# uu_{xy}-u_{x}u_{y}=0

• December 23rd 2010, 06:40 PM
dwsmith
uu_{xy}-u_{x}u_{y}=0
Find the general solution.

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

My PDE book doesn't offer explanations so I have no clue on how to approach this problem.
• December 23rd 2010, 08:36 PM
Prove It
You probably have to use Separation of Variables.
• December 23rd 2010, 08:37 PM
dwsmith
How is that done since there is a term u_{xy}?
• December 24th 2010, 05:47 PM
dwsmith
$\displaystyle uu_{xy}=u_xu_y\Rightarrow \int\frac{u_{xy}}{u_x}dy=\int\frac{u_y}{u}dy\Right arrow ln(u_x)=ln(u)+f(x)$

Should I move $ln(u)+f(x)$ this to left and then exponentiate, exponentiate first, or integrated as is?
• December 24th 2010, 06:18 PM
Ackbeet
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 $u(x,y)=X(x)\,Y(y),$ in which case $u_{x}=X'Y,$ and $u_{y}=XY'.$ Finally, $u_{xy}=X'Y'.$ Plugging all this into the pde $uu_{xy}=u_{x}u_{y}$ yields

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

I'm not sure exactly what this means. Does it mean that any solution that is a product of a function of $x$ and a function of $y$ is a solution to the pde? I'm inclined to think so.
• December 24th 2010, 06:19 PM
dwsmith
It came from here.

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

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

$\displaystyle \frac{\partial u}{\partial y}\left[ln(u_x)\right]=\frac{\partial u}{\partial y}\left[ln(u)+f(x)\right]\Rightarrow \frac{u_{xy}}{u_x}=\frac{u_y}{u}$
• December 24th 2010, 06:49 PM
dwsmith
$\displaystyle ln(u_x)=ln(u)+f(x)\Rightarrow ln(u_x)-ln(u)=f(x)\Rightarrow \frac{u_x}{u}=e^{f(x)}\Rightarrow \frac{u_x}{u}=f_2(x)$

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

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

I think this is how it is done.
• December 25th 2010, 03:45 AM
Ackbeet
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!
• December 25th 2010, 06:54 AM
Jester
You could also have achieved the same with the transformation

$u = e^v$

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

$uu_{xy} - u_x u_y = 1 \; \text{or}\; u u_{xy} - u_x - u_y = 0.$
• December 25th 2010, 01:46 PM
Ackbeet
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 $(t,x,u)$) admit (functional) separable solutions in the form
$F(u) = T(t)X(x)$
$F(u) = T(t) + X(x)$?