Deriving d'Alemberts solution - Boundary conditions

[divB]

Hi,

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

<br /> \frac{\partial^2 u(x,t)}{\partial t^2} = \frac{\partial^2 u(x,t)}{\partial x^2}\\<br /> u(x,0) = f(x) \\<br /> u_t(x,0) = 0<br />

is

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

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

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

And the solution to this differential equation is:

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

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

My approach was: Since u(x,t) was transferred to Fourier domain via x rather than t, u(s,0) = \hat{u}(x,0). 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!

 

[divB]

Got it; finally!
It is so trivial from where I was: A=B was correct, it is a constant regarding t but not s, so A(s)=B(s).

The other "boundary" condition is u(x,0)=f(x), so when transforming back my solution: u(x,t) = A(x+t)+B(x-t) with t=0: u(x,0)=A(x)+B(x) which means that A(x)=B(x)=f(x)/2