# Fourier Transform plus Shift

1. Jun 4, 2006

### cloud18

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

$$u(x+1,t) - 2u(x,t) + u(x-1,t) = u_t$$
$$u(x,0) = f(x)$$

Hint: Use the shift formula

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

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

$$\exp{(-i\omega)} \overline{u} - 2\overline{u} + \exp{(i\omega)}\overline{u} = \overline{u}_t$$

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

Last edited: Jun 4, 2006
2. Jun 4, 2006

### vsage

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

3. Jun 4, 2006

### cloud18

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

$$(\exp{(-i\omega)} - 2 + \exp{(i\omega)})\overline{u} = \overline{u}_t$$

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):

$$\frac{1}{2\pi}\int{e^{-i\omega x-2(1-\cos{\omega})t} \int{e^{i\omega\xi}f(\xi)d\xi}d\omega}$$

Last edited: Jun 4, 2006