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Using Laplace transform to solve this (simple) PDE

  1. Oct 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve [itex]\frac{\partial^2 v}{\partial x^2} = \frac{\partial^2 v}{\partial t^2}, x > 0, t > 0[/itex] subject to
    [itex]v(x,0) = 0[/itex]
    [itex]v_t(x,0) = 0[/itex]
    [itex]v(0,t) = f(t)[/itex]
    and where [itex]v[/itex] is bounded for all [itex]x > 0[/itex], by taking Laplace transforms with respect to [itex]t[/itex].

    2. Relevant equations



    3. The attempt at a solution

    Take Laplace transforms with respect to [itex]t[/itex]:
    [itex]\frac{\partial^2}{\partial x^2} V = s^2 V(x,s) - s v(x,0) - \frac{dv}{dt}(x,0)[/itex]
    [itex]\frac{\partial^2}{\partial x^2} V = s^2 V(x,s)[/itex]
    where [itex]V = \mathcal{L} \{ v \}[/itex].

    Solve this ODE.
    [itex]\lambda^2 - s^2 = 0[/itex]
    [itex]\lambda = \pm s[/itex]

    Here I start to get confused. By Laplace transform definition [itex]s[/itex] is a complex number. So is this ODE solution correct?

    [itex]V(x,s) = A(s) e^{sx} + B(s) e^{-sx}[/itex]

    Because [itex]v[/itex] is bounded the transform [itex]V[/itex] is also bounded for all [itex]x > 0[/itex] hence [itex]A(s) = 0[/itex].

    [itex]V(x,s) = B(s) e^{-sx}[/itex]
    [itex]V(0,s) = F(s) = B(s)[/itex]
    [itex]V(x,s) = F(s) e^{-sx}[/itex]

    Invert using given formula [itex]\mathcal{L} \{ H(t - c) f(t - x) \} = e^{-cs} F(s), c > 0[/itex] where [itex]H[/itex] is the Heaviside function we get
    [itex]v(x,t) = H(t - x)f(t - x)[/itex].

    Is this correct?
     
  2. jcsd
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