UPSC 2026 Maths Optional Paper 2 Q5d — Step-by-Step Solution

10 marks · Section B

Sources, sinks, doublets · Mechanics & Fluid Dynamics · asked 9× in 14 yrs · Read the full method →

Question

A simple source of strength $m$ is fixed at the origin $O$ in a uniform stream of incompressible fluid moving with velocity $U\hat{i}$ . Find out the velocity potential $\phi$ at any point $P$ of the stream where $OP = r$ and $\theta$ is the angle $\overrightarrow{OP}$ makes with the direction $\hat{i}$ . Find the differential equation of the stream lines and show that they lie on the surface $Ur^2\sin^2\theta - 2m\cos\theta = \text{constant}$ .

Technique

This is the Rankine half-body: superpose a three-dimensional simple source on a uniform stream. The flow is axisymmetric about the $x$ -axis and irrotational, so $\phi$ is just the sum of the two elementary potentials — but the sign convention for $\phi$ and the meaning of “strength $m$ ” must be declared, because the printed target $Ur^{2}\sin^{2}\theta-2m\cos\theta$ fixes both. The streamline equation $dr/q_r = r\,d\theta/q_\theta$ is then made exact by the integrating factor $2r\sin\theta$ — which is the same thing as recognising the Stokes stream function $\psi=\tfrac12 Ur^{2}\sin^{2}\theta-m\cos\theta$ .

Solution

Step 0 — Conventions (state them; the answer is not checkable without them).

(a) Sign of the potential. We use the convention standard in the Indian hydrodynamics texts,

\vec q=-\nabla\phi .

(b) Strength of a simple source. A simple (three-dimensional, point) source of strength $m$ at $O$ produces the purely radial field $q_r=m/r^{2}$ . Its total efflux across a sphere of radius $r$ is

\oint \vec q\cdot\hat n\,dS=4\pi r^{2}\cdot\frac{m}{r^{2}}=4\pi m,

so “strength $m$ ” here means an emission of $4\pi m$ units of volume per unit time. Its potential is $\phi_{\text{source}}=m/r$ , since $-\partial_r(m/r)=m/r^{2}$ . The printed answer forces this convention: the alternative reading (efflux $m$ , $\phi=m/4\pi r$ ) would produce $Ur^{2}\sin^{2}\theta-\dfrac{m}{2\pi}\cos\theta=$ const. It is the coefficient $2m$ in the printed target that selects the first reading. Declaring which convention is in force is itself worth a mark, because without it the answer is not checkable at all.

(c) Geometry. Spherical polar coordinates $(r,\theta,\omega)$ with the polar axis along $\hat i$ : $\theta$ is the angle $\overrightarrow{OP}$ makes with $\hat i$ (so $x=r\cos\theta$ ) and $\omega$ is the azimuth. Nothing depends on $\omega$ and there is no swirl, so the flow is axisymmetric: $q_\omega=0$ , $\partial/\partial\omega\equiv0$ .

Step 1 — The velocity potential.

Uniform stream. We need $\vec q=U\hat i$ , i.e. $-\nabla\phi_{\text{stream}}=U\hat i$ , so $\phi_{\text{stream}}=-Ux=-Ur\cos\theta$ .

Source at $O$ . $\phi_{\text{source}}=\dfrac{m}{r}$ .

Both are harmonic ( $\nabla^{2}\phi=0$ away from $O$ ) and the problem is linear, so they superpose:

\boxed{\ \phi(r,\theta)=\frac{m}{r}-Ur\cos\theta\ }\qquad (r>0). \tag{1}

Step 2 — The velocity components (and a check that (1) is right).

q_r=-\frac{\partial\phi}{\partial r}=-\left(-\frac{m}{r^{2}}-U\cos\theta\right)=U\cos\theta+\frac{m}{r^{2}},

q_\theta=-\frac{1}{r}\frac{\partial\phi}{\partial\theta}=-\frac{1}{r}\bigl(Ur\sin\theta\bigr)=-U\sin\theta,\qquad q_\omega=0. \tag{2}

Check. As $r\to\infty$ , $(q_r,q_\theta)\to(U\cos\theta,\,-U\sin\theta)$ , which is exactly $U\hat i$ resolved in spherical polars ✓. As $r\to0$ , $q_r\sim m/r^{2}$ , the source ✓. And the continuity equation for axisymmetric incompressible flow,

\nabla\cdot\vec q=\frac{1}{r^{2}}\frac{\partial}{\partial r}\bigl(r^{2}q_r\bigr)+\frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\bigl(\sin\theta\,q_\theta\bigr),

is satisfied: with $r^{2}q_r=Ur^{2}\cos\theta+m$ and $\sin\theta\,q_\theta=-U\sin^{2}\theta$ ,

\frac{1}{r^{2}}\frac{\partial}{\partial r}\bigl(Ur^{2}\cos\theta+m\bigr)=\frac{2Ur\cos\theta}{r^{2}}=\frac{2U\cos\theta}{r},\qquad \frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}\bigl(-U\sin^{2}\theta\bigr)=\frac{-2U\sin\theta\cos\theta}{r\sin\theta}=-\frac{2U\cos\theta}{r},

whose sum is $0$ for $r>0$ ✓ (the source is a singularity at $r=0$ , where continuity necessarily fails).

Step 3 — The differential equation of the streamlines.

A streamline is a curve everywhere tangent to $\vec q$ . In spherical polars the element of arc has components $(dr,\ r\,d\theta,\ r\sin\theta\,d\omega)$ , so tangency reads

\frac{dr}{q_r}=\frac{r\,d\theta}{q_\theta}=\frac{r\sin\theta\,d\omega}{q_\omega}.

Since $q_\omega=0$ , the last ratio forces $d\omega=0$ : every streamline lies in a meridian plane $\omega=$ const. Within that plane, substituting (2),

\boxed{\ \frac{dr}{\,U\cos\theta+\dfrac{m}{r^{2}}\,}=\frac{r\,d\theta}{-U\sin\theta}\ }\qquad\text{i.e.}\qquad \frac{dr}{d\theta}=-\,\frac{r\bigl(Ur^{2}\cos\theta+m\bigr)}{Ur^{2}\sin\theta}. \tag{3}

Step 4 — Integrate (3).

Cross-multiplying (3),

-U\sin\theta\,dr=r\left(U\cos\theta+\frac{m}{r^{2}}\right)d\theta \quad\Longrightarrow\quad U\sin\theta\,dr+Ur\cos\theta\,d\theta+\frac{m}{r}\,d\theta=0 .

This is not exact as it stands. Multiply by the integrating factor $2r\sin\theta$ :

2Ur\sin^{2}\theta\,dr+2Ur^{2}\sin\theta\cos\theta\,d\theta+2m\sin\theta\,d\theta=0 .

Now recognise the two exact differentials

d\bigl(Ur^{2}\sin^{2}\theta\bigr)=2Ur\sin^{2}\theta\,dr+2Ur^{2}\sin\theta\cos\theta\,d\theta,\qquad d\bigl(-2m\cos\theta\bigr)=2m\sin\theta\,d\theta,

so the equation is

d\Bigl(Ur^{2}\sin^{2}\theta-2m\cos\theta\Bigr)=0 .

Integrating,

\boxed{\ Ur^{2}\sin^{2}\theta-2m\cos\theta=\text{constant}\ } \tag{4}

which is the required result. Because each streamline lies in a meridian plane and (4) is independent of $\omega$ , each constant in (4) determines a surface of revolution about the $x$ -axis, generated by rotating the corresponding streamline; the streamlines therefore lie on these surfaces, exactly as the question phrases it. $\qquad\blacksquare$

Step 5 — The same result via the Stokes stream function (the structural reason, worth two lines).

For axisymmetric incompressible flow the Stokes stream function $\psi(r,\theta)$ is defined by

q_r=\frac{1}{r^{2}\sin\theta}\frac{\partial\psi}{\partial\theta},\qquad q_\theta=-\frac{1}{r\sin\theta}\frac{\partial\psi}{\partial r},

which satisfies continuity identically. From (2),

\frac{\partial\psi}{\partial\theta}=r^{2}\sin\theta\left(U\cos\theta+\frac{m}{r^{2}}\right)=Ur^{2}\sin\theta\cos\theta+m\sin\theta \ \Longrightarrow\ \psi=\tfrac12Ur^{2}\sin^{2}\theta-m\cos\theta+h(r),

and $q_\theta=-\dfrac{1}{r\sin\theta}\bigl(Ur\sin^{2}\theta+h'(r)\bigr)=-U\sin\theta$ forces $h'(r)=0$ . Taking $h\equiv0$ ,

\psi=\tfrac12\,Ur^{2}\sin^{2}\theta-m\cos\theta . \tag{5}

Streamlines are $\psi=$ const, and $2\psi=Ur^{2}\sin^{2}\theta-2m\cos\theta$ : the printed surface is $2\psi=$ const. (The integrating factor $2r\sin\theta$ of Step 4 is precisely the factor that converts the tangency condition into $d\psi=0$ — the two routes are the same computation.)

Step 6 — Stagnation point, and what the surfaces physically are.

Stagnation. $q_\theta=-U\sin\theta=0\Rightarrow\theta=0$ or $\pi$ . With $q_r=U\cos\theta+m/r^{2}$ :

$\theta=0$ : $q_r=U+m/r^{2}>0$ — no stagnation downstream;
$\theta=\pi$ : $q_r=-U+m/r^{2}=0\Rightarrow r=\sqrt{m/U}$ .

\text{Stagnation point } S:\quad \left(r,\theta\right)=\left(\sqrt{\tfrac{m}{U}},\ \pi\right),\quad\text{i.e. }x=-\sqrt{m/U}\ \text{on the upstream axis}.

This is where the source, pushing out radially, exactly balances the oncoming stream.

The dividing surface. Evaluating (5) at $S$ : $\psi_S=\tfrac12U\cdot\tfrac{m}{U}\cdot\sin^{2}\pi-m\cos\pi=m$ . So the streamline through the stagnation point is

Ur^{2}\sin^{2}\theta-2m\cos\theta=2m . \tag{6}

(Consistency: on the upstream axis $\theta=\pi$ the left side of (4) equals $+2m$ for every $r$ — so the whole upstream axis up to $S$ belongs to this same streamline, as it must.)

Surface (6) is the Rankine half-body: it separates the fluid emitted by the source (which stays inside, forever) from the oncoming stream (which stays outside). Being a stream surface, no fluid crosses it, so it may be replaced by a rigid boundary without altering the exterior flow — that is what makes the construction useful. Its asymptotic radius follows from (6) as $\theta\to0$ with $R=r\sin\theta$ held finite: $UR^{2}-2m=2m$ , so

R_\infty=2\sqrt{\frac{m}{U}},

which agrees with a pure flux balance ( $U\pi R_\infty^{2}=4\pi m$ , the source output carried away downstream) — an independent confirmation of the " $4\pi m$ " reading of the strength.

Answer

\boxed{\;\phi=\frac{m}{r}-Ur\cos\theta\ \ (\vec q=-\nabla\phi);\qquad \frac{dr}{U\cos\theta+\dfrac{m}{r^{2}}}=\frac{r\,d\theta}{-U\sin\theta},\ \ d\omega=0;\qquad Ur^{2}\sin^{2}\theta-2m\cos\theta=\text{const}\;}

with $\psi=\tfrac12Ur^{2}\sin^{2}\theta-m\cos\theta$ , stagnation point at $\bigl(\sqrt{m/U},\,\pi\bigr)$ and dividing (Rankine half-body) surface $Ur^{2}\sin^{2}\theta-2m\cos\theta=2m$ .

(Under the opposite convention $\vec q=+\nabla\phi$ the potential is $\phi=Ur\cos\theta-\dfrac{m}{r}$ ; the velocity field, $\psi$ , the streamline equation and the printed surface are all unchanged.)