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Fourier transform of differentials equation

  1. Apr 5, 2010 #1
    hey there.. i really dont know how to start answering this question.. can someone please guide me to solve this question..

    maths.jpg
     
  2. jcsd
  3. Apr 5, 2010 #2

    gabbagabbahey

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    Since the question tells you to use Fourier transforms, you start by taking the FT of both sides of the differential equation...
     
  4. Apr 5, 2010 #3
    still blur... :confused:
     
  5. Apr 5, 2010 #4

    gabbagabbahey

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    [tex]F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+F\left[a^2u\right]=F[f(x)][/itex]


    Call the FT of [itex]f(x)[/itex], [itex]F[f(x)]=\tilde{f}(k)[/itex]. Call the FT of [itex]u(x)[/itex], [itex]F[u(x)]=\tilde{u}(k)[/itex]. Your textbook likely tells you how to express the Fourier transform of [itex]-\frac{d^2u}{dx^2}[/itex] and [itex]a^2u(x)[/itex] in terms of [itex]\tilde{u}(k)[/itex]....do exactly that
     
  6. Apr 5, 2010 #5
    i dont know how to proceed from here...

    [tex]F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+F\left[a^2u\right]=F[f(x)][/tex]

    [tex]F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-\frac{d^2u}{dx^2}\right]+a^2F\left[u\right]=F[f(x)][/tex]

    [tex]F\left[-\frac{d^2u}{dx^2}+a^2u\right]=F\left[-(-k^2\tilde{u}(k,y))\right]+a^2F\left[u\right]=F[f(x)][/tex]
     
  7. Apr 5, 2010 #6

    gabbagabbahey

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    When you take the fourier transform of [itex]-\frac{d^2u}{dx^2}[/itex] (That's what the [itex]F[\ldots][/itex] means) you get [itex]k^2F[/itex], or [itex]k^2\tilde{u}(k)[/itex].

    This gives you an algebraic (instead of differential) equation you can solve for [itex]\tilde{u}[/itex]. To find [itex]u[/itex], you then take the inverse FT.

    This is a very basic application of using FTs to solve DEs and I have a hard time believing that your textbook doesn't provide a similar example. You should really be in the habit of studying the relevant section of your text/notes before attempting homework problems that are based off of it.
     
  8. Apr 6, 2010 #7
    i got the general solution already.. howeverr.. how am i going to prove...
    mathss.jpg
     
  9. Apr 6, 2010 #8

    i've done the inverse..

    F^-1[g^(k)] = g(x) = 1/ √2∏ ∫g^(k)e^ikx dk
    .............................. = 1/ √2∏ ∫ 1/ (k^2 + a^2)e^ikx dk
    ...............................= √2∏ [(e^-a|x|) / 2a]
     
  10. Apr 7, 2010 #9

    gabbagabbahey

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    [tex]\tilde{u}(k)\neq\frac{1}{k^2+a^2}[/tex]

    What happened to the FT of [itex]f(x)[/itex]?
     
  11. Apr 7, 2010 #10
    accidently double post
     
    Last edited: Apr 7, 2010
  12. Apr 7, 2010 #11
    mathsss.jpg
    ok.. this is my full solution...
    am i got it right until here?
     
  13. Apr 7, 2010 #12

    gabbagabbahey

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    That looks good, now use the composition rule to take the inverse FT of [tex]\hat{u}=\hat{g}\hat{f}[/itex]...
     
  14. Apr 7, 2010 #13
    inverse both side...
    i get..
    mathssss.jpg

    is that my general solution?
     
  15. Apr 7, 2010 #14

    gabbagabbahey

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    Does [itex]\frac{1}{2a}\int f(x) e^{-a|x|}dx[/itex] really represent the convolution of [itex]f[/itex] and [itex]g[/itex]?
     
  16. Apr 7, 2010 #15
    by convolution theorem,
    F^−1{F*G} = f ·g
    correct?
    so, i just substitute it..?
    kinda confused :confused:
     
  17. Apr 7, 2010 #16
    ok.. i revised bout it already..
    mathsssss.jpg
     
  18. Apr 7, 2010 #17

    gabbagabbahey

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    That's better!:smile:

    However, that's just part of the general solution. You can also add any function, [itex]u_h[/itex], which satisfies the corresponding homogeneous ODE, [itex]-\frac{d^2 u_h}{dx^2}+a^2u_h=0[/itex], to the [itex]u(x)[/itex] you've found, and the result will still satisfy your ODE. So, you need to find that homogeneous solution and add it to what you've found above in order to obtain the true general solution.
     
  19. Apr 7, 2010 #18
    should i solve the integration u(x) = 1/2a ∫ f(x - y) e^-a|y| dy?
     
  20. Apr 7, 2010 #19

    gabbagabbahey

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    First worry about finding the homogeneous part of the general solution. Solve the Homogeneous equation.
     
  21. Apr 8, 2010 #20
    can u give me more clue.. i really can't get it..
    which homogeneous equation i need to solve?
     
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