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For an incompressible flow, $\nabla \cdot \mathbf{u}=0$. If the flow is also irrotational, then locally $\mathbf{u}=\nabla \varphi$ for a velocity potential, and therefore

The boundary conditions areThe fluid region is not simply connected, so a potential may change by the constant $\kappa$ after one circuit while its gradient remains single-valued. Superposing uniform flow, the cylinder doublet, and the circulation gives the potential flow around a circular cylinder with circulationIts velocity components are

On $r=a$, $u_r=0$ and, with $\lambda =\kappa /(4\pi Ua)$,The Bernoulli equation gives $p=p_{\infty }+\rho (U^2-u_{\theta }^2)/2$. The pressure force per unit length on the cylinder isThe $x$ component vanishes by symmetry, while ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ givesin agreement with the Kutta–Joukowski theorem.

A stagnation point satisfies $u_r=u_{\theta }=0$. On the cylinder this means $\sin \theta =\lambda$. Thus $|\lambda |<1$ gives two surface stagnation points, $|\lambda |=1$ gives one coincident surface point, and $|\lambda |>1$ gives none on the surface. Away from the cylinder, $u_r=0$ requires $\cos \theta =0$. Solving $u_{\theta }=0$ then gives one physical exterior root when $|\lambda |>1$:At $|\lambda |=1$ this root lies on $r=a$ and agrees with the single surface point.