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Use instantaneous polar coordinates centered on the moving circular cylinder, but measure the fluid velocity in the laboratory frame. Impermeability equates the normal components of body and fluid velocity:Since ${\mathbf{e}}_x\cdot {\mathbf{e}}_r=\cos \theta$ and $u_r={\varphi }_r$, the no-penetration boundary condition isThe velocity potential obeys the exterior Laplace equation, by incompressibility. The boundary data have only the first cosine harmonic. A decaying harmonic potential of this form is $\varphi =C\cos \theta /r$. Matching its radial derivative gives $-C/a^2=U$, so $C=-U{a}^2$. The zero circulation condition rules out an angular potential term, and the integrated boundary condition gives zero net volume flux, ruling out a radial logarithmic source term. The remaining decaying harmonics have zero coefficients by the boundary data. HenceUsing the streamfunction in polar coordinates, $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$, givesOne may also obtain both functions as the real and imaginary parts of $w(z)=-U{a}^2/z$, a two-dimensional potential doublet. The velocity decays as $r^{-2}$ and the source of this singular representation lies inside the excluded body. These are the fields used to calculate the added mass of a circular cylinder.

A velocity potential satisfies $\mathbf{u}=\nabla \varphi$. Integration of ${\varphi }_r=Q/(2\pi r)$ and $r^{-1}{\varphi }_{\theta }=0$ givesChoose the streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. Its streamfunction in polar coordinates satisfies $u_r=r^{-1}{\psi }_{\theta }$ and $u_{\theta }=-{\psi }_r$, soAdditive constants are immaterial. This complex potential of a two-dimensional potential flow is defined on a chosen logarithm branch. Its complex velocity $w^{\prime }=Q/(2\pi z)=u-iv$ is single-valued away from the origin. Continuing once around the line source increases $\psi$ by $Q$, consistently with its nonzero volume flux.

Take outward radial volume flux per unit width as positive. The streamfunction in polar coordinates immediately givesFor spanwise width $W$, the volumetric flow rate is $Q=Wq$. The wedge inside radius $r$ has cross-sectional area $\alpha r^2/2$, whose rate of increase is $r^2\dot{\alpha }/2$. Opening therefore draws in exactly the volume needed to fill the expanding wedge; closing expels it. This is also a direct check of conservation of mass for the moving control volume. It follows from the wall boundary conditions and remains true even though the printed interior streamfunction is erroneous.

The coefficient of the growing term is fixed by the far-field complex potential: $A=\beta e^{i\alpha }$. On $z=a{e}^{i\theta }$, the no-penetration boundary condition makes the cylinder a streamline, so the imaginary part of $A{a}^2{e}^{2i\theta }+B{a}^{-2}{e}^{-2i\theta }$ must be constant. Choosing the two terms to be complex conjugates makes that constant zero, givingThis is the zero-circulation solution, also obtained by the Milne-Thomson circle theorem. Impermeability and the leading far field alone permit an additional circulation term $-i\Gamma \log z/(2\pi )$; the requested algebraic form implicitly selects $\Gamma =0$. No no-slip boundary condition is imposed on an inviscid flow.

The streamfunction in polar coordinates isIt gives $v_r(a,\theta )=0$ andThe liquid still exerts normal pressure on each surface element even though tangential viscous traction is absent. For this steady flow with zero vorticity, the Bernoulli equation has one constant $C$ throughout the connected fluid region:There is no finite uniform far-field pressure to substitute for $C$, because this straining velocity field grows with radius. Taking $\mathbf{n}=(\cos \theta ,\sin \theta )$ outward from the cylinder, the fluid force per unit axial width isIndeed $p(\theta +\pi )=p(\theta )$ while $\mathbf{n}(\theta +\pi )=-\mathbf{n}(\theta )$: diametrically opposite elements cancel exactly. Zero resultant force does not mean zero local pressure or zero deformation tendency.

The inner cylinder is $x^2+(y-a{)}^2=a^2$. With $x=r\cos \theta$ and $y=r\sin \theta$, this becomes $r^2-2ar\sin \theta =0$, so away from the touching pointAt this boundary the outward radial vector from the cylinder center is $(\sin 2\theta ,-\cos 2\theta )$, so the counterclockwise unit tangent is $\mathbf{t}=(\cos 2\theta ,\sin 2\theta )$. Since ${\mathbf{e}}_r\cdot \mathbf{t}=\cos \theta$ and ${\mathbf{e}}_{\theta }\cdot \mathbf{t}=\sin \theta$, the no-slip boundary condition givesFor a streamfunction in polar coordinates, no penetration through a stationary material curve makes $\psi$ constant along that curve. Both cylinders touch at the origin, so continuity forces their constants to agree. Since adding a constant does not alter velocity, choose their common value as $\boxed{\psi =0}$.

In the exterior domain $r>a$, the streamfunction obeysUsing the Streamfunction in polar coordinates conventionthe no-slip condition on the stationary disk isThus $\psi$ is constant on the disk and ${\psi }_r=0$ there. Choose the arbitrary constant to be zero:

The imposed shear ${\mathbf{u}}_{\infty }=\gamma y{\mathbf{e}}_x$ has streamfunctionThe far-field condition is therefore

A stagnation point of a fluid flow is a point at which the velocity field vanishes. For a two-dimensional streamfunction, it is a critical point satisfying both first spatial derivatives of the streamfunction equal to zero.