- #1

divB

- 87

- 0

I shall show (using Fourier transform) that the solution to

[tex]

\frac{\partial^2 u(x,t)}{\partial t^2} = \frac{\partial^2 u(x,t)}{\partial x^2}\\

u(x,0) = f(x) \\

u_t(x,0) = 0

[/tex]

is

[tex]

u(x,t) = (f(x+t) + f(x-t))/2

[/tex]

I got it almost: Taking the Fourier transform in the variable x (leaving t constant) and the derivative theorem gives:

[tex]

\hat{u}_{tt}(s,t) + 4\pi^2 s^2 \hat{u}(s,t) = 0

[/tex]

And the solution to this differential equation is:

[tex]

\hat{u}(s,t) = A e^{j2\pi s t} + B e^{-j2\pi s t}

[/tex]

However, now I am stuck in incorporating the boundary conditions and I do not see why A and B should become [itex]f(x)=u(x,0)[/itex].

My approach was: Since u(x,t) was transferred to Fourier domain via x rather than t, [itex]u(s,0) = \hat{u}(x,0)[/itex]. Derivating the solution for t and filling t=0 gives A=B. But I neither see the factor of 1/2 there, nore why A and B should be f(x). Can anyone give me a pointer?

Thanks!