Exercises — Stream function, velocity potential

Two independent tests. WHY two? Because $\varphi$ needs no spin (irrotational) and $\psi$ needs no squishing (incompressible) — different physics.

Test for $\psi$ (incompressibility, the Continuity Equation): $\nabla \cdot \vec{V}=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=\frac{\partial (-y)}{\partial x}+\frac{\partial (x)}{\partial y}=0+0=0.$ Divergence is zero ⇒ $\psi$ exists. ✓

Test for $\varphi$ (irrotationality): in 2D the only vorticity component is $\omega =\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\frac{\partial (x)}{\partial x}-\frac{\partial (-y)}{\partial y}=1-(-1)=2\ne 0.$ Non-zero spin ⇒ $\varphi$ does NOT exist. ✗

This field is solid-body rotation (like a spinning turntable): every particle circles the origin, so there is real spin, and no single "height" scalar can describe it.

Rule from the parent note: $\psi =\text{const}$ are streamlines (the fluid runs along them); $\varphi =\text{const}$ are equipotentials (crossed perpendicularly by the flow).

• $\psi =3x-2y=\text{const}$ → straight lines of slope $\frac{dy}{dx}=\frac{3}{2}$. These are the streamlines.
• $\varphi =2x+3y=\text{const}$ → straight lines of slope $-\frac{2}{3}$. These are the equipotentials.

Slopes multiply to $\frac{3}{2}\cdot (-\frac{2}{3})=-1$: they are perpendicular, exactly as a flow net demands.

Apply the definitions ($u={\psi }_y$, $v=-{\psi }_x$). WHY these and not $\varphi$'s? Because we were handed $\psi$. $u=\frac{\partial }{\partial y}(x^2-y^2)=-2y,v=-\frac{\partial }{\partial x}(x^2-y^2)=-2x.$ Incompressible? By construction any $\psi$ gives $\nabla \cdot \vec{V}=0$. Check: $u_x+v_y=0+0=0$ ✓. Irrotational? $\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}=\frac{\partial (-2x)}{\partial x}-\frac{\partial (-2y)}{\partial y}=-2-(-2)=0.$ Zero spin ⇒ also irrotational. So a $\varphi$ exists too — this is an ideal flow. (It is the classic "flow toward a corner" / stagnation flow.)

We integrate, because $\varphi$ is the anti-derivative of velocity ($u={\varphi }_x$). $\varphi =\int udx=\int (-2y)dx=-2xy+f(y).$ Here $f(y)$ is an unknown function of $y$ only — it vanishes when we differentiate by $x$, so integration cannot see it yet. Pin it down using the other equation $v={\varphi }_y$: ${\varphi }_y=-2x+f^{\prime }(y)\stackrel{!}{=}v=-2x\Rightarrow f^{\prime }(y)=0\Rightarrow f=\text{const}.$ So $\varphi =-2xy$ (constant absorbed). Check: ${\varphi }_x=-2y=u$ ✓, ${\varphi }_y=-2x=v$ ✓.

The two scalars must agree on $u$ and $v$. The bridge is ${\varphi }_x={\psi }_y$ and ${\varphi }_y=-{\psi }_x$. ${\varphi }_x=2x,{\psi }_y=2x✓$ ${\varphi }_y=2y,-{\psi }_x=-2y.$ Second equation: $2y\stackrel{?}{=}-2y$ → only true at $y=0$. Refuted. These do not describe the same flow.

Deeper reason: $\varphi =x^2+y^2$ has ${\nabla }^2\varphi =2+2=4\ne 0$, so it fails the Laplace Equation ${\nabla }^2\varphi =0$ that every valid velocity potential of an incompressible flow must satisfy. It was never a legal potential to begin with.

From $\varphi$: $u={\varphi }_x=2x$, $v={\varphi }_y=-2y$. Now build $\psi$ from $u={\psi }_y$: $\psi =\int 2xdy=2xy+g(x).$ Fix $g$ with $v=-{\psi }_x$: $-{\psi }_x=-(2y+g^{\prime }(x))\stackrel{!}{=}-2y\Rightarrow g^{\prime }(x)=0$. So $\boxed{\psi =2xy}$. Laplace check: ${\nabla }^2\varphi ={\varphi }_{xx}+{\varphi }_{yy}=2+(-2)=0✓,{\nabla }^2\psi ={\psi }_{xx}+{\psi }_{yy}=0+0=0✓.$

WHY superpose? Because the Laplace Equation is linear: if ${\psi }_1$ and ${\psi }_2$ each solve ${\nabla }^2\psi =0$, so does their sum. Streamfunctions add. $\psi =Uy+\frac{m}{2\pi }\theta ,y=r\sin \theta .$

From ${\psi }_2$ to a Cartesian velocity — do it carefully. The source stream function is ${\psi }_2=\frac{m}{2\pi }\theta$. Its velocity in polar coordinates comes from the polar definitions of the stream function, $u_r=\frac{1}{r}\frac{\partial \psi }{\partial \theta },u_{\theta }=-\frac{\partial \psi }{\partial r}.$ So $u_r=\frac{1}{r}\cdot \frac{m}{2\pi }=\frac{m}{2\pi r}$ (pure radial outflow) and $u_{\theta }=0$. WHY convert to Cartesian? Because the stream flows horizontally ($u$-direction), so to add the two flows we need both expressed in the same $(u,v)$ basis. A purely radial velocity of magnitude $u_r$ points along the unit vector $\hat{r}=(x,y)/r$, so its Cartesian components are $u_r\hat{r}=\frac{m}{2\pi r}\cdot \frac{(x,y)}{r}=\frac{m}{2\pi }\frac{(x,y)}{x^2+y^2},\text{since }{r}^2=x^2+y^2.$ That is where the vector form $\frac{m}{2\pi }\frac{(x,y)}{x^2+y^2}$ comes from — it is nothing more than "radial speed times the outward unit vector." Adding the uniform stream $(U,0)$: $u=U+\frac{m}{2\pi }\frac{x}{x^2+y^2},v=\frac{m}{2\pi }\frac{y}{x^2+y^2}.$ Stagnation on the $x$-axis: set $y=0$ (then $v=0$ automatically) and $u=0$: $U+\frac{m}{2\pi }\frac{x}{x^2}=0\Rightarrow U+\frac{m}{2\pi x}=0\Rightarrow x=-\frac{m}{2\pi U}.$ With $U=1,m=2\pi$: $x=-\frac{2\pi }{2\pi \cdot 1}=-1$. Stagnation point at $(-1,0)$. Read it off the figure above: the amber dot sits at $(-1,0)$, upstream (to the left) of the white source dot. The white arrows on the far left are the incoming uniform stream; the cyan curves are streamlines of the combined $\psi$. Exactly at the amber dot the source's outward push (pointing left, in $-x$, on that side) cancels the stream's rightward push — so the flow halts, splits, and wraps around the "half-body." The analytic $x=-1$ is precisely that amber dot.

Key fact from the parent note: flux between two streamlines is just $Q={\psi }_2-{\psi }_1$. WHY: flux $=\int (udy-vdx)=\int d\psi =\Delta \psi$.

Evaluate $\psi =Uy+\frac{m}{2\pi }\theta$ with $U=1,\frac{m}{2\pi }=1$, so $\psi =y+\theta$ (with $\theta \in (-\pi ,\pi ]$).

• At $(-1,0)$: on the negative $x$-axis $\theta =\pi$, $y=0$ ⇒ ${\psi }_1=0+\pi =\pi$.
• At $(0,1)$: $\theta =\pi /2$, $y=1$ ⇒ ${\psi }_2=1+\pi /2$. $Q={\psi }_2-{\psi }_1=\left(1+\frac{\pi }{2}\right)-\pi =1-\frac{\pi }{2}\approx -\mathrm{0.5708.}$ WHY the minus sign — it is orientation, not a mistake. The quantity $Q={\psi }_2-{\psi }_1$ measures flux crossing a curve from streamline $1$ to streamline $2$ with a fixed handedness (positive when the flow crosses left-to-right as you walk from the first streamline to the second). Here $\psi$ actually decreases going from $(-1,0)$ to $(0,1)$, so the fluid crosses in the opposite sense to our chosen walking order — hence the negative sign. Swap the two endpoints (walk from $(0,1)$ to $(-1,0)$) and you get $+\left(\frac{\pi }{2}-1\right)$. The physical amount of fluid is the same either way: its magnitude is $|Q|=\frac{\pi }{2}-1\approx 0.571\text{units (per unit depth).}$ The sign only records which way it flows; the size is what "how much fluid" means.

First, the tool we need: velocity from a potential in polar coordinates. The gradient $\nabla \varphi$ is the vector "steepest-increase of $\varphi$." In Cartesian it is $({\varphi }_x,{\varphi }_y)$. In polar coordinates, a step in the radial direction of length $dr$ changes $\varphi$ by ${\varphi }_{r}dr$, but a step in the angular direction sweeps an arc length $rd\theta$ (not $d\theta$) — because moving through angle $d\theta$ at radius $r$ physically moves you a distance $rd\theta$. Since velocity is change-in-$\varphi$ per unit distance, the tangential component must divide by that arc length: $\nabla \varphi =\left({\varphi }_r,\frac{1}{r}{\varphi }_{\theta }\right)\Rightarrow u_r=\frac{\partial \varphi }{\partial r},u_{\theta }=\frac{1}{r}\frac{\partial \varphi }{\partial \theta }.$ That extra $1/r$ is the whole reason the angular formula looks different from the radial one.

(a) Stream function. In polar form $u_{\theta }=-\frac{\partial \psi }{\partial r}$. So $-\frac{\partial \psi }{\partial r}=\frac{\Gamma }{2\pi r}\Rightarrow \psi =-\frac{\Gamma }{2\pi }\ln r.$ Streamlines $\psi =\text{const}\Rightarrow r=\text{const}$: concentric circles — correct, fluid whirls around.

Velocity potential. Using the polar rule just derived, $u_{\theta }=\frac{1}{r}\frac{\partial \varphi }{\partial \theta }=\frac{\Gamma }{2\pi r}\Rightarrow \frac{\partial \varphi }{\partial \theta }=\frac{\Gamma }{2\pi }\Rightarrow \varphi =\frac{\Gamma }{2\pi }\theta .$ Note this $\varphi$ and $\psi$ are the swapped mirror of the source in the parent note — vortex and source are conjugates.

(b) Irrotational check + single-valuedness. The vorticity of a purely azimuthal flow is ${\omega }_z=\frac{1}{r}\frac{\partial (r{u}_{\theta })}{\partial r}=\frac{1}{r}\frac{\partial }{\partial r}\left(r\cdot \frac{\Gamma }{2\pi r}\right)=\frac{1}{r}\frac{\partial }{\partial r}\left(\frac{\Gamma }{2\pi }\right)=0(r\ne 0).$ So although each particle orbits, it does not spin about its own axis — that is why it is irrotational everywhere except the singular core $r=0$.

But $\varphi =\frac{\Gamma }{2\pi }\theta$ is multivalued. Walk one full loop around the origin: $\theta$ increases by $2\pi$, so $\varphi$ jumps by $\frac{\Gamma }{2\pi }\cdot 2\pi =\Gamma$ even though you returned to the same physical point. A genuine single-valued height cannot do that. This is precisely the L1 caveat in action: the flow region (the plane minus the origin) is not simply-connected — it has a hole at $r=0$, and loops around that hole cannot be shrunk to a point. So "curl $\varphi =0$" holds yet $\varphi$ is only single-valued if we cut the plane along a ray (a branch cut, e.g. the negative $x$-axis) so that no path can encircle the core. The jump $\Gamma$ across the cut is exactly the circulation — the vortex's defining quantity. (The velocity itself stays perfectly single-valued; only the potential carries the ambiguity.)

(c) Bernoulli. For irrotational flow, $p+\frac{1}{2}\rho V^2=\text{const}$ everywhere. Speed $V=u_{\theta }=\frac{\Gamma }{2\pi r}$. With $\Gamma =2\pi ,\rho =1$: $V=\frac{1}{r}$. $p(r)=C-\frac{1}{2}\rho V^2=C-\frac{1}{2r^2}.$ $p(1)-p(2)=\left(C-\frac{1}{2}\right)-\left(C-\frac{1}{8}\right)=-\frac{1}{2}+\frac{1}{8}=-\frac{3}{8}.$ So the pressure is lower near the centre by $\frac{3}{8}$ units — exactly why the middle of a whirlpool dips down.

• Streamlines: $\psi =-\frac{\Gamma }{2\pi }\ln r=\text{const}\Rightarrow r=\text{const}$ → concentric circles.
• Equipotentials: $\varphi =\frac{\Gamma }{2\pi }\theta =\text{const}\Rightarrow \theta =\text{const}$ → radial rays.
• A circle and a radius always cross at $90°$.

Match this to the figure above: the cyan circles are the streamlines ($r=\text{const}$), the amber rays are the equipotentials ($\theta =\text{const}$), and the white dot marks the singular core $r=0$ (where $\varphi$'s branch cut lives). Every place a cyan circle meets an amber ray, they cross at a right angle — that is the flow net, drawn.

Algebraic proof of orthogonality (matches the picture). The gradient of a scalar points across its level curves, so: $\nabla \varphi =(u,v)\text{(velocity)},\nabla \psi =({\psi }_x,{\psi }_y)=(-v,u).$ Their dot product: $\nabla \varphi \cdot \nabla \psi =(u)(-v)+(v)(u)=0.$ Zero dot product ⇒ perpendicular gradients ⇒ perpendicular level curves. So the algebra ($\nabla \varphi \cdot \nabla \psi =0$) is the exact statement of what your eye sees in the figure: circles ⟂ rays.

Existence test ::: $\psi$ needs $\nabla \cdot \vec{V}=0$; $\varphi$ needs $\nabla \times \vec{V}=0$ and a simply-connected (hole-free) region for single-valuedness. Recover $\varphi$ from velocity ::: integrate $u$ over $x$, then fix the leftover $f(y)$ using $v={\varphi }_y$. Superposition rule ::: $\psi$ and $\varphi$ add because Laplace's equation is linear. Polar velocity from a potential ::: $u_r={\varphi }_r,u_{\theta }=\frac{1}{r}{\varphi }_{\theta }$ (the $1/r$ is arc-length weighting). Orbiting vs spinning ::: free vortex ($u_{\theta }\propto 1/r$) is irrotational (but $\varphi$ jumps by $\Gamma$ per loop); solid-body ($u_{\theta }\propto r$) genuinely spins. Flux meaning of $\psi$ ::: $Q={\psi }_2-{\psi }_1$ between two streamlines; its sign records orientation, its magnitude the amount.