Fourier Transform plus Shift

  • Thread starter cloud18
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  • #1
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Find the solution (in integral form) of the equation:

[tex]
u(x+1,t) - 2u(x,t) + u(x-1,t) = u_t
[/tex]
[tex]u(x,0) = f(x)[/tex]

Hint: Use the shift formula

[tex]
F[f(ax-b)] = \frac{\exp{i\omega b/a}}{|a|} \overline{f}(\omega/a)
[/tex]

So I took the Fourier transform of each term using the shift formula:

[tex]
\exp{(-i\omega)} \overline{u} - 2\overline{u} + \exp{(i\omega)}\overline{u} = \overline{u}_t
[/tex]

But I don't think this is correct thus far...
 
Last edited:

Answers and Replies

  • #2
vsage
Why don't you think it is right? Also, that expression is not fully simplified. Maybe you were expecting something with sinusoids?
 
  • #3
8
0
Well I guess I get stuck on how to solve the ODE:

[tex]
(\exp{(-i\omega)} - 2 + \exp{(i\omega)})\overline{u} = \overline{u}_t
[/tex]

So [tex](\exp{(-i\omega)} - 2 + \exp{(i\omega)})[/tex] can be treated as a constant with respect to t. But when I solve I get exponents raised to exponents...


The answer is suppose to be (limits of integration -inf to +inf):

[tex]
\frac{1}{2\pi}\int{e^{-i\omega x-2(1-\cos{\omega})t} \int{e^{i\omega\xi}f(\xi)d\xi}d\omega}
[/tex]
 
Last edited:

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