= Solution
Let $X(t)$ denote the polymer–oil material interface, measured from the injection well, and let $U(t)$ be the <Darcy velocity>. <Incompressibility> and uniform cross-sectional area make $U$ independent of $x$ in this planar model. Integrating <Darcy's law> through the injected layer and the oil layer gives
$$
\Delta P=\frac{U}{k}\left[\mu X+\mu_h(L-X)\right].
$$
The material interface moves at the <pore velocity>, so $X'=U/\phi$. Put $a=\mu-\mu_h$ and $A=\mu_hL$. Then $(A+aX)X'=k\Delta P/\phi$, which integrates from $X(0)=0$ to
$$
AX+\frac a2X^2=\frac{k\Delta P}{\phi}t.
$$
Thus the <fixed-pressure planar Darcy displacement> has
$$
\boxed{\dot X(t)=\frac{k\Delta P}{\phi\sqrt{(\mu_hL)^2+2(\mu-\mu_h)k\Delta P\,t/\phi}},\qquad U(t)=\phi\dot X(t).}
$$
For $a\ne0$, $X=[\sqrt{A^2+2ak\Delta P\,t/\phi}-A]/a$; its continuous $a=0$ limit is $X=k\Delta P\,t/(\phi\mu_hL)$. Positive layer <dynamic viscosities> ensure the resistance stays positive until breakthrough, which occurs at $t_b=\phi L^2(\mu+\mu_h)/(2k\Delta P)$. The formulas stop there, since the two-layer geometry then changes.
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