So the semi-infinite strip means that we are working in the region where $(x,y)\in [0,\infty)\times [0,H]$. We have Laplace's equation, which simply states:
$$\nabla^{2} \Phi = 0 \iff \frac{\partial^{2} \Phi}{\partial x^{2}}+\frac{\partial^{2} \Phi}{\partial y^{2}}=0$$
And we have $\Phi(0,y)=f(y)$ and $\lim_{x\to\infty}\Phi(x,y) = 0$ (from physical conditions) as our boundary conditions, along with:
$$\left.\frac{\partial \Phi}{\partial y}\right|_{(x,y)=(x,0)} = \left.\frac{\partial \Phi}{\partial y}\right|_{(x,y)=(x,H)}=0$$
We note that using the method of characteristics we can assume that: $\Phi(x,y)=X(x)Y(y)$, and therefore we can rewrite the Laplace equation above:
$$Y(y)\frac{\partial^{2} X}{\partial x^{2}}+X(x)\frac{\partial^{2} Y}{\partial y^{2}}=0 \implies \frac{1}{X(x)}\frac{\partial^{2} X}{\partial x^{2}}=-\frac{1}{Y(y)}\frac{\partial^{2} Y}{\partial y^{2}}$$
But $x$ and $y$ are independent, and therefore we must have that:
$$\frac{1}{X(x)}\frac{\partial^{2} X}{\partial x^{2}}=k^{2} \land \frac{1}{Y(y)}\frac{\partial^{2} Y}{\partial y^{2}}=-k^{2}$$
These are ODEs with known solutions:
$$X(x)=\alpha \exp(kx)+\beta\exp(-kx) \land Y(y) = \gamma \sin(ky) + \delta \cos(ky)$$
Where $\exists \alpha,\beta,\gamma,\delta\in\mathbb{R}$. We now wish to find the value of these coefficients based on the initial conditions. We can see that our second boundary condition, $\lim_{x\to\infty}\Phi(x,y)=0$ means that $X(x)$ must have a decaying form, thus $\alpha = 0$. We can now use the differential boundary conditions:
$$\left.\frac{\partial \Phi}{\partial y}\right|_{(x,y)=(x,0)}=0 \implies \gamma k \cos(0) - \delta k \sin(0) = 0$$
Therefore we have $\gamma k = 0$, and therefore $\gamma = 0$ (if we assume $k \neq 0$ for non-static solution). We can now look at the second differential boundary condition:
$$\left.\frac{\partial \Phi}{\partial y}\right|_{(x,y)=(x,H)}=0 \implies \delta k \sin(k H)=0$$
But assuming $\delta \neq 0$ (for non-trivial solution), we must have $k = \frac{n \pi}{H}$, $\exists n \in \mathbb{N}_{0}$.
We finally turn to our last boundary condition $\Phi(0,y)=f(y)$, where $f(y)$ is an arbitrary function. We can write this as:
$$\sum_{n = 0}^{\infty} \delta_{n} \beta_{n} \sin(k_{n} y) = f(y)$$
But WLOG we can assume $\beta_{n} = 1$, $\forall n \in \mathbb{N}_{0}$ therefore we have:
$$\sum_{n=0}^{\infty}\delta_{n} \sin(k_{n} y) = f(y)$$
We can recognize this as a Fourier sine series, thus giving us coefficients:
$$\delta_{n} = \frac{2}{H}\int_{0}^{H} f(y) \sin\left(\frac{n \pi y}{H}\right)\:\mathrm{d}y\tag{1}$$
So we have in general:
$$\Phi(x,y) = \sum_{n=0}^{\infty}\delta_{n}\exp\left(-\frac{n\pi x}{H}\right)\sin\left(\frac{n \pi y}{H}\right)$$
Where $\delta_{n}$ is defined by $(1)$.
I hope this helps clear things up!
Best Answer
Yes, there is one condition missing, as your semi-infinite strip has, very simply put, four boundary segments, at $x \in \{0,1\}$ and at $y \in \{0,\infty\}$. But you have prescribed conditions on only three of the boundary segments so far, which leaves you with arbitrary constants in the solution (you obtain $C_n + D_n = a_n$ where only $a_n$ is known).
The boundedness of the solution as $y \rightarrow \infty$ usually follows from some physical considerations.