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Math Help - 2nd Order PDE hyperbolic

  1. #1
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    2nd Order PDE hyperbolic

    Hi, I have to find the general solution to this second order pde.

    x^2Z_{xx} - y^2Z_{yy} = 0 Where x and y not equal to 0.

    I have found characteristics to be n=xy and v=y/x and I have used these to change the variables to get

    Z_{vn} = 1/(2n) Z_v

    And now i'm stuck, for some reason I just dont know how to solve this with the first order partial derivative on the RHS.

    Please help.
    Thanks Katy
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  2. #2
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    I'm a little unsure. Here goes though. You have:

    \frac{\partial^2 z}{\partial v \partial n}=\frac{1}{2n}\frac{\partial z}{\partial v}

    and if the function is well-behaved, then:

    \frac{\partial^2 z}{\partial n \partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}=\frac{\partial}{\partial n}\frac{\partial z}{\partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}

    Let \frac{\partial z}{\partial v}=g(n,v) then we have \frac{\partial g}{\partial n}=\frac{1}{2n}g which is an ordinary DE in the variable n with v kept constant. Then:

    \frac{dg}{dn}-\frac{1}{2n}g=0

    Solving via integrating factor and noting that the constant of integration is now an arbitrary function of v, I obtain:

    d\left[n^{1/2}g\right]=0

    and so:

    g(n,v)=n^{-1/2} H(v) with  H(v) arbitrary.

    Now:

    \frac{\partial z}{\partial v}=n^{-1/2} H(v)

    so that:

    z(n,v)=n^{-1/2}\int H(v) dv+G(n)
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  3. #3
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    Thats fantastic, thats the answer I'm supposed to get.

    Thank you very much
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  4. #4
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    Quote Originally Posted by shawsend View Post
    I'm a little unsure. Here goes though. You have:

    \frac{\partial^2 z}{\partial v \partial n}=\frac{1}{2n}\frac{\partial z}{\partial v}

    and if the function is well-behaved, then:

    \frac{\partial^2 z}{\partial n \partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}=\frac{\partial}{\partial n}\frac{\partial z}{\partial v}=\frac{1}{2n}\frac{\partial z}{\partial v}

    Let \frac{\partial z}{\partial v}=g(n,v) then we have \frac{\partial g}{\partial n}=\frac{1}{2n}g which is an ordinary DE in the variable n with v kept constant. Then:

    \frac{dg}{dn}-\frac{1}{2n}g=0 (**)

    Solving via integrating factor and noting that the constant of integration is now an arbitrary function of v, I obtain:

    d\left[n^{1/2}g\right]=0

    and so:

    g(n,v)=n^{-1/2} H(v) with  H(v) arbitrary.

    Now:

    \frac{\partial z}{\partial v}=n^{-1/2} H(v)

    so that:

    z(n,v)=n^{-1/2}\int H(v) dv+G(n)
    I agree in the most part except for a negative (**) separates. Also since H(v) is arbitrary we can let \tilde{H} = \int H(v)\,dv

    @ Katy. You could also introduce the coords

     n = \ln x + \ln y,\;\; v = \ln x - \ln y

    to obtain the new PDE

    u_{nv} = \frac{1}{2} u_v - maybe a little easier.
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  5. #5
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    Hi. I made a mistake with the integrating factor. It should be:

    z(n,v)=n^{1/2}\left(\int H(v) dv+G(n)\right)

    or As Danny mentioned:

    z(n,v)=n^{1/2}\left(h(v)+g(n)\right)

    and I checked it with:

    H(v)=v^2+\sin(v)

    G(n)=3n-4 \cos(n)

    which equates to:

    z(x,y)=(xy)^{1/2}\left[\frac{1}{3}\left(\frac{y}{x}\right)^3-\cos(y/x)+3 x y-4 \cos(xy)\right]
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