← 2023 Paper 2
UPSC 2023 Maths Optional Paper 2 Q8c — Step-by-Step Solution 20 marks · Section B
Navier-Stokes equation for a viscous fluid · Mechanics & Fluid Dynamics · asked 4× in 14 yrs · Read the full method →
Question
Determine under what conditions the velocity field u = c ( x 2 − y 2 ) , v = − 2 c x y , w = 0 u=c(x^2-y^2),\,v=-2cxy,\,w=0 u = c ( x 2 − y 2 ) , v = − 2 c x y , w = 0 z z z B x = 0 = B y , B z = − g B_x=0=B_y,\,B_z=-g B x = 0 = B y , B z = − g
Technique
Substitute into continuity (passes); compute Laplacian (vanishes, so flow is harmonic); integrate the three pressure derivatives subject to compatibility.
Solution
Step 1 — Continuity (incompressibility check)
For an incompressible Newtonian fluid, ∇ ⋅ V ⃗ = 0 \nabla\cdot\vec V=0 ∇ ⋅ V = 0
u x + v y + w z = 2 c x + ( − 2 c x ) + 0 = 0 ✓ . u_x+v_y+w_z=2cx+(-2cx)+0=0\quad\checkmark. u x + v y + w z = 2 c x + ( − 2 c x ) + 0 = 0 ✓ . So the flow is automatically incompressible — no condition on the parameters is needed for continuity.
Step 2 — Convective acceleration
u u x + v u y + w u z = c ( x 2 − y 2 ) ( 2 c x ) + ( − 2 c x y ) ( − 2 c y ) + 0 = 2 c 2 x ( x 2 − y 2 ) + 4 c 2 x y 2 = 2 c 2 x ( x 2 + y 2 ) , u\,u_x+v\,u_y+w\,u_z=c(x^{2}-y^{2})(2cx)+(-2cxy)(-2cy)+0=2c^{2}x(x^{2}-y^{2})+4c^{2}xy^{2}=2c^{2}x(x^{2}+y^{2}), u u x + v u y + w u z = c ( x 2 − y 2 ) ( 2 c x ) + ( − 2 c x y ) ( − 2 cy ) + 0 = 2 c 2 x ( x 2 − y 2 ) + 4 c 2 x y 2 = 2 c 2 x ( x 2 + y 2 ) , u v x + v v y + w v z = c ( x 2 − y 2 ) ( − 2 c y ) + ( − 2 c x y ) ( − 2 c x ) + 0 = − 2 c 2 y ( x 2 − y 2 ) + 4 c 2 x 2 y = 2 c 2 y ( x 2 + y 2 ) , u\,v_x+v\,v_y+w\,v_z=c(x^{2}-y^{2})(-2cy)+(-2cxy)(-2cx)+0=-2c^{2}y(x^{2}-y^{2})+4c^{2}x^{2}y=2c^{2}y(x^{2}+y^{2}), u v x + v v y + w v z = c ( x 2 − y 2 ) ( − 2 cy ) + ( − 2 c x y ) ( − 2 c x ) + 0 = − 2 c 2 y ( x 2 − y 2 ) + 4 c 2 x 2 y = 2 c 2 y ( x 2 + y 2 ) , u w x + v w y + w w z = 0. u\,w_x+v\,w_y+w\,w_z=0. u w x + v w y + w w z = 0. Step 3 — Viscous (Laplacian) terms
∇ 2 u = u x x + u y y + u z z = 2 c + ( − 2 c ) + 0 = 0 , \nabla^{2}u=u_{xx}+u_{yy}+u_{zz}=2c+(-2c)+0=0, ∇ 2 u = u xx + u y y + u z z = 2 c + ( − 2 c ) + 0 = 0 , ∇ 2 v = v x x + v y y + v z z = 0 + 0 + 0 = 0 , ∇ 2 w = 0. \nabla^{2}v=v_{xx}+v_{yy}+v_{zz}=0+0+0=0,\quad \nabla^{2}w=0. ∇ 2 v = v xx + v y y + v z z = 0 + 0 + 0 = 0 , ∇ 2 w = 0. Each velocity component is harmonic — viscous stresses vanish identically. This means the NS equations reduce to Euler form :
ρ ( V ⃗ ⋅ ∇ ) V ⃗ = − ∇ p + ρ B ⃗ . \rho\,(\vec V\cdot\nabla)\vec V=-\nabla p+\rho\vec B. ρ ( V ⋅ ∇ ) V = − ∇ p + ρ B . Step 4 — Conditions for being an NS solution
Substituting into the three momentum equations (steady, ∂ t = 0 \partial_t=0 ∂ t = 0
Component NS equation ⇒ \Rightarrow ⇒ x x x ρ ⋅ 2 c 2 x ( x 2 + y 2 ) = − p x \rho\cdot 2c^{2}x(x^{2}+y^{2})=-p_x ρ ⋅ 2 c 2 x ( x 2 + y 2 ) = − p x p x = − 2 ρ c 2 x ( x 2 + y 2 ) p_x=-2\rho c^{2}x(x^{2}+y^{2}) p x = − 2 ρ c 2 x ( x 2 + y 2 ) y y y ρ ⋅ 2 c 2 y ( x 2 + y 2 ) = − p y \rho\cdot 2c^{2}y(x^{2}+y^{2})=-p_y ρ ⋅ 2 c 2 y ( x 2 + y 2 ) = − p y p y = − 2 ρ c 2 y ( x 2 + y 2 ) p_y=-2\rho c^{2}y(x^{2}+y^{2}) p y = − 2 ρ c 2 y ( x 2 + y 2 ) z z z 0 = − p z − ρ g 0=-p_z-\rho g 0 = − p z − ρ g p z = − ρ g p_z=-\rho g p z = − ρ g
The compatibility (mixed-partials) conditions on p p p
∂ y p x = − 4 ρ c 2 x y = ∂ x p y ✓ , \partial_y p_x=-4\rho c^{2}xy=\partial_x p_y\;\checkmark, ∂ y p x = − 4 ρ c 2 x y = ∂ x p y ✓ , ∂ z p x = 0 = ∂ x p z ✓ , ∂ z p y = 0 = ∂ y p z ✓ . \partial_z p_x=0=\partial_x p_z\;\checkmark,\quad \partial_z p_y=0=\partial_y p_z\;\checkmark. ∂ z p x = 0 = ∂ x p z ✓ , ∂ z p y = 0 = ∂ y p z ✓ . All consistent. So the velocity field is a Navier-Stokes solution for any constant c c c μ \mu μ — the only required condition is the fluid be incompressible (which is satisfied) and the integrability of ∇ p \nabla p ∇ p
Condition: ∇ ⋅ V ⃗ = 0 (incompressible) — already satisfied. \boxed{\;\text{Condition: } \nabla\cdot\vec V=0\text{ (incompressible)} \text{ — already satisfied.}\;} Condition: ∇ ⋅ V = 0 (incompressible) — already satisfied. (Equivalently: the flow is irrotational and harmonic, so viscous terms vanish and NS reduces to Euler with Bernoulli.)
Step 5 — Integrate for the pressure
Integrate p x p_x p x x x x y , z y,z y , z
p = − 2 ρ c 2 ∫ x ( x 2 + y 2 ) d x + h ( y , z ) = − 2 ρ c 2 ( x 4 4 + x 2 y 2 2 ) + h ( y , z ) = − ρ c 2 ( x 4 2 + x 2 y 2 ) + h ( y , z ) . p=-2\rho c^{2}\!\int x(x^{2}+y^{2})\,dx+h(y,z)=-2\rho c^{2}\!\left(\frac{x^{4}}{4}+\frac{x^{2}y^{2}}{2}\right)+h(y,z)=-\rho c^{2}\!\left(\tfrac{x^{4}}{2}+x^{2}y^{2}\right)+h(y,z). p = − 2 ρ c 2 ∫ x ( x 2 + y 2 ) d x + h ( y , z ) = − 2 ρ c 2 ( 4 x 4 + 2 x 2 y 2 ) + h ( y , z ) = − ρ c 2 ( 2 x 4 + x 2 y 2 ) + h ( y , z ) . Match p y p_y p y p y = − 2 ρ c 2 x 2 y + h y = ! − 2 ρ c 2 y ( x 2 + y 2 ) = − 2 ρ c 2 x 2 y − 2 ρ c 2 y 3 \;p_y=-2\rho c^{2}x^{2}y+h_y\stackrel{!}{=}-2\rho c^{2}y(x^{2}+y^{2})=-2\rho c^{2}x^{2}y-2\rho c^{2}y^{3} p y = − 2 ρ c 2 x 2 y + h y = ! − 2 ρ c 2 y ( x 2 + y 2 ) = − 2 ρ c 2 x 2 y − 2 ρ c 2 y 3 h y = − 2 ρ c 2 y 3 ⇒ h ( y , z ) = − ρ c 2 y 4 2 + k ( z ) h_y=-2\rho c^{2}y^{3}\;\Rightarrow\;h(y,z)=-\tfrac{\rho c^{2}y^{4}}{2}+k(z) h y = − 2 ρ c 2 y 3 ⇒ h ( y , z ) = − 2 ρ c 2 y 4 + k ( z )
Match p z p_z p z k ′ ( z ) = − ρ g ⇒ k ( z ) = − ρ g z + const \;k'(z)=-\rho g\;\Rightarrow\;k(z)=-\rho g z+\text{const} k ′ ( z ) = − ρ g ⇒ k ( z ) = − ρ g z + const
Collecting:
p = − ρ c 2 ( x 4 2 + x 2 y 2 + y 4 2 ) − ρ g z + C = − ρ c 2 2 ( x 2 + y 2 ) 2 − ρ g z + C . p=-\rho c^{2}\!\left(\tfrac{x^{4}}{2}+x^{2}y^{2}+\tfrac{y^{4}}{2}\right)-\rho g z+C=-\tfrac{\rho c^{2}}{2}(x^{2}+y^{2})^{2}-\rho g z+C. p = − ρ c 2 ( 2 x 4 + x 2 y 2 + 2 y 4 ) − ρ g z + C = − 2 ρ c 2 ( x 2 + y 2 ) 2 − ρ g z + C . The speed of the flow:
∣ V ⃗ ∣ 2 = u 2 + v 2 = c 2 ( x 2 − y 2 ) 2 + 4 c 2 x 2 y 2 = c 2 [ ( x 2 − y 2 ) 2 + 4 x 2 y 2 ] = c 2 ( x 2 + y 2 ) 2 . |\vec V|^{2}=u^{2}+v^{2}=c^{2}(x^{2}-y^{2})^{2}+4c^{2}x^{2}y^{2}=c^{2}\bigl[(x^{2}-y^{2})^{2}+4x^{2}y^{2}\bigr]=c^{2}(x^{2}+y^{2})^{2}. ∣ V ∣ 2 = u 2 + v 2 = c 2 ( x 2 − y 2 ) 2 + 4 c 2 x 2 y 2 = c 2 [ ( x 2 − y 2 ) 2 + 4 x 2 y 2 ] = c 2 ( x 2 + y 2 ) 2 . So the pressure can be written
Answer
p ( x , y , z ) = p 0 − 1 2 ρ ∣ V ⃗ ∣ 2 − ρ g z , \boxed{\;p(x,y,z)=p_0-\tfrac{1}{2}\rho|\vec V|^{2}-\rho g z,\;} p ( x , y , z ) = p 0 − 2 1 ρ ∣ V ∣ 2 − ρ g z ,