Put $c'=\chi(y)\bar c_x$. The balance and reflecting boundary conditions give
$$
D\chi''=\frac Uh y,
\qquad
\chi'(\pm h)=0,
\qquad
\bar\chi=0.
$$
Integration yields
$$
\boxed{
\chi(y)=\frac UD
\left(\frac{y^3}{6h}-\frac{hy}{2}\right)}.
$$
Averaging the full transport equation gives
$$
\bar c_t+\overline{u\chi}\,\bar c_{xx}
=D\bar c_{xx}.
$$
Now
$$
\overline{u\chi}
=\frac{U^2}{D}
\left(\frac{\overline{y^4}}{6h^2}
-\frac{\overline{y^2}}2\right)
=-\frac{2U^2h^2}{15D},
$$
because $\overline{y^2}=h^2/3$ and $\overline{y^4}=h^4/5$. Thus the flow-induced <Taylor dispersion> coefficient and total effective diffusivity are
$$
\boxed{
D_T=\frac{2U^2h^2}{15D}},
\qquad
\boxed{
D_{\rm eff}=D+\frac{2U^2h^2}{15D}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 334 / 2 / b / iii / Solution

The transverse and longitudinal <Péclet numbers> are
$$
\boxed{\operatorname{Pe}_h=\frac{Uh}{D}},
\qquad
\boxed{\operatorname{Pe}_L=\frac{UL}{D}
=\frac Lh\operatorname{Pe}_h}.
$$
The <Taylor dispersion> regime requires transverse diffusion to act within a longitudinal advection time,
$$
\frac{h^2}{D}\ll\frac LU
\quad\Longleftrightarrow\quad
\operatorname{Pe}_h\ll\frac Lh,
$$
while longitudinal molecular diffusion is slow on that advection time,
$$
\operatorname{Pe}_L\gg1.
$$
Together,
$$
\boxed{
\frac hL\ll\operatorname{Pe}_h\ll\frac Lh},
\qquad
\boxed{
1\ll\operatorname{Pe}_L\ll\left(\frac Lh\right)^2}.
$$

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 334 / 2 / b / i / Solution

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{wiki}

Taylor dispersion is enhanced longitudinal spreading produced when transverse molecular diffusion repeatedly moves solute between streamlines of different velocity.


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