An Integral Equation with the Convolution Theorem for Fourier Transforms


by mathmajor314
Tags: convolution, fourier transform, integral equation
mathmajor314
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#1
Apr22-11, 08:29 PM
P: 9
The problem: Solve the integral equation [tex]\int[/tex][tex]\stackrel{\infty}{-\infty}[/tex]exp(-abs(x-y))u(y)dy+u=f(x) for -[tex]\infty<x<\infty[/tex].

The solutions say "Use the convolution theorem to find u(x)=f(x)-[tex]\frac{4}{3}[/tex][tex]\int[/tex]f(t)exp(-3abs(x-t))dt."

The Convolution Theorem in my book states "If the functions f(x), g(x) have Fourier transforms F(u), G(u), respectively, then the Fourier transform of [tex]\int[/tex]g([tex]\xi[/tex])f(x-[tex]\xi\[/tex])d[tex]\xi[/tex] is F(u)G(u)."

Now, I know that the Fourier transform of exp(-abs(x)) is 2/(1+u^2) but I'm not sure what to do next and I have no idea where the 4/3 comes from.

Thank you in advance for any help!
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vela
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#2
Apr23-11, 02:56 AM
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You seem to be using the letter u in two different ways. Just to clarify, is the original equation

[tex]\int_{-\infty}^\infty e^{-\lvert x-y \rvert}u(y)\,dy + u(x) = f(x)[/tex]

and the Fourier transform of g(x)=e-|x|

[tex]G(k) = \int_{-\infty}^\infty e^{-\lvert x \rvert}e^{-ikx}\,dx = \frac{2}{1+k^2}[/tex]

?

Try taking the Fourier transform of the first equation and solve for U(k).
mathmajor314
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#3
Apr23-11, 05:18 PM
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I'm sorry, the original equation should have a 4 in front of the integral sign. Also, in my book, the problem only has u instead of u(x) on the left hand side. And yes, that is the correct Fourier transform of g(x).

Taking the Fourier transform of the left hand side, I know that the Fourier transform of a sum is the sum of the Fourier transforms and I know that the convolution theorem shows that the Fourier transform of the integral is

[tex]\frac{2}{1+k^2}[/tex][tex]\int u(k)e^{-ikx}\\\\,dx[/tex]

but then I'm not sure what to do with the extra u (or maybe it's supposed to be u(x) ) on the left hand side.

After this, I don't know where the 4/3 comes from or how to get a 3 in front of the absolute value of x-t.

Thanks for your help!

vela
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#4
Apr23-11, 05:55 PM
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An Integral Equation with the Convolution Theorem for Fourier Transforms


OK, so if we denote the Fourier transform of u(x) and f(x) as U(k) and F(k) respectively, the equation becomes

[tex]4\left(\frac{2}{1+k^2}\right)U(k) + U(k) = F(k)[/tex]

Now solve for U(k).

I'm sure the second u means u(x). I was confused by the Fourier transform of e-|x| you wrote in the original post. You used u as the variable there, which doesn't make sense.
mathmajor314
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#5
Apr24-11, 12:53 AM
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Clearly, U(k) = [tex]\frac{F(k)}{1+4(\frac{2}{1+k^2})}[/tex].

Next, I think I am supposed to apply the inverse Fourier transform version of the Convolution Theorem, which states "If the functions f(x), g(x) have Fourier transforms F(k), G(k), respectively, then the inverse Fourier transform of [tex]\frac{1}{2\pi}[/tex][tex]\int F(s)G(k-s)ds[/tex] is f(x)g(x)." I'm not sure how to do this since I can't seem to figure out what the inverse Fourier transform of [tex]\frac{1}{1+4(\frac{2}{1+k^2})}[/tex] is. And I still don't know where the 3's in the solution come from... =(
Char. Limit
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#6
Apr24-11, 12:58 AM
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Quote Quote by mathmajor314 View Post
Clearly, U(k) = [tex]\frac{F(k)}{1+4(\frac{2}{1+k^2})}[/tex].

Next, I think I am supposed to apply the inverse Fourier transform version of the Convolution Theorem, which states "If the functions f(x), g(x) have Fourier transforms F(k), G(k), respectively, then the inverse Fourier transform of [tex]\frac{1}{2\pi}[/tex][tex]\int F(s)G(k-s)ds[/tex] is f(x)g(x)." I'm not sure how to do this since I can't seem to figure out what the inverse Fourier transform of [tex]\frac{1}{1+4(\frac{2}{1+k^2})}[/tex] is. And I still don't know where the 3's in the solution come from... =(
Try rearranging it. I believe you can prove that [tex]\frac{1}{1+4\left(\frac{2}{1+k^2}\right)} = \frac{k^2+1}{k^2+9}[/tex]. Might make it easier.
mathmajor314
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#7
Apr24-11, 01:09 AM
P: 9
I rearranged it and found that too but I can't figure out what to do with [tex] \int \frac{F(k) (k^2+9)}{k^2+1} dk [/tex].
vela
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#8
Apr24-11, 02:13 AM
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First, find and fix your algebra mistake.

The solution is u(x) = f(x) + stuff, right? So you need an F(k) by itself that'll inverse transform to f(x). Can you see how you might manipulate what you have to get that?
mathmajor314
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#9
Apr24-11, 02:38 PM
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Oops, it should be [tex]\frac{F(k)(1+k^2)}{9+k^2}[/tex].

I know that the inverse transform of [tex]\frac{1}{9+k^2}[/tex] is [tex]\frac{exp(-3|k|}{6}[/tex] so I guess that's where that 3 comes from. I'm still stuck on getting F(k) by itself though.
vela
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#10
Apr24-11, 02:52 PM
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Try using k2+1 = (k2+9)-8.
mathmajor314
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#11
Apr24-11, 02:58 PM
P: 9
Ugh!!! Why didn't I think of that!? Now I've got it. Thank you for all of your help; I appreciate it!


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