= Solution
Take eastward $x$, upward $z$, positive <wavenumbers> $k,K$, and a real instantaneous amplitude $A$. A common growing factor does not affect the <phase> geometry. Put $C=\cosh(Kz)$ and $S=\sinh(Kz)$, using the <hyperbolic cosine> and <hyperbolic sine>. The prescribed <streamfunction> is
$$
\psi'=A[C\cos(kx)-\alpha S\sin(kx)]
=AR_\psi\cos[kx+\delta_\psi(z)],
$$
where
$$
R_\psi^2=C^2+\alpha^2S^2,\qquad
\delta_\psi=\arctan[\alpha\tanh(Kz)].
$$
Following a fixed <phase> gives
$$
\frac{dx_\psi}{dz}=-\frac{\delta_{\psi,z}}k
=-\frac{\alpha K}{k(C^2+\alpha^2S^2)}<0.
$$
Therefore the <streamfunction> pattern \b[tilts westward with height]. The lower boundary leads the upper boundary eastward.
In the conventional <quasi-geostrophic streamfunction> normalization, <hydrostatic balance> gives $b'=f_0\psi'_z$. Therefore
$$
b'=f_0AK[S\cos(kx)-\alpha C\sin(kx)].
$$
For the <phase> calculation, multiplication by the constant $f_0$ is immaterial. Define a continuous <phase> $\delta_b=\operatorname{atan2}(\alpha C,S)$; this avoids a spurious discontinuity at $z=0$. Then
$$
R_b^2=S^2+\alpha^2C^2,\qquad
\delta_{b,z}=-\frac{\alpha K}{S^2+\alpha^2C^2},
\qquad
\boxed{\frac{dx_b}{dz}=\frac{\alpha K}{k(S^2+\alpha^2C^2)}>0.}
$$
The <buoyancy perturbation> pattern \b[tilts eastward with height], opposite to the <streamfunction> pattern. At $z=0$, $\psi'=A\cos(kx)$ while $b'=-f_0AK\alpha\sin(kx)$, so the two fields are a quarter <wavelength> apart there.
Both perturbation amplitudes are largest at the boundaries. Indeed,
$$
R_\psi^2=1+(1+\alpha^2)\sinh^2(Kz),\qquad
R_b^2=\alpha^2+(1+\alpha^2)\sinh^2(Kz),
$$
which increase with $|z|$. The physical reason is particularly clear through <stratified quasi-geostrophic inversion>: the <Eady model> has no interior background <potential-vorticity gradient>, and a growing <normal mode> has zero interior <potential vorticity> anomaly. The independent thermal anomalies reside at the rigid upper and lower boundaries, where meridional motion advects the basic horizontal <buoyancy gradient>. Each <Eady edge wave> induces an interior balanced flow that decays away from its boundary on scale $K^{-1}$. Their interacting, phase-locked combination produces the growing <baroclinic instability>. Boundary confinement of the thermal sources, rather than a rigid-wall requirement $\psi'=0$, explains the amplitude pattern.
For a sketch with a quarter-<wavelength> difference between the boundary <streamfunction> patterns, choose
$$
2\arctan[\alpha\tanh(KD)]=\frac\pi2,
\qquad \alpha\tanh(KD)=1.
$$
An upper positive <streamfunction> crest is then $\lambda_x/4$ west of its lower counterpart, where $\lambda_x=2\pi/k$. The <buoyancy perturbation> crests slope the other way. The quarter-<wavelength> boundary assumption is used for the <streamfunction>; differentiating in $z$ fixes the <buoyancy perturbation> and does not generally give it the same boundary <phase> separation.
\Image[/past-exam-of-the-geophysical-fluid-dynamics-course-of-the-university-of-oxford/2023-baroclinic-wave-structure.png]
{title=Original <Eady model> sketch: <streamfunction> crests tilt westward and <buoyancy perturbation> crests eastward. Dashed curves follow positive maxima; the <streamfunction> boundary <phase> difference is a quarter <wavelength>.}
The connection to poleward <heat transport> can be checked without relying on the sketch. The meridional <geostrophic flow> and <buoyancy perturbation> are
$$
v'=\psi'_x=-Ak[C\sin(kx)+\alpha S\cos(kx)],
\qquad
b'=f_0AK[S\cos(kx)-\alpha C\sin(kx)].
$$
Average over one zonal <wavelength>, using equal averages $1/2$ for squared <sine> and <cosine> and zero for their product:
$$
\boxed{\overline{v'b'}=\frac{f_0A^2kK\alpha}{2}(C^2-S^2)
=\frac{f_0A^2kK\alpha}{2}.}
$$
In the Northern Hemisphere this is positive: warm anomalies preferentially move northward and cold anomalies southward. For the same eastward-increasing basic <thermal wind> in the Southern Hemisphere, $f_0<0$ and the basic <buoyancy gradient> reverses; the negative northward <eddy buoyancy flux> is again poleward. For <thermal expansion coefficient> $\alpha_T>0$ and <specific heat capacity at constant pressure> $c_p$, the corresponding <sensible heat> flux in a thermal <Boussinesq approximation> is $\rho_0c_p\overline{v'T'}=\rho_0c_p\overline{v'b'}/(g\alpha_T)$. The growing amplitude makes this flux grow as $A^2$. The tilt supplies the correlation absent in an untilted disturbance and reduces the background <available potential energy>.
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