Partial differential equations $u_{tt}=k$ I'm trying to make sure I am understanding things correctly, which is probably not the case. But I am playing with this toy example. Let's say that I just want to find the solution the following system.
$$u_{tt}=k$$
Where $k$ is a constant. There are some initial conditions but for now lets just say the initial conditions are arbitrary functions of $x$: $u|_{t=0}=\phi(x)$ and $u_t|_{t=0}=\psi(x)$.  This is the idea of a constant external acceleration, something like that. Anyways I am pretty sure that the idea is to integrate twice, and you end up with a function like this, which is considered a solution.
$$u(t,x)=\int \int k \ dw\ dz$$
So everything depends of the bounds of this integration, which is what I am really not understanding. Is this where we need to use the initial conditions? Of course in this context we could have a function instead of $k$ but I wanted to try something very simple. Am I on the right track or is there something big I'm missing?
Thanks!
 A: You are absolutely on the right track.  To clarify, all you really need to do is remember that you should be using definite integrals here in order to precisely wield the fundamental theorem of calculus.  Indeed, if you have a solution then you can fix an $x$ and integrate in time from $0$ to $t$ to see that (using the FTC)
$$
u_t(t,x) = u_t(0,x) + \int_0^t u_{tt}(s,x) ds = \psi(x) + \int_0^t k ds = \psi(x) + tk.
$$
Hence, we now know that $u_t(t,x) = \psi(x) + tk$ for all $t,x$. Then we integrate again and use FTC again to see that
$$
u(t,x) = u(0,x) + \int_0^t u_t(s,x) dx = \phi(x) + \int_0^t (\psi(x) + tk) ds = \phi(x) + t \psi(x) + \frac{kt^2}{2}.
$$
We thus arrive at exactly the expression you would get from the constant acceleration ODE $\ddot{z}(t) = k$ upon integrating in time, namely
$$
z(t) = z(0) + \dot{z}(0) t + \frac{kt^2}{2},
$$
with the slight twist that the data are now varying with respect to the spatial variable $x$.  Of course, in both situations we assume we have a solution and then derive this form, but we can then check that this form actually does solve the equations.
