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With the complex velocity convention $w^{\prime }(z)=u-iv$, differentiation gives $w^{\prime }=-i\Gamma /(2\pi z)$. Thus $u=-\Gamma y/(2\pi r^2)$ and $v=\Gamma x/(2\pi r^2)$. Resolving into polar components gives $\boxed{u_r=0,u_{\theta }=\Gamma /(2\pi r)}$.

Superposition for the opposite-signed pair givesup to a constant and branch choices. A vortex moves with the regular velocity generated by the other vortex, excluding its own singular field. At $z=ia$, the other contribution is $i\Gamma /[2\pi (2ia)]=\Gamma /(4\pi a)$; at $z=-ia$ it is the same real number. Hence the pair keeps its separation and translates parallel to the real axis withfor $a>0$.

For a vortex at $z_1=x_1+i{y}_1$ in the first quadrant, reflect it across either wall with opposite strength, then across both walls with its original strength. Thus the images are ${\bar{z}}_1$ and $-{\bar{z}}_1$ of strength $-\Gamma$, and $-z_1$ of strength $+\Gamma$. The potential isOn either real or imaginary axis, $z^2$ is real and the numerator and denominator of $(z^2-z_1^2)/(z^2-{\bar{z}}_1^2)$ are conjugates. Its modulus is one, making the stream function constant on both walls and enforcing no normal flow. A regular entire background flow could be added only with additional far-field data; the displayed image solution is the one generated by the specified vortex and its walls.

Write the complex potential of a two-dimensional potential flow as $w=\varphi +i\psi$. Its complex velocity is $u-iv=dw/dz$. Hereso the far-field motion has velocity $(U,0)$ and speed $U$. In polar coordinates the stream function, shifted by a harmless constant, isIt is zero on $r=a$, proving the cylinder is a streamline.

Stagnation points are zeros of $dw/dz$, hence roots of $U{z}^2-i\Gamma z/(2\pi )-U{a}^2=0$. Put $\gamma =\Gamma /(4\pi Ua)\in (0,1)$. The two roots areBoth have modulus $a$ and are distinct. Their angles satisfy $\sin \theta =\gamma$, namely $\arcsin \gamma$ and $\pi -\arcsin \gamma$. The tangential velocity on the cylinder is $-2U\sin \theta +\Gamma /(2\pi a)$, confirming the same zeros. Streamlines are level curves of $\psi$; the positive circulation places both stagnation points on the upper side and increases the flow speed below the cylinder.

The stream function gives $u=2\alpha xy$ and $v=-\alpha y^2$. Its scalar vorticity isFor $\alpha \ne 0$ this does not vanish on any nonempty open region of the plane. A holomorphic complex potential of a two-dimensional potential flow would produce an irrotational flow there. Equivalently, its imaginary part would be a harmonic function, whereas $\Delta \psi =2\alpha x$. Thus no complex potential exists on any nonempty open region for this flow; vanishing vorticity only along the vertical line does not provide a two-dimensional domain.

For a twice continuously differentiable stream function, equality of mixed derivatives givesThis is the incompressible flow condition in two dimensions. A streamline is a curve whose tangent at each regular point is parallel to the instantaneous velocity field. Parameterize it by ${\gamma }^{\prime }(s)=c(s)(u,v)$. The chain rule yieldsTherefore the stream function is constant on each streamline. This proof includes arbitrary regular parametrizations; at a stagnation point the velocity vanishes and the usual tangent direction is not determined.

Start from the uniform-flow complex potential of a two-dimensional potential flow $Uz$. Milne-Thomson's circle theorem adds the image doublet $U{a}^2/z$, yielding constant stream function on $|z|=a$. Add a line vortex with positive counterclockwise circulation $\Gamma$:Indeed, on $z=a{e}^{i\theta }$ the uniform-flow and doublet terms sum to the real number $2Ua\cos \theta$, while the vortex has imaginary part $-\Gamma \log a/(2\pi )$, also constant. Thus the cylinder is a streamline and satisfies the impermeability condition. The vortex velocity is $\Gamma /(2\pi r)$ in the counterclockwise direction. Although the potential is multivalued when $\Gamma \ne 0$, its velocity is single-valued.

Differentiate and locate the coefficient of $z^{-1}$ in the square:For the counterclockwise contour in the printed Blasius force theorem, the residue theorem gives ${\oint }_C(w^{\prime }{)}^{2}dz=2U\Gamma$. ConsequentlyThese are force components per unit cylinder length; with the stated circulation convention the lift is downward for positive $U\Gamma$.

Use the stream function convention $u={\psi }_y$, $v=-{\psi }_x$. It gives $u=-\sin x\sin y$, $v=-\cos x\cos y$. Thenso the flow is incompressible.

For a planar incompressible flow, a stream function satisfies $u={\psi }_y$, $v=-{\psi }_x$. A velocity potential satisfies $u={\varphi }_x$, $v={\varphi }_y$ where the flow is irrotational; such a potential exists locally, and globally on a simply connected nonsingular region. The irrotational flow condition is $v_x-u_y=0$. Where both exist, the Cauchy-Riemann equations make the complex potential $w=\varphi +i\psi$ analytic with $w^{\prime }=u-iv$.

For the logarithmic potential, write $z-z_0=r{e}^{i\theta }$. ThenThe outward volume flux per unit span across a circle is $2\pi r{u}_r=m$. Hence it is a line source of strength $m$. The potential may use local logarithm branches; its velocity is single-valued away from the source.

By superposition, the full flow hasFor positive $m$ and $U$, the stagnation points areThey are real, one upstream of the source and one downstream of the sink. Away from the singularities, streamlines are level curves ofThe zero level through the two stagnation points contains an upper and a lower separating curve, symmetric about the real axis and about $x=1/2$. Together they bound a Rankine oval. Flow inside goes from source to sink; the external stream divides around this oval and tends to horizontal flow far away. This also specifies the sketch for every positive parameter ratio; the figure illustrates one ratio.

Use the convention $w=\varphi +i\psi$, with $dw/dz=u_x-i{u}_y$. ThenIndeed $w^{\prime }(z)=z=x+iy$, giving velocity components $(x,-y)$. The streamlines are the level sets of the stream function:For $c\ne 0$ these are rectangular hyperbolas, restricted to $y>0$. The level $c=0$ consists of the vertical axis and the wall $y=0$. Directly on the wall the normal velocity is $-y=0$, so it is a streamline. The arrows point down toward the wall and then away from the stagnation point along the wall.

A point vortex of circulation $\Gamma$ at $z_0$ contributes the complex potential of a two-dimensional potential flow $-i\Gamma \log (z-z_0)/(2\pi )$ with the stated convention. The method of images places a vortex of the opposite circulation at ${\overline{z}}_0$ below the wall, outside the fluid domain. Superposition with the stagnation flow givesFor real $z=x$, the distances $|x-z_0|$ and $|x-{\overline{z}}_0|$ agree. The two vortex contributions to the stream function are $-\Gamma \log |x-z_0|/(2\pi )$ and $+\Gamma \log |x-{\overline{z}}_0|/(2\pi )$, which cancel. The background stream function is also zero there, so $y=0$ remains an impermeable streamline. The image changes the boundary condition without introducing another singularity in the upper fluid region.

Orient the body boundary $C$ counterclockwise. Its outward normal vector satisfies $\mathbf{n}ds=(dy,-dx)$, so the pressure force on the body has complex componentFor steady incompressible flow and irrotational flow, Bernoulli's principle gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant-pressure integral vanishes. Write $q=w^{\prime }=u-iv$. The impermeability condition makes velocity tangent to $C$, implying $qdz=\bar{q}d\bar{z}$. Thus $|q{|}^{2}d\bar{z}=q^{2}dz$ along the boundary, andThis proves the Blasius force theorem, with force measured per unit length perpendicular to the flow plane.

For a circle of radius $a$, Milne-Thomson's circle theorem replaces an incident analytic potential $f(z)$ by $f(z)+\overline{f(a^2/\bar{z})}$, making the stream function constant on the circle. Its reflected singularities must lie inside the circle; here the physical source lies outside. With $c=Q/(2\pi )$ and $d=a^2/b$, this yields, up to additive constants,The interior image source and sink have equal and opposite strengths, so there is no extra net source inside the body. No independent vortex term is added. The potential may have logarithmic branches, but the velocity and pressure are single-valued.

Only $d$ and $0$ lie inside $C$. The residues of $(w^{\prime }{)}^2$ areTheir sum is $-2c^2{a}^2/[b(b^2-a^2)]$. The residue theorem and the preceding force formula therefore giveThe force points towards the external source. Reflection symmetry across the horizontal axis independently checks the vanishing vertical component.

Orient the body boundary counterclockwise. The outward normal from the body satisfies $nds=(dy,-dx)$, so pressure givesFor steady irrotational flow, Bernoulli's equation makes $p+\rho |u{|}^2/2$ constant in the connected fluid region. Its constant contribution integrates to zero. Write $q=w^{\prime }=u-iv$. Impermeability says the velocity is tangent to the boundary, so $qdz=\bar{q}d\bar{z}$ there. Consequentlywhich proves Blasius' theorem with this contour convention.

The background stream function is the imaginary part of $w_0(z)=\alpha (z-c{)}^3$. Since its coefficients are real, the circle theorem givesIts derivative isIn $q^2$ the coefficient of $z^{-1}$ comes from the exponent pairs $(2,-3)$ and $(1,-2)$, including both orders. It is $36{\alpha }^{2}c{a}^4+36{\alpha }^2{c}^3{a}^2$. The residue theorem then givesThe vanishing transverse force also agrees with reflection symmetry about the real axis.

The Milne-Thomson circle theorem says that, when $f$ is a complex potential analytic inside and on $|z|=R$ and all its flow singularities are outside, the potential outside an impermeable circle isThe reflected expression is analytic in the exterior: conjugating a local power series turns its argument into $R^2/z$. It creates only interior image singularities. On $|z|=R$ we have $R^2/\bar{z}=z$, so $W=f+\bar{f}$ is real. The stream function is therefore constant on the circle, which is the impermeability condition. At infinity the reflected term is $\overline{f(0)}+O(z^{-1})$, so it changes the velocity by $O(z^{-2})$ and leaves the imposed far flow unchanged. These observations prove the theorem, up to the irrelevant additive potential constant and with no extra circulation introduced.

With $a=z_1$, $|a|>1$, and unit radius, it givesThe image dipole is at $1/\bar{a}$ with moment $-\bar{m}/{\bar{a}}^2$; writing the image term this way changes only an additive constant.

Reflection in the line $y=-x$ is $Rz=\overline{iz}=-i\bar{z}$. The printed expression has two conjugation bars: $w(z)+\overline{w(\overline{iz})}$. On the line $Rz=z$, this is $w(z)+\overline{w(z)}$, which is real; hence its stream function vanishes there. Applying this reflection to the dipole givesConsequentlyThe image lies at $-i\bar{a}$, with moment $-i\bar{m}$, on the opposite side of the wall.

The Milne-Thomson circle theorem says that, when $f$ is a complex potential analytic inside and on $|z|=R$ and all its flow singularities are outside, the potential outside an impermeable circle isThe reflected expression is analytic in the exterior: conjugating a local power series turns its argument into $R^2/z$. It creates only interior image singularities. On $|z|=R$ we have $R^2/\bar{z}=z$, so $W=f+\bar{f}$ is real. The stream function is therefore constant on the circle, which is the impermeability condition. At infinity the reflected term is $\overline{f(0)}+O(z^{-1})$, so it changes the velocity by $O(z^{-2})$ and leaves the imposed far flow unchanged. These observations prove the theorem, up to the irrelevant additive potential constant and with no extra circulation introduced.

With $a=z_1$, $|a|>1$, and unit radius, it givesThe image dipole is at $1/\bar{a}$ with moment $-\bar{m}/{\bar{a}}^2$; writing the image term this way changes only an additive constant.

Reflection in the line $y=-x$ is $Rz=\overline{iz}=-i\bar{z}$. The printed expression has two conjugation bars: $w(z)+\overline{w(\overline{iz})}$. On the line $Rz=z$, this is $w(z)+\overline{w(z)}$, which is real; hence its stream function vanishes there. Applying this reflection to the dipole givesConsequentlyThe image lies at $-i\bar{a}$, with moment $-i\bar{m}$, on the opposite side of the wall.

The Milne-Thomson circle theorem says that if a complex potential $f(z)$ has no singularities on or inside $|z|=a$, then a circular impermeable boundary can be introduced by takingThe new singularities are the reflected ones inside the circle. On $|z|=a$, $a^2/z=\overline{z}$, so $W(z)=f(z)+\overline{f(z)}$ is real; the stream function is constant and there is no normal velocity. At infinity the added term tends to $f^{\ast }(0)$, a harmless potential constant, so it preserves the far-field flow. This proves the theorem with zero added net circulation about the cylinder. A logarithmic term may then specify a different cylinder circulation.

For $f(z)=-i\Gamma \log (z-b)/(2\pi )$, put $c=a^2/b$. The reflected term iswith locally consistent logarithm branches. The circle-theorem solution has zero cylinder circulation: its image vortex at $c$ and its central vortex cancel. To impose circulation $-\Gamma$, add $i\Gamma \log z/(2\pi )$. The result isIt consists of the physical vortex and an opposite image vortex inside the obstacle. On the circle, $|z-b|=(b/a)|z-c|$, so its stream function is indeed constant.

The complex velocity isInside the positively oriented obstacle contour, $q^2$ has only the double pole at $c$. Its residue isThe residue theorem and the supplied Blasius force theorem therefore giveConsequentlyThe force points toward the held vortex, independent of the sign of its circulation.

The residue calculation is the force on a circular cylinder from a held exterior vortex.

Orient $C$ counterclockwise around the solid body and let $\mathbf{n}$ point from the solid into the fluid. The pressure force is $-{\int }_{C}p\mathbf{n}ds$. Since $\mathbf{n}ds=(dy,-dx)$,For steady irrotational flow, Bernoulli's equation gives $p=p_0-\rho |w^{\prime }{|}^2/2$. The constant term integrates to zero. The boundary is a streamline, so $dw=w^{\prime }dz$ is real there, giving $w^{\prime }dz=\overline{w^{\prime }}d\overline{z}$. Thereforewhich is Blasius force theorem with the specified orientation.

The prescribed stream function is the imaginary part of $w_0(z)=\alpha (z-c{)}^3$. The zero-circulation solution from the Milne-Thomson circle theorem isOn $|z|=a$ the two terms are conjugates, so the boundary stream function is constant. The added term has no singularities in the exterior and does not change the prescribed growing far-field part. Additive constants are immaterial.

Differentiating givesThe coefficient of $z^{-1}$ in $(w^{\prime }{)}^2$ is $36{\alpha }^{2}c{a}^2(a^2+c^2)$. The residue theorem and Blasius force theorem giveThe zero-circulation convention is relevant: if an independently imposed circulation $\Gamma$ is allowed, adding $\Gamma \log z/(2\pi i)$ changes $F_y$ to $-3\rho \alpha \Gamma c^2$. Thus the claimed zero transverse force belongs to the standard circle-theorem construction, or to a single-valued exterior complex potential of a two-dimensional potential flow, which excludes that extra circulation.

Let the thin transverse length be $R{e}^{-1/2}$ times the longitudinal scale and write $Y=R{e}^{1/2}y$, $V=R{e}^{1/2}v$. Continuity then remains $u_x+V_Y=0$. Transverse momentum gives $p_Y=0$ to leading order; the quiescent exterior has constant pressure, hence $p_x=0$. Axial viscous diffusion in $x$ is smaller by $R{e}^{-1}$, leavingSymmetry about the jet centreline requires $V=0$ and $u_Y=0$ at $Y=0$. At large positive $Y$, require $u\to 0$ and $u_Y\to 0$. The transverse velocity can approach a finite nonzero scaled entrainment velocity; imposing $V\to 0$ as well would rule out the jet. The physical transverse velocity remains smaller by $R{e}^{-1/2}$ and matches an entraining outer flow.

Using continuity, the momentum equation is ${\partial }_x(u^2)+{\partial }_Y(uV-u_Y)=0$. Integrating across the jet, the boundary terms vanish because $u,u_Y\to 0$ while $V$ is finite. ThusApart from the dimensional density and velocity/length factors, this is the conserved axial momentum flux supplied by the nozzle. Entrainment increases volume flux without adding axial momentum from the resting exterior.

For $u=x^{\alpha }{f}^{\prime }(\eta )$ with $\eta =Y/x^{\beta }$, use the stream function $\psi =x^{\alpha +\beta }f(\eta )$ so that $u={\psi }_Y$ and $V=-{\psi }_x$. Momentum conservation requires $2\alpha +\beta =0$. Inertia and diffusion scale as $x^{2\alpha -1}$ and $x^{\alpha -2\beta }$, so $\alpha +2\beta =1$. Therefore $\boxed{\alpha =-1/3,\beta =2/3}$. In particularSubstitution reduces momentum to $3f^{(3)}+f{f}^{\prime \prime }+(f^{\prime }{)}^2=0$, with $f(0)=f^{\prime \prime }(0)=0$ and $f^{\prime }(\infty )=0$. Integrating first gives $3f^{\prime \prime }+f{f}^{\prime }=0$, the integration constant vanishing at the centreline. A second integration gives $3f^{\prime }+f^2/2=C$. Parameterize its positive constant as $C=18A^2$ and solve the resulting Riccati equation with $f(0)=0$:The momentum integral fixes its scale, since ${\int }_{-\infty }^{\infty }{\mathrm{sech}}^{4}zdz=4/3$:The upper-half entrainment is $V\to -2A{x}^{-2/3}$; the negative side has the opposite sign, drawing ambient fluid inward. This is the Bickley jet, valid in the slender far-downstream laminar regime, rather than in the immediate nozzle region.

Two-dimensional incompressibility permits $u={\psi }_y$, $v=-{\psi }_x$. Taking the curl of the Stokes momentum equation eliminates pressure and gives ${\nabla }^2(v_x-u_y)=0$. Since $v_x-u_y=-{\nabla }^2\psi$, the stream function obeys $\boxed{{\nabla }^4\psi =0}$.

Scale radius by $a$, velocity by $U$ and stream function by $Ua$. In polar coordinates $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$. No penetration makes $\psi$ constant on the cylinder; choose that constant zero. No slip also requires ${\psi }_r=0$ at $r=1$. The uniform stream has $\psi =r\sin \theta$, specifying the far-field condition.

The boundary forcing has angular mode $\sin \theta$, so its corresponding mode must have $\psi =f(r)\sin \theta$. Define $L_1=d^2/d{r}^2+r^{-1}d/dr-r^{-2}$. Then ${\nabla }^4\psi =(L_1^{2}f)\sin \theta$, while $L_1{r}^s=(s^2-1)r^{s-2}$. Thus the indicial equation of $L_1^2$ is $(s^2-1)[(s-2{)}^2-1]=(s-3)(s-1{)}^2(s+1)=0$, givingThe repeated exponent one supplies the logarithmic term. The condition $f^{\prime }\to 1$ first forces $A=B=0$ and $C=1$. Then $f(1)=0$ forces $D=-1$, but $f^{\prime }(1)=C-D=2\ne 0$. No additional Fourier mode can repair the missing boundary condition in this angular mode. Therefore the unbounded-domain Stokes problem has no solution with both no slip and a nonzero uniform far stream.

This is Stokes' paradox. It identifies a nonuniform approximation, rather than the absence of physical flow. For any small but nonzero Reynolds number, convective inertia becomes important sufficiently far away. An inner Stokes region must be matched to an outer Oseen or Navier-Stokes region that retains that inertia. Alternatively, finite external boundaries introduce a different well-posed physical problem; they cannot be sent to infinity while keeping the naive Stokes solution unchanged.

The Von Mises boundary-layer transformation uses $\Psi$ as transverse coordinate. With $u=U(x,\Psi )$ and ${\Psi }_x=-V$, ${\Psi }_Y=u=U$, the chain rule givesThe streamwise advection terms reduce to $u{u}_x+V{u}_Y=U{U}_x$, so, for $Y>0$ where $U>0$, division by $U$ givesContinuity was already enforced by the stream function. Substitute $U=g(\eta )$ with $\eta =\Psi /\sqrt{x}$. Then $U_x=-\eta g^{\prime }/(2x)$ and ${\partial }_{\Psi }(U{U}_{\Psi })=(g{g}^{\prime }{)}^{\prime }/x$, yieldingThe wall coordinate is zero because $\Psi$ was set to zero there, and the far condition matches unit external speed. This is the transformed Blasius similarity problem. One must not additionally prescribe a finite $g^{\prime }(0)$: the coordinate change degenerates at the no-slip wall, and the regular physical shear corresponds to $g\propto \sqrt{\eta }$ near zero. This similarity reduction is obtained for the attached monotone boundary layer assumed in the question.

Existence of a positive normalized profile can also be seen directly. Put $q=g{g}^{\prime }$ and choose $q(0)=a>0$. In $g$ coordinates the regular initial-value system is $dq/dg=-\eta /2$, $d\eta /dg=g/q$, with $\eta (0)=0$. It starts as $g\sim \sqrt{2a\eta }$. For finite positive $\eta$, $q^{\prime }/q=-\eta /(2g)$ keeps $q$ positive and $q\le a$, while $(g^2{)}^{\prime }=2q$ gives $g\le \sqrt{2a\eta }$. Consequently $q^{\prime }/q\le -\sqrt{\eta }/(2\sqrt{2a})$, so $q$ is integrable at infinity. It follows that $g$ increases to a finite positive limit $M$. The rescaled profile $\tilde{g}(\eta )=M^{-1}g(\sqrt{M}\eta )$ satisfies the same equation and has limit one. This supplies the requested similarity solution with both boundary conditions, rather than only its formal differential equation.