= Solution
Integrating the incompressible <continuity equation> $u_x+v_y=0$ across the moving gap, with $v(x,0)=0$ and $v(x,h)=-V$, gives the <Reynolds lubrication equation>
$$
\frac{dQ}{dx}=V.
$$
Hence
$$
Q=Vx+C
$$
for a constant $C$, and the flux formula from part (b) gives
$$
p_x=\frac{6\mu U}{h^2}
-\frac{12\mu(Vx+C)}{h^3}.
$$
Define the spatial mean
$$
\langle f\rangle=\frac1L\int_0^Lf(x)\,dx.
$$
Because both ends meet fluid at pressure $p_0$, the <pressure recovery condition in lubrication flow> is
$$
0=p(L)-p(0)=\int_0^Lp_x\,dx.
$$
It determines
$$
C=\frac U2
\frac{\langle h^{-2}\rangle}{\langle h^{-3}\rangle}
-V\frac{\langle xh^{-3}\rangle}{\langle h^{-3}\rangle}.
$$
Therefore the requested <pressure gradient in a translating and squeezing finite gap> is
$$
\boxed{
p_x=\frac{6\mu U}{h^2}
-\frac{12\mu}{h^3}
\left[
Vx+\frac U2
\frac{\langle h^{-2}\rangle}{\langle h^{-3}\rangle}
-V\frac{\langle xh^{-3}\rangle}{\langle h^{-3}\rangle}
\right].}
$$
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