Expand $g$ in its <Fourier sine series>
$$
g(y)=\sum_{n=1}^{\infty}g_n\sin(k_ny),
\qquad
k_n=\frac{n\pi}{a},
\qquad
g_n=\frac2a\int_0^a g(y)\sin(k_ny)\,dy.
$$
By <separation of variables>, the decaying solution of <Laplace equation> in each half-strip has $n$th mode proportional to $e^{-k_n|x|}\sin(k_ny)$. Continuity at $x=0$ and integration of the equation across the <Dirac delta function> require the normal-derivative jump
$$
\phi_x(0^+,y)-\phi_x(0^-,y)=g(y).
$$
The jump of $C_ne^{-k_n|x|}$ is $-2k_nC_n$, so
$$
\boxed{\phi(x,y)=-\sum_{n=1}^{\infty}\frac{g_n}{2k_n}e^{-k_n|x|}\sin(k_ny)}.
$$

For $g(y)=\delta(y-c)$, its Fourier coefficients are
$$
g_n=\frac2a\sin\frac{n\pi c}{a}.
$$
Thus
$$
\phi(x,y)=-\frac1\pi\sum_{n=1}^{\infty}
\frac1n e^{-n\pi|x|/a}
\sin\frac{n\pi y}{a}\sin\frac{n\pi c}{a}.
$$
Translation in $x$ gives the <Dirichlet Green function for an infinite strip> with source $(b,c)$:
$$
\boxed{G(x,y;b,c)=-\frac1\pi\sum_{n=1}^{\infty}
\frac1n e^{-n\pi|x-b|/a}
\sin\frac{n\pi y}{a}\sin\frac{n\pi c}{a}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ib / Paper 1 / 14B / Solution

On $\mathbb R\times(0,a)$, separation into sine modes gives the Dirichlet Green function for the Laplacian
$$
G(x,y;b,c)=-\frac1\pi\sum_{n=1}^\infty
\frac{e^{-n\pi|x-b|/a}}n
\sin\frac{n\pi y}{a}\sin\frac{n\pi c}{a}.
$$


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 Incoming links: Dirichlet Green function for an infinite strip