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Thread: uu_{xy}-u_{x}u_{y}=0

  1. #1
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    uu_{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.
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  2. #2
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    You probably have to use Separation of Variables.
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  3. #3
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    How is that done since there is a term u_{xy}?
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  4. #4
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    $\displaystyle \displaystyle uu_{xy}=u_xu_y\Rightarrow \int\frac{u_{xy}}{u_x}dy=\int\frac{u_y}{u}dy\Right arrow ln(u_x)=ln(u)+f(x)$

    Should I move $\displaystyle ln(u)+f(x)$ this to left and then exponentiate, exponentiate first, or integrated as is?
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  5. #5
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    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.
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  6. #6
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    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.

    $\displaystyle \displaystyle \frac{\partial u}{\partial y}\left[ln(u_x)\right]=\frac{\partial u}{\partial y}\left[ln(u)+f(x)\right]\Rightarrow \frac{u_{xy}}{u_x}=\frac{u_y}{u}$
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  7. #7
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    $\displaystyle \displaystyle ln(u_x)=ln(u)+f(x)\Rightarrow ln(u_x)-ln(u)=f(x)\Rightarrow \frac{u_x}{u}=e^{f(x)}\Rightarrow \frac{u_x}{u}=f_2(x)$

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

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

    I think this is how it is done.
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  8. #8
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    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!
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  9. #9
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    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

    $\displaystyle uu_{xy} - u_x u_y = 1 \; \text{or}\; u u_{xy} - u_x - u_y = 0.$
    Last edited by Jester; Dec 25th 2010 at 06:22 AM.
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  10. #10
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    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.
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  11. #11
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    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

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