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