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Math Help - fourier transform of DE

  1. #1
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    fourier transform of DE

    hey there.. i really dont know how to start answering this question.. can someone please guide me to solve this question..


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  2. #2
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    Hi. Start taking Fourier Transforms. In general:

    \mathcal{F}\left\{y^{(n)}\right\}=i^n s^n \mathcal{F}\left\{y\right\}

    I get:

    \begin{aligned}<br />
\widehat{y}&=\frac{\widehat{f}}{s^2+a^2}\\<br />
&=\widehat{g}\widehat{f}<br />
\end{aligned}<br />

    Now take the inverse transform by expressing the right side as a convolution.
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  3. #3
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    i dont know how to proceed from here...

    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)]

    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)]

    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)]


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  4. #4
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    Quote Originally Posted by shawsend View Post
    Hi. Start taking Fourier Transforms. In general:

    \mathcal{F}\left\{y^{(n)}\right\}=i^n s^n \mathcal{F}\left\{y\right\}

    I get:

    \begin{aligned}<br />
\widehat{y}&=\frac{\widehat{f}}{s^2+a^2}\\<br />
&=\widehat{g}\widehat{f}<br />
\end{aligned}<br />

    Now take the inverse transform by expressing the right side as a convolution.
    i get the general solution already.. but.. how am i going to prove..
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  5. #5
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    First note (from a table or Mathematica):

    \mathcal{F}\left\{e^{-a|x|}\right\}=a\sqrt{2/\pi} \frac{1}{a^2+s^2}
    then:
    \mathcal{F}\left\{1/a \sqrt{\pi/a}e^{-a|x|}\right\}=\frac{1}{a^2+s^2}

    From above, we had:

    \widehat{y}=\widehat{g}\widehat{f}

    so use the convolution theorem (look it up and review it):

    \widehat{g}(s)\widehat{f}(s)=\frac{1}{\sqrt{2\pi}}  \widehat{(g\star f)}(s)

    and therefore:

    y(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\  pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt

    and there are various versions of the transform using different coefficients so you need to check the constants I have up there. Now, look at the integral, for positive a, that's going to converge for f that has an order less than e^{kx}. Polynomials do but not e^{x^2}.
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  6. #6
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    Quote Originally Posted by shawsend View Post
    Now take the inverse transform by expressing the right side as a convolution.
    so.. y^ = g^f^ is not my general solution yet?
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  7. #7
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    You're gonna' have to invert that expression as the y-hats represent the Fourier transforms of the function. Also, I didn't include the complimentary components of the solution arising from the homogeneous equation. Add that part to the solution for the general solution. Also, I may not have done the Fourier transform/inverse transform correctly. I'm expecting you to go over it and make sure it's correct and in the process, learn for you self how to work problems this way. You can verify everything by just solving a particular case numerically and comparing it to the solution obtained symbolically.
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  8. #8
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    by inverse g^, i get....

    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]

    correct?
    so.. what's the next step?
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  9. #9
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    let's look at a particular example:

    u''-4u=-\sin(x)=g(x)

    Taking Fourier Transforms of both sides:

    \mathcal{F}\left\{u''-4u=g(x)\right\}

    -s^2\widehat{u}-4\widehat{u}=\widehat{g}

    so:

    \widehat{u}=-\widehat{g} \frac{1}{s^2+4}

    but I know that:

    \mathcal{F}\left\{\sqrt{\pi/8} e^{-2|x|}\right\}=\frac{1}{s^2+4}

    so I let:

    \widehat{h}=\frac{1}{s^2+4}

    that means

    \widehat{u}=-\widehat{g}\widehat{h}

    I now take the inverse Fourier Transform of both sides:

    \mathcal{F}^{-1}\left\{ \widehat{u}\right\}=\mathcal{F}^{-1}\left\{-\widehat{g}\widehat{h}\right\}

    and the inversse Fourier Transform of a product of transforms is a convolution integral:

    u(x)=-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}-\sin(x-y)\sqrt{\pi/8} e^{-2|y|}dy

    u(x)=1/4\left(\int_{-\infty}^0 \sin(x-y)e^{2y}dy+\int_0^{\infty} \sin(x-y)e^{-2y}dy\right)

    simplifying that I get:

    u(x)=1/5\sin(x)

    so that the general solution is:

    u(x)=c_1 e^{-2x}+c_2 e^{2x}+1/5 \sin(x)
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  10. #10
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    ok.. this is my full solution...


    [img]http://i99.photobucket.com/albums/l313/nameck_5/mathsss.jpg[\img]


    am i right until there?
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  11. #11
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    ok.. this is my full solution...
    am i got it right until here?
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  12. #12
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    continue.. i get...
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  13. #13
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    i your example, your g(x) is given.. but, my question, isn't given..

    how am i going to solve integration of f(x - y)?
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  14. #14
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    now.. to answer my question, u'' + a^2u = x^5
    after solving everything, i manage to get this equation..


    [tex] u(x)= ∫- ∞ to 0
    1/2a(x-y)^{5}e^{ay}dy+ ∫0 to ∞
    (x-y)^{5}e^{-ay}dy

    should i solve it to get the general solution?
    because, i need to do integration by parts 5 times to solve it..


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  15. #15
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    Quote Originally Posted by shawsend View Post
    First note (from a table or Mathematica):

    \mathcal{F}\left\{e^{-a|x|}\right\}=a\sqrt{2/\pi} \frac{1}{a^2+s^2}
    then:
    \mathcal{F}\left\{1/a \sqrt{\pi/a}e^{-a|x|}\right\}=\frac{1}{a^2+s^2}

    From above, we had:

    \widehat{y}=\widehat{g}\widehat{f}

    so use the convolution theorem (look it up and review it):

    \widehat{g}(s)\widehat{f}(s)=\frac{1}{\sqrt{2\pi}}  \widehat{(g\star f)}(s)

    and therefore:

    y(x)=\frac{1}{\sqrt{2\pi}}\frac{1}{a}\sqrt{\frac{\  pi}{a}}\int_{-\infty}^{\infty} e^{-a|x|}f(x-t)dt

    and there are various versions of the transform using different coefficients so you need to check the constants I have up there. Now, look at the integral, for positive a, that's going to converge for f that has an order less than e^{kx}. Polynomials do but not e^{x^2}.
    should i use convergence test to justify the answer?
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