← 2020 Paper 2

UPSC 2020 Maths Optional Paper 2 Q7c — Step-by-Step Solution

15 marks · Section B

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

Question

A velocity potential in a two-dimensional fluid flow is given by ϕ(x,y)=xy+x2−y2\phi(x,y)=xy+x^2-y^2. Find the stream function for this flow.

Technique

u⃗=∇ϕ\vec u=\nabla\phi; Cauchy–Riemann relations ϕx=ψy, ϕy=−ψx\phi_x=\psi_y,\ \phi_y=-\psi_x; integrate to recover ψ\psi up to an additive constant.

Solution

Step 1 — Velocity components

With velocity potential ϕ\phi (so u⃗=∇ϕ\vec u=\nabla\phi),

u=∂ϕ∂x=y+2x,v=∂ϕ∂y=x−2y.u=\frac{\partial\phi}{\partial x}=y+2x,\qquad v=\frac{\partial\phi}{\partial y}=x-2y.

(Sanity: the flow must be incompressible. ∂u∂x+∂v∂y=2+(−2)=0\dfrac{\partial u}{\partial x}+\dfrac{\partial v}{\partial y}=2+(-2)=0, so ϕ\phi is harmonic and a stream function exists.)

Step 2 — Cauchy–Riemann relations for the conjugate ψ\psi

The stream function ψ\psi satisfies

u=∂ϕ∂x=∂ψ∂y,v=∂ϕ∂y=−∂ψ∂x.u=\frac{\partial\phi}{\partial x}=\frac{\partial\psi}{\partial y},\qquad v=\frac{\partial\phi}{\partial y}=-\frac{\partial\psi}{\partial x}.

Step 3 — Integrate

From ∂ψ∂y=u=2x+y\dfrac{\partial\psi}{\partial y}=u=2x+y:

ψ=∫(2x+y) dy=2xy+y22+g(x).\psi=\int(2x+y)\,dy=2xy+\frac{y^2}{2}+g(x).

Differentiate in xx and use ∂ψ∂x=−v=−(x−2y)=−x+2y\dfrac{\partial\psi}{\partial x}=-v=-(x-2y)=-x+2y:

∂ψ∂x=2y+g′(x)=2y−x ⇒ g′(x)=−x ⇒ g(x)=−x22+C.\frac{\partial\psi}{\partial x}=2y+g'(x)=2y-x\ \Rightarrow\ g'(x)=-x\ \Rightarrow\ g(x)=-\frac{x^2}{2}+C.

Step 4 — Stream function

Answer

  ψ(x,y)=2xy+12(y2−x2)+C.  \boxed{\;\psi(x,y)=2xy+\frac{1}{2}\big(y^2-x^2\big)+C.\;}
We post more of this — worked solutions, CSAT trap breakdowns, guide chapters — a few times a week on Telegram. Free, no sign-in. Join

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.