= Solution
For a stationary film of uniform thickness, <thermal conduction> gives $T_{zz}=0$, so $T=T_0+cz$. The cooling boundary condition determines $c$:
$$
-\kappa c=\alpha(c h+\Delta T),\qquad
\boxed{T(z)=T_0-\frac{\alpha\Delta T}{\kappa+\alpha h}z.}
$$
Consequently the surface <temperature> is
$$
T_s=T_0-\Delta T\frac{\alpha h}{\kappa+\alpha h}
=T_0-\Delta T\frac{\alpha h}{\kappa}
+O\!\left(\Delta T\left(\frac{\alpha h}{\kappa}\right)^2\right).
$$
This is the <quasistatic temperature of a conductively cooled film>, with a weak-cooling expansion when $\alpha h/\kappa\ll1$.
Let the thickness and longitudinal scales be $h_*,L$, with $\epsilon=h_*/L\ll1$. <Incompressibility> gives transverse speed of order $\epsilon U$, and film evolution has the kinematic time scale $L/U$. For the <temperature> variation scale $\Theta$, longitudinal <diffusion> divided by transverse <diffusion> is $h_*^2/L^2=\epsilon^2$. Each advective term, and the time derivative on that evolution scale, divided by transverse <diffusion> is
$$
\frac{U\Theta/L}{\kappa\Theta/h_*^2}
=\frac{Uh_*^2}{\kappa L}=Pe.
$$
Thus $\epsilon^2\ll1$ and $Pe\ll1$ make the leading thermal profile locally linear and quasistatic. Retaining the weak-cooling condition gives the same surface-temperature approximation with the local $h(x,t)$. For any independently imposed faster time variation one would also require its thermal relaxation ratio $h_*^2/(\kappa t_{\mathrm{evol}})$ to be small.
Define the positive surface-tension gradient coefficient
$$
B=-\frac{\gamma'\Delta T\alpha}{\kappa}>0.
$$
The weak-cooling approximation gives $\gamma_s\simeq\gamma_0+Bh$, so the <Marangoni stress> is $\gamma_{s,x}=Bh_x$. The <Young–Laplace equation> gives the leading liquid <pressure> $p=p_a-\gamma_0h_{xx}$, with constant ambient <pressure>. In <lubrication theory>, $p_z=0$ and $\mu u_{zz}=p_x$. The <no-slip boundary condition> at the wall and the tangential surface condition $\mu u_z(h)=\gamma_{s,x}$ give
$$
u(z)=\frac{p_x}{\mu}\left(\frac{z^2}{2}-hz\right)
+\frac{\gamma_{s,x}}\mu z.
$$
Hence the <volume flux> is
$$
Q=\int_0^hu\,dz=-\frac{h^3p_x}{3\mu}+
\frac{h^2\gamma_{s,x}}{2\mu}
=\frac{\gamma_0h^3h_{xxx}}{3\mu}+
\frac{Bh^2h_x}{2\mu}.
$$
<Conservation of mass> gives the <thermocapillary thin-film equation>
$$
\boxed{h_t+\partial_x\left(
\frac{\gamma_0h^3h_{xxx}}{3\mu}
-\frac{\gamma'\Delta T\alpha}{2\kappa\mu}h^2h_x\right)=0.}
$$
The replacement of <surface tension> by $\gamma_0$ in the capillary term needs $Bh_*/\gamma_0\ll1$. For comparable capillary and thermocapillary fluxes,
$$
\frac{Q_{\mathrm M}}{Q_{\mathrm C}}
\sim\frac{Bh_*}{\gamma_0\epsilon^2}=O(1),
$$
so the fractional tension change is only $O(\epsilon^2)$. Both the tension correction to $\gamma h_{xxx}$ and the differentiated-tension contribution $\gamma_xh_{xx}$ are then higher-order. Weak cooling alone would not justify that replacement if $|\gamma'|\Delta T/\gamma_0$ were allowed to grow without bound.
The capillary term smooths the film: a crest has negative <curvature> and higher liquid <pressure>, causing flow away from it. The thermal term has the opposite effect. A thicker region has a colder surface and therefore higher <surface tension>; <Marangoni stress> pulls surface fluid toward that region, reinforcing the thickness disturbance. These mechanisms give fourth-order damping and second-order growth respectively.
Use $H=h/\widehat h$, $X=\epsilon x/\widehat h$, and $\tau=t/\widehat t$. The two dimensionless flux-divergence coefficients are $B\epsilon^2\widehat t/(2\mu)$ and $\gamma_0\epsilon^4\widehat t/(3\mu\widehat h)$. Setting both equal to one gives the <balanced scales for a thermocapillary film>:
$$
\boxed{\epsilon^2=\frac{3B\widehat h}{2\gamma_0}
=-\frac{3\gamma'\Delta T\alpha\widehat h}{2\kappa\gamma_0},\qquad
\widehat t=\frac{4\mu\gamma_0}{3B^2\widehat h}
=\frac{4\mu\gamma_0\kappa^2}{3(\gamma'\Delta T\alpha)^2\widehat h}.}
$$
These choices yield $H_\tau+(H^2H_X)_X+(H^3H_{XXX})_X=0$. The predicted $\epsilon$ must be small for the assumed long-wave approximation.
Linearize about $H=1$ using a <normal mode> of amplitude $\delta$. The resulting equation is $s-k^2+k^4=0$, so the <dispersion relation> is
$$
\boxed{s(k)=k^2-k^4.}
$$
Modes with $0<|k|<1$ grow, while $|k|>1$ modes decay. Differentiating $s$ gives the most unstable wavenumbers
$$
\boxed{|k|=\frac1{\sqrt2},\qquad s_{\max}=\frac14.}
$$
The physical fastest wavenumber is $\epsilon/(\sqrt2\widehat h)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-44-dispersion.png]
{title=Thermocapillary film growth rate showing the unstable band and fastest-growing wavenumber}
For a positive steady film with zero <volume flux>, $H^2H_X+H^3H_{XXX}=0$. Divide by $H^3$ and integrate once:
$$
H_{XXX}=-\frac{H_X}{H},\qquad H_{XX}+\ln H=C.
$$
Choosing the thickness scale so that $H=1$ when $H_{XX}=0$ sets $C=0$. Multiply $H_{XX}=-\ln H$ by $H_X$ and integrate to obtain the <zero-flux thermocapillary film profiles> relation
$$
\boxed{\tfrac12H_X^2+H(\ln H-1)=E.}
$$
Thus the stated potential is $V(H)=H(\ln H-1)$ and the integration constant $E$ is the corresponding first integral.
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