Worked examples — Stream function, velocity potential

This page is the worked-example workout for the parent topic. We will not re-derive the theory — instead we hunt down every kind of situation the topic can throw at you and solve one representative problem for each. If a symbol looks unfamiliar, the parent note builds it; here we use it and always say why.

Before anything, a one-line reminder of the two toolkits we will keep reaching for:

The scenario matrix

Here is the full landscape of cases. Every cell below is covered by an example further down.

We work them in order. Every numeric result is machine-checked at the bottom.

Ex 1 — Cell A: both scalars, all signs positive

Forecast: guess first — will the streamlines be horizontal or vertical? Will $\varphi =$const lines be the same set or perpendicular?

1. Check divergence and curl. $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0+0=0$ and $\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=0-0=0$. Why this step? Both scalars are only legal when their existence conditions hold. Divergence-free ⇒ $\psi$ allowed; curl-free ⇒ $\varphi$ allowed. Both are zero, so both exist.
2. Build $\varphi$. From $u={\varphi }_x=3$, integrate in $x$: $\varphi =3x$ (constant absorbed). Why this step? $\varphi$ is defined by $u={\varphi }_x$; undoing the derivative means integrating.
3. Build $\psi$. From $u={\psi }_y=3$, integrate in $y$: $\psi =3y$. Check $v=-{\psi }_x=0$ ✓. Why this step? $\psi$ uses the cross index ($u={\psi }_y$), so integrate in $y$, not $x$.

Verify: $\psi =3y=$const ⇒ horizontal lines (streamlines along the flow). $\varphi =3x=$const ⇒ vertical lines. They cross at $90^{\circ }$. Recompute $u={\varphi }_x=3$ ✓, $v={\varphi }_y=0$ ✓.

Ex 2 — Cell B: negative velocity (sign flip)

Forecast: the streamlines are still horizontal — but which way does the flux count? Positive or negative?

1. Build $\psi$. $u={\psi }_y=-2\Rightarrow \psi =-2y$. Why this step? Same recipe; the sign of $u$ carries straight into $\psi$.
2. Flux between two streamlines $={\psi }_2-{\psi }_1$ with $y_2=4,y_1=0$: $\psi (4)-\psi (0)=-8-0=-8$. Why this step? $\psi$ measures volume flow between streamlines; the difference is the flux (per unit depth).

Verify: magnitude $|-8|=|u|\cdot \Delta y=2\times 4=8$ ✓. The minus sign correctly reports that fluid crosses from high-$y$ to low-$y$ as seen against the standard orientation — a sign flip in $u$ flips the flux sign, exactly as it must.

Ex 3 — Cell C: flow at an angle (both components nonzero)

Forecast: streamlines should tilt. What slope do you predict from $v/u$?

1. Streamline slope. $\frac{dy}{dx}=\frac{v}{u}=\frac{1}{2}$. Why this step? Velocity is tangent to streamlines, so the slope is $v/u$.
2. Build $\psi$. $u={\psi }_y=2\Rightarrow \psi =2y+g(x)$. Then $v=-{\psi }_x=-g^{\prime }(x)=1\Rightarrow g(x)=-x$. So $\psi =2y-x$. Why this step? When both components live, integrate one, then use the other equation to pin the integration "constant" $g(x)$.
3. Build $\varphi$. $u={\varphi }_x=2\Rightarrow \varphi =2x+h(y)$; $v={\varphi }_y=h^{\prime }(y)=1\Rightarrow h=y$. So $\varphi =2x+y$.

Verify: streamline $\psi =2y-x=c\Rightarrow y=\frac{1}{2}x+\frac{c}{2}$, slope $\frac{1}{2}$ ✓ matches step 1. Orthogonality: $\nabla \varphi =(2,1)$, $\nabla \psi =(-v,u)=(-1,2)$; dot product $=2(-1)+1(2)=0$ ✓ — perpendicular grid.

Ex 4 — Cell D: rotational flow (degenerate for $\varphi$)

Forecast: whirlpool alarm — will the potential $\varphi$ survive?

1. Curl test. $\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\omega -(-\omega )=2\omega =2\ne 0$. Why this step? $\varphi$ exists only if curl is zero. Here it is $2\omega$ — the flow spins — so no single-valued $\varphi$. Cell D is genuinely degenerate.
2. Divergence test. $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0+0=0$. Why this step? $\psi$ needs only incompressibility, which does hold. So $\psi$ still exists.
3. Build $\psi$. $u={\psi }_y=-\omega y\Rightarrow \psi =-\frac{\omega }{2}{y}^2+g(x)$; $v=-{\psi }_x=-g^{\prime }(x)=\omega x\Rightarrow g=-\frac{\omega }{2}{x}^2$. So $\psi =-\frac{\omega }{2}(x^2+y^2)$.

Verify: at $\omega =1$, $\psi =-\frac{1}{2}(x^2+y^2)$; streamlines $\psi =$const ⇒ circles $x^2+y^2=$const ✓ (spinning fluid moves in circles). Recompute $u={\psi }_y=-y$ ✓, $v=-{\psi }_x=x$ ✓. Lesson: not every flow has a $\varphi$ — the first common mistake, made concrete.

Ex 5 — Cell E: is $\psi$ even allowed? (divergence gate)

Forecast: will the divergence be zero?

1. Divergence. $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=1+1=2\ne 0$. Why this step? No $\psi$ with the flux property exists unless divergence is zero. Here it's $2$ — fluid is being created everywhere. $\psi$ illegal.
2. Fix the sign of one component: try $u=x$, $v=-y$. Now divergence $=1+(-1)=0$ ✓. Why this step? Flipping $v$ makes what flows out in $x$ get sucked in through $y$ — a valid stagnation-point flow.
3. Build $\psi$ for the fixed field. $u={\psi }_y=x\Rightarrow \psi =xy+g(x)$; $v=-{\psi }_x=-(y+g^{\prime }(x))=-y\Rightarrow g^{\prime }=0$. So $\psi =xy$.

Verify: streamlines $\psi =xy=$const ⇒ hyperbolas — the classic corner/stagnation flow. Recompute $u={\psi }_y=x$ ✓, $v=-{\psi }_x=-y$ ✓, divergence $0$ ✓. Moral: always run the divergence gate first.

Ex 6 — Cell F: polar source, singular origin, all quadrants

Read the figure first: the magenta arrows are the streamlines — they shoot straight out from the origin like the spokes of a wheel, one for each angle $\theta$. The violet dashed circles are the equipotentials $\varphi =$const. Notice that every spoke pierces every circle at a perfect right angle: that crossing pattern is the flow net you should extract. The orange dot at the centre is the singular source point.

Forecast: what happens to the speed as $r\to 0$? Are streamlines circles or spokes?

1. Build $\varphi$. $u_r=\frac{\partial \varphi }{\partial r}=\frac{m}{2\pi r}\Rightarrow \varphi =\frac{m}{2\pi }\ln r$. Why this step? In polar coordinates $\nabla \varphi$ has radial component $\partial \varphi /\partial r$; integrate it.
2. Build $\psi$. The polar Cauchy–Riemann relation ties the radial velocity to the angular derivative of $\psi$: $u_r=\frac{1}{r}\frac{\partial \psi }{\partial \theta }$. Setting this equal to $\frac{m}{2\pi r}$ gives $\frac{\partial \psi }{\partial \theta }=\frac{m}{2\pi }\Rightarrow \psi =\frac{m}{2\pi }\theta$. Why this step? We use the polar definition of $\psi$ (its $\theta$-derivative supplies $u_r$), not the generic "$\psi$ is constant along a streamline." Here each streamline is a ray of fixed $\theta$, so $\psi$ actually labels rays by their angle — it changes from ray to ray, and is constant only along one given spoke.
3. All-quadrant behaviour (from the figure). $\psi =\frac{m}{2\pi }\theta$ increases smoothly through quadrant I ($0\le \theta <\frac{\pi }{2}$), II, III, IV as you sweep the spokes anticlockwise; $\varphi$ ignores $\theta$ so its equipotential circles $r=$const are valid in all four quadrants. Singular origin: as $r\to 0$, $u_r=\frac{m}{2\pi r}\to \infty$ and $\varphi =\frac{m}{2\pi }\ln r\to -\infty$ — the source point is a genuine singularity, excluded from the domain.
4. The branch-cut (multi-valuedness of $\psi$). Because $\theta$ itself jumps by $2\pi$ every time you cross a chosen ray (say the negative-$x$ axis), $\psi =\frac{m}{2\pi }\theta$ jumps by $\frac{m}{2\pi }\cdot 2\pi =m$ across that ray. So $\psi$ is not single-valued on a loop around the origin — it is defined up to a branch cut, a line we agree not to cross so that $\theta$ (and hence $\psi$) stays single-valued on the rest of the plane. Why this step? Ignoring the cut would let you "prove" $\psi$ returns to itself after a full loop, contradicting the flux $=m$ result below.
5. Flux over a full loop. $Q=\psi (\theta =2\pi )-\psi (0)=\frac{m}{2\pi }\cdot 2\pi =m$. Why this step? Flux between streamlines is the $\psi$ difference; over a whole circle that difference is exactly the source strength $m$ — and it is nonzero precisely because $\psi$ jumps by $m$ across the cut.

Verify: with $m=4\pi$: $\varphi =2\ln r$, $\psi =2\theta$, and $Q=4\pi$. Check $u_r=\partial \varphi /\partial r=2/r=\frac{4\pi }{2\pi r}$ ✓. Circles (equipotentials) meet radial spokes (streamlines) at $90^{\circ }$ — orthogonality holds in every quadrant, and the $\psi$-jump of $m$ across the cut matches the total outflow.

Ex 7 — Cell G: irrotational vortex, limiting behaviour

Read the figure first: now the magenta circles are the streamlines — fluid swirls round and round — and the violet dashed spokes are the equipotentials. It is the exact mirror image of the source: what were streamlines there are equipotentials here, and vice versa. The small arrows on the circles show the swirl sense; the orange core at the centre is where the speed blows up.

Forecast: it swirls — surely it has vorticity? (Watch the surprise.)

1. Build $\varphi$. Here $u_{\theta }=\frac{1}{r}\frac{\partial \varphi }{\partial \theta }=\frac{\Gamma }{2\pi r}\Rightarrow \varphi =\frac{\Gamma }{2\pi }\theta$. Why this step? Now the angular velocity component is nonzero, so $\varphi$ depends on $\theta$.
2. Build $\psi$. $u_{\theta }=-\frac{\partial \psi }{\partial r}=\frac{\Gamma }{2\pi r}\Rightarrow \psi =-\frac{\Gamma }{2\pi }\ln r$. Why this step? The polar $\psi$ relation for the swirl component carries a minus, mirroring $v=-{\psi }_x$.
3. Vorticity — the twist (shown, not asserted). In polar coordinates the out-of-plane vorticity of a purely azimuthal field is ${\omega }_z=\frac{1}{r}\frac{\partial (r{u}_{\theta })}{\partial r}-\frac{1}{r}\frac{\partial u_r}{\partial \theta }.$ With $u_r=0$ and $r{u}_{\theta }=r\cdot \frac{\Gamma }{2\pi r}=\frac{\Gamma }{2\pi }$ (a constant in $r$), we get $\frac{\partial }{\partial r}\left(\frac{\Gamma }{2\pi }\right)=0$, so ${\omega }_z=0$ for every $r>0$. It is irrotational everywhere except the origin, where all the circulation is concentrated. This is the opposite of Ex 4's solid-body rotation, which was rotational everywhere. Why this step? It exposes the subtlety: swirling flow can still be irrotational — and now we've earned it with the polar curl formula.
4. The branch-cut (multi-valuedness of $\varphi$). Here it is $\varphi =\frac{\Gamma }{2\pi }\theta$ that carries the $\theta$: it jumps by $\frac{\Gamma }{2\pi }\cdot 2\pi =\Gamma$ across the chosen cut. So the potential is multi-valued, defined up to a branch cut, exactly mirroring $\psi$ in the source case. That jump is the circulation $\Gamma$ around the origin.
5. Limits (from the figure). As $r\to \infty$, the circles grow huge and $u_{\theta }\to 0$ (fluid barely moves far out). As $r\to 0^{+}$, $u_{\theta }\to \infty$ — the singular core marked in orange.

Verify: with $\Gamma =4\pi$: $\varphi =2\theta$, $\psi =-2\ln r$. Circulation around any loop enclosing the origin $=\oint u_{\theta }rd\theta =\frac{\Gamma }{2\pi }\cdot 2\pi =\Gamma =4\pi$ ✓, matching the $\varphi$-jump across the cut. Note equipotentials $\varphi =$const are radial spokes and streamlines $\psi =$const are circles — exactly swapped vs the source (Ex 6). This source↔vortex swap is a favourite flow-net duality.

Ex 8 — Cell H: real-world flux word problem

Forecast: guess the number from "speed × width" before using $\psi$.

1. Identify the speed. $u={\psi }_y=5\text{m/s}$, $v=-{\psi }_x=0$. Why this step? Differentiate $\psi$ to recover the actual velocity — that's what the water is doing.
2. Flux $={\psi }_2-{\psi }_1$. $Q=\psi (3)-\psi (1)=15-5=10{\text{m}}^2/\text{s}$ (per unit depth). Why this step? The whole point of $\psi$: the difference between two streamline values is the volumetric flow between them — no integration needed.

Verify (units + physics): $Q=u\times \Delta y=5\frac{\text{m}}{\text{s}}\times 2\text{m}=10\frac{{\text{m}}^2}{\text{s}}$ ✓. Units are area/time (2D "flow rate per unit depth") — consistent. Multiply by a channel depth of, say, $2\text{m}$ to get a true volume rate $20{\text{m}}^3/\text{s}$.

Ex 9 — Cell I: exam twist — recover the other scalar & prove orthogonality

Forecast: does a $\varphi$ even exist here? What velocity does $\psi =x^2-y^2$ hide?

1. Extract velocity. $u={\psi }_y=-2y$, $v=-{\psi }_x=-2x$. Why this step? $\psi$ always encodes velocity via the cross-with-minus recipe.
2. Check both conditions. Divergence $={\partial }_x(-2y)+{\partial }_y(-2x)=0$ ✓ ($\psi$ guarantees this). Curl $={\partial }_{x}v-{\partial }_{y}u=-2-(-2)=0$ ✓ ⇒ $\varphi$ exists. Why this step? You must earn the right to a $\varphi$ before hunting for it.
3. Build $\varphi$. $u={\varphi }_x=-2y\Rightarrow \varphi =-2xy+h(y)$; $v={\varphi }_y=-2x+h^{\prime }(y)=-2x\Rightarrow h^{\prime }=0$. So $\varphi =-2xy$. Why this step? Integrate one C–R equation, fix the leftover with the other.
4. Orthogonality at $(2,1)$. $\nabla \varphi =({\varphi }_x,{\varphi }_y)=(-2y,-2x)=(-2,-4)$; $\nabla \psi =({\psi }_x,{\psi }_y)=(2x,-2y)=(4,-2)$. Dot product $=(-2)(4)+(-4)(-2)=-8+8=0$ ✓. Why this step? Zero dot product ⇒ the $\varphi$-line and $\psi$-line cross at $90^{\circ }$ there.

Verify: recompute from $\varphi$: $u={\varphi }_x=-2y$ ✓, $v={\varphi }_y=-2x$ ✓ — matches step 1. Both scalars satisfy Laplace: ${\nabla }^2\psi =2-2=0$, ${\nabla }^2\varphi =0+0=0$ ✓. Every cell of the matrix is now covered.

Answer these before moving on: