= Solution
Let $u=\phi_1-\phi_2$. It is a <harmonic function> in the bounded region and vanishes on its entire boundary. For smooth solutions and boundary, the <divergence theorem> applied to $u\nabla u$ gives the energy identity
$$
\int_R|\nabla u|^2dV
=\int_Su\,\partial_nu\,dS-\int_Ru\nabla^2u\,dV=0.
$$
The integrand is nonnegative, so $\nabla u=0$ in $R$. Thus $u$ is constant on each connected component, and its zero <Dirichlet boundary condition> makes every constant zero. Hence
$$
\boxed{\phi_1=\phi_2}.
$$
This proves the <Uniqueness of the Dirichlet problem> rather than assuming it.
For the separated expression, write $\phi=F(x)G(y)$. The hyperbolic factor obeys $F''=\lambda^2F$ and the trigonometric factor obeys $G''=-\lambda^2G$, so
$$
\phi_{xx}+\phi_{yy}=F''G+FG''=0
$$
for all the constants, including $\lambda=0$. In the half-strip take $a>0$ so the region has positive width. For every nonzero integer $n$, the decaying <harmonic sine mode in a half-strip> is
$$
\boxed{\phi(x,y)=e^{-|n|\pi x/a}\sin\left(\frac{n\pi y}{a}\right)}.
$$
Its second derivatives cancel, it vanishes at $y=0,a$, and it has the prescribed sine trace at $x=0$. The absolute value in the exponential ensures decay for negative as well as positive $n$. For $n=0$, the zero function is the solution. Thus all integer cases are covered, and the decay is uniform across the strip.
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