I've spent the last day trying to work out a solution for this question:
If u = f(x+ct) + g(x-ct) where f,g are twice-differentiable functions of 1 variable and c is a constant, show that $\partial^2 u / \partial t^2$ = $c^2 \partial ^2u / \partial x^2$
I end up getting: $\partial^2 u / \partial t^2$ = $c^2 (\partial ^2u / \partial g^2)(g'(x-ct))^2 -2c^2(\partial ^2u/\partial g \partial f) * f'(x+ct)*g'(x-ct) + c^2(\partial ^2u/\partial f^2)*(f'(x+ct))^2 $
and
$c^2\partial ^2u/\partial x^2 = c^2(\partial ^2u/\partial g^2)*(g'(x-ct))^2 + 2c^2(\partial ^2u/\partial g \partial f)*(f'(x+ct))*(g'(x-ct)) + c^2(\partial ^2u/\partial f^2)*(f'(x+ct))^2 $
the minus term in Utt seems to prevent the two sides from cancelling out. Only way I can see it working is if $\partial ^2u/\partial g \partial f = 0 $.
Anyone have any ideas where I went wrong?
Thanks
Your differentiaon is not correct in convetional way.
Let $u(x, t) = f(x + c t) + g(x -ct)$, then $$\frac{\partial u}{\partial x} = f'(x +ct) + g'(x -ct)$$ $$\frac{\partial^2 u}{\partial x^2} = f''(x +ct) + g''(x - ct)$$ $$\frac{\partial u}{\partial t} = cf'(x +ct) - c g'(x - ct)$$ $$\frac{\partial^2 u}{\partial t^2} = c^2f''(x +ct) + c^2 g''(x - ct) = c^2 \frac{\partial^2 u}{\partial x^2}$$