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Math Help - ODE (ar laplace transform)

  1. #1
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    ODE (ar laplace transform)

    where L[u(x,t)] = U(x,s) with respect to t
    Given u(x,o)=x u(0.t)=o

    It is an imhonogeous equation. It need to find out general soland particular sol.

    if it is a honogeous equation, just very easy to do.Am I correct in finding general sol.?
    what is the yp(particular sol) in order to find U(x,s)


    Finally, How to do inverse laplace transform to find u(x,t).
    Thank
    Attached Thumbnails Attached Thumbnails ODE (ar laplace transform)-1.jpg  
    Last edited by cyw1984; August 5th 2007 at 06:14 PM.
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  2. #2
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    Give us the partial differencial equation with the boundary and initial value problems. The way you have it just makes no sense.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Give us the partial differencial equation with the boundary and initial value problems. The way you have it just makes no sense.
    updated
    bivp is u(x,0)=x u(0,t)=0
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  4. #4
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    That does not help. Give me the original equation without taking the Laplace Transform. The one you have is after you taken the Laplace Transformation.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    That does not help. Give me the original equation without taking the Laplace Transform. The one you have is after you taken the Laplace Transformation.
    O..sorry

    The original equation is

    Uxx= Ut (-inf < x < + inf , t>0)
    u(x,0)=x u(0,t)=0

    Use the foruier transform and laplace transform to solve this PDE.

    Thank, The above one I have is after I taken the Laplace Transformation.
    Last edited by cyw1984; August 5th 2007 at 07:32 PM.
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    \mathcal{L}\left\{ \frac{\partial^2 u}{\partial x^2} \right\} = \mathcal{L} \left\{ \frac{\partial^2 u}{\partial t} \right\}

    \frac{d^2 U(x,s)}{dx^2} = sU(x,s) - u(x,0) = sU(x,s)-x

    \frac{d^2 U(x,s)}{dx^2} - sU(x,s) = x

    U(x,s) = Ae^{x\sqrt{s}}+Be^{-x\sqrt{s}} - \frac{x}{s}

    Now solve for A,B with the conditions given.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    \mathcal{L}\left\{ \frac{\partial^2 u}{\partial x^2} \right\} = \mathcal{L} \left\{ \frac{\partial^2 u}{\partial t} \right\}

    \frac{d^2 U(x,s)}{dx^2} = sU(x,s) - u(x,0) = sU(x,s)-x

    \frac{d^2 U(x,s)}{dx^2} - sU(x,s) = x

    U(x,s) = Ae^{x\sqrt{s}}+Be^{-x\sqrt{s}} - \frac{x}{s}

    Now solve for A,B with the conditions given.

    Now I have solve A=-B
     A = \frac{L}{s} \frac{1}{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}

     U(x,s) = \frac{L}{s} \frac{1}{e^{L\sqrt{s}}-e^{-L\sqrt{s}}} [{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}] - \frac{x}{s}
    Am I correct?
    Then how to do inverse laplace transform. The U(x,s) is very complex.....
    Last edited by cyw1984; August 6th 2007 at 09:57 AM.
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  8. #8
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    Quote Originally Posted by cyw1984 View Post
    Now I have solve A=-B
     A = \frac{L}{s} \frac{1}{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}

     U(x,s) = \frac{L}{s} \frac{1}{e^{x\sqrt{s}}-e^{-x\sqrt{s}}} [{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}] - \frac{x}{s}
    Am I correct?
    Then how to do inverse laplace transform. The U(x,s) is very complex.....
    I just realized that perhaps we need another point. You provided u(0,t)=0. Is there supposed to be another one? Because you cannot solve for A and B yet because there is 1 equation and two unknowns, if you provide one more conditions then we have 2 equations and 2 unknowns and it should solve.
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  9. #9
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    Quote Originally Posted by ThePerfectHacker View Post
    I just realized that perhaps we need another point. You provided u(0,t)=0. Is there supposed to be another one? Because you cannot solve for A and B yet because there is 1 equation and two unknowns, if you provide one more conditions then we have 2 equations and 2 unknowns and it should solve.
    now if u(0,t)=0 u(L,t) =0


     U(x,s) = \frac{L}{s} \frac{1}{e^{L\sqrt{s}}-e^{-L\sqrt{s}}} [{e^{x\sqrt{s}}-e^{-x\sqrt{s}}}] - \frac{x}{s}

    Is my answer correct???
    Thank
    and how to do inverse laplace.it is so complex
    Last edited by cyw1984; August 6th 2007 at 09:58 AM.
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