← 2021 Paper 2

UPSC 2021 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

What sources/sinks have velocity potential $w=\log_e(z-a^2/z)$? Sketch streamlines; show that two streamlines subdivide into circle $r=a$ and $y$-axis.

Technique

Factor $z-a^2/z$ to identify sources/sinks; streamline $\psi=$ const; verify $\psi$ constant on circle and on $y$-axis.

Solution

Step 1 — Simplify $w$

$z-\dfrac{a^2}{z}=\dfrac{z^2-a^2}{z}=\dfrac{(z-a)(z+a)}{z}$.

$w=\log\!\left[\dfrac{(z-a)(z+a)}{z}\right]=\log(z-a)+\log(z+a)-\log z$.

Step 2 — Identify sources/sinks

Each $\log(z-z_0)$ term corresponds to a source (positive strength) or sink (negative) at $z_0$.

$+\log(z-a)$: source at $z=a$. $+\log(z+a)$: source at $z=-a$. $-\log z$: sink at $z=0$.

(In 2D potential flow, a source of strength $k$ has potential $k\log r$. The complex potential $\log(z-z_0)$ corresponds to a source of unit strength at $z_0$.)

Step 3 — Velocity field

$dw/dz=\dfrac{1}{z-a}+\dfrac{1}{z+a}-\dfrac{1}{z}$.

Sum the fractions: $\dfrac{(z+a)z+(z-a)z-(z-a)(z+a)}{z(z-a)(z+a)}=\dfrac{z^2+az+z^2-az-(z^2-a^2)}{z(z^2-a^2)}=\dfrac{z^2+a^2}{z(z^2-a^2)}$.

Step 4 — Streamlines $\psi=$ const

$\psi=\Im w=\arg(z-a)+\arg(z+a)-\arg z$.

Streamline $\psi=$ const means $\theta_1+\theta_2-\theta=$ const, where $\theta_1=\arg(z-a)$, $\theta_2=\arg(z+a)$, $\theta=\arg z$.

Step 5 — Stagnation points

$dw/dz=0\Rightarrow z^2+a^2=0\Rightarrow z=\pm ia$.

Stagnation at $(0,\pm a)$ on $y$-axis.

Step 6 — The two special streamlines

$y$-axis ($\theta=\pi/2$ or $-\pi/2$): $z=iy$.

$\arg(z-a)=\arg(iy-a)$, $\arg(z+a)=\arg(iy+a)$, $\arg z=\pi/2$ (for $y>0$).

By symmetry of $-a,+a$ about origin, $\arg(iy-a)+\arg(iy+a)=$ angle subtended by interval $[-a,a]$ at $(0,y)$ from above. For pure imaginary $z$, this is $\pi$ (the angles add to $\pi$).

Wait, for $y>0$: $\arg(iy-a)$ is in $(\pi/2,\pi)$; $\arg(iy+a)$ is in $(0,\pi/2)$; sum is $\pi/2+\pi/2=\pi$. So $\psi=\pi-\pi/2=\pi/2$ on positive $y$-axis. Const ⇒ streamline.

Circle $r=a$: $z=ae^{i\phi}$. Then $z-a=a(e^{i\phi}-1)$, $z+a=a(e^{i\phi}+1)$.

$\arg(z-a)$ and $\arg(z+a)$: standard “inscribed angle” — sum equals… let me think.

For point $z$ on circle $|z|=a$, the angles $\arg(z-a)$ and $\arg(z+a)$ subtended at the points $a$ and $-a$ on the diameter. Inscribed-angle theorem: angle subtended by diameter at any point on circle is $\pi/2$.

Hmm more precisely: $\arg(z-a)-\arg(z+a)$ is the angle subtended by the chord from $-a$ to $a$ at point $z$. For $z$ on the circle, this is the inscribed angle subtending the diameter — which is $\pi/2$.

And $\arg z=\phi$.

Hmm: $\psi=\arg(z-a)+\arg(z+a)-\arg z$ — sum, not difference.

Let me check at $z=ai$ (top of circle): $z-a=-a+ai$, $\arg=3\pi/4$. $z+a=a+ai$, $\arg=\pi/4$. $\arg z=\pi/2$. Sum: $3\pi/4+\pi/4-\pi/2=\pi-\pi/2=\pi/2$. Same value as on $y$-axis ✓.

At $z=a$ (rightmost, but this is a singularity). Skip.

At $z=ae^{i\phi}$ for $\phi=\pi/4$: $z=a(\cos\pi/4+i\sin\pi/4)=a/\sqrt 2(1+i)$.

$z-a=a/\sqrt 2(1+i)-a=a(1/\sqrt 2-1+i/\sqrt 2)$. Real part $\approx -0.293a$, imag $\approx 0.707a$. $\arg\approx\arctan(0.707/-0.293)$ in second quadrant: $\arctan(-2.414)+\pi\approx -1.178+\pi\approx 1.963$.

$z+a=a/\sqrt 2(1+i)+a=a(1+1/\sqrt 2,1/\sqrt 2)\approx a(1.707,0.707)$. $\arg\approx\arctan(0.707/1.707)\approx 0.393$.

Sum: $1.963+0.393=2.356\approx 3\pi/4$.

$\arg z=\pi/4\approx 0.785$.

$\psi=3\pi/4-\pi/4=\pi/2$ ✓.

So on the circle $|z|=a$, $\psi=\pi/2$ is constant — it’s a streamline.

Step 7 — Sketch

• Sources at $\pm a$ on $x$-axis emit flow radially.
• Sink at origin absorbs flow.
• Two stagnation points at $(0,\pm a)$.
• The circle $r=a$ and the $y$-axis are streamlines $\psi=\pi/2$ (which “branch” at the stagnation points).

The two streamlines literally subdivide along the circle and the $y$-axis.

Answer