Fourier Transform and Shifting in Solving ODEs?

[cloud18]

Find the solution (in integral form) of the equation:

<br /> u(x+1,t) - 2u(x,t) + u(x-1,t) = u_t <br />
u(x,0) = f(x)

Hint: Use the shift formula

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

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

<br /> \exp{(-i\omega)} \overline{u} - 2\overline{u} + \exp{(i\omega)}\overline{u} = \overline{u}_t<br />

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

 

[vsage]

Why don't you think it is right? Also, that expression is not fully simplified. Maybe you were expecting something with sinusoids?

 

[cloud18]

Well I guess I get stuck on how to solve the ODE:

<br /> (\exp{(-i\omega)} - 2 + \exp{(i\omega)})\overline{u} = \overline{u}_t<br />

So (\exp{(-i\omega)} - 2 + \exp{(i\omega)}) 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):

<br /> \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}<br />