Conformal Mapping & Plane Potential Theory

Derivation

Conformal Mapping & Plane Potential Theory

D-189 Home PU-205 Threads fields Depends on Cauchy-Riemann Equations & Harmonic Conjugates, laplace-equation-separation

Statement

Let \(w=f(z)\) be analytic and non-degenerate (\(f'(z)\neq 0\)) on a domain \(D\) of the complex plane, with \(z=x+iy\) and \(w=u+iv\). Then \(f\) is conformal: it preserves the angle and orientation between smooth curves crossing at any point of \(D\). Moreover, if \(\phi(u,v)\) is harmonic (\(\Delta_w\phi=0\)) on the image \(f(D)\), then the pullback \(\Phi(x,y)=\phi\big(u(x,y),v(x,y)\big)\) is harmonic on \(D\). Consequently a plane potential problem \(\Delta\Phi=0\) with Dirichlet or Neumann data on \(\partial D\) can be transported to a simpler domain, solved there, and mapped back, with the transformation law \(\Delta_z\Phi=|f'(z)|^{2}\,\Delta_w\phi\).

Why it matters

Two-dimensional electrostatics and incompressible, irrotational (ideal) fluid flow are both governed by Laplace's equation for a scalar potential. Analytic functions form an enormous library of maps that turn awkward boundaries — a charged strip, a wedge, an aerofoil, the region outside a cylinder — into trivially simple ones such as a half-plane or the interior of a circle, where the solution is already known. Solve once on the simple domain, transport back, and the hard geometry is dispatched exactly, not numerically.

The method is the analytic engine behind the Joukowski aerofoil, the edge fields of capacitors, the capacitance of oddly shaped conductors, and the Schwarz–Christoffel treatment of polygonal boundaries. It is the concrete pay-off of the Cauchy–Riemann equations: complex differentiability is exactly the condition that geometry and potential theory travel together under the map.

Assumptions

\(f\) is analytic (holomorphic) on \(D\).Without complex differentiability the Cauchy–Riemann equations fail, \(u\) and \(v\) need not be harmonic, and neither angle preservation nor the Laplacian transformation law holds.

\(f'(z)\neq 0\) throughout \(D\).At a critical point \(f'(z_0)=0\) the map is not conformal there: if \(f'\) vanishes to order \(k\), angles are multiplied by \(k+1\), so a right angle can open into a straight line and the geometry is distorted.

The problem is genuinely two-dimensional (no \(z_3\)-dependence, invariance along one Cartesian axis).Analytic-function methods live in the plane; a field with real three-dimensional structure has no complex-potential representation and the whole apparatus is unavailable.

The domain and boundary data are compatible with a single-valued potential (or the multivaluedness is physically intended, e.g. circulation).Maps like \(\log z\) are multivalued; if branch structure is ignored the recovered potential is inconsistent, though a controlled multivalued part legitimately encodes net charge per length or circulation.

Derivation

\[ f(z_0+\delta z)=f(z_0)+f'(z_0)\,\delta z+o(|\delta z|) \]

Analyticity gives a genuine complex derivative, so near \(z_0\) the map is a linear map \(\delta w\approx f'(z_0)\,\delta z\) plus higher order. A

\[ f'(z_0)=\rho\,e^{i\alpha},\qquad \delta w \approx \rho\,e^{i\alpha}\,\delta z \]

Write the (non-zero) derivative in polar form. Multiplication by \(f'(z_0)\) is a rotation by \(\alpha=\arg f'(z_0)\) and a scaling by \(\rho=|f'(z_0)|\). A

\[ \arg(\delta w_2)-\arg(\delta w_1)=\big(\alpha+\arg\delta z_2\big)-\big(\alpha+\arg\delta z_1\big)=\arg(\delta z_2)-\arg(\delta z_1) \]

Two curves through \(z_0\) with tangent increments \(\delta z_1,\delta z_2\) both acquire the same rotation \(\alpha\), which cancels in their difference. The angle between them — in magnitude and sense — is preserved. This is conformality. A

\[ u_x=v_y,\qquad u_y=-v_x \]

The Cauchy–Riemann equations (prior result) hold because \(f\) is analytic; they are the coordinate form of "\(f'\) exists as a complex number". B

\[ u_{xx}=v_{yx},\qquad u_{yy}=-v_{xy}\;\Rightarrow\; u_{xx}+u_{yy}=v_{yx}-v_{xy}=0 \]

Differentiate the first CR relation by \(x\) and the second by \(y\), add, and use equality of mixed partials \(v_{yx}=v_{xy}\). Hence \(u\) is harmonic; identically \(v\) is harmonic. Real and imaginary parts of an analytic function are harmonic conjugates. B

\[ \Phi(x,y)=\phi\big(u(x,y),v(x,y)\big) \]

Define the pullback of a target-plane field \(\phi(u,v)\) through the map. We now compute its Laplacian in \(z\). B

\[ \Phi_x=\phi_u u_x+\phi_v v_x,\qquad \Phi_y=\phi_u u_y+\phi_v v_y \]

Chain rule in two variables. B

\[ \Phi_{xx}+\Phi_{yy}=\phi_{uu}\big(u_x^2+u_y^2\big)+\phi_{vv}\big(v_x^2+v_y^2\big)+2\phi_{uv}\big(u_xv_x+u_yv_y\big)+\phi_u\Delta u+\phi_v\Delta v \]

Differentiate again, collecting the pure second derivatives of \(\phi\), the cross term, and the terms multiplying \(\Delta u,\Delta v\). C

\[ \Delta u=\Delta v=0,\qquad u_xv_x+u_yv_y = u_x(-u_y)+u_y(u_x)=0 \]

By steps 4–5 both harmonics vanish; the cross term vanishes by the CR equations (\(v_x=-u_y,\ v_y=u_x\)). The mixed \(\phi_{uv}\) term drops out entirely. C

\[ u_x^2+u_y^2=u_x^2+v_x^2=|f'(z)|^{2},\qquad v_x^2+v_y^2=v_x^2+u_x^2=|f'(z)|^{2} \]

Use \(u_y=-v_x\) and \(v_y=u_x\), and \(f'(z)=u_x+iv_x\), so \(|f'|^2=u_x^2+v_x^2\). Both prefactors equal the common Jacobian factor. C

\[ \boxed{\;\Delta_z\Phi=\big(\phi_{uu}+\phi_{vv}\big)\,|f'(z)|^{2}=|f'(z)|^{2}\,\Delta_w\phi\;} \]

Substitute steps 9–10 into step 8. Since \(|f'(z)|^2\neq 0\) on \(D\), \(\Delta_z\Phi=0\iff\Delta_w\phi=0\): harmonicity is preserved exactly. C

Result

\[ f'(z)\neq 0\ \text{analytic}\ \Longrightarrow\ \text{angles preserved, and}\quad \Delta_z\Phi=|f'(z)|^{2}\,\Delta_w\phi \]

Reading. An analytic map with non-vanishing derivative rotates-and-scales every infinitesimal neighbourhood by the single complex number \(f'(z)\), so it preserves angles (conformal). The Laplacian merely picks up the positive real factor \(|f'(z)|^2\); therefore a harmonic potential stays harmonic under the map. To solve \(\Delta\Phi=0\) on an awkward domain \(D\), map \(D\) conformally to a simple domain, carry the boundary values along, solve the easy Laplace problem there, and pull the solution back.

Units check. \(z,w\) are dimensionless position ratios (or carry length; the map is a pure geometric relation), so \(f'(z)\) is dimensionless and \(|f'|^2\) is a dimensionless multiplier. If \(\phi\) is a potential in volts (V), \(\Phi\) is in volts too, and both \(\Delta_z\Phi\) and \(|f'|^2\Delta_w\phi\) carry units \(\mathrm{V\,m^{-2}}\); the identity is dimensionally consistent.

Limiting cases

\(f(z)=az+b\) (\(a\neq0\)): a similarity — rigid rotation, uniform scale, translation. \(|f'|^2=|a|^2\) constant; every harmonic function maps to a harmonic function with no local distortion.
\(f(z)=z^{n}\): conformal except at the critical point \(z=0\), where \(f'=0\) and a wedge of opening \(\theta\) is stretched to \(n\theta\) — the standard tool for corner and edge fields.
\(f(z)=1/z\): inversion in the unit circle composed with reflection; maps circles-and-lines to circles-and-lines, the workhorse for exterior-of-cylinder problems.
Near any non-critical point the map degenerates to the pure similarity \(\delta w\approx f'(z_0)\delta z\); conformality is fundamentally a local, first-order statement.
Identity map \(f(z)=z\): \(|f'|^2=1\), \(\Delta_z\Phi=\Delta_w\phi\) — the transformation law reduces to nothing, as it must.

Breaks when

Critical points, \(f'(z_0)=0\). Conformality fails at \(z_0\); angles are multiplied by the order of the first non-vanishing derivative plus one. Field lines can develop stagnation points or corners there (e.g. the trailing edge of a Joukowski aerofoil), and the Laplacian factor \(|f'|^2\) collapses to zero.
Three-dimensional problems. The entire complex-potential machinery is planar. A point charge, a finite rod, or any field with true \(z_3\)-dependence has no analytic-function representation; attempting a 2D conformal solve silently answers a different (infinite-cylinder) problem.
Non-Laplacian physics. The moment the governing equation is not \(\Delta\Phi=0\) — space charge (Poisson \(\Delta\Phi=-\rho/\varepsilon_0\)), compressible or viscous flow, screening (Helmholtz/Yukawa), or nonlinear media — harmonicity is not the invariant and mapping no longer transports the solution.
Branch cuts and multivaluedness mishandled. With logarithmic or fractional-power maps the potential is multivalued; ignoring the branch structure (net line charge, circulation) yields a discontinuous, unphysical result.

Failure modes

Dropping the \(|f'|^2\) factor when transforming source terms. Harmonicity is preserved without it, but a Poisson right-hand side, a charge density, or a field-energy density does not transform trivially — the Jacobian \(|f'|^2\) must be carried.
Assuming field magnitudes are preserved. Only angles are. \(|\mathbf E|\) in the \(z\)-plane equals \(|f'(z)|\) times its value in the \(w\)-plane; students routinely read the target-plane field strength as the physical one.
Using a map that is not injective on \(D\). If \(f\) is not one-to-one, the "inverse image" is ambiguous and boundary data get double-counted; conformality is local, global bijectivity is a separate requirement (Riemann mapping theorem guarantees it only for simply connected proper subdomains).
Confusing the potential \(\Phi=\operatorname{Re}\Omega\) with the stream function \(\Psi=\operatorname{Im}\Omega\). Both are harmonic and conjugate; swapping them exchanges equipotentials and field lines and gets boundary conditions backwards.
Ignoring critical points in the interior. Solving as if \(f\) were everywhere conformal misplaces stagnation points and mis-predicts corner-field singularities.
Applying the method to a doubly connected region without accounting for the topological invariant (the modulus / conformal type), then being surprised the map to a disc does not exist.

Discussion

The deep content is that a single algebraic fact — the derivative of an analytic function is one complex number \(f'(z)\), a rotation-scaling — simultaneously delivers a geometric theorem (angle preservation) and an analytic theorem (harmonicity preservation). These are not two coincidences; they are the same statement seen through the Cauchy–Riemann equations. The CR equations say the Jacobian of the map is a scalar times a rotation, which is precisely what makes it angle-preserving, and precisely what makes the Laplacian transform by a scalar with no cross term.

The natural object is the complex potential \(\Omega(z)=\Phi(x,y)+i\Psi(x,y)\), itself analytic. In electrostatics \(\Phi\) is the electric potential and \(\Psi\) the flux function; in ideal flow \(\Phi\) is the velocity potential and \(\Psi\) the stream function. Because \(\Omega\) is analytic, the physical field is read straight off the derivative: \(\overline{E_x+iE_y}=-\,\Omega'(z)\) in electrostatics, and \(u-iv=\Omega'(z)\) (velocity components) in flow. Equipotentials \(\Phi=\text{const}\) and field lines \(\Psi=\text{const}\) form an orthogonal net — orthogonal because \(\Phi,\Psi\) are harmonic conjugates and their gradients are perpendicular. Conformal mapping simply carries this orthogonal net, undistorted in angle, from a simple domain onto the real one.

The rigorous backbone is the Riemann mapping theorem: any simply connected domain that is not the whole plane is conformally equivalent to the open unit disc, and the map is essentially unique once three real boundary parameters are fixed. This guarantees that "map to something simple" is not wishful thinking but a theorem — with the important caveats that multiply connected domains carry conformal invariants (a topological modulus) obstructing a map to the disc, and that boundary regularity of the map (Carathéodory's theorem) is what lets Dirichlet data be transported continuously. Schwarz–Christoffel construction realises maps onto arbitrary polygons explicitly, turning the abstract existence into a computable integral \(f'(z)=C\prod_k (z-a_k)^{\beta_k}\).

Common misconceptions. Conformal maps are not shape-preserving — they preserve angles locally, not lengths, areas, or global shape; a square grid maps to a curvilinear net that meets at right angles but is stretched non-uniformly. "Conformal" is a first-order, pointwise property, so a map can badly deform finite figures while being perfectly conformal at every point (except critical points). And the method solves \(\Delta\Phi=0\) only: it is a tool for source-free potential theory in the plane, not a universal PDE solver.

Worked examples

Field in a conducting right-angle corner via \(w=z^{2}\)

Two semi-infinite grounded conducting plates meet at a right angle occupying the first quadrant \(0\le\arg z\le\pi/2\); far away a uniform field is imposed. Find the potential in the corner by mapping to a half-plane. B

\[ w=z^{2},\qquad z=x+iy,\ \ w=u+iv=(x^2-y^2)+2ixy \]

The map \(w=z^2\) opens the quarter-plane wedge (opening \(\pi/2\)) into the upper half-plane (opening \(\pi\)): angles at the corner double, \(2\times(\pi/2)=\pi\). B

\[ \phi(u,v)=A\,v\quad(\text{half-plane: potential} = A\times\text{height above real axis, grounded on }v=0) \]

In the half-plane the grounded boundary is the whole real axis \(v=0\); the simplest harmonic solution vanishing there is linear in \(v\), with \(A\) fixing the field strength. \(\phi\) is harmonic, \(\phi=0\) on \(v=0\). B

\[ \Phi(x,y)=\phi(u,v)=A\,v=2A\,xy \]

Pull back: \(v=2xy\). By the theorem \(\Phi=2Axy\) is automatically harmonic (\(\Phi_{xx}+\Phi_{yy}=0-0=0\)) and vanishes on both plates \(x=0\) and \(y=0\). B

\[ \mathbf E=-\nabla\Phi=-2A\,(y,\,x),\qquad |\mathbf E|=2A\sqrt{x^2+y^2}=2A\,r \]

Field grows linearly with distance from the corner and is tangent to neither plate except at the vertex, where it vanishes — the corner is a field null for this interior geometry. C

\[ \Phi(x,y)=2A\,xy,\qquad |\mathbf E|=2A\,r \]

Reading. The right-angle corner problem is solved exactly by doubling angles with \(z^2\). Take \(A=100\ \mathrm{V\,m^{-2}}\): at the point \((x,y)=(0.03,0.04)\ \mathrm{m}\), \(\Phi=2(100)(0.03)(0.04)=0.24\ \mathrm{V}\) and \(|\mathbf E|=2(100)(0.05)=10\ \mathrm{V\,m^{-1}}\).

Units check. \(A\) in \(\mathrm{V\,m^{-2}}\), \(xy\) in \(\mathrm{m^2}\) gives \(\Phi\) in V; \(2Ar\) has \(\mathrm{V\,m^{-2}\cdot m}=\mathrm{V\,m^{-1}}\), correct for a field.

Ideal flow past a cylinder via \(w=z+\frac{a^{2}}{z}\)

Uniform stream of speed \(U\) approaches a circular cylinder of radius \(a\). Find the velocity field and the stagnation points using the Joukowski-type complex potential. B

\[ \Omega(z)=U\!\left(z+\frac{a^{2}}{z}\right) \]

This complex potential is analytic outside \(|z|=a\). On the circle \(z=a e^{i\theta}\), \(\Omega=U(ae^{i\theta}+ae^{-i\theta})=2Ua\cos\theta\), which is real, so \(\Psi=\operatorname{Im}\Omega=0\): the circle is a streamline — a valid solid boundary. B

\[ \Omega'(z)=U\!\left(1-\frac{a^{2}}{z^{2}}\right)=u-iv\quad(\text{complex velocity}) \]

The velocity components come straight from the analytic derivative, \(\Omega'(z)=u-iv\). As \(|z|\to\infty\), \(\Omega'\to U\): recovers the uniform stream at infinity. B

\[ \Omega'(z)=0\ \Rightarrow\ 1-\frac{a^{2}}{z^{2}}=0\ \Rightarrow\ z=\pm a \]

Stagnation points where the velocity vanishes: the leading and trailing points of the cylinder on the flow axis. These are exactly the critical points \(f'=0\) at which conformality is lost — physically, flow splits and rejoins. C

\[ \text{On the surface } z=ae^{i\theta}:\quad \Omega'=U\big(1-e^{-2i\theta}\big),\qquad |\mathbf v|=2U|\sin\theta| \]

Surface speed: \(1-e^{-2i\theta}=e^{-i\theta}(e^{i\theta}-e^{-i\theta})=2i\sin\theta\,e^{-i\theta}\), magnitude \(2|\sin\theta|\). Maximum \(2U\) at the top and bottom (\(\theta=\pm\pi/2\)), zero at \(\theta=0,\pi\). C

\[ |\mathbf v|_{\text{surface}}=2U|\sin\theta|,\qquad \text{stagnation at }z=\pm a \]

Reading. The exact plane flow past a cylinder is read off an analytic complex potential; peak surface speed is twice the free-stream speed. For \(U=3\ \mathrm{m\,s^{-1}}\) and \(a=0.10\ \mathrm{m}\): peak surface speed \(2U=6\ \mathrm{m\,s^{-1}}\) at the shoulders, and (by Bernoulli, \(p+\tfrac12\rho v^2=\text{const}\), \(\rho=1000\ \mathrm{kg\,m^{-3}}\)) the shoulder pressure drop below free stream is \(\tfrac12\rho\big((2U)^2-U^2\big)=\tfrac12(1000)(36-9)=1.35\times10^{4}\ \mathrm{Pa}\).

Units check. \(2U|\sin\theta|\) has units of \(U\), i.e. \(\mathrm{m\,s^{-1}}\); the pressure drop \(\tfrac12\rho v^2\) has \(\mathrm{kg\,m^{-3}\cdot m^2\,s^{-2}}=\mathrm{Pa}\), correct.

Problems

(A) Angle multiplication. A wedge of interior opening \(\pi/3\) is mapped by \(w=z^{n}\) onto a half-plane. Find \(n\).
Solution
The map \(z^n\) multiplies angles at the origin by \(n\). To open \(\pi/3\) to \(\pi\) requires \(n\cdot(\pi/3)=\pi\), so \(n=3\). The map is \(w=z^3\).
(B) Harmonicity of a pullback. Verify directly that \(\Phi(x,y)=x^2-y^2\), the real part of \(f(z)=z^2\), is harmonic, and identify its harmonic conjugate.
Solution
\(\Phi_{xx}=2\), \(\Phi_{yy}=-2\), so \(\Phi_{xx}+\Phi_{yy}=0\): harmonic. The conjugate \(\Psi\) satisfies CR: \(\Psi_y=\Phi_x=2x\Rightarrow\Psi=2xy+g(x)\); \(\Psi_x=-\Phi_y=2y\Rightarrow 2y+g'(x)=2y\Rightarrow g'=0\). Thus \(\Psi=2xy\), and \(f=z^2=(x^2-y^2)+2ixy\), consistent.
(C) Field strength scaling. Under a conformal map \(w=f(z)\), a target-plane field has magnitude \(|\mathbf E_w|=5\ \mathrm{V\,m^{-1}}\) at a point where \(|f'(z)|=0.4\). Find the physical field magnitude \(|\mathbf E_z|\) at the corresponding \(z\).
Solution
Since \(\mathbf E=-\nabla\Phi\) and \(\Phi=\phi\circ f\), the gradient picks up a factor \(|f'(z)|\): \(|\mathbf E_z|=|f'(z)|\,|\mathbf E_w|\). Numerically \(|\mathbf E_z|=0.4\times5=2\ \mathrm{V\,m^{-1}}\). (Angles preserved, magnitude scaled by \(|f'|\).)
(B) Line charge to uniform field. Using \(\Omega(z)=-\dfrac{\lambda}{2\pi\varepsilon_0}\ln z\) for a line charge \(\lambda\) at the origin, show the equipotentials are circles and find the field magnitude at radius \(r\).
Solution
Write \(z=re^{i\theta}\), \(\ln z=\ln r+i\theta\). Then \(\Phi=\operatorname{Re}\Omega=-\frac{\lambda}{2\pi\varepsilon_0}\ln r\), constant on circles \(r=\text{const}\): equipotentials are concentric circles. Field: \(|\mathbf E|=|\Omega'(z)|=\frac{\lambda}{2\pi\varepsilon_0}\left|\frac1z\right|=\frac{\lambda}{2\pi\varepsilon_0 r}\), the standard line-charge field. For \(\lambda=1\ \mathrm{nC\,m^{-1}}\) at \(r=0.05\ \mathrm{m}\): \(|\mathbf E|=\frac{10^{-9}}{2\pi(8.85\times10^{-12})(0.05)}\approx 360\ \mathrm{V\,m^{-1}}\).
(C) Critical point and stagnation. For flow past a cylinder, \(\Omega(z)=U(z+a^2/z)\) with \(U=2\ \mathrm{m\,s^{-1}}\), \(a=0.2\ \mathrm{m}\). Locate the stagnation points and compute the surface speed at \(\theta=30^\circ\).
Solution
Stagnation where \(\Omega'(z)=U(1-a^2/z^2)=0\Rightarrow z=\pm a=\pm0.2\ \mathrm{m}\) (fore and aft on the axis). Surface speed \(|\mathbf v|=2U|\sin\theta|=2(2)|\sin30^\circ|=2(2)(0.5)=2\ \mathrm{m\,s^{-1}}\). At the shoulders \(\theta=90^\circ\), \(|\mathbf v|=2U=4\ \mathrm{m\,s^{-1}}\), the maximum. The stagnation points are exactly the critical points \(\Omega'=0\), where conformality of the associated map fails and the flow divides.