Q: Let f,g: R->R where f and g are twice differentiable. Show that u(c,t)=f(x-at)+g(x+at) is a solution of the wave equation: utt = a^2*uxx, where utt is the derivative interms of t twice and same to x.
My attempt of solving this question is to first find the first derivative of u interms of t, as follows:
Let w=x-at and v=x+at.
ut (x,t) = (df/dw)(?w/?t) + (dg/dv)(?v/?t) = f'(w)(-a) + g'(v)(a).
utt (x,t)= (?ut/?w)(?w/?t) + (?ut/?v)(?v/?t) = [f''(w)(-a)+g'(v)(a)](-a) + [f'(w)(-a) + g''(v)(a)](a) = a^2 * f''(w) - a^2 g'(v) - a^2 * f'(w) + a^2 * g''(v) = a^2 * [f''(w)-g'(v)-f'(w)+g''(v)]
ux (x,t) = (df/dw)(?w/?x) + (dg/dv)(?v/?x) = f'(w)(1) + g'(v)(1)
uxx (x,t) = (?ux/?w)(?w/?x) + (?ux/?v)(?v/?x) = [f''(w) + g'(v)](1) + [f'(w) + g''(v)](1) = f''(w)+g'(v)+f'(w)+g''(v)
But f''(w)-g'(v)-f'(w)+g''(v) ? f''(w)+g'(v)+f'(w)+g''(v), how do go from here to utt = a^2*uxx.
Thank you.
My attempt of solving this question is to first find the first derivative of u interms of t, as follows:
Let w=x-at and v=x+at.
ut (x,t) = (df/dw)(?w/?t) + (dg/dv)(?v/?t) = f'(w)(-a) + g'(v)(a).
utt (x,t)= (?ut/?w)(?w/?t) + (?ut/?v)(?v/?t) = [f''(w)(-a)+g'(v)(a)](-a) + [f'(w)(-a) + g''(v)(a)](a) = a^2 * f''(w) - a^2 g'(v) - a^2 * f'(w) + a^2 * g''(v) = a^2 * [f''(w)-g'(v)-f'(w)+g''(v)]
ux (x,t) = (df/dw)(?w/?x) + (dg/dv)(?v/?x) = f'(w)(1) + g'(v)(1)
uxx (x,t) = (?ux/?w)(?w/?x) + (?ux/?v)(?v/?x) = [f''(w) + g'(v)](1) + [f'(w) + g''(v)](1) = f''(w)+g'(v)+f'(w)+g''(v)
But f''(w)-g'(v)-f'(w)+g''(v) ? f''(w)+g'(v)+f'(w)+g''(v), how do go from here to utt = a^2*uxx.
Thank you.