Potential flow

At a Glance

Why This Chapter Matters

Potential flow has appeared in 9 of the last 14 years with a heavy skew toward 15- and 20-mark questions — the highest average mark of any T1 fluid dynamics atom. The questions always resolve into one of five patterns: find the stream function from a given potential, verify incompressibility and find the potential, compute streamlines via the streamline ODE, analyse a complex potential, or apply unsteady Bernoulli. Mastering the Cauchy–Riemann relations between ϕ\phi and ψ\psi and the two-step streamline integration covers four of the five patterns.

Minimum Theory

Velocity potential and stream function. For an irrotational incompressible 2D flow, there exists a velocity potential ϕ(x,y)\phi(x,y) with q⃗=∇ϕ\vec q = \nabla\phi, i.e. u=ϕx,  v=ϕyu=\phi_x,\;v=\phi_y. Incompressibility (∇⋅q⃗=0\nabla\cdot\vec q=0) forces ϕ\phi to be harmonic: ∇2ϕ=ϕxx+ϕyy=0\nabla^2\phi=\phi_{xx}+\phi_{yy}=0. The conjugate harmonic ψ\psi (the stream function) satisfies the Cauchy–Riemann relations u=ψy,  v=−ψxu=\psi_y,\;v=-\psi_x. Streamlines are ψ=const\psi=\text{const} and are everywhere perpendicular to the equipotentials ϕ=const\phi=\text{const}. The complex potential w=ϕ+iψw=\phi+i\psi is analytic in z=x+iyz=x+iy, and the complex velocity is dw/dz=u−ivdw/dz=u-iv.

Sign note. Some authors write q⃗=−∇ϕ\vec q=-\nabla\phi. The 2017 and 2016 UPSC solutions use this convention; the 2018, 2020, 2021 solutions use q⃗=+∇ϕ\vec q=+\nabla\phi. Either is acceptable — state it clearly at the start of every answer.

Testing motion possibility and finding ψ\psi. A velocity potential ϕ\phi represents a possible (incompressible irrotational) flow iff ∇2ϕ=0\nabla^2\phi=0. To find ψ\psi: integrate ∂ψ/∂y=u\partial\psi/\partial y=u w.r.t. yy to get ψ=∫u dy+g(x)\psi=\int u\,dy+g(x); then differentiate and match −∂ψ/∂x=v-\partial\psi/\partial x=v to determine g(x)g(x).

Streamlines in 3D and special flows. Streamlines satisfy the ODE dx/u=dy/v=dz/wdx/u=dy/v=dz/w; integrate pairwise to find two independent first integrals. Key building blocks: uniform flow (ϕ=Ux\phi=Ux, streamlines horizontal); source/sink (ϕ=mln⁡r\phi=m\ln r in 2D, ϕ=−m/r\phi=-m/r in 3D); point vortex (ϕ=Γθ/2π\phi=\Gamma\theta/2\pi, streamlines = concentric circles, multi-valued); doublet/dipole in 3D (ϕ=−Mcos⁡θ/r2\phi=-M\cos\theta/r^2, streamlines sin⁡2θ/r=const\sin^2\theta/r=\text{const}). Unsteady irrotational flow satisfies Bernoulli’s equation: p/ρ+∂ϕ/∂t+12∣q⃗∣2=f(t)p/\rho+\partial\phi/\partial t+\tfrac12|\vec q|^2=f(t), with f(t)=p∞/ρf(t)=p_\infty/\rho when the fluid is at rest at infinity.

Potential flow: relationships between \phi, \psi, complex potential, and velocity

Question Archetypes

Five patterns cover every potential flow question in the corpus.

ArchetypeYou are seeing this when…
potential-stream-functiongiven ϕ\phi, find ψ\psi; or given q⃗\vec q, show flow is possible, find ϕ\phi, find streamlines
streamlines-from-potentialgiven ϕ\phi only, find and identify the streamlines geometrically
complex-potential-analysisa complex potential w(z)w(z) is given; find streamlines, equipotentials, velocity, singularities
potential-boundary-valueflow between boundaries (concentric spheres, etc.); find ϕ\phi matching normal-velocity BCs
unsteady-bernoullia sphere or body moves unsteadily; derive surface pressure using ∂ϕ/∂t\partial\phi/\partial t

potential-stream-function (5 question(s); 2017, 2018, 2020, 2021, 2022)

Recognition Cues

Solution Template

Given ϕ\phi, find u,v,ψu,v,\psi:

  1. Compute u=ϕx,  v=ϕyu=\phi_x,\;v=\phi_y.
  2. Check ∇2ϕ=ux+vy=0\nabla^2\phi=u_x+v_y=0 — if yes, “possible flow.”
  3. Integrate ψy=u\psi_y=u w.r.t. yy: ψ=∫u dy+g(x)\psi=\int u\,dy+g(x).
  4. Differentiate: ψx=−v\psi_x=-v. Determine g′(x)g'(x), hence g(x)g(x). Write ψ\psi.

Given q⃗=(u,v,w)\vec q=(u,v,w), verify and find ϕ\phi:

  1. Check ∇⋅q⃗=ux+vy+wz=0\nabla\cdot\vec q=u_x+v_y+w_z=0.
  2. Integrate u=−ϕxu=-\phi_x (or +ϕx+\phi_x) to get ϕ\phi up to a function of the remaining variables; match with v,wv,w.
  3. Find streamlines from dx/u=dy/v=dz/wdx/u=dy/v=dz/w.

Worked Example(s)

2018 Paper 2, 2018-P2-Q8b (15 marks)

For ϕ=x2y−xy2+13(x3−y3)\phi=x^2y-xy^2+\tfrac13(x^3-y^3), find velocity components, stream function, and check whether ϕ\phi represents a possible flow.

Velocity. u=ϕx=2xy−y2+x2u=\phi_x=2xy-y^2+x^2; v=ϕy=x2−2xy−y2v=\phi_y=x^2-2xy-y^2.

Possibility check. ux=2x+2yu_x=2x+2y; vy=−2x−2yv_y=-2x-2y; ∇2ϕ=ux+vy=0\nabla^2\phi=u_x+v_y=0. Possible flow.

Stream function. Integrate ψy=u=x2+2xy−y2\psi_y=u=x^2+2xy-y^2: ψ=x2y+xy2−y33+g(x).\psi=x^2y+xy^2-\frac{y^3}{3}+g(x). Match ψx=−v=−(x2−2xy−y2)\psi_x=-v=-(x^2-2xy-y^2): 2xy+y2+g′(x)=−x2+2xy+y2  ⟹  g′(x)=−x2  ⟹  g=−x33.2xy+y^2+g'(x)=-x^2+2xy+y^2\;\Longrightarrow\;g'(x)=-x^2\;\Longrightarrow\;g=-\tfrac{x^3}{3}.   ψ=x2y+xy2−13(x3+y3).  \boxed{\;\psi=x^2y+xy^2-\tfrac13(x^3+y^3).\;}


2020 Paper 2, 2020-P2-Q7c (15 marks)

Velocity potential ϕ=xy+x2−y2\phi=xy+x^2-y^2. Find the stream function.

Velocity. u=ϕx=y+2xu=\phi_x=y+2x; v=ϕy=x−2yv=\phi_y=x-2y. Check: ux+vy=2+(−2)=0u_x+v_y=2+(-2)=0 ✓.

Stream function. Integrate ψy=u=2x+y\psi_y=u=2x+y: ψ=2xy+y2/2+g(x)\psi=2xy+y^2/2+g(x). Match ψx=−v=2y−x\psi_x=-v=2y-x: 2y+g′(x)=2y−x  ⟹  g′(x)=−x  ⟹  g=−x2/2.2y+g'(x)=2y-x\;\Longrightarrow\;g'(x)=-x\;\Longrightarrow\;g=-x^2/2.   ψ=2xy+12(y2−x2)+C.  \boxed{\;\psi=2xy+\tfrac12(y^2-x^2)+C.\;}


2021 Paper 2, 2021-P2-Q7c (20 marks)

Show q⃗=λ(−y,x)/(x2+y2)\vec q=\lambda(-y,x)/(x^2+y^2) is a possible incompressible flow. Find streamlines; is motion potential; find ϕ\phi.

Incompressibility. u=−λy/r2u=-\lambda y/r^2, v=λx/r2v=\lambda x/r^2 where r2=x2+y2r^2=x^2+y^2. ux=2λxyr4,vy=−2λxyr4;∇⋅q⃗=0. ✓u_x=\frac{2\lambda xy}{r^4},\quad v_y=\frac{-2\lambda xy}{r^4};\quad\nabla\cdot\vec q=0.\ \checkmark

Streamlines. dx/(−y)=dy/x  ⇒  x dx+y dy=0  ⇒  x2+y2=Cdx/(-y)=dy/x\;\Rightarrow\;x\,dx+y\,dy=0\;\Rightarrow\;x^2+y^2=C. Concentric circles.

Irrotationality. ωz=vx−uy=λ(y2−x2)/r4−λ(y2−x2)/r4=0\omega_z=v_x-u_y=\lambda(y^2-x^2)/r^4-\lambda(y^2-x^2)/r^4=0. Irrotational except at origin.

Velocity potential. ∇(λarctan⁡(y/x))=λ(−y/r2, x/r2)=q⃗\nabla(\lambda\arctan(y/x))=\lambda(-y/r^2,\,x/r^2)=\vec q.   ϕ=λarctan⁡(y/x)=λθ(multi-valued: point vortex).  \boxed{\;\phi=\lambda\arctan(y/x)=\lambda\theta\quad(\text{multi-valued: point vortex}).\;}


2017 Paper 2, 2017-P2-Q8c (15 marks)

Velocity q⃗=(3xz/r5, 3yz/r5, (3z2−r2)/r5)\vec q=(3xz/r^5,\,3yz/r^5,\,(3z^2-r^2)/r^5). Show motion possible, potential is z/r3z/r^3, find streamlines.

Incompressibility (∇⋅q⃗=0\nabla\cdot\vec q=0). Compute ux+vy+wzu_x+v_y+w_z; use x2+y2=r2−z2x^2+y^2=r^2-z^2 to collapse the terms: ∇⋅q⃗=10zr5−15z(r2−z2)+15z3−5zr2r7=10zr5−10zr5=0. ✓\nabla\cdot\vec q=\frac{10z}{r^5}-\frac{15z(r^2-z^2)+15z^3-5zr^2}{r^7}=\frac{10z}{r^5}-\frac{10z}{r^5}=0.\ \checkmark

Velocity potential (convention q⃗=−∇ϕ\vec q=-\nabla\phi). Verify ϕ=z/r3\phi=z/r^3: −ϕx=3xz/r5=u,−ϕy=3yz/r5=v,−ϕz=(3z2−r2)/r5=w. ✓-\phi_x=3xz/r^5=u,\quad-\phi_y=3yz/r^5=v,\quad-\phi_z=(3z^2-r^2)/r^5=w.\ \checkmark

Streamlines. From dx/x=dy/ydx/x=dy/y: meridian planes y/x=C1y/x=C_1. In each plane, Stokes stream function ψ=ρ2/r3\psi=\rho^2/r^3 (with ρ=x2+y2\rho=\sqrt{x^2+y^2}) gives:   yx=C1,x2+y2(x2+y2+z2)3/2=C2.  \boxed{\;\frac{y}{x}=C_1,\qquad\frac{x^2+y^2}{(x^2+y^2+z^2)^{3/2}}=C_2.\;} (Classical doublet/dipole streamlines — sin⁡2θ/r=C2\sin^2\theta/r=C_2 in spherical coordinates.)


2022 Paper 2, 2022-P2-Q5e (10 marks)

Spherical velocity: vr=2Mr−3cos⁡θv_r=2Mr^{-3}\cos\theta, vθ=Mr−3sin⁡θv_\theta=Mr^{-3}\sin\theta (corrected exponent). Show potential flow; find ϕ\phi and streamlines.

Irrotationality (spherical ψ\psi-component of curl): (∇×v⃗)ϕ=1r ⁣[∂(rvθ)∂r−∂vr∂θ]=1r ⁣[−2Mr−3sin⁡θ−(−2Mr−3sin⁡θ)]=0. ✓(\nabla\times\vec v)_\phi=\frac{1}{r}\!\left[\frac{\partial(rv_\theta)}{\partial r}-\frac{\partial v_r}{\partial\theta}\right]=\frac1r\!\left[-2Mr^{-3}\sin\theta-(-2Mr^{-3}\sin\theta)\right]=0.\ \checkmark

Velocity potential. vr=∂ϕ/∂r=2Mcos⁡θ/r3v_r=\partial\phi/\partial r=2M\cos\theta/r^3; integrate: ϕ=−Mcos⁡θ/r2+g(θ)\phi=-M\cos\theta/r^2+g(\theta). Match vθ=(1/r)∂ϕ/∂θ=Msin⁡θ/r3v_\theta=(1/r)\partial\phi/\partial\theta=M\sin\theta/r^3: g′(θ)=0g'(\theta)=0.   ϕ=−Mcos⁡θr2(3D doublet/dipole).  \boxed{\;\phi=-\frac{M\cos\theta}{r^2}\quad(\text{3D doublet/dipole}).\;}

Streamlines via Stokes stream function: ψ=Msin⁡2θ/r\psi=M\sin^2\theta/r, so streamlines are sin⁡2θ/r=const\sin^2\theta/r=\text{const}.

Common Traps


streamlines-from-potential (2 question(s); 2014, 2024)

Recognition Cues

Solution Template

  1. Compute q⃗=∇ϕ\vec q=\nabla\phi to get u,vu,v (2D) or u,v,wu,v,w (3D).
  2. 2D: use the complex potential w=ϕ+iψw=\phi+i\psi; read ψ=Im⁡(w)\psi=\operatorname{Im}(w); streamlines are ψ=const\psi=\text{const}.
  3. 3D: write the streamline ODEs dx/u=dy/v=dz/wdx/u=dy/v=dz/w; integrate pairwise to get two independent first integrals f(x,y,z)=C1f(x,y,z)=C_1 and g(x,y,z)=C2g(x,y,z)=C_2.
  4. Identify the geometric shape of the level curves (circles, planes, hyperboloids, etc.).

Streamlines (\psi= const, dashed) are perpendicular to equipotentials (\phi= const, solid); velocity \vec q is tangent to streamlines

Worked Example(s)

2014 Paper 2, 2014-P2-Q7c (20 marks)

ϕ=12log⁡ ⁣[(x+a)2+y2(x−a)2+y2]\phi=\tfrac12\log\!\left[\tfrac{(x+a)^2+y^2}{(x-a)^2+y^2}\right]. Determine the streamlines.

Identify the flow. Write ϕ=ln⁡r1−ln⁡r2\phi=\ln r_1-\ln r_2 where r1,2=(x±a)2+y2r_{1,2}=\sqrt{(x\pm a)^2+y^2}. This is a source at (−a,0)(-a,0) and sink at (a,0)(a,0) of unit strength.

Complex potential. w=log⁡(z+a)−log⁡(z−a)=log⁡ ⁣z+az−aw=\log(z+a)-\log(z-a)=\log\!\dfrac{z+a}{z-a}.

Stream function. ψ=Im⁡(w)=arg⁡z+az−a=θ1−θ2\psi=\operatorname{Im}(w)=\arg\dfrac{z+a}{z-a}=\theta_1-\theta_2, where θ1,θ2\theta_1,\theta_2 are the angles from (∓a,0)(\mp a,0) to (x,y)(x,y).

Streamlines ψ=k\psi=k. Setting θ1−θ2=k\theta_1-\theta_2=k and using tan⁡(θ1−θ2)=−2ay/(x2+y2−a2)\tan(\theta_1-\theta_2)=-2ay/(x^2+y^2-a^2): x2+y2+2aky−a2=0  ⟹  x2+ ⁣(y+ak)2=a2(k2+1)k2.x^2+y^2+\frac{2a}{k}y-a^2=0\;\Longrightarrow\;x^2+\!\left(y+\frac ak\right)^2=\frac{a^2(k^2+1)}{k^2}.

  Streamlines are circles centred on the y-axis, each passing through (±a,0).  \boxed{\;\text{Streamlines are circles centred on the }y\text{-axis, each passing through }(\pm a,0).\;}

(Inscribed-angle theorem: constant subtended angle ⟺ circular arc through the chord endpoints.)


2024 Paper 2, 2024-P2-Q5e (10 marks)

ϕ=12(x2+y2−2z2)\phi=\tfrac12(x^2+y^2-2z^2). Determine the streamlines.

Velocity. u=xu=x, v=yv=y, w=−2zw=-2z. (Check: ∇2ϕ=1+1−2=0\nabla^2\phi=1+1-2=0 ✓.)

Streamline ODEs. dx/x=dy/y=dz/(−2z)dx/x=dy/y=dz/(-2z).

Pair 1 (dx/x=dy/ydx/x=dy/y): y/x=C1y/x=C_1.

Pair 2 (dx/x=dz/(−2z)dx/x=dz/(-2z)): 2ln⁡∣x∣+ln⁡∣z∣=const2\ln|x|+\ln|z|=\text{const}, i.e. x2z=C2x^2z=C_2.

  yx=C1andx2z=C2.  \boxed{\;\frac{y}{x}=C_1\quad\text{and}\quad x^2z=C_2.\;}

(Intersections of meridian planes y=C1xy=C_1 x with hyperbolic surfaces z=C2/x2z=C_2/x^2. These are not circles — check ∇2ϕ=0\nabla^2\phi=0 to confirm this is genuinely a potential flow.)

Common Traps


complex-potential-analysis (1 question(s); 2021)

Recognition Cues

Solution Template

  1. Write w=ϕ+iψw=\phi+i\psi by expressing ww in terms of z=x+iyz=x+iy.
  2. For ww involving log⁡\log or arctan⁡\arctan: use log⁡(ReiΘ)=ln⁡R+iΘ\log(Re^{i\Theta})=\ln R+i\Theta to identify ϕ=Θ/2\phi=\Theta/2 or ϕ=ln⁡R\phi=\ln R, etc.
  3. Streamlines: set ψ=const\psi=\text{const} and identify the geometric locus (often a circle or line).
  4. Equipotentials: set ϕ=const\phi=\text{const} — these are the orthogonal family.
  5. Velocity: dw/dz=u−ivdw/dz=u-iv; compute this derivative.
  6. Singularities: where dw/dzdw/dz is undefined or where ww has a branch point.

Worked Example(s)

2021 Paper 2, 2021-P2-Q8c (20 marks)

Complex potential w=tan⁡−1zw=\tan^{-1}z. Show streamlines and equipotential curves are circles; find velocity; check singularities at z=±iz=\pm i.

Logarithmic form. w=tan⁡−1z=12iln⁡ ⁣1+iz1−izw=\tan^{-1}z=\dfrac{1}{2i}\ln\!\dfrac{1+iz}{1-iz}.

Write 1+iz1−iz=ReiΘ\dfrac{1+iz}{1-iz}=Re^{i\Theta}. Since Re⁡(w)=Θ/2=ϕ\operatorname{Re}(w)=\Theta/2=\phi and Im⁡(w)=−ln⁡R/2=ψ\operatorname{Im}(w)=-\ln R/2=\psi.

Compute: R=∣1+iz∣∣1−iz∣=(1−y)2+x2(1+y)2+x2R=\dfrac{|1+iz|}{|1-iz|}=\dfrac{\sqrt{(1-y)^2+x^2}}{\sqrt{(1+y)^2+x^2}}.

Streamlines (ψ=\psi= const, i.e. R=R= const): x2+(1−y)2x2+(1+y)2=k  ⟹  x2+ ⁣(y−1+k1−k)2=4k(1−k)2.\frac{x^2+(1-y)^2}{x^2+(1+y)^2}=k\;\Longrightarrow\;x^2+\!\left(y-\frac{1+k}{1-k}\right)^2=\frac{4k}{(1-k)^2}. These are circles with centres on the yy-axis passing through (0,±1)(0,\pm1).

Equipotentials (ϕ=\phi= const, Θ=\Theta= const): the angle subtended at (x,y)(x,y) by the segment [−i,+i][-i,+i] is constant. By the inscribed-angle theorem, equipotentials are also circles through (0,±1)(0,\pm1), orthogonal to the streamlines.

Velocity. dw/dz=11+z2dw/dz=\dfrac{1}{1+z^2}.

Singularities. 1+z2=0⇒z=±i1+z^2=0\Rightarrow z=\pm i: logarithmic branch points of ww; velocity diverges there.

  Both families are circles through (0,±1);V=1/(1+z2);branch points at z=±i.  \boxed{\;\text{Both families are circles through }(0,\pm1);\quad V=1/(1+z^2);\quad\text{branch points at }z=\pm i.\;}

Common Traps


potential-boundary-value (1 question(s); 2016)

Recognition Cues

Solution Template

  1. Write ϕ\phi as a linear combination of the elementary harmonic functions appropriate to the geometry (e.g. x,  x/r3x,\;x/r^3 for each Cartesian direction in 3D spherical geometry).
  2. Compute the normal velocity ∂ϕ/∂r\partial\phi/\partial r on each shell.
  3. Match to the specified boundary velocity component in the normal direction — this gives one equation per shell per direction.
  4. Solve the linear system for the coefficients.
  5. Compute q⃗=−∇ϕ\vec q=-\nabla\phi to find the velocity at any interior point.

Worked Example(s)

2016 Paper 2, 2016-P2-Q7b (20 marks)

Liquid between concentric spheres r=ar=a (moving Ui^U\hat i) and r=br=b (moving Vj^V\hat j). Show the potential is ϕ=[a3U(1+12b3r−3)x−b3V(1+12a3r−3)y]/(b3−a3)\phi=\bigl[a^3U(1+\tfrac12 b^3r^{-3})x-b^3V(1+\tfrac12 a^3r^{-3})y\bigr]/(b^3-a^3) and find the velocity.

Harmonic ansatz. The natural harmonics linear in a direction are xx and x/r3x/r^3 (each satisfies ∇2=0\nabla^2=0). Take ϕ=(Ax+Bx/r3)+(Cy+Ey/r3)\phi=(Ax+Bx/r^3)+(Cy+Ey/r^3), with the xx-group carrying the inner-sphere UU-condition and the yy-group the outer-sphere VV-condition.

Boundary conditions (normal velocity on each sphere, q⃗=−∇ϕ\vec q=-\nabla\phi, n^=r/r\hat n=\mathbf r/r):

Coefficients: A=a3Ub3−a3,B=a3b3U2(b3−a3),C=−b3Vb3−a3,E=−a3b3V2(b3−a3).A=\frac{a^3U}{b^3-a^3},\quad B=\frac{a^3b^3U}{2(b^3-a^3)},\quad C=\frac{-b^3V}{b^3-a^3},\quad E=\frac{-a^3b^3V}{2(b^3-a^3)}.

Substituting gives the stated ϕ\phi. ■\blacksquare

Velocity. Differentiating ϕ\phi using ∂x(x/r3)=r−3−3x2r−5\partial_x(x/r^3)=r^{-3}-3x^2r^{-5} etc.:   q⃗=−a3Ui^+b3Vj^b3−a3+a3b32r3(b3−a3) ⁣[3(d⋅r^)r^⋅rr−d],d=Ui^−Vj^.  \boxed{\;\vec q=\frac{-a^3U\hat i+b^3V\hat j}{b^3-a^3}+\frac{a^3b^3}{2r^3(b^3-a^3)}\!\left[\frac{3(\mathbf d\cdot\hat r)\hat r\cdot r}{r}-\mathbf d\right],\quad\mathbf d=U\hat i-V\hat j.\;}

Common Traps


unsteady-bernoulli (1 question(s); 2019)

Recognition Cues

Solution Template

  1. Write ϕ=−F(t)/r\phi=-F(t)/r (radial harmonic for incompressible irrotational flow at infinity, decaying). From u=∂ϕ/∂r=F/r2u=\partial\phi/\partial r=F/r^2 and the BC u(R,t)=R˙u(R,t)=\dot R: determine F(t)=R2R˙F(t)=R^2\dot R.
  2. Compute ∂ϕ/∂t\partial\phi/\partial t at r=Rr=R; compute u2u^2 at r=Rr=R.
  3. Apply unsteady Bernoulli: p=Π−ρ(∂ϕ/∂t+12u2)p=\Pi-\rho(\partial\phi/\partial t+\tfrac12 u^2) (where Π\Pi = pressure at infinity).
  4. Substitute and simplify. Common result: pR=Π+ρ(RR¨+32R˙2)p_R=\Pi+\rho(R\ddot R+\tfrac32\dot R^2).
  5. If required, rewrite using d2(R2)/dt2=2R˙2+2RR¨d^2(R^2)/dt^2=2\dot R^2+2R\ddot R to match the stated form.

Worked Example(s)

2019 Paper 2, 2019-P2-Q8b (15 marks)

Sphere radius R(t)R(t) vibrates radially in infinite incompressible fluid. Establish the surface pressure Π+12ρ{d2(R2)/dt2+(R˙)2}\Pi+\tfrac12\rho\{d^2(R^2)/dt^2+(\dot R)^2\}.

Step 1 — Velocity potential. Radial symmetry: ϕ=−F(t)/r\phi=-F(t)/r, u=F/r2u=F/r^2. BC at r=Rr=R: F=R2R˙F=R^2\dot R. So ϕ=−R2R˙/r\phi=-R^2\dot R/r.

Step 2 — Terms at r=Rr=R. ∂ϕ∂t∣R=−ddt(R2R˙)=−(2RR˙2+R2R¨),u2∣R=R˙2.\left.\frac{\partial\phi}{\partial t}\right|_R=-\frac{d}{dt}(R^2\dot R)=-(2R\dot R^2+R^2\ddot R),\qquad u^2|_R=\dot R^2.

Step 3 — Unsteady Bernoulli. pR=Π−ρ ⁣[−(2RR˙2+R2R¨)+12R˙2]=Π+ρ ⁣(RR¨+32R˙2).p_R=\Pi-\rho\!\left[-(2R\dot R^2+R^2\ddot R)+\tfrac12\dot R^2\right]=\Pi+\rho\!\left(R\ddot R+\tfrac32\dot R^2\right).

Step 4 — Reconcile with printed form. 12ρ{d2(R2)/dt2+R˙2}=12ρ{(2R˙2+2RR¨)+R˙2}=ρ(RR¨+32R˙2)\tfrac12\rho\{d^2(R^2)/dt^2+\dot R^2\}=\tfrac12\rho\{(2\dot R^2+2R\ddot R)+\dot R^2\}=\rho(R\ddot R+\tfrac32\dot R^2). ✓

  pR=Π+12ρ ⁣{d2(R2)dt2+ ⁣(dRdt)2}.  ■\boxed{\;p_R=\Pi+\tfrac12\rho\!\left\{\tfrac{d^2(R^2)}{dt^2}+\!\left(\tfrac{dR}{dt}\right)^2\right\}.\;}\quad\blacksquare

Common Traps


Marks-Aware Writing

10-mark questions (2022-Q5e, 2024-Q5e): One to two sentences of setup (state ϕ\phi or the flow type), then execute the algorithm. For streamlines: write the two ODEs, integrate in pairs, box the result. No lengthy preamble.

15-mark questions (2017, 2018, 2019, 2020): Three to four clear steps. For the stream function: verify possibility (∇2ϕ=0\nabla^2\phi=0), then integrate step by step. For unsteady Bernoulli: label Steps 1–3 (potential, evaluate at r=Rr=R, Bernoulli). The verification step at the end earns the final method marks.

20-mark questions (2014, 2016, 2021-Q7c, 2021-Q8c): Full proof format. For boundary-value problems: state the harmonic ansatz, write both BCs, solve the coefficient system, then differentiate for velocity. For complex potentials: extract ϕ,ψ\phi,\psi explicitly, then show each geometric claim separately (streamlines circles, equipotentials circles, velocity, singularities).

Practice Set

YearPaper/QMarksArchetypeOne-line hint
2022P2-Q5e10potential-stream-functionSpherical dipole; check curl-free in spherical; ϕ=−Mcos⁡θ/r2\phi=-M\cos\theta/r^2
2024P2-Q5e10streamlines-from-potentialq⃗=(x,y,−2z)\vec q=(x,y,-2z); pair 1: y/x=C1y/x=C_1; pair 2: x2z=C2x^2z=C_2
2020P2-Q7c15potential-stream-functionϕ=xy+x2−y2\phi=xy+x^2-y^2; integrate ψy=y+2x\psi_y=y+2x then match ψx\psi_x
2018P2-Q8b15potential-stream-functionCheck ∇2ϕ=0\nabla^2\phi=0; integrate CR relations to get ψ=x2y+xy2−13(x3+y3)\psi=x^2y+xy^2-\tfrac13(x^3+y^3)
2017P2-Q8c15potential-stream-functionDivergence cancellation uses x2+y2=r2−z2x^2+y^2=r^2-z^2; ϕ=z/r3\phi=z/r^3 with q⃗=−∇ϕ\vec q=-\nabla\phi
2019P2-Q8b15unsteady-bernoulliϕ=−R2R˙/r\phi=-R^2\dot R/r; $\phi_t
2021P2-Q7c20potential-stream-functionDiv=0; curl=0; streamlines x2+y2=Cx^2+y^2=C; ϕ=λθ\phi=\lambda\theta (point vortex, multi-valued)
2021P2-Q8c20complex-potential-analysisw=tan⁡−1zw=\tan^{-1}z; write as log⁡\log; $R=
2016P2-Q7b20potential-boundary-valueAnsatz Ax+Bx/r3+Cy+Ey/r3Ax+Bx/r^3+Cy+Ey/r^3; decouple x,yx,y systems; solve for 4 coefficients
2014P2-Q7c20streamlines-from-potentialSource-sink complex potential; ψ=θ1−θ2\psi=\theta_1-\theta_2; inscribed-angle → circles through (±a,0)(\pm a,0)

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