- #1
mathmajor314
- 9
- 0
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!
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!