= Solution
Locally, zero curl permits $\mathbf v=\nabla\Phi$; for a global single-valued potential the <circulation> around every noncontractible loop must also vanish. <incompressibility> gives $\nabla\cdot\mathbf v=\nabla^2\Phi=0$. The exterior of a cylinder is multiply connected, so local irrotationality alone does not exclude a circulating solution.
For inviscid constant-density flow under conservative gravity, Euler's equation is $\partial_t\mathbf v+(\mathbf v\cdot\nabla)\mathbf v=-\nabla(p/\rho+\phi_g)$. The identity $(\mathbf v\cdot\nabla)\mathbf v=\nabla(v^2/2)-\mathbf v\times(\nabla\times\mathbf v)$ gives, for <irrotational flow>,
$$
\partial_t\Phi+\frac p\rho+\frac{v^2}{2}+\phi_g=C(t).
$$
For <steady flow>, absorb the potential's time-only gauge term and obtain \b[$p+\rho v^2/2+\rho\phi_g=\mathrm{constant}$] throughout the connected fluid region. Without irrotationality the steady constant need only be common along each <streamline>.
Use $U$ for the uniform upstream <speed> and take the flow along $+x$. The <boundary conditions> are $v_r(a,\theta)=0$ and $\mathbf v\to U\mathbf e_x$ at infinity. A perfect inviscid model imposes no penetration, not no slip. For the intended no-circulation solution, require a globally single-valued potential, so the coefficient of $\theta$ is zero. The logarithmic term would be a net source and violates no penetration, so its coefficient is also zero. Growing terms with $n\geq2$ violate the uniform far field. For each remaining $n\geq2$, the surface normal condition then <forces> its decaying coefficient to vanish. The $n=1$ terms satisfy $C_1=U$ and $D_1=Ua^2$:
$$
\boxed{\Phi=U(r+a^2/r)\cos\theta,\qquad
v_r=U(1-a^2/r^2)\cos\theta,\quad
v_\theta=-U(1+a^2/r^2)\sin\theta.}
$$
An irrelevant additive potential constant is understood. At the same elevation, Bernoulli balance with the upstream stream gives
$$
\boxed{p(a,\theta)=p_\infty+\frac12\rho U^2(1-4\sin^2\theta).}
$$
The <force> per unit height is
$$
\mathbf F'=-a\int_0^{2\pi}p(a,\theta)(\cos\theta,\sin\theta)\,d\theta
=\boxed{\mathbf0.}
$$
<pressure> <symmetry> gives zero <drag> and zero <lift> in this zero-circulation idealization. <D'Alembert's paradox> is the failure of this inviscid model to represent real separated viscous wakes and their <drag>, not a claim that a real pole feels no <force>.
The zero-circulation assumption matters: $\Phi\to\Phi+\Gamma\theta/(2\pi)$ leaves the stated <velocity> <boundary conditions> intact but adds $\Gamma/(2\pi a)$ to the surface tangential <speed>. It produces $F'_y=-\rho U\Gamma$ for positive counterclockwise <circulation>. Thus the <boundary conditions> alone do not justify excluding the angular term; an initially noncirculating inviscid start, or an explicitly single-valued potential, supplies the missing condition.<Flow Past a Cylindrical Obstacle by Richard Fitzpatrick>[equations 5.73–5.81 and discussion after 5.83: circulation, force and the zero-circulation initial condition]{ref}
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2013/paper-1-cylinder-flow.png]
{title=Original calculated zero-circulation cylinder streamlines and surface pressure coefficient}
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