← 2021 Paper 2

UPSC 2021 Maths Optional Paper 2 Q7c — Step-by-Step Solution

20 marks · Section B

Potential flow · Mechanics & Fluid Dynamics · asked 10× in 13 yrs · Read the full method →

Question

Show $\vec q=\dfrac{\lambda(-y\hat\imath+x\hat\jmath)}{x^2+y^2}$ is a possible incompressible flow. Find streamlines. Is the motion potential? If yes, find velocity potential.

Technique

Verify $\nabla\cdot\vec q=0$ (incompressibility); find streamlines from $dx/u=dy/v$; verify $\nabla\times\vec q=0$ (locally) for potential; identify $\phi=\lambda\theta$.

Solution

Step 1 — Incompressibility

$\nabla\cdot\vec q=\partial_x\!\left(\dfrac{-\lambda y}{x^2+y^2}\right)+\partial_y\!\left(\dfrac{\lambda x}{x^2+y^2}\right)$.

$\partial_x(-\lambda y/(x^2+y^2))=-\lambda y\cdot\partial_x(1/(x^2+y^2))=-\lambda y\cdot(-2x/(x^2+y^2)^2)=\dfrac{2\lambda xy}{(x^2+y^2)^2}$.

$\partial_y(\lambda x/(x^2+y^2))=\lambda x\cdot(-2y/(x^2+y^2)^2)=\dfrac{-2\lambda xy}{(x^2+y^2)^2}$.

Sum: $0$. ✓

So $\nabla\cdot\vec q=0$ — incompressible.

Step 2 — Streamlines

$\dfrac{dx}{u}=\dfrac{dy}{v}$ where $u=-\lambda y/(x^2+y^2)$, $v=\lambda x/(x^2+y^2)$.

$\dfrac{dx}{-y}=\dfrac{dy}{x}$ (cancel $\lambda/(x^2+y^2)$).

$x\,dx+y\,dy=0\Rightarrow d(x^2+y^2)=0\Rightarrow x^2+y^2=$ const.

Streamlines are concentric circles centred at origin.

Step 3 — Test for potential flow

Curl of $\vec q$ (in 2D, the $z$-component): $\omega_z=\partial_x v-\partial_y u$.

$\partial_x v=\partial_x(\lambda x/(x^2+y^2))=\lambda\cdot\dfrac{(x^2+y^2)-x\cdot 2x}{(x^2+y^2)^2}=\dfrac{\lambda(y^2-x^2)}{(x^2+y^2)^2}$.

$\partial_y u=\partial_y(-\lambda y/(x^2+y^2))=-\lambda\cdot\dfrac{(x^2+y^2)-y\cdot 2y}{(x^2+y^2)^2}=\dfrac{-\lambda(x^2-y^2)}{(x^2+y^2)^2}=\dfrac{\lambda(y^2-x^2)}{(x^2+y^2)^2}$.

$\omega_z=\dfrac{\lambda(y^2-x^2)}{(x^2+y^2)^2}-\dfrac{\lambda(y^2-x^2)}{(x^2+y^2)^2}=0$.

So $\vec q$ is curl-free (except at the origin, where it’s singular).

Step 4 — Velocity potential

Since curl-free on $\mathbb R^2\setminus\{0\}$, locally there exists $\phi$ with $\vec q=\nabla\phi$.

$\partial\phi/\partial x=u=-\lambda y/(x^2+y^2)$.

$\partial\phi/\partial y=v=\lambda x/(x^2+y^2)$.

Compare with $\partial_y\arctan(y/x)=\dfrac{1/x}{1+y^2/x^2}=\dfrac{x}{x^2+y^2}$.

$\partial_x\arctan(y/x)=\dfrac{-y/x^2}{1+y^2/x^2}=\dfrac{-y}{x^2+y^2}$.

So $\nabla(\lambda\arctan(y/x))=\lambda\!\left(\dfrac{-y}{x^2+y^2},\dfrac{x}{x^2+y^2}\right)=\vec q$ ✓.

Velocity potential: $\phi=\lambda\arctan(y/x)=\lambda\theta$ (the polar angle).

Answer