← 2015 Paper 2
UPSC 2015 Maths Optional Paper 2 Q8c — Step-by-Step Solution 20 marks · Section B
Two-Dimensional and Axisymmetric Flow · Mechanics & Fluid Dynamics · Read the full method →
Question
In an axisymmetric motion, show that stream function exists due to equation of continuity. Express the velocity components in terms of the stream function. Find the equation satisfied by the stream function if the flow is irrotational.
Technique
Continuity in spherical coordinates ⇒ integrability ⇒ Stokes stream function ψ \psi ψ ψ \psi ψ ω ϕ \omega_\phi ω ϕ ψ \psi ψ E 2 ψ = 0 E^2\psi=0 E 2 ψ = 0
Solution
Setup. Use cylindrical coordinates ( R , ϕ , z ) (R,\phi,z) ( R , ϕ , z ) ( r , θ , ϕ ) (r,\theta,\phi) ( r , θ , ϕ ) ϕ \phi ϕ ϕ \phi ϕ
For axisymmetric flow, v ⃗ = v r r ^ + v θ θ ^ \vec v=v_r\hat r+v_\theta\hat\theta v = v r r ^ + v θ θ ^ v ϕ v_\phi v ϕ ∂ / ∂ ϕ = 0 \partial/\partial\phi=0 ∂ / ∂ ϕ = 0
Step 1 — Continuity equation (incompressible)
∇ ⋅ v ⃗ = 1 r 2 ∂ ∂ r ( r 2 v r ) + 1 r sin θ ∂ ∂ θ ( v θ sin θ ) = 0. \nabla\cdot\vec v=\dfrac{1}{r^2}\dfrac{\partial}{\partial r}(r^2 v_r)+\dfrac{1}{r\sin\theta}\dfrac{\partial}{\partial\theta}(v_\theta\sin\theta)=0. ∇ ⋅ v = r 2 1 ∂ r ∂ ( r 2 v r ) + r sin θ 1 ∂ θ ∂ ( v θ sin θ ) = 0. (Using the spherical divergence formula, dropping ϕ \phi ϕ
Multiply by r 2 sin θ r^2\sin\theta r 2 sin θ
sin θ ∂ ∂ r ( r 2 v r ) + r ∂ ∂ θ ( v θ sin θ ) = 0. \sin\theta\dfrac{\partial}{\partial r}(r^2 v_r)+r\dfrac{\partial}{\partial\theta}(v_\theta\sin\theta)=0. sin θ ∂ r ∂ ( r 2 v r ) + r ∂ θ ∂ ( v θ sin θ ) = 0. This is the integrability condition for the existence of a function ψ ( r , θ ) \psi(r,\theta) ψ ( r , θ ) Stokes stream function — such that
r 2 sin θ ⋅ v r = ∂ ψ ∂ θ , r sin θ ⋅ v θ = − ∂ ψ ∂ r . ( ⋆ ) r^2\sin\theta\cdot v_r=\dfrac{\partial\psi}{\partial\theta},\quad r\sin\theta\cdot v_\theta=-\dfrac{\partial\psi}{\partial r}.\qquad(\star) r 2 sin θ ⋅ v r = ∂ θ ∂ ψ , r sin θ ⋅ v θ = − ∂ r ∂ ψ . ( ⋆ ) (With these definitions, continuity is automatically satisfied: ∂ 2 ψ / ( ∂ r ∂ θ ) = ∂ 2 ψ / ( ∂ θ ∂ r ) \partial^2\psi/(\partial r\partial\theta)=\partial^2\psi/(\partial\theta\partial r) ∂ 2 ψ / ( ∂ r ∂ θ ) = ∂ 2 ψ / ( ∂ θ ∂ r )
Step 2 — Velocity components in terms of ψ \psi ψ
From ( ⋆ ) (\star) ( ⋆ )
v r = 1 r 2 sin θ ∂ ψ ∂ θ , v θ = − 1 r sin θ ∂ ψ ∂ r . \boxed{\;v_r=\dfrac{1}{r^2\sin\theta}\dfrac{\partial\psi}{\partial\theta},\quad v_\theta=-\dfrac{1}{r\sin\theta}\dfrac{\partial\psi}{\partial r}.\;} v r = r 2 sin θ 1 ∂ θ ∂ ψ , v θ = − r sin θ 1 ∂ r ∂ ψ . Step 3 — Verify continuity is satisfied
1 r 2 ∂ r ( r 2 v r ) = 1 r 2 ∂ r ( r 2 ⋅ 1 r 2 sin θ ∂ θ ψ ) = 1 r 2 ∂ r ( ∂ θ ψ sin θ ) = 1 r 2 sin θ ∂ r ∂ θ ψ \dfrac{1}{r^2}\partial_r(r^2 v_r)=\dfrac{1}{r^2}\partial_r\left(r^2\cdot\dfrac{1}{r^2\sin\theta}\partial_\theta\psi\right)=\dfrac{1}{r^2}\partial_r\left(\dfrac{\partial_\theta\psi}{\sin\theta}\right)=\dfrac{1}{r^2\sin\theta}\partial_r\partial_\theta\psi r 2 1 ∂ r ( r 2 v r ) = r 2 1 ∂ r ( r 2 ⋅ r 2 sin θ 1 ∂ θ ψ ) = r 2 1 ∂ r ( sin θ ∂ θ ψ ) = r 2 sin θ 1 ∂ r ∂ θ ψ
1 r sin θ ∂ θ ( v θ sin θ ) = 1 r sin θ ∂ θ ( sin θ ⋅ − 1 r sin θ ∂ r ψ ) = 1 r sin θ ∂ θ ( − 1 r ∂ r ψ ) = − 1 r 2 sin θ ∂ θ ∂ r ψ \dfrac{1}{r\sin\theta}\partial_\theta(v_\theta\sin\theta)=\dfrac{1}{r\sin\theta}\partial_\theta\left(\sin\theta\cdot\dfrac{-1}{r\sin\theta}\partial_r\psi\right)=\dfrac{1}{r\sin\theta}\partial_\theta\left(-\dfrac{1}{r}\partial_r\psi\right)=\dfrac{-1}{r^2\sin\theta}\partial_\theta\partial_r\psi r sin θ 1 ∂ θ ( v θ sin θ ) = r sin θ 1 ∂ θ ( sin θ ⋅ r sin θ − 1 ∂ r ψ ) = r sin θ 1 ∂ θ ( − r 1 ∂ r ψ ) = r 2 sin θ − 1 ∂ θ ∂ r ψ
Sum: 1 r 2 sin θ ∂ r ∂ θ ψ − 1 r 2 sin θ ∂ θ ∂ r ψ = 0 \dfrac{1}{r^2\sin\theta}\partial_r\partial_\theta\psi-\dfrac{1}{r^2\sin\theta}\partial_\theta\partial_r\psi=0 r 2 sin θ 1 ∂ r ∂ θ ψ − r 2 sin θ 1 ∂ θ ∂ r ψ = 0
Step 4 — Irrotational condition
For axisymmetric flow without ϕ \phi ϕ ω ϕ \omega_\phi ω ϕ ϕ ^ \hat\phi ϕ ^ ∇ × v ⃗ \nabla\times\vec v ∇ × v
ω ϕ = 1 r [ ∂ ∂ r ( r v θ ) − ∂ v r ∂ θ ] . \omega_\phi=\dfrac{1}{r}\left[\dfrac{\partial}{\partial r}(rv_\theta)-\dfrac{\partial v_r}{\partial\theta}\right]. ω ϕ = r 1 [ ∂ r ∂ ( r v θ ) − ∂ θ ∂ v r ] . For irrotational flow, ω ϕ = 0 \omega_\phi=0 ω ϕ = 0
Step 5 — Express ω ϕ \omega_\phi ω ϕ ψ \psi ψ
Compute r v θ = r ⋅ − 1 r sin θ ∂ r ψ = − ∂ r ψ sin θ rv_\theta=r\cdot\dfrac{-1}{r\sin\theta}\partial_r\psi=\dfrac{-\partial_r\psi}{\sin\theta} r v θ = r ⋅ r sin θ − 1 ∂ r ψ = sin θ − ∂ r ψ
∂ r ( r v θ ) = − ∂ r r ψ sin θ \partial_r(rv_\theta)=-\dfrac{\partial_{rr}\psi}{\sin\theta} ∂ r ( r v θ ) = − sin θ ∂ r r ψ
∂ θ v r = ∂ θ ( ∂ θ ψ r 2 sin θ ) = 1 r 2 ∂ θ ( ∂ θ ψ sin θ ) = 1 r 2 [ ∂ θ θ ψ sin θ − cos θ ∂ θ ψ sin 2 θ ] \partial_\theta v_r=\partial_\theta\!\left(\dfrac{\partial_\theta\psi}{r^2\sin\theta}\right)=\dfrac{1}{r^2}\partial_\theta\!\left(\dfrac{\partial_\theta\psi}{\sin\theta}\right)=\dfrac{1}{r^2}\left[\dfrac{\partial_{\theta\theta}\psi}{\sin\theta}-\dfrac{\cos\theta\partial_\theta\psi}{\sin^2\theta}\right] ∂ θ v r = ∂ θ ( r 2 sin θ ∂ θ ψ ) = r 2 1 ∂ θ ( sin θ ∂ θ ψ ) = r 2 1 [ sin θ ∂ θ θ ψ − sin 2 θ cos θ ∂ θ ψ ]
So
ω ϕ = 1 r [ − ∂ r r ψ sin θ − 1 r 2 ( ∂ θ θ ψ sin θ − cos θ ∂ θ ψ sin 2 θ ) ] \omega_\phi=\dfrac{1}{r}\left[-\dfrac{\partial_{rr}\psi}{\sin\theta}-\dfrac{1}{r^2}\!\left(\dfrac{\partial_{\theta\theta}\psi}{\sin\theta}-\dfrac{\cos\theta\partial_\theta\psi}{\sin^2\theta}\right)\right] ω ϕ = r 1 [ − sin θ ∂ r r ψ − r 2 1 ( sin θ ∂ θ θ ψ − sin 2 θ cos θ ∂ θ ψ ) ] = − 1 r sin θ [ ∂ r r ψ + 1 r 2 ∂ θ θ ψ − cos θ r 2 sin θ ∂ θ ψ ] =-\dfrac{1}{r\sin\theta}\left[\partial_{rr}\psi+\dfrac{1}{r^2}\partial_{\theta\theta}\psi-\dfrac{\cos\theta}{r^2\sin\theta}\partial_\theta\psi\right] = − r sin θ 1 [ ∂ r r ψ + r 2 1 ∂ θ θ ψ − r 2 sin θ cos θ ∂ θ ψ ] = − 1 r sin θ E 2 ψ , =-\dfrac{1}{r\sin\theta}\,E^2\psi, = − r sin θ 1 E 2 ψ , where the Stokes operator is
E 2 ≡ ∂ 2 ∂ r 2 + sin θ r 2 ∂ ∂ θ ( 1 sin θ ∂ ∂ θ ) = ∂ 2 ∂ r 2 + 1 r 2 ∂ 2 ∂ θ 2 − cos θ r 2 sin θ ∂ ∂ θ . E^2\equiv\dfrac{\partial^2}{\partial r^2}+\dfrac{\sin\theta}{r^2}\dfrac{\partial}{\partial\theta}\!\left(\dfrac{1}{\sin\theta}\dfrac{\partial}{\partial\theta}\right)=\dfrac{\partial^2}{\partial r^2}+\dfrac{1}{r^2}\dfrac{\partial^2}{\partial\theta^2}-\dfrac{\cos\theta}{r^2\sin\theta}\dfrac{\partial}{\partial\theta}. E 2 ≡ ∂ r 2 ∂ 2 + r 2 sin θ ∂ θ ∂ ( sin θ 1 ∂ θ ∂ ) = ∂ r 2 ∂ 2 + r 2 1 ∂ θ 2 ∂ 2 − r 2 sin θ cos θ ∂ θ ∂ . Setting ω ϕ = 0 \omega_\phi=0 ω ϕ = 0
Answer
E 2 ψ = ∂ 2 ψ ∂ r 2 + sin θ r 2 ∂ ∂ θ ( 1 sin θ ∂ ψ ∂ θ ) = 0. \boxed{\;E^2\psi=\dfrac{\partial^2\psi}{\partial r^2}+\dfrac{\sin\theta}{r^2}\dfrac{\partial}{\partial\theta}\!\left(\dfrac{1}{\sin\theta}\dfrac{\partial\psi}{\partial\theta}\right)=0.\;} E 2 ψ = ∂ r 2 ∂ 2 ψ + r 2 sin θ ∂ θ ∂ ( sin θ 1 ∂ θ ∂ ψ ) = 0.