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Homework Help: Fourier Transform of Differential Equation

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A differential equation [*] is given by:

    [tex] \frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t} [/tex]

    By first Fourier transforming the equation (*) with respect to x, show by substitution that:

    [tex] u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t} [/tex]

    is the Fourier transform of the solution of [*] , where A(k) is an unknown function of k.

    2. Relevant equations

    Derivative property of Fourier transforms (with respect to x):

    [tex] F\left[ \frac{\partial f}{\partial x} \right] = ikF[f] [/tex]

    Also know:

    [tex] F(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x)e^{ikx}dx [/tex]

    [tex] F(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} F(k)e^{ikx}dk [/tex]

    3. The attempt at a solution

    .. Fourier transform [*] with respect to x, treating t as a parameter. k used as a variable.

    Firstly rearrange [*] to get just [tex] \frac{\partial^{3}u}{\partial x^{3}} [/tex] on the LHS

    Then the LHS:

    [tex] F\left[ \frac{\partial^{3}u}{\partial x^{3}} \right] = F\left[ \frac{\partial^{2}}{\partial x^{2}}\left( \frac{\partial u}{\partial x} \right)\ right] = -k^{2}\left( \frac{\partial^{2}}{\partial x^{2}} \right)F [/tex]

    Then the RHS:

    RHS = [tex] \frac{\partial u}{\partial t} - 2 \left( \frac{\partial u}{\partial x} \right) [/tex]

    can take the t term outside the integral as it is just a constant parameter, therefore:

    [tex] F\left[ \left(\frac{\partial u}{\partial t}\right)-\left(\frac{2\partial u}{\partial x}\right) \right] = \frac{1}{\sqrt{2\pi}}\left(\frac{\partial}{\partial t} \right)\int_{-\infty}^{\infty}\frac{-2\partial u}{\partial x}e^{-ikx} dx [/tex]

    .. then obviously some more steps, not sure really where to go next though.

    Not sure if I'm even going about this in vaguely the right way :grumpy: .. so help will definately be appreciated!
     
  2. jcsd
  3. Feb 26, 2010 #2

    Mapes

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    Can you proofread the left-hand side? It's not clear what you're doing. You want to be replacing every spatial derivative with ik.

    The integral definition of F(u) isn't needed in this problem.
     
  4. Feb 26, 2010 #3
    This is the LHS calculations again, in full:

    [tex] F\left[ \frac{\partial^{3}u}{\partial x^{3}} \right] = F\left[ \frac{\partial^{2}}{\partial x^{2}}\left( \frac{\partial u}{\partial x} \right)\right]

    = ikF\left[\frac{\partial u^{2}}{\partial x} \right] = (ik)^{2}F\left[u\right] = -k^{2}\left( \frac{\partial^{2}u^{2}}{\partial x^{2}} \right)[/tex]

    .. hopefully that's more clear now (and somewhat correct!)?!?
     
  5. Feb 26, 2010 #4

    Mapes

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    I agree up to the second equals sign. Then you seem to lose a derivative, and later lose F altogether. Why not just use the relationship for a spatial derivative three times?
     
  6. Feb 26, 2010 #5
    The term being [tex] \frac{\partial^{3}u}{\partial x^{3}} [/tex] and not just a second deriviative has thrown me somewhat, though I know it shouldn't!

    .. the relationship for spacial derivative 3 times? Sorry to sound stupid! At the moment I'm trying to adapt a second derivative example to this question, so I thought would be some errors.

    I was using this relationship for a property of Fourier transform derivatives:

    [tex] F\left[ \frac{\partial f}{\partial x} \right] = ikF[f] [/tex]

    Hence that's where the derivative went for F (from 3rd to 2nd).

    .. bit confused now with all these derivatives! :|
     
  7. Feb 26, 2010 #6

    vela

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    Use the fact that

    [tex]\frac{\partial^{3}u}{\partial x^{3}} = \frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)[/tex]

    so

    [tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = F\left[\frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)\right] = ?[/tex]
     
  8. Feb 26, 2010 #7
    .. :umm: nope I don't really get it

    ..

    [tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = F\left[\frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)\right] = ik[/tex] ??

    (since you said earlier I need to replace every spacial derivative with ik)

    .. apologies for not being very clever at the moment with this!
     
  9. Feb 26, 2010 #8

    vela

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    Compare what you wrote here:
    with what you have here:
    What is f(x) in this case?
    We're backing up a bit here since you didn't see what Mapes was saying earlier.
     
  10. Feb 26, 2010 #9
    .. I did see Mapes comment and my comment after was an attempt to show how I had got that answer, allbeit incorrect, because I wasn't sure where I'd gone wrong as weren't completely sure how to do the calculation.

    Right, so as far as where we are now..

    [tex]f(x) = \frac{\partial^2 u}{\partial x^{2}}[/tex]

    ??
     
  11. Feb 26, 2010 #10

    vela

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    Yes, so you get

    [tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = F\left[\frac{\partial}{\partial x}\left(\frac{\partial^{2}u}{\partial x^{2}}\right)\right] = ikF\left[\frac{\partial^2 u}{\partial x^{2}}\right][/tex]

    Do you see now where this is going?
     
  12. Feb 27, 2010 #11
    [tex]= -k^{2}F\left[\frac{\partial u}{\partial x}\right][/tex]

    ??
     
  13. Feb 27, 2010 #12

    vela

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    Yes, and you want to do it one more time.
     
  14. Feb 27, 2010 #13
    ..

    [tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = -ik^{3}\left[u\right][/tex]

    ??
     
  15. Feb 27, 2010 #14

    Mapes

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    Looks good. Perhaps you can see where this is starting to match elements in the solution. :smile:
     
  16. Feb 27, 2010 #15

    vela

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    And perhaps you can see what Mapes meant when suggesting you replace every spatial derivative with ik.

    [tex]
    F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = (ik)^{3}F
    [/tex]
     
  17. Feb 27, 2010 #16
    Yep! It's obvious now!

    Righto, now know these expressions then:

    [tex]F\left[\frac{\partial^{3}u}{\partial x^{3}}\right] = (ik)^{3}F[/tex]

    [tex] F\left[ \frac{\partial u}{\partial x} \right] = ikF [/tex]

    so presumably:

    [tex] \frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = (ik)^{3}F + 2ikF[/tex]

    correct??

    .. and then the other side of the original equation:

    [tex] F\left[ \left(\frac{\partial u}{\partial t}\right)\right] = \frac{1}{\sqrt{2\pi}}\left(\frac{\partial}{\partial t} \right)\int_{-\infty}^{\infty} u e^{-ikx} dx [/tex]

    also correct??

    Hopefully getting somewhere with this now!
     
  18. Feb 28, 2010 #17
    .. had a look at this again, and I don't really get how to now 'sum up' these results in order to get through substitution the expression required.

    Also, another question following on from this..

    "Find [tex]A(k)[/tex] for the case where [tex]u(x,t)[/tex] at [tex]t=0[/tex] is given by [tex]u(x,0) = U_{0}\delta(x-a)[/tex] where [tex]U_{0}[/tex] is a constant."

    I think this follows from the answer to the first question then, but obviously I don't know how to do this at the moment :confused:
     
  19. Feb 28, 2010 #18

    ideasrule

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    Don't get confused like I was by the question. The "u" in this equation:

    [tex]
    \frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t}
    [/tex]

    is NOT the same as the "u" in this equation:

    [tex]
    u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}
    [/tex]

    The latter "u" is meant to be the Fourier transform of the former "u".

    To proceed, "substitution" just means putting in [tex]
    u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}
    [/tex] for F and showing that both sides of the differential equation are equal.
     
  20. Feb 28, 2010 #19
    .. so input this:

    [tex]u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t} [/tex]

    into this:

    [tex]\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t} [/tex]

    ??

    How do I do this seeing how there are both x and t dependant differentials, and the first equation doesn't distinguish whether it it dx or dt!? Unless differentiate it with respect to x, to get a result, and then differentiate (original equation) to get a different result for dt?!
     
  21. Feb 28, 2010 #20

    ideasrule

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    No, reread my last post. I said that the "u" given by [itex]
    u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}[/itex] is NOT the same as the "u" given by [itex]\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = \frac{\partial u}{\partial t} [/itex].

    You did a lot of work to get [itex]\frac{\partial^{3}u}{\partial x^{3}} + 2 \left( \frac{\partial u}{\partial x} \right) = (ik)^{3}F + 2ikF[/itex], so use it. F is actually [itex]u(k,t) = A(k)e^{-i \left( k^{3}-2k \right) t}[/itex], as the question states.
     
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