= Solution
Here write $\sigma$ for the <buoyancy perturbation> about the resting constant-$N$ <stratification>; the total <buoyancy> also includes its background <gradient> $N^2$. Use kinematic perturbation <pressure> $p$. The weak-forcing <Linearized Boussinesq equations> are
$$
u_t=-p_x+F,\qquad w_t=-p_z+\sigma,\qquad\sigma_t=-N^2w,\qquad u_x+w_z=0.
$$
Weakness means that perturbation advection is small compared with the retained linear terms. With the specified <streamfunction> convention, the transverse <vorticity> is
$$
\eta=u_z-w_x=\psi_{zz}+\psi_{xx}=\nabla^2\psi.
$$
Differentiate the horizontal <momentum> equation in $z$ and subtract the $x$ derivative of vertical <momentum>. The <pressure> derivatives cancel, giving
$$
\boxed{\nabla^2\psi_t=F_z-\sigma_x,\qquad\nabla^2\psi_{tt}+N^2\psi_{xx}=F_{zt}.}
$$
The second equation uses $\sigma_t=N^2\psi_x$ and is the forced internal-wave equation.
For free waves at <frequency> $|\omega|\ll N$, the <internal gravity wave> dispersion relation gives $|k|/|m|\simeq|\omega|/N\ll1$. Thus the horizontal response scale is much larger than the vertical scale. Vertical acceleration is small in the vertical <momentum> equation, leaving the <hydrostatic approximation> $p_z=\sigma$. Correspondingly the transverse <vorticity> is predominantly $u_z$: neglect $-w_x=\psi_{xx}$ relative to $\psi_{zz}$ in the inertial <vorticity> term. The restoring term $N^2\psi_{xx}$ must still be retained, since it balances the small time derivatives.
For $m\ne0$ put $\psi=e^{imz}P(x,t)$, $\sigma=e^{imz}B(x,t)$ and interpret physical fields as real parts. The reduced equation is
$$
\boxed{P_{tt}-c^2P_{xx}=-\frac{i}{m}f_t,\qquad c=\frac N{|m|}.}
$$
One can solve it while obtaining the <buoyancy> simultaneously. Define $U=imP$ and $H=iB/(cm)$. The first-order hydrostatic equations become
$$
U_t=f+cH_x,\qquad H_t=cU_x.
$$
Hence $R_+=U+H$ and $R_-=U-H$ obey
$$
(\partial_t-c\partial_x)R_+=f,\qquad(\partial_t+c\partial_x)R_-=f.
$$
Apply the <method of characteristics> with zero fields in the past. With
$$
I_\pm(x,t)=\int_{-\infty}^t f\bigl(x\mp c(s-t),s\bigr)\,ds,
$$
the solution is $R_\pm=I_\pm$. Consequently the <hydrostatic response to a localized horizontal force> is
$$
\boxed{\psi=-\frac{ie^{imz}}{2m}(I_++I_-),\qquad
\sigma=-\frac{icme^{imz}}2(I_+-I_-).}
$$
These are exactly the required two characteristic integrals, derived with their relative sign fixed by the <buoyancy> equation. Quiet-past initial conditions exclude additional freely propagating solutions.
For the consistency check, set $\tau=s-t$, so both integrals run over the fixed range $(-\infty,0)$. Differentiation at fixed $x,z$ now gives
$$
\boxed{w_t=\frac{ie^{imz}}{2m}\int_{-\infty}^0
\left[f_{xt}(x-c\tau,t+\tau)+f_{xt}(x+c\tau,t+\tau)\right]d\tau.}
$$
This change of variable is important: it includes the time dependence of both the sampled forcing and the moving integration geometry without omitting a term. Each straight path intersects $|x|\leq a$ for a time interval of length at most $2a/c$. If $L=\sup|f_{xt}|$, then
$$
|w_t|\leq\frac{2a}{|m|c}L=\frac{2a}{N}L.
$$
The characteristic size of the hydrostatic <buoyancy> and pressure-gradient balance is $|m|af_{\max}$, as follows from the analogous bound on $B$ in its integral formula. Thus the omitted acceleration has relative scale at most a constant times
$$
\frac{L}{|m|Nf_{\max}}\ll1.
$$
Together with the low-frequency assumption, this verifies the requested hydrostatic consistency. It is a comparison of characteristic magnitudes, rather than division by a <buoyancy> value at a node where both terms may vanish. The capital $N$ here is the <buoyancy> <frequency>, as printed in the PDF.
For $t\geq0$, only $0\leq s\leq t$ contributes. If $|x|>ct+a$, every point $x\mp c(s-t)$ lies outside the source interval, so both integrals vanish:
$$
\boxed{\psi=\sigma=0\quad\text{for }|x|>ct+a.}
$$
This finite propagation cone belongs to the reduced hydrostatic wave equation.
The hypotheses do not require an even forcing, so no single symmetric graph follows for every permitted $f$. For an illustrative smooth even nonnegative profile, $U$ is even and $H$ is odd. At late time and fixed $x$, if $A_f=\int_{-a}^af(r,\infty)\,dr$, then
$$
U\longrightarrow\frac{A_f}{2c},\qquad
H\longrightarrow\frac1{2c}\left[\int_x^\infty f(r,\infty)\,dr-\int_{-\infty}^x f(r,\infty)\,dr\right].
$$
Thus the <streamfunction> develops a broad nearly level interior response with outgoing fronts, while the <buoyancy> changes sign across the source and approaches opposite interior plateaus to its left and right. Both vanish beyond the fronts. For $m>0$, a vertical phase with $\sin(mz)=1$ gives real fields $\psi=U/m$ and $\sigma=cmH$. The figure plots those profiles up to positive normalization factors, for an even compact forcing increasing slowly from zero.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-77-forced-response.png]
{title=Causal streamfunction and buoyancy profiles for a slowly established even localized horizontal force}
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