Stream function, velocity potential: Question bank

Before we start, a quick refresher on the three symbols so every line below is readable from zero:

True or false — justify

Whether

ψ

exists depends on incompressibility, not on spin

True —

ψ

needs only

\nabla \cdot V = 0

(2D). A flow can spin like a whirlpool and still have a perfectly good stream function.

Whether

ϕ

exists depends on irrotationality, not on incompressibility

True —

ϕ

needs only

\nabla \times V = 0

. A compressible-but-irrotational flow still has a

ϕ

; its

ϕ

just won't obey Laplace.

Every flow that has a

ψ

also has a

ϕ

False —

ψ

needs incompressible,

ϕ

needs irrotational. A rotating incompressible flow (solid-body rotation) has

ψ

but no single-valued

ϕ

If both

ϕ

and

ψ

exist, both must satisfy Laplace's equation

True — both existing means incompressible AND irrotational, and those two conditions are exactly what force [[Laplace Equation|

\nabla^{2} ϕ = 0

and

\nabla^{2} ψ = 0

]].

ψ

satisfies Laplace's equation because the flow is incompressible

False —

ψ

obeys Laplace because the flow is irrotational; incompressibility is what let

ψ

exist in the first place. The roles are swapped from

ϕ

ϕ

satisfies Laplace's equation because the flow is incompressible

True — plug

V = \nabla ϕ

into

\nabla \cdot V = 0

and you get

\nabla^{2} ϕ = 0

. Here incompressibility is the source of Laplace.

Streamlines and equipotential lines always cross at right angles

False as stated — only when BOTH

ψ

and

ϕ

exist (ideal flow) is

\nabla ψ \cdot \nabla ϕ = 0

. In a purely incompressible-but-rotational flow there is no

ϕ

to be perpendicular to.

The difference

ψ_{2} - ψ_{1}

equals the volume flow rate between those two streamlines

True — flux

= \int (u d y - v d x) = \int d ψ = ψ_{2} - ψ_{1}

per unit depth. This is

ψ

's defining physical meaning.

Adding a constant to

ψ

changes the flow

False — velocities are derivatives of

ψ

, and the derivative of a constant is zero. Only differences of

ψ

(fluxes) are physical.

Adding a constant to

ϕ

changes the velocity

False —

V = \nabla ϕ

, and

\nabla (ϕ + c) = \nabla ϕ

. The zero-level of a potential is always arbitrary, exactly like Electrostatic Potential.

Spot the error

"Continuity gives

u_{x} + v_{y} = 0

, so I set

u = ψ_{x}, v = ψ_{y}

Wrong sign/index pattern: this gives

ψ_{xx} + ψ_{y y}

, not zero. The correct choice

u = ψ_{y}, v = - ψ_{x}

makes continuity read

ψ_{y x} - ψ_{x y} = 0

automatically.

"The minus sign in

v = - ψ_{x}

is just a convention you can flip freely."

The sign is what cancels the mixed partials so continuity holds identically. Flip it and

ψ

no longer guarantees incompressibility — it stops being a valid stream function.

ϕ =

const are the streamlines because fluid follows level curves of a scalar."

ϕ =

const are equipotentials, perpendicular to the flow; fluid crosses them.

ψ =

const are the streamlines the fluid actually travels along.

\nabla \times (\nabla ϕ) = 0

proves every flow is irrotational."

It only proves that flows written as

\nabla ϕ

are irrotational. The physics is the converse: you may write

V = \nabla ϕ

only after checking the flow is irrotational.

"For the source,

ϕ = \frac{m}{2 π} ln r

, and since

ln

blows up at

r = 0

, the model is wrong."

The singularity at

r = 0

is the source point itself — an idealised injection of fluid. The model is valid everywhere except that one point, which is deliberate. See Bernoulli Equation usage of such point singularities.

"Cauchy–Riemann relations only matter in complex analysis, not fluids."

The fluid pair

ϕ_{x} = ψ_{y}, ϕ_{y} = - ψ_{x}

are the Cauchy-Riemann Equations; that's why

ϕ + i ψ

is an analytic function and complex methods solve ideal flows.

"A source has

ψ = \frac{m}{2 π} θ

, which is single-valued."

It is multi-valued: going once around (

θ \to θ + 2 π

) increases

ψ

m

. That jump is exactly the flux

m

leaving the source, so the multivaluedness is physical, not a bug.

Why questions

Why can two velocity components be replaced by one scalar for incompressible flow?

Incompressibility is one equation tying

u

and

v

together, removing one degree of freedom — so a single function (

ψ

) can encode both while satisfying that constraint by construction.

Why does velocity point along

\nabla ϕ

but across

\nabla

's of

ψ

-lines' tangent?

\nabla ϕ = (u, v)

is the velocity, so it points where

ϕ

increases fastest. The streamline direction is

(u, v)

too, which is why

\nabla ψ = (- v, u)

is perpendicular to it.

Why are equipotentials and streamlines perpendicular in ideal flow?

\nabla ϕ = (u, v)

and

\nabla ψ = (- v, u)

; their dot product is

- uv + uv = 0

, so their level curves cross at right angles, forming a flow net.

Why does an irrotational flow have no single-valued

ϕ

around a vortex?

Circulation around the vortex is nonzero, so integrating

V \cdot d r

once around gains a fixed amount —

ϕ

jumps each loop, so it can't be single-valued even though the flow is irrotational away from the centre.

Why is

ψ

constant along a streamline?

Moving along the flow,

d ψ = ψ_{x} d x + ψ_{y} d y = - v d x + u d y

, and on a streamline

u d y = v d x

, so

d ψ = 0

. No fluid crosses a streamline, which is why the "lane number" doesn't change.

Why does the flux between streamlines not depend on the path chosen between them?

Because

\int d ψ

only sees the endpoints'

ψ

values,

Q = ψ_{2} - ψ_{1}

regardless of the connecting curve — the same reason work in a potential field is path-independent.

Edge cases

Solid-body rotation

V = (- Ω y, Ω x)

: does

ψ

exist? Does

ϕ

\nabla \cdot V = 0

ψ = - \frac{Ω}{2} (x^{2} + y^{2})

exists; but

\nabla \times V = 2Ω \neq = 0

so it is rotational and

ϕ

does not exist.

A free vortex

u_{θ} = \frac{Γ}{2 π r}

: does

ϕ

exist? Is it single-valued?

It is irrotational everywhere except

r = 0

, so locally

ϕ = \frac{Γ}{2 π} θ

exists — but it increases by

Γ

per loop, so it is multi-valued globally. A classic irrotational-yet-no-single-

ϕ

case.

Uniform flow at rest (

V = 0

): what are

ϕ

and

ψ

Both are constants (any constants), since all their derivatives vanish. Degenerate but consistent — no velocity means no meaningful level curves.

Compressible steady flow: can you still define a stream function?

Not the ordinary

ψ

; you need a density-weighted version

ρ u = ψ_{y}, ρ v = - ψ_{x}

so that

(ρ u)_{x} + (ρ v)_{y} = 0

(mass, not volume, is conserved).

At a stagnation point where

u = v = 0

, what happens to the flow net?

\nabla ϕ = 0

and

\nabla ψ = 0

there, so the orthogonality argument degenerates — streamlines can cross at that single point (e.g. the front of a cylinder), the one place right angles need not hold.

What does

ψ

mean at the exact centre of a source?

It is undefined there (

θ

has no value at

r = 0

); the point is a singularity where fluid is created, so no finite lane number applies.

If a flow is irrotational but compressible, does

ϕ

satisfy Laplace's equation?

No —

ϕ

exists (irrotational) but

\nabla \cdot V \neq = 0

, so

\nabla^{2} ϕ = \nabla \cdot V \neq = 0

. Laplace needs incompressibility too.

Recall One-line self-test

Cover everything above and answer: which condition gives $ψ$ , which gives $ϕ$ , and which gives Laplace for each? $ψ$ ← incompressible; $ϕ$ ← irrotational; Laplace for $ϕ$ ← incompressible; Laplace for $ψ$ ← irrotational. The Laplace conditions are the cross-over — that's the whole trap.