Consider the boundary-value problem for the Laplacian $\nabla^2\phi(x,y)=0$ within a semi-infinite strip $0<x<a$, $0<y<\infty$, with the following boundary conditions
$$\partial_x\phi(0,y)=\partial_x\phi(a,y)=0$$
$$\phi(x,0)=f(x)$$
$$\lim_{y\rightarrow\infty}\phi(x,y)=0$$
The solution from separation of variables is
$$\phi(x,y)=\sum_{n=0}^\infty c_n \exp(-n\pi y/a)\cos(n\pi x/a)$$
With the $c_n$ determined from the Fourier cosine series of $f$
$$c_{n \neq 0}=\frac{2}{a}\int_0^a dx\ f(x)\cos(n\pi x/a)$$
$$c_0=\frac{1}{a}\int_0^a dx\ f(x)$$
For the special case $f(x)=const=\phi_0$, only the $n=0$ term is nonzero, but then $\phi(x,y)=\phi_0$ doesn't satisfy all the boundary conditions, so it seems the BVP is ill-posed.
Question: We have specified $\phi$ or $\partial_x \phi$ over the boundary of the region. In a purely Dirichlet or Neumann problem, that would be enough. In this case, the boundary conditions are mixed, and it appears that there are additional conditions that must be enforced (on the boundary data) to make the problem well posed. What are these conditions?
The same problem in a rectangle $0<x<a$, $0<y<b$, and BC $\phi(x,b)=0$ with $f(x)=\phi_0$ gives the solution that $\phi(x,y)=\phi_0 (1-y/b)$. In the sense that we are taking $\lim_{b\rightarrow\infty}$ for this problem, I see how end up with the solution to the first problem, yet this is not entirely satisfactory since that solution doesn't satisfy the BCs.
Best Answer
A necessary and sufficient condition for the well-posedness is $$ \int_{0}^{a}f(x)\, dx=0. $$ This is equivalent to $c_0=0$, where $c_0$ is one of the Fourier coefficients defined in the question. Note that the condition $c_0=0$ is needed to ensure that $\lim_{y\to \infty}\phi(x,y)=0$. In particular, this excludes the case $f=\text{const.}$ mentioned by Sal and A rural reader. The reason why it is sufficient is because your $\phi$ solves the problem.
Now, let me explain some background to this which is related to what Bob Terrell wrote. For this purpose, let us for the moment consider a Neumann boundary condition $$ (\partial_y \phi)(x,0)=h(x) $$ instead of the Dirichlet one. Then it is well-known (see https://doi.org/10.1016/S0065-2156(08)70244-8 for example) that the problem is well-posed (unique existence of the solution decaying as $y\to \infty$) if and only if $h(x)$ is "self-equibrated", that is, $$ \int_{0}^{a}h(x)\, dx=0. $$ This condition resolves the essential concern raised by Bob Terrell about the flux. [N.B.: The term "self-equibrated" is used in the field of elasticity. You might also be interested in what is called "Saint-Venant's principle" (https://en.wikipedia.org/wiki/Saint-Venant's_principle).] Now, let us return to the Dirichlet case. Denote by $h(x)=(\partial_y \phi)(x,0)$ the (local) flux at the bottom. Then from your expression, we have $$ h(x)=-\frac{\pi}{a}\sum_{n=1}^{\infty}nc_n \cos\left( \frac{n\pi x}{a} \right). $$ This gives an expression of the Neumann data in terms of the Dirichlet data, i.e., the Fourier coefficients $(c_n)_{n=1}^{\infty}$ of $f$. Then the (total) flux at the bottom is automatically zero: $$ \int_{0}^{a}h(x)\, dx=0. $$ This is because $\int_{0}^{a}\cos(n\pi x/a)\, dx=0$. As we mentioned above, this ensures that the problem is well-posed.
However, if the region is not rectangular, it becomes difficult to express the Neumann data $h(x)$ in terms of the Dirichlet data $f(x)$. And in such cases, the condition $\int_{0}^{a}f(x)\, dx=0$ might not be sufficient (although I cannot give any example right now), whereas the condition corresponding to $\int_{0}^{a}h(x)\, dx=0$ ensures that the problem is well-posed even for regions not necessarily rectangular.