= Solution
Separate the uniform stream and an arbitrary time-dependent gauge from each <velocity potential>: $\phi_j=U_jx+f_j(t)+\phi'_j$. Only <gradients> of the <velocity potential> determine the <velocity field>. The remote fluid must remain in its unperturbed uniform stream, so
$$
\boxed{\nabla\phi_1\to(U_1,0)\quad(z\to+\infty),\qquad
\nabla\phi_2\to(U_2,0)\quad(z\to-\infty).}
$$
After fixing the harmless gauge, the nonzero-<wavenumber> perturbation <velocity potentials> themselves vanish at the corresponding infinities. <Incompressible flow> and <irrotational flow> give the <Laplace equation> for each perturbation. For a <normal mode> with $k>0$, the vertical amplitude solves $\Phi_j''-k^2\Phi_j=0$, hence
$$
\phi'_1=A_1e^{-kz}e^{ik(x-ct)},\qquad
\phi'_2=A_2e^{kz}e^{ik(x-ct)}.
$$
The opposite exponentials are excluded because they make the perturbation grow at infinity. It is the perturbation, rather than the total potential $U_jx+f_j(t)$, that decays.
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