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Find the Fourier series solution to the differential equation

  1. Dec 8, 2013 #1
    Find the Fourier series solution to the differential equation x"+x=t

    It's given that x(0)=x(1)=0

    So, I'm trying to find a Fourier serie to x(t) and f(t)=t, and I'm know it must a serie of sin...



    So here's my question...the limits of integration to the Bn, how do I define them? Will it be like 0 to L to both series? And about the x", after a I find the f(x) Fourier series I must just derive it and replace in the x"??
     
  2. jcsd
  3. Dec 8, 2013 #2
    t = Ʃ Tn sin (n∏x/L)

    where f(t) = t = 2/L ∫ f(t) sin(n∏x/L)

    If the period 2L = 2, my limits on the integral will be 0 to 1?
     
  4. Dec 9, 2013 #3

    vanhees71

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    Since the right-hand side of the equation is not periodic, you have to use a Fourier integral rather than a Fourier series. Further one has to regularize the right-hand side, because it's not a Fourier-transformable function. I guess that the idea is that the external force is switched on at [itex]t=0[/itex]. So I'd write
    [tex]t \rightarrow \Theta(t) t \exp(-\epsilon t).[/tex]
    Then you can evaluate the Fourier transform of both the left-hand side and the right-hand side of the equation. At the end of the calculation, after transforming back to the time domain, you can take [itex]\epsilon \rightarrow 0^+[/itex].

    I also don't understand, why you have given boundary conditions at t=0 and t=1 rather than an initial condition [itex]x(t=0)=x_0[/itex], [itex]\dot{x}(t=0)=v_0[/itex]. This you could solve by first finding a particular solution of the inhomogeneous equation (using the Fourier-integral ansatz) and then add the general solution of the homogeneous equation.
     
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