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Laplace's Equation (Potential Flow)

Equation

\[\boxed{\,\nabla^2\phi = 0\,} \qquad \text{and in 2-D also}\quad \nabla^2\psi = 0\]

Physical meaning

The governing equation of potential flow: a velocity field that is both irrotational (\(\mathbf{v} = \nabla\phi\)) and incompressible (\(\nabla\cdot\mathbf{v} = 0\)) must have a harmonic potential. The same equation rules electrostatics and Newtonian gravity — one PDE, three physical theories, and every solution technique transfers.

Variables

Symbol Meaning SI unit
\(\phi\) velocity potential (\(\mathbf{v} = \nabla\phi\)) m² s⁻¹
\(\psi\) stream function (\(u = \partial_y\psi\), \(v = -\partial_x\psi\)) m² s⁻¹

Assumptions

Irrotational (\(\nabla\times\mathbf{v}=0\)) + incompressible (\(\nabla\cdot\mathbf{v}=0\)). Away from boundaries and wakes at high Reynolds number, these hold remarkably well.

Derivation

\[\nabla\cdot\mathbf{v} = \nabla\cdot(\nabla\phi) = \nabla^2\phi = 0\]

For \(\psi\): the definition satisfies continuity identically, and irrotationality \(\partial_x v - \partial_y u = 0\) gives \(\nabla^2\psi = 0\).

Properties that do all the work

  • Linearity → superposition: build flows from elementary solutions (uniform, source, vortex, doublet)
  • Maximum principle: no interior velocity extrema — stagnation points live on boundaries
  • Uniqueness: boundary conditions determine the flow
  • Conformal invariance: solutions map to solutions — Joukowski airfoils

Applications

Airfoil lift (with circulation), flow around cylinders/half-bodies, groundwater flow, surface-wave theory (the perturbation potentials solve Laplace), and the Poisson-equation solvers of computational physics (Laplace = Poisson with zero source).

Limitations

No viscosity ⇒ no drag, no boundary layers, no separation. Wakes and lift require injecting circulation/vortex sheets by hand.

Quiz

Q1 (conceptual). Why does \(\phi\) exist only for irrotational flows, while \(\psi\) (2-D) exists for any incompressible flow?

Answer

\(\mathbf{v} = \nabla\phi\) forces \(\nabla\times\mathbf{v} = 0\), so a rotational flow can't have a \(\phi\). The stream function only encodes \(\nabla\cdot\mathbf{v} = 0\); \(\psi\) exists for shear flows too — it just isn't harmonic unless the flow is also irrotational.

Q2 (computational). Verify that \(\phi = \frac{Q}{2\pi}\ln r\) solves \(\nabla^2\phi = 0\) for \(r \neq 0\).

Answer

In polar coordinates \(\nabla^2\phi = \frac{1}{r}\partial_r(r\,\partial_r\phi) = \frac{1}{r}\partial_r\left(r\cdot\frac{Q}{2\pi r}\right) = \frac{1}{r}\partial_r\left(\frac{Q}{2\pi}\right) = 0\). The singularity at \(r=0\) is the source itself.