- #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...

[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...

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