Complex potential of a two-dimensional potential flow ($W=\Phi +i\Psi ,W^{\prime }=u-iv$)

For an incompressible irrotational two-dimensional potential flow, an analytic complex potential of a two-dimensional potential flow combines the velocity potential $\Phi$ and streamfunction $\Psi$. The Cauchy-Riemann equations give $u={\Phi }_x={\Psi }_y$ and $v={\Phi }_y=-{\Psi }_x$, hence $W^{\prime }(z)=u-iv$. Curves of constant $\Psi$ are streamlines. An analytic coordinate change $z=f(\zeta )$ transforms the complex velocity by $dW/dz=(dW/d\zeta )/f^{\prime }(\zeta )$ wherever $f^{\prime }\ne 0$.

A planar potential dipole at $a$ with complex moment $m$ has complex potential of a two-dimensional potential flow $w(z)=m/(z-a)$. Its complex velocity is $w^{\prime }(z)=-m/(z-a{)}^2$, interpreted as $v_x-i{v}_y$. Unlike the three-dimensional potential dipole, this is a two-dimensional singularity of the logarithmic-source potential. For example, two opposite source coefficients $s$ at $a\pm \delta /2$ have potential $s[\log (z-a-\delta /2)-\log (z-a+\delta /2)]$, whose leading term is $-s\delta /(z-a)$ as the separation tends to zero with $s\delta$ fixed. This gives the dipole as a limiting source–sink pair, with the complex sign and orientation included in $m$.

A transported planar potential dipole follows the surrounding regular velocity, so its position satisfies the displayed equation after its own singular term is removed. The conjugation comes from the convention $w^{\prime }=v_x-i{v}_y$. Reflected image parameters are held fixed when taking the spatial derivative, then evaluated at the instantaneous dipole position.

For a unit circular wall, the Milne-Thomson circle theorem gives image potential $\bar{m}z/(1-\bar{a}z)$ and therefore $\dot{\bar{a}}=\bar{m}/(1-|a{|}^2{)}^2$. For a horizontal straight wall the image is $\bar{m}/(z-\bar{a})$, giving $\dot{\bar{a}}=-\bar{m}/(a-\bar{a}{)}^2$. For the line $y=-x$, reflected conjugation gives image $-i\bar{m}/(z+i\bar{a})$ and hence $\dot{\bar{a}}=i\bar{m}/(a+i\bar{a}{)}^2$. In each case the image lies outside the fluid region and makes the wall a streamline; differentiation of the regular image proves the motion formula. Several compatible straight-wall reflections can be summed before the same operation is applied.

For a complex potential $w(z)=m/(z-a)$, reflection in the straight wall through zero at angle $\theta$ is $Rz=e^{2i\theta }\bar{z}$. The image construction $W(z)=w(z)+\overline{w(Rz)}$ givesOn the wall $Rz=z$, this is real, so its stream function is constant there. This proves the impermeability condition. Thus the image position is $a^{\prime }=e^{2i\theta }\bar{a}$ and its moment is $m^{\prime }=e^{2i\theta }\bar{m}$. If the point dipole is advected by the regular image flow, the complex velocity convention $W^{\prime }=u-iv$ giveswhere $d=\mathrm{Im}(e^{-i\theta }a)$ is its signed distance from the wall. The last equality follows from $a-a^{\prime }=2i{e}^{i\theta }d$. It applies when $d\ne 0$.

A two-dimensional point sink of strength $Q>0$ removes volume flux $Q$ per unit depth at $z_0$. Its complex potential of a two-dimensional potential flow is the displayed expression, so $u-iv=-Q/[2\pi (z-z_0)]$. The radial velocity is $u_r=-Q/(2\pi r)$: integrating over a circle gives outward flux $2\pi r{u}_r=-Q$. This is the planar cross-section of a line sink. A passive particle satisfies $d(r^2)/dt=-Q/\pi$ and reaches the ideal point singularity in finite time. This is a flow singularity, distinct from a stable sink equilibrium of a regular dynamical system.

For a two-dimensional potential doublet of fixed complex strength $M$ outside a stationary circle of radius $a$, the Milne-Thomson circle theorem gives the image contribution $\bar{M}z/(a^2-\bar{b}z)$. If the singularity is stipulated to follow the regular fluid velocity at its position, differentiating this term and using the complex velocity convention gives the displayed motion law. Thus $\mathrm{Im}(b\bar{M})$ is constant: the trajectory is a straight line parallel to $M$. With $M=m{e}^{i\alpha }$ and initial location $b_0>a$ real, its polar coordinates satisfy $r\sin (\theta -\alpha )=-b_0\sin \alpha$. This statement assumes fixed orientation; the translational rule alone supplies no equation for orientation evolution.

For a steady two-dimensional incompressible flow and irrotational flow around an impermeable body, the displayed contour formula gives the force on the body per unit span. Here $W^{\prime }=u-iv$ is the complex velocity, $\rho$ is the constant mass density, and $C$ runs counterclockwise around the body, excluding other fluid singularities. To verify the sign, Bernoulli's principle gives $p=p_0-\rho |q{|}^2/2$. On a counterclockwise boundary, the complex pressure force is $d(X-iY)=-ipd\bar{z}$. Tangency gives $qdz=\bar{q}d\bar{z}$, and therefore $|q{|}^{2}d\bar{z}=q^{2}dz$. Integrating removes the constant $p_0$ term and gives the formula. The theorem is named after Heinrich Blasius.

The Milne-Thomson circle theorem places an image dipole inside a cylinder at the inverse point of an external dipole. In the residue of the squared complex velocity, only the cross term between the analytic external velocity and the image double pole contributes. The product of their conjugate dipole strengths is independent of orientation, so the cylinder force is radial toward the external dipole.

For a two-dimensional steady ideal flow outside a body, let $q=w^{\prime }$ be the complex velocity. Suppose $q$ is holomorphic in the exterior and $q^2$ has a Laurent series near infinity, with coefficient $a_{-1}$ of $z^{-1}$. Orient the body contour counterclockwise and assume the usual impermeability and Bernoulli hypotheses of the Blasius force theorem. Deforming the contour to a large circle gives $\oint q^{2}dz=2\pi i{a}_{-1}$. Substitution in Blasius' formula proves $F_x-i{F}_y=-\pi \rho a_{-1}$.

In particular, $q(z)=q_{\infty }+O(z^{-2})$ makes the force zero: its square has no reciprocal term. This criterion handles branch-cut obstacles by applying the contour in their analytic exterior, rather than separately assigning forces to endpoint singularities.

Place a held point vortex of circulation $\Gamma$ at real position $b>a$ outside a circular cylinder $|z|=a$, and prescribe circulation $-\Gamma$ about the cylinder. An opposite image at $c=a^2/b$ gives complex velocity $q(z)=-i\Gamma (1/(z-b)-1/(z-c))/(2\pi )$. On the positively oriented cylinder contour, the only pole of $q^2$ inside is at $c$, with residue $-{\Gamma }^2/(2{\pi }^2(b-c))$. The residue theorem and Blasius force theorem give $F_x-i{F}_y=\rho {\Gamma }^2/(2\pi (b-c))$. Thus $F_y=0$ and $F_x=\rho {\Gamma }^{2}b/(2\pi (b^2-a^2))>0$: the force on the cylinder points toward the vortex. The specified cylinder circulation is part of the hypothesis; a zero-circulation cylinder has an additional central image term.

For steady incompressible flow and irrotational flow of constant mass density, with no body force, this formula gives the hydrodynamic torque about the origin per unit span on an impermeable body. Orient its boundary counterclockwise, and let positive torque be counterclockwise. The proof uses $M=\mathrm{Re}{\oint }_{C}pzd\bar{z}$, the Bernoulli equation $p=P_{\ast }-\rho |w^{\prime }{|}^2/2$, and the tangency identity $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. The real part of the constant-pressure term vanishes because $\mathrm{Re}(zd\bar{z})=d|z{|}^2/2$. The full complex integral of $zd\bar{z}$ need not vanish.

In a planar potential flow, superpose a positive-$x$ uniform flow of speed $U$ and a line source of strength $Q=2\pi m$ at the real point $b>1$. Introduce an impermeable unit circular cylinder with zero circulation. The Milne-Thomson circle theorem gives the complex potential of a two-dimensional potential flowThe displayed horizontal force per unit length acts on the cylinder, and $F_y=0$ by reflection symmetry. To prove it, let $q=w^{\prime }$ be the complex velocity. At infinity, $q=U+m/z+O(z^{-2})$, so the sum of all finite residues of $q^2$ is $2Um$. At the external pole $b$, write $q=m/(z-b)+H(z)$; thenThe residue theorem and the counterclockwise Blasius force theorem give $F_x-i{F}_y=2\pi \rho m[H(b)-U]$, proving the formula. With $m>0$ and $U>0$, its sign changes at $b=(m+\sqrt{m^2+4U^2})/(2U)$.

The Milne-Thomson circle theorem constructs a circular impermeable boundary in a two-dimensional potential flow. Suppose an incident complex potential of a two-dimensional potential flow $f$ is analytic on and inside $|z|=a$, with its physical singularities outside that disk. Then the displayed $W$ has constant stream function on $|z|=a$. The proof is the identity $a^2/\bar{z}=z$ on the circle: there $W=f(z)+\overline{f(z)}$ is real. Hence the circle is a streamline and satisfies the no-penetration boundary condition. The reflected term has its image singularities inside the obstacle. Additional imposed circulation must be specified separately. The theorem is named after Louis Melville Milne-Thomson.

A two-dimensional potential doublet is a localized potential flow singularity with complex potential of a two-dimensional potential flow $W=M/(z-b)$, where $M=m{e}^{i\alpha }$ specifies strength and orientation. It is the limiting field of an equal point source and sink as their separation tends to zero with strength times separation fixed, with the sign absorbed in $M$. Its complex velocity is $-M/(z-b{)}^2$, so the velocity decays as inverse distance squared. The total volumetric flow rate around the singularity is zero. Its two-dimensional decay differs from that of the three-dimensional source dipole.

For a complex potential of a two-dimensional potential flow $W=\varphi +i\psi$ in the coordinate $z=x+iy$, the Cauchy-Riemann equations give $W^{\prime }={\varphi }_x-i{\varphi }_y=u-iv$. A material point with physical velocity $(u,v)$ therefore satisfies $\dot{z}=\overline{W^{\prime }(z)}$. Complex conjugation matters when deriving a pathline from a complex potential of a two-dimensional potential flow.

The hodograph plane records velocity rather than position. For a two-dimensional potential flow, the complex coordinate $q=u-iv$ is the complex velocity. A straight impermeable wall fixes the velocity direction and hence maps to a radial line in this plane. A constant-speed free streamline maps to a circle centered at zero. Thus a free-boundary problem in physical space can become a fixed geometric domain in velocity space. Recovering the physical map then uses $dz/dw=1/q(w)$, wherever $q$ is nonzero.

A unit-speed free layer turning downward around a reentrant wall corner has complex velocity $q=u-iv$ in the first-quadrant exterior of the unit disc. Upstream $q=1$, downstream $q=i$, and the corner has infinite speed because its fluid angle is $3\pi /2$. Squaring $q$, applying $(s-1)/(s+1)$, and squaring again maps this domain to the upper half-plane with upstream, corner and downstream at $0,1,\infty$. The potential strip map $e^{\pi w}$ has the same normalization, proving the formula. The sign of the turn and the fluid corner angle determine whether the relevant hodograph is an exterior or interior quarter disc.

A semi-infinite horizontal plate at unit height and a floor at zero give a velocity potential strip of height $a$ and an upper unit semicircle in the complex velocity $q$. Their matching half-plane maps give $(1+e^{\pi w/a}{)}^{-1}=((q-1)/(q+1){)}^2$. Along the fluid free boundary put $q=e^{-i\theta }$, $-\pi <\theta <0$. Then $dz/d\theta =-(a/\pi )e^{i\theta }\cot (\theta /2)$ and $y=1-(a/\pi )(\theta +\pi +\sin \theta )$. Its downstream limit is $1-a$, which must equal the downstream depth $a$. Thus the contraction is exactly one half.

For a planar reservoir draining below a vertical lip of height one into a unit-speed stream, the potential strip $0<\mathrm{Im}w<a$ and first-quadrant unit velocity disk satisfyOn the fluid free boundary put $q=e^{-i\theta }$, $-\pi /2\le \theta <0$. Then $dz/d\theta =-2a{e}^{i\theta }\cot \theta /\pi$, so the downstream height is $1-2a/\pi$. Equating it to $a$ gives $a=\pi /(\pi +2)$.

For an inviscid irrotational unit-speed layer turning through a right angle, the complex potential of a two-dimensional potential flow occupies the strip $0<\mathrm{Im}w<1$ and its complex velocity $q$ occupies a fourth-quadrant quarter disk. The maps $r=q^2$, $s=(r-1)/(r+1)$ and $s^2$ take this disk to the upper half-plane, sending upstream velocity one, corner stagnation zero, and downstream velocity $-i$ to $0,1,\infty$. The map $e^{\pi w}$ has the same normalization, proving the displayed relation.

The conformal map of a vertical half-strip by sine sends $-\pi /2<\mathrm{Re}z<\pi /2$, $\mathrm{Im}z>0$ to the upper half-plane. The complex potential of a two-dimensional potential flow $W(z)=\log (a+\sin z)$ therefore pulls back the radial logarithmic flow originating at $\zeta =-a$. Its complex velocity is $W^{\prime }=\cos z/(a+\sin z)$, with a boundary singularity at $z=-\pi /2+i\mathrm{arcosh}a$. For $z=x+iy$, define $D=(a+\sin x\cosh y{)}^2+{\cos }^{2}x{\sinh }^{2}y$. ThenThe bottom boundary carries rightward flow, the right boundary upward flow, and the left boundary flow away from the singular point. The velocity tends to $(0,1)$ at large height. The finite bottom corners are stagnation points, not singularities of $W$.