Partial Diff qn

  1. 1. The problem statement, all variables and given/known data
    Does the following differential equation for u(x,y) have solutions which have the form of a product of functions of each independent variable:

    [tex]\partial[/tex]2u/[tex]\partial[/tex]x[tex]\partial[/tex]y=u


    3. The attempt at a solution[/]

    =>[tex]\int[/tex][tex]\int[/tex]u dx dy =uxy=f(x)f(y) ???
     
    Last edited: May 21, 2009
  2. jcsd
  3. im not sure if that even makes sense, but if u(x,y)=f(x)f(y) and d*f(x)/dx=u/constant and d*f(y)/dy=constant... i think, i don't know im confused..
     
  4. Dick

    Dick 25,887
    Science Advisor
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    Set u(x,y)=f(x)*g(y). What is your partial derivative in terms of f(x), f'(x), g(y), and g'(y)?
     
  5. so [tex]\partial[/tex]2u/[tex]\partial[/tex]x[tex]\partial[/tex]y=f(x)g(y)??
    so in terms of f'(x) and g'(y) im not sure..
     
  6. Cyosis

    Cyosis 1,495
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    You have an equation that equates the derivative of u to u itself. You can already guess what u will be, but lets not guess.

    Setting u(x,y)=f(x)g(y), then the equation becomes [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}=f(x)g(y)[/tex]. Now what Dick asks you to do is to calculate [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}[/tex].

    Hint:

    [tex]
    \frac{d}{dx}f(x)= ?
    [/tex]

    In terms of f'(x)
     
    Last edited: May 21, 2009
  7. Dick

    Dick 25,887
    Science Advisor
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    I actually meant u(x,y)=f(x)g(y). There's no reason why the function of x would have to be the same as the function of y. And yes, evaluate the derivative.
     
  8. Cyosis

    Cyosis 1,495
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    Sorry, that was sloppy of me I actually meant f(x)g(x). I will fix it.
     
  9. hmm d/dx of f(x).. isn't that just f'(x)??.... is it something like the integral of f'(x)g'(y)??
     
  10. so... [tex]
    \frac{\partial^2 f(x)g(y)}{\partial x \partial y}
    [/tex] =[tex]\int[/tex][tex]\int[/tex]f'(x)g'(y) dx dy??
     
  11. Cyosis

    Cyosis 1,495
    Homework Helper

    What you've written down now is the equivalent to [itex]\frac{d}{dx}f(x)=\int f'(x) dx[/itex]. Which is wrong. Lets test it, f(x)=x then [itex]\frac{d}{dx}x=1 \neq \int 1 dx=x+c[/itex].

    This is right, so if you evaluate [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}[/tex] you get....?
     
  12. u get f'(x)g'(y)?? Im confused becasue when u take partial of f(x), u get d/dy(d/dx*f(x))=0??
     
  13. Cyosis

    Cyosis 1,495
    Homework Helper

    Yes that's correct, but you're not taking the partial of f(x), you're talking the partial derivative of f(x)g(y).

    [tex]
    \frac{\partial^2 f(x)g(y)} {\partial x \partial y}=\frac{\partial}{\partial x} f(x) \frac{\partial}{\partial y}g(y)=f'(x)g'(y)
    [/tex]
     
  14. ok so since f(x)g(y)=f'(x)g(y)=u(x,y)=??
     
  15. Cyosis

    Cyosis 1,495
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    You forgot a prime it should be [itex]f'(x)g'(y)=f(x)g(y)=u(x,y)[/itex]. You can instantly see which function satisfies the equation, but you can also separate the variables.
     
  16. e^x+y?? so f(x)=e^x and g(y)=e^y same as the derivativeS?
     
  17. Cyosis

    Cyosis 1,495
    Homework Helper

    That would be a solution indeed, you're on the right track. it is not the general solution however. The general solution will involve integration constants and a constant in the argument of the exponential. Can you see how to integrate the differential equation?
     
  18. e^xy?? so f(x)=e^x and g(y)=e^y same as the derrivativeS?

    srry this was an accident
     
  19. Cyosis

    Cyosis 1,495
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    Don't randomly guess, [itex]f'(x)g'(y)=y e^{xy}xe^{xy} \neq e^{xy}[/itex]. Separate the variables first, that is all the x dependent stuff on one side and all the y dependent stuff on the other side.
     
  20. na im confused now... like f(x)=e^c1 g(y)=e^c2 and u(x,y)=e^c1+c2??
     
  21. Cyosis

    Cyosis 1,495
    Homework Helper

    If c are constants then f'(x)=g'(y)=0, which doesn't fit either. Can you please separate the variables first.
     
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