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UPSC 2025 Maths Optional Paper 2 Q5e — Step-by-Step Solution

10 marks · Section B

Sources, sinks, doublets · Mechanics & Fluid Dynamics · asked 8× in 13 yrs · Read the full method →

Question

A source and a sink of equal strength are placed at points (±a2,0)\left(\pm\dfrac{a}{2}, 0\right) within a fixed circular boundary x2+y2=a2x^2 + y^2 = a^2. Show that the streamlines are given by (r2−a24)(r2−4a2)−4a2y2=ky(r2−a2)\left(r^2 - \dfrac{a^2}{4}\right)(r^2 - 4a^2) - 4a^2 y^2 = ky(r^2 - a^2) where kk is a constant and r2=x2+y2r^2 = x^2 + y^2.

Technique

Use two-dimensional ideal flow with the circle theorem (method of images): each interior source/sink is imaged in ∣z∣=a|z|=a by a like singularity at the inverse point together with an opposite one at the centre. The complex potential is a sum of logarithms; the stream function is its imaginary part, and streamlines are ψ=const\psi = \text{const}.

Solution

Image system (circle theorem)

Use the complex variable z=x+iyz = x + iy. Put a source of strength mm at z1=a/2z_1 = a/2 and a sink of strength mm (a source of strength −m-m) at z2=−a/2z_2 = -a/2. Both lie inside ∣z∣=a|z| = a.

By the circle theorem (Milne–Thomson), a source mm at a real interior point bb is imaged in ∣z∣=a|z|=a by a source mm at the inverse point a2/ba^2/b and a sink mm at the origin.

The two central images (−m-m and +m+m at the origin) cancel. The full singularity system is therefore:

+m at a2,+m at 2a,−m at −a2,−m at −2a.+m \text{ at } \tfrac{a}{2},\quad +m \text{ at } 2a,\quad -m \text{ at } -\tfrac{a}{2},\quad -m \text{ at } -2a.

Complex potential and stream function

With source strength normalised so a single source contributes m2πln⁡(z−z0)\dfrac{m}{2\pi}\ln(z - z_0),

w=m2π[ln⁡ ⁣(z−a2)+ln⁡(z−2a)−ln⁡ ⁣(z+a2)−ln⁡(z+2a)]=m2πln⁡(z−a2)(z−2a)(z+a2)(z+2a).w = \frac{m}{2\pi}\Big[\ln\!\big(z - \tfrac{a}{2}\big) + \ln(z - 2a) - \ln\!\big(z + \tfrac{a}{2}\big) - \ln(z + 2a)\Big] = \frac{m}{2\pi}\ln\frac{\big(z - \tfrac{a}{2}\big)(z - 2a)}{\big(z + \tfrac{a}{2}\big)(z + 2a)}.

The stream function is ψ=Im⁡w=m2πarg⁡ND\psi = \operatorname{Im} w = \dfrac{m}{2\pi}\arg\dfrac{N}{D}, where

N=(z−a2)(z−2a),D=(z+a2)(z+2a).N = \big(z - \tfrac{a}{2}\big)(z - 2a),\qquad D = \big(z + \tfrac{a}{2}\big)(z + 2a).

Streamlines ψ=const\psi = \text{const} mean arg⁡(N/D)=const\arg(N/D) = \text{const}, i.e. arg⁡N−arg⁡D=c′\arg N - \arg D = c'. Equivalently, the locus where

Im⁡(NDˉ)Re⁡(NDˉ)=tan⁡c′=const⟺Re⁡(NDˉ)=λ Im⁡(NDˉ)\frac{\operatorname{Im}(N\bar D)}{\operatorname{Re}(N\bar D)} = \tan c' = \text{const} \quad\Longleftrightarrow\quad \operatorname{Re}(N\bar D) = \lambda\,\operatorname{Im}(N\bar D)

for a constant λ\lambda.

Compute Re⁡(NDˉ)\operatorname{Re}(N\bar D) and Im⁡(NDˉ)\operatorname{Im}(N\bar D)

With z=x+iyz = x + iy and r2=x2+y2r^2 = x^2 + y^2, a direct expansion gives:

Im⁡(NDˉ)=5a y (x2+y2−a2)=5a y (r2−a2),\operatorname{Im}(N\bar D) = 5a\,y\,(x^2 + y^2 - a^2) = 5a\,y\,(r^2 - a^2), Re⁡(NDˉ)=a4−174a2x2−334a2y2+x4+2x2y2+y4.\operatorname{Re}(N\bar D) = a^4 - \tfrac{17}{4}a^2 x^2 - \tfrac{33}{4}a^2 y^2 + x^4 + 2x^2y^2 + y^4.

The real part factorises exactly as the target left-hand side. Expanding

(r2−a24)(r2−4a2)−4a2y2=r4−174a2r2+a4−4a2y2,\left(r^2 - \tfrac{a^2}{4}\right)(r^2 - 4a^2) - 4a^2 y^2 = r^4 - \tfrac{17}{4}a^2 r^2 + a^4 - 4a^2 y^2,

and using r4=x4+2x2y2+y4r^4 = x^4 + 2x^2y^2 + y^4,  r2=x2+y2\,r^2 = x^2 + y^2:

=x4+2x2y2+y4−174a2x2−174a2y2+a4−4a2y2=a4−174a2x2−334a2y2+x4+2x2y2+y4,= x^4 + 2x^2y^2 + y^4 - \tfrac{17}{4}a^2 x^2 - \tfrac{17}{4}a^2 y^2 + a^4 - 4a^2 y^2 = a^4 - \tfrac{17}{4}a^2 x^2 - \tfrac{33}{4}a^2 y^2 + x^4 + 2x^2y^2 + y^4,

which is identical to Re⁡(NDˉ)\operatorname{Re}(N\bar D).

Streamline equation

The streamline condition Re⁡(NDˉ)=λ Im⁡(NDˉ)\operatorname{Re}(N\bar D) = \lambda\,\operatorname{Im}(N\bar D) becomes

(r2−a24)(r2−4a2)−4a2y2=λ⋅5a y (r2−a2).\left(r^2 - \tfrac{a^2}{4}\right)(r^2 - 4a^2) - 4a^2 y^2 = \lambda\cdot 5a\,y\,(r^2 - a^2).

Writing k=5aλk = 5a\lambda (an arbitrary constant), this is exactly

(r2−a24)(r2−4a2)−4a2y2=k y (r2−a2).\boxed{\left(r^2 - \tfrac{a^2}{4}\right)(r^2 - 4a^2) - 4a^2 y^2 = k\,y\,(r^2 - a^2).}

Answer

The image system places like singularities at ±2a\pm 2a (the inverse points), giving the streamlines

(r2−a24)(r2−4a2)−4a2y2=k y (r2−a2),k constant.\left(r^2 - \tfrac{a^2}{4}\right)(r^2 - 4a^2) - 4a^2 y^2 = k\,y\,(r^2 - a^2),\quad k \text{ constant}.
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