= Solution
Using $\gamma=\lambda$ and the <Pythagorean trigonometric identity>,
$$
|\mathbf u|^2=u_r^2+u_\theta^2
=A^2\gamma^2r^{2\gamma-2},
$$
so the speed is independent of $\theta$. Since $\gamma=\pi/\alpha>1$,
$$
|\mathbf u(r)|=u_0\left(\frac r{r_0}\right)^{\gamma-1}
$$
and the speed tends to zero at the vertex. The steady irrotational <Bernoulli equation> gives
$$
p+\frac12\rho|\mathbf u|^2
=p_0+\frac12\rho u_0^2.
$$
Thus the <fluid pressure> is also independent of $\theta$, with
$$
p(r)=p_0+\frac12\rho u_0^2
\left[1-\left(\frac r{r_0}\right)^{2\gamma-2}\right].
$$
In particular, the pressure difference between the vertex and $r_0$ is
$$
\boxed{p(0)-p_0=\frac12\rho u_0^2}.
$$
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