Velocity potential and Laplace's equation

Why it matters

Outside the thin boundary layer, the flow around a wing at low speed is very nearly incompressible and irrotational. Under those two assumptions the full Navier–Stokes problem collapses to one linear equation for one scalar — Laplace's equation for the velocity potential. Every classical result that follows (cylinder flow, Kutta–Joukowski lift, thin airfoil theory, panel methods) is a solution of this equation.

Key ideas

Vorticity and irrotational flow. Vorticity ω = ∇ × V is twice the local angular velocity of a fluid element. A flow with ω = 0 everywhere is irrotational. A flow starting from rest or from a uniform stream, with no viscous action, stays irrotational (Kelvin's theorem, later topic) — so the outer flow around a streamlined body is irrotational to a very good approximation.

Velocity potential. For any irrotational flow there exists a scalar φ with V = ∇φ, because the curl of a gradient is always zero. Irrotationality alone is the condition — φ exists in compressible irrotational flow too. In 2-D Cartesian coordinates u = ∂φ/∂x, v = ∂φ/∂y; in polar coordinates V_r = ∂φ/∂r, V_θ = (1/r)∂φ/∂θ. Lines of constant φ are equipotential lines.

Laplace's equation. Incompressible continuity is ∇·V = 0. Substituting V = ∇φ gives ∇²φ = 0. So:

• irrotational ⇒ φ exists;
• irrotational + incompressible ⇒ φ satisfies Laplace's equation. A candidate φ that does not satisfy ∇²φ = 0 can still describe an irrotational flow, but not an incompressible one.

Stream function (2-D). Incompressible 2-D continuity is satisfied identically by ψ with u = ∂ψ/∂y, v = −∂ψ/∂x. Lines of constant ψ are streamlines and the difference ψ₂ − ψ₁ is the volume flow rate per unit depth between them. If the flow is also irrotational, ∇²ψ = 0. Equipotentials and streamlines intersect at right angles (except at stagnation points).

Linearity and superposition. Laplace's equation is linear: the sum of any two solutions is a solution. Complicated flows are therefore built by adding elementary flows (uniform stream, sources, doublets, vortices). Note that the pressure is not linear in φ — compute velocities by superposition first, then the pressure.

Boundary conditions.

• Far field: the disturbance dies out, ∇φ → V∞.
• Solid surface (flow tangency, no penetration): ∂φ/∂n = 0, i.e. the normal velocity is zero. Equivalently, the body surface is a streamline (ψ = constant).
• The no-slip condition cannot be imposed — that is the price of dropping viscosity. Real tangential slip at the wall is resolved by the boundary layer.
• For a lifting body the solution is not unique until the circulation is fixed (the Kutta condition).

Pressure. In irrotational, inviscid, steady incompressible flow Bernoulli's equation p + ½ρV² = constant holds between any two points, not just along one streamline. Once V is known everywhere from φ, the pressure follows.

Circulation. Γ = ∮ V·ds around a closed curve. By Stokes' theorem it equals the flux of vorticity through the curve, so Γ = 0 around any curve enclosing only irrotational fluid (in a simply connected region).

Formulas

• V = ∇φ; 2-D: u = ∂φ/∂x, v = ∂φ/∂y; polar: V_r = ∂φ/∂r, V_θ = (1/r)·∂φ/∂θ — φ in m²/s, velocities in m/s.
• ∇²φ = ∂²φ/∂x² + ∂²φ/∂y² + ∂²φ/∂z² = 0 — incompressible, irrotational flow.
• u = ∂ψ/∂y, v = −∂ψ/∂x; polar: V_r = (1/r)·∂ψ/∂θ, V_θ = −∂ψ/∂r — 2-D incompressible; ψ in m²/s.
• ω_z = ∂v/∂x − ∂u/∂y — 2-D vorticity (1/s); irrotational if zero.
• ∂u/∂x + ∂v/∂y = 0 — 2-D incompressible continuity.
• Γ = ∮ V·ds — circulation (m²/s).
• p₁ + ½ρV₁² = p₂ + ½ρV₂² — any two points in steady, inviscid, irrotational, incompressible flow.

Worked examples

Example 1 (standard). A 2-D flow has φ = 2(x² − y²) m²/s with x, y in metres, in air of ρ = 1.225 kg/m³. Check that it is a valid incompressible potential flow, find the velocity at (1, 2) m, the stream function, and the pressure difference between the origin and (1, 2).

1. ∂²φ/∂x² + ∂²φ/∂y² = 4 − 4 = 0 ✓ — Laplace's equation holds.
2. u = ∂φ/∂x = 4x = 4 m/s, v = ∂φ/∂y = −4y = −8 m/s; |V| = √(16 + 64) = 8.94 m/s.
3. ∂ψ/∂y = u = 4x ⇒ ψ = 4xy + f(x); −∂ψ/∂x = −4y = v ⇒ f′ = 0. So ψ = 4xy (flow in a right-angled corner).
4. The origin is a stagnation point (u = v = 0). Bernoulli: p₀ − p = ½ρV² = 0.5 × 1.225 × 80 = 49 Pa.

Answer: V = (4, −8) m/s, |V| = 8.94 m/s, ψ = 4xy, p_origin − p(1,2) = 49 Pa.

Example 2 (GATE level). A 2-D water flow (ρ = 1000 kg/m³) has u = x² − y², v = −2xy (m/s, x and y in m). (a) Is it incompressible? (b) Is it irrotational? (c) Find φ. (d) Find p(1, 0) − p(2, 1).

1. Continuity: ∂u/∂x + ∂v/∂y = 2x − 2x = 0 ✓ incompressible.
2. Vorticity: ∂v/∂x − ∂u/∂y = −2y − (−2y) = 0 ✓ irrotational.
3. ∂φ/∂x = x² − y² ⇒ φ = x³/3 − xy² + g(y); ∂φ/∂y = −2xy + g′ = −2xy ⇒ g′ = 0. So φ = x³/3 − xy².
4. At (2, 1): u = 3, v = −4, V² = 25. At (1, 0): u = 1, v = 0, V² = 1.
5. Bernoulli (valid between any two points since the flow is irrotational): p(1,0) − p(2,1) = ½ρ(V₂² − V₁²) = 0.5 × 1000 × (25 − 1) = 12 000 Pa.

Answer: incompressible and irrotational; φ = x³/3 − xy²; p(1,0) − p(2,1) = 12.0 kPa.

Common mistakes

• Thinking φ needs incompressibility to exist. It needs only irrotationality; incompressibility then gives ∇²φ = 0.
• Mixing up the sign conventions: u = +∂ψ/∂y, v = −∂ψ/∂x (some texts reverse the sign of ψ; stay with one book).
• Forgetting the 1/r factor in V_θ = (1/r)∂φ/∂θ.
• Applying Bernoulli across streamlines in a rotational flow.
• Superposing pressures instead of velocities.
• Trying to impose no-slip in potential flow; only flow tangency can be satisfied.

For GATE AE

Common items: check whether a given φ or ψ satisfies Laplace's equation; find u, v at a point; derive ψ from φ or the reverse; test a velocity field for continuity and irrotationality; compute vorticity or circulation; and use Bernoulli to get a pressure difference. Practise with polar-coordinate potentials, since the elementary flows in the next topics are all written in r and θ.

Quick check

1. Does φ = x² + y² represent an incompressible flow?
2. For φ = 3xy, find u and v at (1, 2).
3. What boundary condition does potential flow satisfy at a solid wall?
4. Why can solutions of Laplace's equation be superposed?

Answers: 1. No — ∇²φ = 4 ≠ 0 (it is irrotational but has a net source). 2. u = 3y = 6, v = 3x = 3 (units of φ/length). 3. Zero normal velocity, ∂φ/∂n = 0. 4. Because the equation is linear.