Find the general solution.

$\displaystyle uu_{xy}-u_{x}u_{y}=0$

My PDE book doesn't offer explanations so I have no clue on how to approach this problem.

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- Dec 23rd 2010, 05:40 PMdwsmithuu_{xy}-u_{x}u_{y}=0
Find the general solution.

$\displaystyle uu_{xy}-u_{x}u_{y}=0$

My PDE book doesn't offer explanations so I have no clue on how to approach this problem. - Dec 23rd 2010, 07:36 PMProve It
You probably have to use Separation of Variables.

- Dec 23rd 2010, 07:37 PMdwsmith
How is that done since there is a term u_{xy}?

- Dec 24th 2010, 04:47 PMdwsmith
$\displaystyle \displaystyle uu_{xy}=u_xu_y\Rightarrow \int\frac{u_{xy}}{u_x}dy=\int\frac{u_y}{u}dy\Right arrow ln(u_x)=ln(u)+f(x)$

Should I move $\displaystyle ln(u)+f(x)$ this to left and then exponentiate, exponentiate first, or integrated as is? - Dec 24th 2010, 05:18 PMAckbeet
I'm not sure I think your last steps there are valid. In pde's, separation of variables doesn't look like it does in ode's. You would assume that $\displaystyle u(x,y)=X(x)\,Y(y),$ in which case $\displaystyle u_{x}=X'Y,$ and $\displaystyle u_{y}=XY'.$ Finally, $\displaystyle u_{xy}=X'Y'.$ Plugging all this into the pde $\displaystyle uu_{xy}=u_{x}u_{y}$ yields

$\displaystyle XYX'Y'=X'YXY',$ an identity.

I'm not sure exactly what this means. Does it mean that any solution that is a product of a function of $\displaystyle x$ and a function of $\displaystyle y$ is a solution to the pde? I'm inclined to think so. - Dec 24th 2010, 05:19 PMdwsmith
It came from here.

$\displaystyle uu_{xy}-u_{x}u_{y}=0$

Also, if you differentiate with respect to y, you will obtain this equation.

$\displaystyle \displaystyle \frac{\partial u}{\partial y}\left[ln(u_x)\right]=\frac{\partial u}{\partial y}\left[ln(u)+f(x)\right]\Rightarrow \frac{u_{xy}}{u_x}=\frac{u_y}{u}$ - Dec 24th 2010, 05:49 PMdwsmith
$\displaystyle \displaystyle ln(u_x)=ln(u)+f(x)\Rightarrow ln(u_x)-ln(u)=f(x)\Rightarrow \frac{u_x}{u}=e^{f(x)}\Rightarrow \frac{u_x}{u}=f_2(x)$

$\displaystyle \displaystyle \int\frac{u_x}{u}dx=\int f_2(x)dx\Rightarrow ln(u)=f_2(x)+g(y)\Rightarrow e^{ln(u)}=e^{f_2(x)+g(y)}$

$\displaystyle \Rightarrow u(x,y)=e^{f_2(x)}e^{g(y)}=f_3(x)g_2(y)$

I think this is how it is done. - Dec 25th 2010, 02:45 AMAckbeet
In looking over your solution, I think your steps are valid. I would point out, however, that your final destination is the same place the usual pde separation of variables got you in post # 5.

Looks good! - Dec 25th 2010, 05:54 AMJester
You could also have achieved the same with the transformation

$\displaystyle u = e^v$

Just to add, most of the time separation of variables doesn't work for nonlinear PDE's. For example try

$\displaystyle uu_{xy} - u_x u_y = 1 \; \text{or}\; u u_{xy} - u_x - u_y = 0.$ - Dec 25th 2010, 12:46 PMAckbeet
Reply to Danny:

Right. Dym's equation was the only nonlinear pde I'd ever come across, before this post, that succumbed to separation of variables. - Dec 25th 2010, 01:35 PMJester
There are a few other's. I really like the separation of variables of the Sine-Gordon equation as presented in Debnath's book (rather clever). A very good question is - when does a nonlinear PDE (in variables $\displaystyle (t,x,u) $) admit (functional) separable solutions in the form

$\displaystyle F(u) = T(t)X(x)$

or

$\displaystyle F(u) = T(t) + X(x)$?