UPSC Maths 2025 Paper 2 Q5e — Solution

10 marks · Section B

Question

A source and a sink of equal strength are placed at points $\left(\pm\dfrac{a}{2}, 0\right)$ within a fixed circular boundary $x^2 + y^2 = a^2$ . Show that the streamlines are given by $\left(r^2 - \dfrac{a^2}{4}\right)(r^2 - 4a^2) - 4a^2 y^2 = ky(r^2 - a^2)$ where $k$ is a constant and $r^2 = x^2 + y^2$ .

Technique

Use two-dimensional ideal flow with the circle theorem (method of images): each interior source/sink is imaged in $|z|=a$ 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 $\psi = \text{const}$ .

Solution

Image system (circle theorem)

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

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

Source $+m$ at $a/2$ → image source $+m$ at $a^2/(a/2) = 2a$ , and sink $-m$ at $0$ .
Sink $-m$ at $-a/2$ → image sink $-m$ at $a^2/(-a/2) = -2a$ , and source $+m$ at $0$ .

The two central images ( $-m$ and $+m$ at the origin) cancel. The full singularity system is therefore: $+m \text{ at } \tfrac{a}{2},\quad +m \text{ at } 2a,\quad -m \text{ at } -\tfrac{a}{2},\quad -m \text{ at } -2a.$

Complex potential and stream function

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

= \frac{m}{2\pi}\ln\frac{\big(z - \tfrac{a}{2}\big)(z - 2a)}{\big(z + \tfrac{a}{2}\big)(z + 2a)}.$$ The stream function is $\psi = \operatorname{Im} w = \dfrac{m}{2\pi}\arg\dfrac{N}{D}$, where $$N = \big(z - \tfrac{a}{2}\big)(z - 2a),\qquad D = \big(z + \tfrac{a}{2}\big)(z + 2a).$$ Streamlines $\psi = \text{const}$ mean $\arg(N/D) = \text{const}$, i.e. $\arg N - \arg D = c'$. Equivalently, the locus where $$\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)$$ for a constant $\lambda$. ### Compute $\operatorname{Re}(N\bar D)$ and $\operatorname{Im}(N\bar D)$ With $z = x + iy$ and $r^2 = x^2 + y^2$, a direct expansion gives: $$\operatorname{Im}(N\bar D) = 5a\,y\,(x^2 + y^2 - a^2) = 5a\,y\,(r^2 - a^2),$$ $$\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.$$ The real part factorises exactly as the target left-hand side. Expanding $$\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,$$ and using $r^4 = x^4 + 2x^2y^2 + y^4$, $\,r^2 = x^2 + y^2$: $$= 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,$$ which is **identical** to $\operatorname{Re}(N\bar D)$. ### Streamline equation The streamline condition $\operatorname{Re}(N\bar D) = \lambda\,\operatorname{Im}(N\bar D)$ becomes $$\left(r^2 - \tfrac{a^2}{4}\right)(r^2 - 4a^2) - 4a^2 y^2 = \lambda\cdot 5a\,y\,(r^2 - a^2).$$ Writing $k = 5a\lambda$ (an arbitrary constant), this is exactly $$\boxed{\left(r^2 - \tfrac{a^2}{4}\right)(r^2 - 4a^2) - 4a^2 y^2 = k\,y\,(r^2 - a^2).}$$ ## Answer The image system places like singularities at $\pm 2a$ (the inverse points), giving the streamlines $$\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}.$$