Just to avoid typing Greek letters, let u(x,0)=p(x), ut(x,0) = q(x).
The proposed solution u(x,t)=p(x)+t·q(x) doesn't check, since utt(x,t) ≡ 0, while uxx(x,t) = p’’(x)+t·q’’(x).
The equation utt = uxx is the wave equation with velocity 1, so its general solution is u(x,t) = af(x+t) + bg(x-t) for arbitrary functions f and g, and constants a, b.
Add the initial conditions:
p(x) = u(x,0) = af(x) + bg(x), (1)
q(x) = ut(x,0) = af’(x) - bg’(x). (2)
Derivative of (1): p’(x) = af’(x) + bg’(x). solve with (2) to get
af’(x) = (1/2) [p’(x) + q(x)],
bg’(x) = (1/2) [p’(x) - q(x)].
Let h(x) be the integral of q(x). so that h’(x) = q(x). Then
af(x) = (1/2) [p(x) + h(x)],
bg(x) = (1/2) [p(x) - h(x)].
Then
u(x,t) = af(x+t) + bg(x-t) = (1/2) [p(x+t) + h(x+t) + p(x-t) - h(x-t)], where h'(x) = q(x).
Check:
u(x,0) = (1/2) [p(x) + h(x) + p(x) - h(x)] = p(x).
ut(x,t) = (1/2) [p’(x+t) + h’(x+t) – p’(x-t) + h’(x-t)],
ut(x,0) = (1/2) [p’(x) + h’(x) – p’(x) + h’(x)] = h’(x) = q(x).
Lou
