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Math Help - Partial differential equation

  1. #1
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    Partial differential equation

    Hello
    I have example to resolve:
    yu_{x}(x,y)+xu_{y}(x,y)=u(x,y)
    where u(1,y)=y+1
    Anyone to help me?
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  2. #2
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    Lightbulb Re: Partial differential equation

    You can use these two beauties:

    u[x,y]\to -\frac{\left(-1+\sqrt{1-x^2+y^2}\right) \left(1+\sqrt{1-x^2+y^2}\right)}{x+\sqrt{y^2}} = \frac{x^2-y^2}{x+\sqrt{y^2}}

    u[x,y]\to -\frac{\left(x+\sqrt{y^2}\right) \left(-1+\sqrt{1-x^2+y^2}\right)}{1+\sqrt{1-x^2+y^2}}
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  3. #3
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    Re: Partial differential equation

    Hi !

    Changing the cartesian coordinates system to hyperbolic system leads to a very simple PDE. So, the general solution is obtained (in attachment).
    Then the condition u(1,y)=1+y will be easy to apply in order to express the particular solution.
    Attached Thumbnails Attached Thumbnails Partial differential equation-pde.jpg  
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  4. #4
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    Re: Partial differential equation

    I don't understand...Maybe this is not true.
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  5. #5
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    Re: Partial differential equation

    JJacquelin how you get x and y?
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    Re: Partial differential equation

    I quite not understand your question
    x = rho*cosh(theta) and y = rho*sinh(theta)
    The inverse is : rho = sqrt(x-y) and theta = argtanh(y/x)
    This is the usual relationship between cartesian coordinates and hyperbolic coordinates.
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  7. #7
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    Re: Partial differential equation

    By the way, a much simpler method consists in :
    Let u(x,y) = (x+y)*v(x,y)
    Binging back into the PDE leads to y*(dv/dx)+x*(dv/dy) = 0
    which is very easy to solve.
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  8. #8
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    Re: Partial differential equation

    yu_{x}(x,y)+xu_{y}(x,y)=0
    \frac{dx}{y}=\frac{dy}{x}
    C=\frac{1}{2}( x^{2}-y^{2})- shooting fish in a barrel
    I case study new method, which seems easier. I've done this example, but I got the another answer.
    \frac{dx}{dt}=y\\ \frac{dy}{dt}=x\\ \frac{dz}{dt}=z\\
    x(0)=1\\y(0)=L\\z(0)=L+1\\
    x(t)=yt+1\\y(t)=xt+L\\z(t)Le^{t}+1
    After that I talled t and L
    \\t=(x-1)/y\\L=y-\frac{x(x-1)}{y}
    \\u(x,y)=e^{\frac{x-1}{y}}(y-\frac{x(x-1)}{y})+1\\
    u(1,y)=y+1
    If you understand this method, plese give me the answer it is correct?
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  9. #9
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    Re: Partial differential equation

    Sorry, in fact, I didn't read your method.
    I just bing back your result u(x,y) into the PDE. It doesn't aggree.

    Clue : The final solution of your problem is very simple : u(x,y) = x+y
    Last edited by JJacquelin; August 28th 2012 at 05:01 AM.
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  10. #10
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    Re: Partial differential equation

    Maybe I give in....or could you give me advice in which book I find this method or where (link)?
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  11. #11
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    Re: Partial differential equation

    I understand your solution, thanks for your help, but I have one question, when I have to make a parametrization?
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  12. #12
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    Re: Partial differential equation

    I couldn't give a general answer.
    As many methods : try and see if it allows simplifications or not.
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  13. #13
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    Re: Partial differential equation

    Ok, thank you
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