= Solution
Let $Y(t)$ be the <porous thermal front> and $X(t)$ the polymer front. With the literal stated convention for <Darcy velocity>,
$$
Y'=\Gamma U,\qquad X'=U/\phi,\qquad Y(0)=X(0)=0.
$$
Therefore $Y=\beta X$ with $\beta=\Gamma\phi$. This is <thermal-front retardation in Darcy flow>. The region $0<x<Y$ is cold injected fluid of <dynamic viscosity> $\mu$; $Y<x<X$ is heated injected fluid of <dynamic viscosity> $b\mu$; $X<x<L$ is oil of <dynamic viscosity> $\mu_h$. The <pressure> drop is consequently
$$
\Delta P=\frac{U}{k}\left[\mu Y+b\mu(X-Y)+\mu_h(L-X)\right]=\frac{U}{k}\left[\mu_hL+(\mu_{\rm eff}-\mu_h)X\right],
$$
where $\mu_{\rm eff}=\mu[\beta+b(1-\beta)]$. Apply the preceding <fixed-pressure planar Darcy displacement> calculation with $\mu$ replaced by this effective <dynamic viscosity>. The \b[polymer-front speed] is
$$
\boxed{\dot X=\frac{k\Delta P}{\phi\sqrt{(\mu_hL)^2+2(\mu_{\rm eff}-\mu_h)k\Delta P\,t/\phi}},\qquad\mu_{\rm eff}=\mu[b-(b-1)\Gamma\phi].}
$$
Its position solves $\mu_hLX+(\mu_{\rm eff}-\mu_h)X^2/2=k\Delta P\,t/\phi$, and $Y=\Gamma\phi X$. These expressions hold while both interfaces remain in the field, the <porous thermal front> is sharp and stable, and $0\leq\Gamma\phi\leq1$. For $b=1$ the original constant-viscosity answer is recovered.
Sometimes a thermal retardation factor is defined relative to the <pore velocity> instead. Under that alternative definition one substitutes $\beta=\Gamma$ throughout. The factor of $\phi$ in the boxed answer follows specifically from the PDF's phrase “fraction of the Darcy speed.”
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