Unknown time-dependent boundary data for a BVP involving the wave equation Suppose on we have the following boundary value problem on $(0,1) \in \mathbb{R}$:
\begin{align}
&\nabla^2 u = u_{tt} \ \ \text{(wave equation),}  \\[0.6em] &u(0,t) = u(1,t) = f(t),  \\[0.6em] &u(x,0) = f(0) > 0,  \\[0.6em] &u_t(x,0) = c,
\end{align}
where $f(t)$ is an unknown function of time, and $c$ is a known arbitrary constant. I am interested in understanding conditions on when a solution to this system can be solved in general, when $f(t)$ is unknown a priori.
My suspicion is that the BVP can only be solved only when $f(t) = f$ is constant, for one can then solve this problem by setting $v = u-f$ and performing an eigenbasis expansion, separating variables, et cetera. However when $f$ depends on time, this method doesn't work due to not being able to determine the time dependent coefficients in the basis expansion.
I would be grateful if someone could help me further understand my suspicion, or possibly inform me that this system belongs to a class of BVP that are ill-posed, or something rather.
 A: The problem is well posed, and is discussed for example
in F. John's PDE book, Springer, p.40.
One way of briefly thinking of the solution is as follows.
Write $u(x,t) = f(0)+ct+v(x,t)$. Then $v$ satisfies the wave
equation with zero initial values, and at the ends
$$
 v(0,t) = v(1,t) = f(t)-f(0)-ct.
$$
Lets denote that as $f_1(t)$.
We can describe $v$ as a sum
of traveling waves.
Since the initial values of $v$
are zero, $v$ remains zero until the influence of the boundary
values is felt, that is in the triangle with vertices
$(x,t) = (0,0), (1,0), (1/2,1/2)$.
We let the left boundary value $f_1(t)$
propagate to the right as
$$
 f_1(t-x),
$$
and by symmetry we can add the term
$$
 f_1(t-1+x)
$$
that propagates the right boundary value toward the left.
The sum of these,
$$
 v(x,t) = f_1(t-x)+f_1(t-1+x),
$$
together with zero in the stated triangle,
solves the problem within a certain region. Namely
all the points with $t < x+1$ and $t < 2-x$. You can't go higher
with it because the wave $f_1(t-x)$ has the wrong boundary value
at $x = 1$. It gives $f_1(t-1)$, and what you wanted was
$f_1(t)$. To correct this you use the linearity of the wave
equation to add a new term propagating to the left from
the boundary value
$$
 f_1(t)-f_1(t-1)
$$
in the interval $1 < t < 2$.
Similarly add a reflection of this propagating from the left
end toward the right. Now your solution works for one more
$t$ step and you can continue in this manner for all time.
It looks like the solution will be continuous since $f_1(0) = 0$.
And will be differentiable if $f$ is differentiable and $f'(0) = c$. See F. John for
discussion of weak solutions and a more complete discussion
of the problem.
