Let's solve the problem step by step. --- # 1. **Translation of a Sphere Through a Fluid at Rest** Given: - Sphere of radius \( R \) moving with velocity \( U \) in an unbounded, initially stationary fluid. - Stream function: \( \psi(r, \theta) \). - **Boundary conditions:** - At \( r = R \): \[ \psi = -\frac{1}{2} U R^2 \sin^2\theta \] \[ \frac{\partial \psi}{\partial r} = -U R \sin^2\theta \] - As \( r \to \infty \): \[ \frac{\psi}{r^2} \to \] --- ## (a) **Justify the boundary conditions** **At \( r = R \):** - **No slip boundary condition:** The fluid at the surface of the sphere must move with the sphere. - **Radial velocity \( v_r \):** On the surface, fluid cannot penetrate the sphere, so \( v_r = \) at \( r=R \). - **Tangential velocity \( v_\theta \):** Must match the sphere's tangential velocity. The stream function for axisymmetric flow relates to velocity components as: \[ v_r = \frac{1}{r^2 \sin\theta} \frac{\partial \psi}{\partial \theta} \] \[ v_\theta = -\frac{1}{r \sin\theta} \frac{\partial \psi}{\partial r} \] Plugging in the given boundary conditions into these relations ensures the correct velocities at the surface (no-slip and no-penetration). **As \( r \to \infty \):** - Far away from the sphere, the disturbance vanishes, so the stream function should decay faster than \( r^2 \). --- ## (b) **Solve the stream function equation** The governing equation is: \[ E^4 \psi = E^2(E^2\psi) = \] where \( E^2 \) is the axisymmetric Stokes operator. ### **General Solution Form** The general solution for the axisymmetric Stokes flow around a sphere is: \[ \psi(r, \theta) = \left( Ar + \frac{B}{r} + Cr^3 + \frac{D}{r^2} \right)\sin^2\theta \] ### **Apply Boundary Conditions** 1. **At \( r = R \):** \[ \psi(R, \theta) = -\frac{1}{2} U R^2 \sin^2\theta \] \[ \left( AR + \frac{B}{R} + CR^3 + \frac{D}{R^2} \right)\sin^2\theta = -\frac{1}{2} U R^2 \sin^2\theta \] 2. **At \( r = R \):** \[ \frac{\partial \psi}{\partial r}\Big|_{r=R} = -U R \sin^2\theta \] \[ \left( A - \frac{B}{r^2} + 3Cr^2 - \frac{2D}{r^3} \right)\sin^2\theta \Big|_{r=R} = -U R \sin^2\theta \] 3. **As \( r \to \infty \):** - Terms growing faster than \( r^2 \) must vanish: \( C = \). - \( \psi/r^2 \to \) implies \( A = \). So, the solution reduces to: \[ \psi(r, \theta) = \left( \frac{B}{r} + \frac{D}{r^2} \right) \sin^2\theta \] ### **Find \( B \) and \( D \):** Plug into conditions at \( r = R \): 1. \( \psi(R, \theta) = -\frac{1}{2} U R^2 \sin^2\theta \): \[ \left( \frac{B}{R} + \frac{D}{R^2} \right) = -\frac{1}{2} U R^2 \] 2. \( \frac{\partial \psi}{\partial r}\Big|_{r=R} = -U R \sin^2\theta \): \[ \frac{\partial \psi}{\partial r} = \left( -\frac{B}{r^2} - \frac{2D}{r^3} \right) \sin^2\theta \] \[ -\frac{B}{R^2} - \frac{2D}{R^3} = -U R \] Now solve the system: \[ \frac{B}{R} + \frac{D}{R^2} = -\frac{1}{2} U R^2 \tag{1} \] \[ -\frac{B}{R^2} - \frac{2D}{R^3} = -U R \tag{2} \] Multiply (1) by \( R \): \[ B + \frac{D}{R} = -\frac{1}{2} U R^3 \tag{3} \] Multiply (2) by \( -R^2 \): \[ B + 2\frac{D}{R} = U R^3 \tag{4} \] Subtract (3) from (4): \[ \left(B + 2\frac{D}{R}\right) - \left(B + \frac{D}{R}\right) = U R^3 - \left(-\frac{1}{2} U R^3\right) \] \[ \frac{D}{R} = \frac{3}{2} U R^3 \implies D = \frac{3}{2} U R^4 \] Plug \( D \) back into (3): \[ B + \frac{3}{2} U R^3 = -\frac{1}{2} U R^3 \implies B = -2 U R^3 \] **Final Stream Function:** \[ \boxed{ \psi(r, \theta) = \left( -\frac{2 U R^3}{r} + \frac{3 U R^4}{2 r^2} \right) \sin^2\theta } \] --- ## (c) **Fluid velocity components \( v_r \) and \( v_\theta \)** Recall: \[ v_r = \frac{1}{r^2 \sin\theta} \frac{\partial \psi}{\partial \theta} \] \[ v_\theta = -\frac{1}{r \sin\theta} \frac{\partial \psi}{\partial r} \] ### **Compute derivatives:** \[ \psi(r, \theta) = f(r) \sin^2\theta \] \[ \frac{\partial \psi}{\partial \theta} = 2 f(r) \sin\theta \cos\theta \] So, \[ v_r = \frac{1}{r^2 \sin\theta} \cdot 2 f(r) \sin\theta \cos\theta = \frac{2 f(r) \cos\theta}{r^2} \] Now, \[ \frac{\partial \psi}{\partial r} = f'(r) \sin^2\theta \] \[ v_\theta = -\frac{1}{r \sin\theta} f'(r) \sin^2\theta = -\frac{f'(r) \sin\theta}{r} \] Where: \[ f(r) = -\frac{2 U R^3}{r} + \frac{3 U R^4}{2 r^2} \] \[ f'(r) = 2 U R^3 \frac{1}{r^2} - 3 U R^4 \frac{1}{r^3} \] So, \[ v_r = \frac{2 \cos\theta}{r^2} \left( -\frac{2 U R^3}{r} + \frac{3 U R^4}{2 r^2} \right) = \left( -\frac{4 U R^3}{r^3} + \frac{3 U R^4}{r^4} \right) \cos\theta \] \[ v_\theta = -\frac{\sin\theta}{r} \left( 2 U R^3 \frac{1}{r^2} - 3 U R^4 \frac{1}{r^3} \right) = -2 U R^3 \frac{\sin\theta}{r^3} + 3 U R^4 \frac{\sin\theta}{r^4} \] Therefore, \[ \boxed{ v_r = \left( -\frac{4 U R^3}{r^3} + \frac{3 U R^4}{r^4} \right) \cos\theta } \] \[ \boxed{ v_\theta = \left( -2 U R^3 \frac{1}{r^3} + 3 U R^4 \frac{1}{r^4} \right) \sin\theta } \] --- ## (d) **Equation for the pressure field** For Stokes flow, the pressure gradient is related to the velocity by: \[ \nabla p = \mu \nabla^2 \vec{v} \] For axisymmetric flow, the pressure field can be derived (after some algebra, see Stokes' solution): \[ p = P_\infty - \frac{3 \mu U R}{2 r^2} \cos\theta \] **Where:** - \( P_\infty \) is the far-field pressure, - \( \mu \) is the fluid viscosity. \[ \boxed{ p(r, \theta) = P_\infty - \frac{3 \mu U R}{2 r^2} \cos\theta } \] --- ## **Summary of Answers** **(a) Boundary conditions justified by no-slip, no-penetration, and decay at infinity.** **(b) Stream function:** \[ \boxed{ \psi(r, \theta) = \left( -\frac{2 U R^3}{r} + \frac{3 U R^4}{2 r^2} \right) \sin^2\theta } \] **(c) Velocity components:** \[ \boxed{ v_r = \left( -\frac{4 U R^3}{r^3} + \frac{3 U R^4}{r^4} \right) \cos\theta } \] \[ \boxed{ v_\theta = \left( -2 U R^3 \frac{1}{r^3} + 3 U R^4 \frac{1}{r^4} \right) \sin\theta } \] **(d) Pressure field:** \[ \boxed{ p(r, \theta) = P_\infty - \frac{3 \mu U R}{2 r^2} \cos\theta } \] Let me know if you need any step explained more deeply!