitution do I need to make?

Thanks.
Dear dwsmith,

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

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

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

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

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

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

(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.

7. By making that substitution, we don't obtain the same solutions.

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

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

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

Any thoughts?

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

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

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

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

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

Any thoughts?

I could say $\displaystyle \frac{1}{\lambda}=\lambda_2$ 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,

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

You will have to do it all over again.....

9. 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,

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

You will have to do it all over again.....
Thanks. I am not redoing the problem though haha.