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Solving a partial DE

  1. May 26, 2004 #1
    hi all
    i have been trying to solve to following problem,

    [tex]
    \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2}
    + 2\frac{\partial u}{\partial x} + u = 0
    [/tex]

    [tex]
    u=u(x,y)
    [/tex]

    after a bit of work using the change of variables

    [tex]
    \zeta=\zeta(x,y)=y-x
    [/tex]

    and

    [tex]
    \eta=\eta(x,y)=y+x
    [/tex]

    i obtain

    [tex]- 4\frac{\partial u}{\partial \zeta \partial \eta} = 2\frac{\partial u}{\partial\zeta} - 2\frac{\partial u}{\partial\eta}-u[/tex]

    but i am unsure how to solve this, i used maple to solve this problem and it gave out a fairly harmless answer so i am pretty sure there would be any easy way to solve the above equation.

    could someone kindly show me how to obtain a solution to this problem
     
    Last edited: May 26, 2004
  2. jcsd
  3. May 26, 2004 #2

    arildno

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    I would suggest a separation of variables solution,
    u(x,y)=f(x)g(y)
     
  4. May 28, 2004 #3
    Hi;
    The first substitution to use is "running wave" substitution:
    [tex]
    z=x-ct
    [/tex]
    and you will have:
    [tex]
    \frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2}
    + 2\frac{\partial u}{\partial x} + u = 0
    [/tex]
    [tex]
    u_{xx}-u_{yy}+2u_x+u=0
    [/tex]
    [tex]
    z=x-ct
    [/tex]
    [tex]
    u_{zz}-c^2u_{zz}+2u_z+u=0
    [/tex]
    [tex]
    (1-c^2)u_{zz}+2u_z+u=0
    [/tex]
    The last equation is a linear ODE of the 2nd order while c<>+1 and c<>-1. Otherwise we have the first order ODE
    [tex]
    2u_z+u=0
    [/tex]
    Anyways both equations are simple and can be solved easily.
    Still, this is not the only way of reduction of the given PDE (there could be another substitutions v=v(u,x,y), z=z(u,x,y) that will give us ordinary DEs). If you need some other forms for v and z, let me know, I can find some for you.
    Best of luck,
    Max.
     
  5. May 28, 2004 #4

    arildno

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    Here's a rather general solution using separation of variables:

    1.
    [tex]u(x,y)=f(x)g(y)\to\frac{f''+2f'+f}{f}=\frac{g''}{g}[/tex]

    2. Since LHS is a function of x, wheras RHS is a function of y, equality is only possible if they are equal to some constant k.
    I set a subscript k on the functions at this moment:
    [tex]\frac{f_{k}''+2f_{k}'+f_{k}}{f_{k}}=k\to{f}_{k}(x)=A_{k}e^{-x-\sqrt{k}x}+B_{k}e^{-x+\sqrt{k}x}[/tex]
    [tex]\frac{g_{k}''}{g_{k}}=k\to{g}_{k}(x)=C_{k}e^{-\sqrt{k}y}+D_{k}e^{\sqrt{k}y}[/tex]

    3. Since the original differential equation is linear, a sum of two solutions (for example, for different k-values) is also a solution, and we may write:

    [tex]u(x,y)=[/tex]
    [tex]e^{-x}\int_{-\infty}^{\infty}\mathcal{C}_{(k,+)}Cosh(\sqrt{k}(x+y))+\mathcal{C}_{(k,-)}Cosh(\sqrt{k}(x-y))dk+[/tex]
    [tex]e^{-x}\int_{-\infty}^{\infty}\mathcal{S}_{(k,+)}Sinh(\sqrt{k}(x+y))+\mathcal{S}_{(k,-)}Sinh(\sqrt{k}(x-y))dk[/tex]

    4. Negative values of k will yield trigonometric functions; the unknown functions of k,
    [tex]\mathcal{C}_{(k,+)},\mathcal{C}_{(k,-)},\mathcal{S}_{(k,+)},\mathcal{S}_{(k,-)}[/tex]
    may be suited to satisfy boundary conditions.
     
    Last edited: May 28, 2004
  6. May 28, 2004 #5
    thanks for your help guys
    i have the problem solved
     
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