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UPSC Maths 2025 Paper 2 Q7c — Solution

20 marks · Section B

Question

Show that for an incompressible steady flow with constant viscosity, the velocity components $u(y) = \left(\dfrac{U}{h}\right)y - \dfrac{hy}{2\mu}\dfrac{dp}{dx}\left(1 - \dfrac{y}{h}\right)$, $v = 0 = w$, with $p = p(x)$, satisfy the equation of motion in the absence of body force. Given that $U$, $h$ and $\dfrac{dp}{dx}$ are constants.

Technique

Substitute the given (generalised Couette–Poiseuille) velocity field into the incompressible Navier–Stokes equations and the continuity equation, and show every equation is satisfied identically — verifying an exact solution of the equations of motion.

Solution

The incompressible Navier–Stokes equations (constant $\mu$, no body force) are $\nabla\cdot\mathbf{V} = 0,$ $\rho\Big(\frac{\partial \mathbf{V}}{\partial t} + (\mathbf{V}\cdot\nabla)\mathbf{V}\Big) = -\nabla p + \mu\nabla^2\mathbf{V}.$

The field is $\mathbf{V} = (u(y),0,0)$, steady ($\partial_t = 0$), with $p = p(x)$ and $\dfrac{dp}{dx} =$ const $\equiv G$.

Continuity

$\nabla\cdot\mathbf{V} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0 + 0 + 0 = 0,$ since $u$ depends only on $y$, and $v = w = 0$. Continuity is satisfied.

Convective (inertia) terms vanish

For the $x$-momentum equation, $(\mathbf{V}\cdot\nabla)u = u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} + w\frac{\partial u}{\partial z} = u\cdot 0 + 0\cdot u'(y) + 0 = 0.$ So the inertia term is zero — the flow is dynamically equivalent to creeping (Stokes) flow.

$x$-momentum equation

It reduces to $0 = -\frac{\partial p}{\partial x} + \mu\Big(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\Big) = -\frac{dp}{dx} + \mu\frac{d^2 u}{dy^2}.$

Compute $u''(y)$. Expand the given $u$: $u(y) = \frac{U}{h}y - \frac{G h}{2\mu}y + \frac{G}{2\mu}y^2,\qquad G \equiv \frac{dp}{dx}.$ Then $u'(y) = \frac{U}{h} - \frac{G h}{2\mu} + \frac{G}{\mu}y,\qquad u''(y) = \frac{G}{\mu}.$ Hence $\mu\frac{d^2u}{dy^2} = \mu\cdot\frac{G}{\mu} = G = \frac{dp}{dx},$ so $-\frac{dp}{dx} + \mu\frac{d^2u}{dy^2} = -G + G = 0.$ The $x$-momentum equation is satisfied identically.

$y$- and $z$-momentum equations

With $v = w = 0$ and the field independent of $z$: $y\text{-momentum: } 0 = -\frac{\partial p}{\partial y} + \mu\nabla^2 v = -0 + 0 = 0,$ since $p = p(x)$ gives $\partial p/\partial y = 0$, and $v\equiv 0$. $z\text{-momentum: } 0 = -\frac{\partial p}{\partial z} + \mu\nabla^2 w = 0,$ since $\partial p/\partial z = 0$ and $w\equiv 0$. Both are satisfied.

Conclusion

All three momentum equations and continuity hold identically, so the stated field is an exact solution of the equations of motion. The key balance is the viscous–pressure relation $\mu\frac{d^2u}{dy^2} = \frac{dp}{dx},$ whose integration (with $u(0)=0$, $u(h)=U$) reproduces exactly the given $u(y)$ — the plane Couette–Poiseuille profile (linear Couette part $\tfrac{U}{h}y$ plus parabolic Poiseuille part driven by $dp/dx$).

Answer

The field $u(y) = \tfrac{U}{h}y - \tfrac{hy}{2\mu}\tfrac{dp}{dx}\big(1-\tfrac{y}{h}\big)$, $v=w=0$, $p=p(x)$, satisfies continuity and all Navier–Stokes momentum equations with no body force, because the inertia terms vanish and $\mu u'' = dp/dx$ balances the pressure gradient.