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Wave equation

  1. Jul 29, 2008 #1
    We are supposed to work this using Laplace transforms

    [tex]U_{tt}=9U_{xx}; -infty<x<infty[/tex]

    [tex]U(x,0)=sinx[/tex]

    [tex]U_t(x,0)=0[/tex]

    The attempt at a solution

    [tex]Let L=\hat{U}[/tex]

    [tex]L[U_{tt}]=s^2\hat{U}-s(sinx)[/tex]

    [tex]L[9U_{xx}]=9\hat{U}_{xx}[/tex]

    [tex]s^2\hat{U}-s(sinx)=9\hat{U}_{xx}[/tex]

    [tex]\hat{U}_{xx}-\frac{s^2}{9}\hat{U}=-\frac{s}{9}sinx[/tex]

    This is a second order nonhomogeneous equation where the homogeneous solution is

    [tex]\hat{U}_h=A(s)e^{\frac{s}{3}x}+B(s)e^{-\frac{s}{3}x}[/tex]

    and the particular solution is

    [tex]\hat{U}_p=\frac{s}{s^2+9}sinx[/tex]

    Then we have

    [tex]\hat{U}=A(s)e^{\frac{s}{3}x}+B(s)e^{-\frac{s}{3}x}+\frac{s}{s^2+9}sinx[/tex]

    We know that

    [tex]\lim_{x\rightarrow\infty}U=0[/tex]

    [tex]\lim_{x\rightarrow{-\infty}}U=0[/tex]

    That second one should say limit as x approaches negative infinity, cant figure out how to get latex to do that

    [tex]\lim_{x\rightarrow\infty}\hat{U}=0[/tex]

    [tex]\lim_{x\rightarrow\infty}\hat{U}=0[/tex]

    Second one negative infinity again

    If we plug these in we get

    [tex]A(s)=\frac{\lim_{x\rightarrow\infty}\frac{s}{s^2+9}sinx}{\lim_{x\rightarrow\infty}e^{-\frac{s}{3}x}}}[/tex]

    [tex]A(s)=0[/tex]

    I'm not sure if that last step is legal, but if it is you can use the same reasoning for B(s)
    Any ideas?
     
  2. jcsd
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