Potential Flow and Laplace's Equation

Derivation

Potential Flow and Laplace's Equation

D-323 Home PU-307 Threads fields Depends on The Continuity Equation, Kelvin's Circulation Theorem

Statement

For the steady or unsteady flow of an incompressible fluid whose velocity field \( \mathbf{u}(\mathbf{x},t) \) is irrotational, \( \nabla \times \mathbf{u} = \mathbf{0} \), there exists a scalar velocity potential \( \phi \) with \( \mathbf{u} = \nabla \phi \), and mass conservation \( \nabla \cdot \mathbf{u} = 0 \) forces this potential to satisfy Laplace's equation \( \nabla^2 \phi = 0 \). The nonlinear dynamics of the flow field thereby collapse onto a single linear, elliptic partial differential equation for one scalar.

Why it matters

The Navier-Stokes and even the inviscid Euler equations are nonlinear in \( \mathbf{u} \) through the advective term \( (\mathbf{u}\cdot\nabla)\mathbf{u} \). Potential flow removes that nonlinearity entirely: the kinematics reduce to Laplace's equation, which is linear, so solutions superpose. Uniform streams, sources, sinks, doublets and vortices can be added at will to build flows past aerofoils and bluff bodies, and the whole machinery of harmonic-function theory, complex potentials and conformal mapping becomes available.

Physically it is the backbone of classical aerodynamics: lift, added mass, wave-making resistance and acoustics all begin here. The dynamics (pressure) then follow separately and algebraically from the unsteady Bernoulli equation once \( \phi \) is known, so the hard part is solved by one boundary-value problem.

Assumptions

Incompressible flow.If density varies with pressure the continuity equation reads \( \partial_t \rho + \nabla\cdot(\rho\mathbf{u}) = 0 \) and \( \nabla\cdot\mathbf{u} = -\tfrac{1}{\rho}\tfrac{D\rho}{Dt} \neq 0 \); the potential obeys a variable-coefficient or wave-type equation, not Laplace's equation.

Irrotational velocity field.If \( \boldsymbol{\omega} = \nabla\times\mathbf{u} \neq \mathbf{0} \) no single-valued potential \( \phi \) exists; one must retain vorticity and solve the full vortical dynamics or introduce a vector streamfunction.

Simply connected domain (or specified circulation).In a multiply connected region (flow around a closed body in 2D) \( \mathbf{u}=\nabla\phi \) may make \( \phi \) multivalued; the potential jumps by the circulation \( \Gamma \) per circuit, which must be prescribed as an extra condition (Kutta condition) to fix the flow.

Barotropic, conservative body forces so Kelvin's theorem applies.If baroclinic torque \( \nabla\rho\times\nabla p \neq \mathbf{0} \) or non-conservative forces act, vorticity is generated from an initially irrotational state and the irrotational assumption is not preserved in time.

Derivation

\[ \nabla \times \mathbf{u} = \mathbf{0} \quad \text{in a simply connected region} \]

Starting hypothesis: the flow is irrotational. By Kelvin's circulation theorem an inviscid, incompressible, barotropic flow started from rest stays irrotational, so this is dynamically self-consistent. A

\[ \mathbf{u} = \nabla \phi \]

A curl-free vector field on a simply connected domain is the gradient of a single-valued scalar potential (Poincaré lemma): since \( \oint \mathbf{u}\cdot d\boldsymbol{\ell}=0 \) for every closed loop, the line integral \( \phi(\mathbf{x}) = \int_{\mathbf{x}_0}^{\mathbf{x}} \mathbf{u}\cdot d\boldsymbol{\ell} \) is path-independent and defines \( \phi \). B

\[ \nabla \times (\nabla \phi) \equiv \mathbf{0} \]

Consistency check: the representation automatically satisfies irrotationality because the curl of any gradient vanishes identically. No information is lost in passing from \( \mathbf{u} \) to \( \phi \). C

\[ \nabla \cdot \mathbf{u} = 0 \]

Continuity for an incompressible fluid: mass conservation \( \partial_t\rho + \nabla\cdot(\rho\mathbf{u})=0 \) with \( \rho = \text{const} \) reduces to a divergence-free velocity field. A

\[ \nabla \cdot (\nabla \phi) = 0 \]

Substitute the potential representation from Step 2 into the incompressibility condition of Step 4. This is the single legal substitution that couples the two hypotheses. A

\[ \nabla \cdot \nabla \phi = \nabla^2 \phi = \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} \]

The divergence of the gradient is by definition the Laplacian operator \( \nabla^2 = \Delta \). Written in Cartesian components it is the sum of unmixed second partials. B

\[ \boxed{\;\nabla^2 \phi = 0\;} \]

Combining Steps 5 and 6: the velocity potential of an incompressible irrotational flow is a harmonic function. The problem is now linear and elliptic. A

Result

\[ \mathbf{u} = \nabla\phi, \qquad \nabla^2\phi = 0 \]

Reading. Incompressible + irrotational is exactly the kinematic content of "harmonic potential". The velocity is read off as the gradient of a scalar that obeys Laplace's equation; boundary conditions (typically \( \partial\phi/\partial n \) prescribed on solid walls, the no-penetration condition) close the problem. Because Laplace's equation is linear, elementary solutions superpose, and because it is elliptic, \( \phi \) is smooth in the interior and determined everywhere by data on the boundary.

Units check. From \( \mathbf{u}=\nabla\phi \), \( [\phi] = [\mathbf{u}][L] = (\mathrm{m\,s^{-1}})(\mathrm{m}) = \mathrm{m^2\,s^{-1}} \). Then \( [\nabla^2\phi] = \mathrm{m^2\,s^{-1}}/\mathrm{m^2} = \mathrm{s^{-1}} \), the same units as \( \nabla\cdot\mathbf{u} \) (a volumetric strain rate). Setting it to zero is dimensionally consistent: \( \mathrm{s^{-1}} = 0 \).

Limiting cases

Two dimensions. \( \nabla^2\phi = \partial_{xx}\phi + \partial_{yy}\phi = 0 \); \( \phi \) pairs with a streamfunction \( \psi \) as the real and imaginary parts of an analytic complex potential \( w(z)=\phi+i\psi \), so complex analysis solves the flow.
Uniform stream. \( \phi = U x \) gives \( \mathbf{u} = (U,0,0) \) and \( \nabla^2\phi = 0 \) trivially — the simplest harmonic building block.
Point source (3D). \( \phi = -\dfrac{Q}{4\pi r} \) is harmonic for \( r>0 \); the singularity at \( r=0 \) carries the volume flux \( Q \).
Compressible but subsonic. Linearising about a uniform stream at Mach \( M \) gives the Prandtl–Glauert equation \( (1-M^2)\phi_{xx}+\phi_{yy}=0 \), which returns to Laplace's equation as \( M\to 0 \).

Breaks when

Rotational / vortical flow. Boundary layers, wakes and separated flows carry vorticity \( \boldsymbol\omega\neq\mathbf 0 \); no single-valued \( \phi \) exists there, and potential theory famously predicts zero drag on a body (d'Alembert's paradox) precisely because it omits these regions.
Compressible / high-speed flow. Once \( \mathbf u \) is comparable to the sound speed, \( \nabla\cdot\mathbf u\neq0 \); near and above \( M=1 \) the governing equation changes type (elliptic → hyperbolic) and shocks appear, so Laplace's equation is invalid.
Multiply connected domains without a Kutta condition. Around a lifting 2D body \( \phi \) is multivalued; circulation \( \Gamma \) is undetermined by \( \nabla^2\phi=0 \) alone and must be fixed physically, otherwise the "solution" is non-unique.
Free surfaces and gravity waves. Although \( \nabla^2\phi=0 \) still holds in the bulk, the nonlinear dynamic boundary condition on a moving free surface makes the overall problem nonlinear again.

Failure modes

Confusing \( \phi \) with the streamfunction \( \psi \). \( \mathbf u=\nabla\phi \) (potential) but \( \mathbf u = \nabla\times(\psi\hat{\mathbf z}) \) in 2D (streamfunction). Both are harmonic here, but \( \phi=\text{const} \) lines are orthogonal to streamlines, not along them.
Assuming incompressible \( \Rightarrow \) irrotational. They are independent conditions; a divergence-free flow can be highly rotational (e.g. solid-body rotation). Both are needed for Laplace's equation.
Dropping a sign in the source potential. Writing \( \phi=+Q/4\pi r \) reverses the radial velocity; a source (outflow) needs \( u_r=+Q/4\pi r^2>0 \), so \( \phi=-Q/4\pi r \).
Using Bernoulli's steady form for unsteady flow. With time-dependent \( \phi \) one must keep the \( \partial\phi/\partial t \) term in \( \tfrac{\partial\phi}{\partial t}+\tfrac12|\nabla\phi|^2+\tfrac{p}{\rho}+gz=\text{const} \).
Forgetting the boundary condition is on the normal derivative. Solid walls impose \( \partial\phi/\partial n = 0 \) (Neumann), not \( \phi=0 \) (Dirichlet); imposing the wrong type gives an unphysical flow.

Discussion

The remarkable feature is linearisation without approximation. Nothing has been neglected: given genuine incompressibility and irrotationality, Laplace's equation is exact. The nonlinearity of the Euler equations has not been thrown away — it has been quarantined into the pressure. Once \( \phi \) is found from the linear kinematic problem, the unsteady Bernoulli equation returns the pressure field algebraically, and the momentum balance is then satisfied identically. This clean split between kinematics (linear, elliptic) and dynamics (a postprocessing step) is what makes potential flow so powerful.

Ellipticity has strong physical consequences. Laplace's equation admits a maximum principle: \( \phi \) attains its extrema on the boundary, so there are no interior "surprises", and information propagates instantaneously throughout the domain — a disturbance anywhere is felt everywhere at once. This is the mathematical signature of incompressibility, where the effective sound speed is infinite. It also means potential flow has no memory and no waves in the bulk: the field is slaved entirely to its instantaneous boundary data.

The link to Kelvin's circulation theorem is what makes the irrotational assumption dynamically stable rather than a lucky initial condition. Kelvin showed that for an inviscid, incompressible, barotropic fluid under conservative forces the circulation \( \Gamma = \oint_C \mathbf u\cdot d\boldsymbol\ell \) around any material loop is conserved, hence \( \boldsymbol\omega=\mathbf 0 \) is preserved for all time if it holds initially. In multiply connected domains the same theorem quantises the allowed multivaluedness of \( \phi \): the jump in \( \phi \) around a body equals its circulation, and the Kutta condition selects the physical value by demanding smooth flow off a sharp trailing edge — the mechanism behind aerodynamic lift.

Common misconceptions. Potential flow is often dismissed as "the physics with no drag", but this understates it: it correctly predicts lift, added mass, pressure distributions away from separation, and the entire outer flow that boundary-layer theory is matched to. The absence of drag (d'Alembert's paradox) is not a flaw in the mathematics but an honest report that viscosity and vorticity have been excluded — the very things that produce drag.

Worked examples

Verify a stagnation-point flow is a valid potential flow and find its speed at a point.

Take the two-dimensional plane stagnation flow with potential \( \phi = \tfrac12 A\,(x^2 - y^2) \), where \( A \) is a strain rate. A

\[ \mathbf u = \nabla\phi = \left(\frac{\partial\phi}{\partial x},\,\frac{\partial\phi}{\partial y}\right) = (A x,\,-A y) \]

Differentiate the potential. The flow streams toward the wall along \( y \) and away along \( x \), stagnating at the origin. A

\[ \nabla^2\phi = \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} = A + (-A) = 0 \]

Laplace's equation is satisfied identically, so this is an admissible incompressible irrotational flow. A

\[ |\mathbf u| = \sqrt{(Ax)^2 + (Ay)^2} = A\sqrt{x^2+y^2} \]

Symbolic speed before numbers. Now insert \( A = 2.0\ \mathrm{s^{-1}} \) at the point \( (x,y)=(0.30,\,0.40)\ \mathrm{m} \). A

\[ |\mathbf u| = (2.0\ \mathrm{s^{-1}})\sqrt{(0.30)^2+(0.40)^2}\ \mathrm{m} = 2.0\times0.50 = 1.0\ \mathrm{m\,s^{-1}} \]

Evaluate. Components: \( \mathbf u = (0.60,\,-0.80)\ \mathrm{m\,s^{-1}} \). A

\[ \nabla^2\phi = 0,\qquad |\mathbf u|(0.3,0.4) = 1.0\ \mathrm{m\,s^{-1}} \]

Reading. The potential is harmonic, confirming a valid potential flow; the local speed is \( 1.0\ \mathrm{m/s} \), directed up-and-out from the stagnation point at the origin. Units check. \( \mathrm{s^{-1}}\times\mathrm{m}=\mathrm{m\,s^{-1}} \), a speed. Good.

A three-dimensional point source: check harmonicity and compute the radial velocity.

A source of volume flux \( Q \) has potential \( \phi = -\dfrac{Q}{4\pi r} \) with \( r=\sqrt{x^2+y^2+z^2} \). B

\[ u_r = \frac{\partial\phi}{\partial r} = -\frac{Q}{4\pi}\frac{d}{dr}\!\left(\frac1r\right) = -\frac{Q}{4\pi}\left(-\frac1{r^2}\right) = \frac{Q}{4\pi r^2} \]

Radial velocity from the gradient in spherical symmetry. Positive \( u_r \) confirms outflow. A

\[ \nabla^2\phi = \frac{1}{r^2}\frac{d}{dr}\!\left(r^2\frac{d\phi}{dr}\right) = \frac{1}{r^2}\frac{d}{dr}\!\left(r^2\cdot\frac{Q}{4\pi r^2}\right) = \frac{1}{r^2}\frac{d}{dr}\!\left(\frac{Q}{4\pi}\right) = 0 \]

Using the spherical radial Laplacian. The bracket is constant, so its derivative vanishes for all \( r>0 \); the source is harmonic everywhere except its singular centre. C

\[ u_r = \frac{Q}{4\pi r^2} \]

Symbolic result before numbers. Take \( Q = 0.020\ \mathrm{m^3\,s^{-1}} \) at radius \( r = 0.50\ \mathrm{m} \). A

\[ u_r = \frac{0.020\ \mathrm{m^3\,s^{-1}}}{4\pi\,(0.50\ \mathrm{m})^2} = \frac{0.020}{4\pi\times0.25}\ \mathrm{m\,s^{-1}} = \frac{0.020}{3.1416}\approx 6.4\times10^{-3}\ \mathrm{m\,s^{-1}} \]

Evaluate the flux over the sphere of area \( 4\pi r^2 = \pi\ \mathrm{m^2} \). A

\[ \nabla^2\phi = 0\ (r>0),\qquad u_r(0.5\,\mathrm{m}) \approx 6.4\times10^{-3}\ \mathrm{m\,s^{-1}} \]

Reading. The source potential solves Laplace's equation away from its centre; at \( 0.5\ \mathrm m \) the radial outflow is about \( 6.4\ \mathrm{mm/s} \), falling as \( 1/r^2 \). Units check. \( \mathrm{m^3\,s^{-1}}/\mathrm{m^2}=\mathrm{m\,s^{-1}} \), a speed. Good.

Problems

Show that the 2D potential \( \phi = U x + \dfrac{m}{2\pi}\ln r \) (uniform stream plus line source, \( r=\sqrt{x^2+y^2} \)) satisfies Laplace's equation, and locate the stagnation point for \( U=1.0\ \mathrm{m/s} \), \( m=2\pi\ \mathrm{m^2/s} \).
Solution
Each term is harmonic: \( \nabla^2(Ux)=0 \) trivially, and \( \nabla^2\ln r = 0 \) for \( r>0 \) (the 2D source is harmonic away from the origin); by linearity the sum is harmonic. Velocity: \( u_x = U + \dfrac{m}{2\pi}\dfrac{x}{r^2} \), \( u_y = \dfrac{m}{2\pi}\dfrac{y}{r^2} \). Stagnation requires \( \mathbf u=0 \): \( u_y=0 \Rightarrow y=0 \); then \( U + \dfrac{m}{2\pi}\dfrac{1}{x}=0 \Rightarrow x = -\dfrac{m}{2\pi U} \). With \( m/2\pi = 1.0\ \mathrm{m^2/s}\cdot\mathrm{(per\ 2\pi)}=1.0\ \mathrm{m^2/s} \) and \( U=1.0\ \mathrm{m/s} \): \( x = -1.0\ \mathrm m \). Stagnation point at \( (-1.0,\,0)\ \mathrm m \).
A velocity field is given by \( \mathbf u = (2xy,\,x^2 - y^2,\,0)\ \mathrm{s^{-1}}\!\cdot\mathrm m \). Is it incompressible? Is it irrotational? If both, find a potential \( \phi \).
Solution
Incompressible: \( \nabla\cdot\mathbf u = \partial_x(2xy)+\partial_y(x^2-y^2) = 2y + (-2y) = 0 \). Yes. Irrotational (\( z \)-component of curl): \( \partial_x(x^2-y^2)-\partial_y(2xy) = 2x - 2x = 0 \). Yes. Potential: \( \partial_x\phi = 2xy \Rightarrow \phi = x^2 y + f(y) \); then \( \partial_y\phi = x^2 + f'(y) = x^2 - y^2 \Rightarrow f'(y)=-y^2 \Rightarrow f=-y^3/3 \). Thus \( \phi = x^2 y - \tfrac13 y^3 \) (up to a constant), and one checks \( \nabla^2\phi = 2y - 2y = 0 \).
For the 3D source of Worked Example 2 with \( Q = 0.020\ \mathrm{m^3/s} \), compute the volume flux through a sphere of radius \( r=0.80\ \mathrm m \) and confirm it is independent of \( r \).
Solution
Flux \( = u_r \cdot 4\pi r^2 = \dfrac{Q}{4\pi r^2}\cdot 4\pi r^2 = Q \). The \( r^2 \) factors cancel, so the flux equals \( Q = 0.020\ \mathrm{m^3/s} \) through any enclosing sphere — this is just conservation of mass, and it is why \( \phi \) is harmonic away from the singularity (no sources in the bulk). Numerically \( u_r(0.8) = 0.020/(4\pi\cdot0.64)=2.49\times10^{-3}\ \mathrm{m/s} \), and \( 2.49\times10^{-3}\times4\pi\times0.64 = 0.020\ \mathrm{m^3/s} \).
A doublet in 2D has \( \phi = -\dfrac{\mu\cos\theta}{r} \) in polar coordinates. Show it is harmonic for \( r>0 \). (Use \( \nabla^2 = \tfrac1r\partial_r(r\partial_r) + \tfrac1{r^2}\partial_{\theta\theta} \).)
Solution
Write \( \phi = -\mu\cos\theta\, r^{-1} \). Radial part: \( \partial_r\phi = \mu\cos\theta\,r^{-2} \); \( r\partial_r\phi = \mu\cos\theta\,r^{-1} \); \( \partial_r(r\partial_r\phi) = -\mu\cos\theta\,r^{-2} \); \( \tfrac1r\partial_r(r\partial_r\phi) = -\mu\cos\theta\,r^{-3} \). Angular part: \( \partial_{\theta\theta}\phi = -\mu(-\cos\theta)r^{-1}\cdot(-1)\)… carefully: \( \partial_\theta\phi = \mu\sin\theta\,r^{-1} \), \( \partial_{\theta\theta}\phi = \mu\cos\theta\,r^{-1} \); \( \tfrac1{r^2}\partial_{\theta\theta}\phi = \mu\cos\theta\,r^{-3} \). Sum: \( -\mu\cos\theta\,r^{-3} + \mu\cos\theta\,r^{-3} = 0 \). Harmonic for \( r>0 \).
Uniform flow past a cylinder of radius \( a \) has \( \phi = U\!\left(r + \dfrac{a^2}{r}\right)\cos\theta \). Verify the no-penetration condition \( u_r = 0 \) on \( r=a \), and give the surface speed.
Solution
Radial velocity: \( u_r = \partial_r\phi = U\!\left(1 - \dfrac{a^2}{r^2}\right)\cos\theta \). At \( r=a \): \( u_r = U(1-1)\cos\theta = 0 \) for all \( \theta \) — the cylinder surface is a streamline, as required. Tangential velocity: \( u_\theta = \tfrac1r\partial_\theta\phi = -U\!\left(1 + \dfrac{a^2}{r^2}\right)\sin\theta \); at \( r=a \), \( u_\theta = -2U\sin\theta \), so the surface speed is \( |u_\theta| = 2U|\sin\theta| \), peaking at \( 2U \) on the sides (\( \theta=\pm\pi/2 \)) and vanishing at the front and rear stagnation points (\( \theta=0,\pi \)). For \( U=3.0\ \mathrm{m/s} \) the maximum surface speed is \( 6.0\ \mathrm{m/s} \).