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Solving ODE using Fourier Transform

  1. Apr 11, 2010 #1
    i have found the general solution which is,

    u(x)= (C1 + C2x)e^ax + (1/2a)[tex]\int f(x-y) e^\left|y\right| dy

    is this correct??
    now, i just want your help to guide me for justifying f(x)=x^5....

    is that wrong if i solve the integration and just substitute the integral which is the range( infinity to negative infinity)??

    thank you...
     

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  3. Apr 11, 2010 #2

    D H

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    That doesn't look correct. Could you show the work that led you to this result? Also, since there is no one agreed-upon definition of the Fourier transform, tell us what formalism you are using.

    I know just enough about Fourier transforms to be dangerous. I've asked other homework helpers who are better versed than am I in Fourier transforms to jump in and help. Until then, show some work and try to think about the second part of the question.

    A hint for that: Are those functions square integrable?
     
  4. Apr 11, 2010 #3
    I Fourier transform both sides and I get this:

    u~(k) = 1/(k^2 + a^2) . f~(k)

    From table,

    1/(k^2 + a^2) = √(∏/2) (e^-a|x|/a)----> denotes this as g~(k)

    u~(k) = g~(k)*f~(k) ----> applied convolution

    After convolution, I get 1/2a ∫f(x-y) e^-a|y| dy ; range -∞<y<∞

    Is it correct up until here??
     
  5. Apr 11, 2010 #4
    can you explain in detail of what you mean by that??
     
  6. Apr 11, 2010 #5

    gabbagabbahey

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    You've correctly found the particular solution, [itex]u_p(x)=\frac{1}{2a}\int_{-\infty}^{\infty}f(x-y)e^{-a|y|}dy[/itex], (assuming [itex]a>0[/itex]) via Fourier Transform methods. But [itex](c_1+c_2x)e^{ax}[/itex] does not satisfy the homogeneous equation [itex]-\frac{d^2 u}{dx^2}+a^2u=0[/itex], and so it is not the correct homogeneous solution.
     
  7. Apr 11, 2010 #6
    -d^2 u/dx^2 + a^u = 0

    2nd order homogeneous eq.

    -λ^2(e^ λx)+a^2e ^λx = 0
    - λ^2+a^2=0
    a^2= λ^2
    a= λ ------- y=(c1+c2x)e^ax

    is this wrong??
     
  8. Apr 11, 2010 #7

    gabbagabbahey

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    [tex]a^2=\lambda^2\implies a=\pm\lambda[/tex]

    You have two distinct roots, not one double root.
     
  9. Apr 11, 2010 #8
    oh issit that way?? i thought

    a=(λ^2)^1/2

    and will give us a=λ because we cancel the 2x(1/2)??
     
  10. Apr 11, 2010 #9

    gabbagabbahey

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    No, [itex](-\lambda)^2=\lambda^2[/itex]. You can't only take the positive root, and then claim that it is a repeated root. It's the exact same concept as solving the equation [itex]x^2=c[/itex]; both [itex]x=\sqrt{c}[/itex] and [itex]x=-\sqrt{c}[/itex] are solutions.
     
  11. Apr 11, 2010 #10
    oh sorry! my mistake!

    i get it already!..thank you..

    anyway Im aware that for the second part f(x)=x^5,

    1/2a ∫(x-y)^5 e^-a|y| dy -∞<y<∞

    right??

    I am aware that i have to use convergence test..isn't it the same?

    i have to integrate it and substitue the range?
     
  12. Apr 11, 2010 #11
    please correct me if im wrong,

    for f(x)=x^5

    let say i take -∞<y<0 (first)

    then i substitute inside---->∫(x-y)^5 e^-a|y| dy

    y=0---> (x)^5 [(e^0)=1]

    which give us when y=0 x^5

    will this solution conclude that f(x)=x^5 is convergence??
     
  13. Apr 11, 2010 #12
    If u don’t mind, can I ask one more question..

    2 d²u/dx² - x du/dx + u =0

    For this question, I already fourier transform both sides which give me,

    u~(k)[-2k² - xik + 1]=0… but it is impossible if just u~(k)=0

    do you have any idea of solving it?
     
  14. Apr 11, 2010 #13

    gabbagabbahey

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    That doesn't look like any convergence test I've ever seen.
     
  15. Apr 11, 2010 #14

    gabbagabbahey

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    [tex]F\left[x\frac{du}{dx}\right]\neq ikx\tilde{u}(k)[/tex]

    There is a rule that tells you how to take the FT of the product of [itex]x^n[/itex] with any function...use that rule.
     
  16. Apr 11, 2010 #15
    u~[-2k² + 1] = i/2∏ [d/dk u~(k)]

    I transfer, xu’ to the right side

    It this correct??

    doesnt [d/dk u~(k)] =iku~(k)

    is the same?
     
    Last edited: Apr 11, 2010
  17. Apr 11, 2010 #16

    gabbagabbahey

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    You're missing a factor of [itex]k[/itex] and you don't need the [itex]1/2\pi[/itex]:

    [tex]F\left[x\frac{du}{dx}\right]=i\frac{d}{dk}\left(F\left[\frac{du}{dx}\right]\right)=i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]=-k\frac{d\tilde{u}}{dk}[/tex]
     
  18. Apr 11, 2010 #17
    ok thank you!, so now i rearrange it,

    u~(k)=(-k/-2k²+1)(du~/dk)
    is this correct if i diffrentiate towards k on the right side?

    that will give me u~(k)=2k³-4k²-k/(-2k²+1)²???
     
    Last edited: Apr 11, 2010
  19. Apr 11, 2010 #18

    gabbagabbahey

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    Actually, I made an error in my previous post:

    [tex]i\frac{d}{dk}\left[(ik)\tilde{u}(k)\right]\neq-k\frac{d\tilde{u}}{dk}[/itex]

    You will need to use the product rule to carry out the derivative.
     
  20. Apr 11, 2010 #19
    :confused: error???

    error in which part??

    yes i got the answer of 2k³-4k²-k/(-2k²+1)² using....(u'v-uv')/v² am i right??
     
  21. Apr 11, 2010 #20

    gabbagabbahey

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    [tex]\frac{d}{dk}\left(k\tilde{u}\right)\neq k\frac{d\tilde{u}}{dk}[/tex]
     
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