I've been tasked with solving the following IBVP for arbitrary $\alpha\in\mathbb{R}$: $$ \begin{cases} \displaystyle\frac{\partial^2 u}{\partial t^2} = c^2\frac{\partial^2 u}{\partial x^2} - g & 0 < x < \pi, t > 0 \\ u(0, t) = u(\pi, t) = 0\quad & t\ge 0 \\ u(x, 0) = \alpha x(\pi - x) & 0 \le x \le \pi \\ \displaystyle\frac{\partial u}{\partial t}(x, 0) = 0 & 0 \le x \le \pi \end{cases} $$ Now, through a string of homogenizing the initial conditions and applying Duhamel's principle I was able to solve for $u$ (hopefully correctly) as $$ u(x, t) = \alpha x(\pi - x) - \frac{8}{\pi}\left(2\alpha + gc^{-2}\right)\sum_{k\ge1,\text{ odd}}\frac{\sin(kx)\sin^2(kct/2)}{k^3} $$
Now, the part I'm struggling with is I've been asked to determine whether or not the IBVP is well posed for $\alpha = \frac{g}{2c^2}$. For this value of $\alpha$, which we will denote $\alpha_0$, it's apparent that the sum in the above formula for $u(x, t)$ is annihilated by the factor of $2\alpha_0 + gc^{-2} = 0$, and what we're left with is something which doesn't depend on $t$ (the equilibrium solution).
In order for the problem to be well-posed, a solution must exist (which one does), it must be unique (ummm...), and it must depend continuously upon the initial conditions (what?).
Given the nature of the course (for which this question is a homework assignment, but I'm not asking for someone else to do my homework, merely some direction), I suspect it's unlikely we would be required to show that the solution depends continuously upon the initial conditions, so my guess is that the problem is not well posed. However, I can't really figure out why?
The way I arrived at the solution $u(x, t)$ above, it wasn't "quite" a string of logical implications leading from "suppose $u$ solves this IBVP" to "this is the formula for $u$", so I'm not positive about uniqueness. That said, for $\alpha = \alpha_0$, I can't figure out any other solutions $u$. If push comes to shove, I suppose I could try an argument where if $u_1, u_2$ solve the IBVP, then $\psi = u_1 - u_2$ solves a slightly different IBVP with homogeneous boundary conditions, and try to use this to show that $\psi = 0$, giving us uniqueness, but I didn't want to dive into such a long argument without exhausting my options first.
Supposing the solution is unique, where do I even begin showing that it depends continuously upon the initial conditions? Is the implication simply that the solution I derived above is "continuous in $\alpha$" (small changes in $\alpha$ correspond to small changes in $u$ with respect to, say, the $L_2$ norm or the $L_\infty$ norm)? Or does "depends continuously on the initial conditions" require I vary the other boundary conditions/initial conditions as well? In this case I would have to let them be arbitrary functions $f(t)$ and $g(x)$, which seems like overkill.
Am I missing something? Is the solution even unique for $\alpha = \alpha_0$? If so, how would I show that the solution does/doesn't depend continuously on the initial conditions?