Along $y=0$, put $x=r$ and $U(x)=-Q/x$. If $\delta(x)$ is the layer thickness, then
$$
u u_x=O\!\left(\frac{Q^2}{x^3}\right),
\qquad
\nu u_{yy}=O\!\left(\frac{\nu Q}{x\delta^2}\right).
$$
The <boundary-layer thickness from an advection-diffusion balance> therefore gives
$$
\frac{Q^2}{x^3}\sim\frac{\nu Q}{x\delta^2},
$$
and hence
$$
\boxed{\delta=x\sqrt{\frac\nu Q}}
$$
up to a constant factor absorbed into the similarity coordinate.

Solution

Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 38B / b / ii / Solution

For streamwise length $L$, outer speed $U$, and layer thickness $\delta$, balancing $u u_x\sim U^2/L$ with $\nu u_{yy}\sim\nu U/\delta^2$ gives
$$
\frac\delta L\sim\operatorname{Re}^{-1/2},
\qquad
\operatorname{Re}=\frac{UL}{\nu}.
$$


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