Paper II — Q8
(a) Two-dimensional harmonic equation in plane polar coordinates (r, θ) takes the form : ∂²V/∂r² + (1/r) ∂V/∂r + (1/r²) ∂²V/∂θ²…
Two-dimensional harmonic equation in plane polar coordinates (r, θ) takes the form :
∂²V/∂r² + (1/r) ∂V/∂r + (1/r²) ∂²V/∂θ² = 0
Deduce that it has solutions of the form (Arⁿ + Br⁻ⁿ) e^±inθ, where A, B, n are constants. Determine V if it satisfies the two-dimensional harmonic equation in the region 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π and satisfies the conditions :
V remains finite as r → 0,
V = Σₙ cₙ cos(nθ) on r = a. 15 marks
Consider the initial value problem :
dy/dx = 1 + y/x, y(1) = 1, h = 1 (Step-size)
Apply the classical fourth order Runge-Kutta method to approximate the solution y(2).
Use Euler's method to find y(2) and y(3). 15 marks
A viscous incompressible fluid is filled between two concentric cylinders of radius a, b (b > a). The flow is steady and no body forces are taken into consideration. Discuss and formulate the velocity of the fluid if : the inner cylinder is given angular velocity Ω while the outer one is held at rest;
the outer cylinder is rotated with angular velocity Ω while the inner one is held at rest. 20 marks
हिंदी में प्रश्न पढ़ें
समतलीय ध्रुवीय निर्देशांकों (r, θ) में द्विविमीय प्रसंवादी (हार्मोनिक) समीकरण निम्न रूप लेता है :
∂²V/∂r² + (1/r) ∂V/∂r + (1/r²) ∂²V/∂θ² = 0
निगमन कीजिए कि इसके हल का रूप (Arⁿ + Br⁻ⁿ) e^±inθ है, जहाँ A, B, n अचर हैं । V ज्ञात कीजिए यदि यह क्षेत्र 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π में द्विविमीय प्रसंवादी समीकरण को संतुष्ट करता है और शर्तों :
r → 0 पर V परिमित रहता है,
r = a पर V = Σₙ cₙ cos(nθ) है
को संतुष्ट करता है । (15 अंक)
प्रारंभिक मान समस्या dy/dx = 1 + y/x, y(1) = 1, h = 1 (पग-माप) का विचार कीजिए ।
y(2) का सन्निकट मान प्राप्त करने के लिए चिरप्रतिष्ठित चतुष्कोटि रूंगे-कुट्टा विधि का उपयोग कीजिए ।
y(2) और y(3) के मान ज्ञात करने के लिए ऑयलर विधि का उपयोग कीजिए । (15 अंक)
त्रिज्या a, b (b > a) के दो संकेंद्री बेलनों के बीच एक श्यान असंपीड्य तरल भरा है । प्रवाह अपरिवर्ती है और कोई भी पिंड बल विचाराधीन नहीं है । तरल के वेग की विवेचना और इसे सूत्रित कीजिए यदि : आंतरिक बेलन को कोणीय वेग Ω दिया जाता है जबकि बाहरी बेलन को विरामावस्था में रखा जाता है;
बाहरी बेलन को कोणीय वेग Ω से घुमाया जाता है जबकि आंतरिक बेलन को विरामावस्था में रखा जाता है । (20 अंक)
Model answer
Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.
(a) Two-dimensional harmonic equation (15 marks)
Assume V(r, θ) = R(r)Θ(θ). Substituting in ∂²V/∂r² + (1/r)∂V/∂r + (1/r²)∂²V/∂θ² = 0 and multiplying by r²/(RΘ) gives (r²R″ + rR′)/R = -Θ″/Θ = n², where n² is the separation constant. The angular equation Θ″ + n²Θ = 0 has solutions e^(±inθ); since V must be single-valued and 2π-periodic, n is an integer. The radial equation r²R″ + rR′ - n²R = 0 is an Euler-Cauchy equation; putting r = eᵗ turns it into d²R/dt² = n²R, so R = Arⁿ + Br⁻ⁿ (for n = 0, R = A + B ln r). Hence the solutions are (Arⁿ + Br⁻ⁿ)e^(±inθ).
In the disc 0 ≤ r ≤ a, condition (i) requires V to stay finite as r → 0, so every r⁻ⁿ term and the ln r term must vanish: B = 0 for all n. The general finite solution is therefore V = Σ Aₙ rⁿ cos(nθ) (with sine terms allowed in general). Condition (ii), V = Σ cₙ cos(nθ) on r = a, contains only cosines, so the sine terms are absent and matching coefficients (orthogonality of the Fourier modes) gives Aₙaⁿ = cₙ, that is Aₙ = cₙ/aⁿ. Therefore
V(r, θ) = c₀ + Σₙ≥₁ cₙ (r/a)ⁿ cos(nθ).
(b) Runge-Kutta and Euler methods (15 marks)
Here f(x, y) = 1 + y/x, x₀ = 1, y₀ = 1, h = 1.
(i) Classical fourth-order Runge-Kutta, y₁ = y₀ + (1/6)(k₁ + 2k₂ + 2k₃ + k₄): k₁ = h f(x₀, y₀) = 1 + 1/1 = 2 k₂ = h f(x₀ + h/2, y₀ + k₁/2) = 1 + 2/1.5 = 2.3333 k₃ = h f(x₀ + h/2, y₀ + k₂/2) = 1 + 2.1667/1.5 = 2.4444 k₄ = h f(x₀ + h, y₀ + k₃) = 1 + 3.4444/2 = 2.7222 y(2) ≈ 1 + (2 + 4.6667 + 4.8889 + 2.7222)/6 = 3.3796.
(ii) Euler's method, yₙ₊₁ = yₙ + h f(xₙ, yₙ): y(2) = 1 + 1·(1 + 1/1) = 3.00; then y(3) = 3 + 1·(1 + 3/2) = 5.50.
The exact solution (integrating factor 1/x) is y = x(1 + ln x), giving y(2) = 3.3863 and y(3) = 6.2958. The RK4 error at x = 2 is about 0.007, while the Euler errors are about 0.386 at x = 2 and 0.796 at x = 3.
(c) Flow between concentric cylinders (20 marks)
For steady flow with no body forces the velocity is purely circumferential, u_θ = v(r), with no dependence on θ or z, so the convective terms vanish and the θ-momentum equation reduces to d²v/dr² + (1/r)dv/dr - v/r² = 0, that is d/dr[(1/r)d(rv)/dr] = 0. Its general solution is v(r) = Ar + B/r: a rigid-rotation part plus a free-vortex part. The radial momentum equation dp/dr = ρv²/r then gives the pressure.
(i) Inner cylinder rotating, outer at rest: v(a) = Ωa, v(b) = 0. These give A = -Ωa²/(b² - a²) and B = Ωa²b²/(b² - a²), so v(r) = Ωa²(b² - r²) / [r(b² - a²)].
(ii) Outer cylinder rotating, inner at rest: v(a) = 0, v(b) = Ωb. These give A = Ωb²/(b² - a²) and B = -Ωa²b²/(b² - a²), so v(r) = Ωb²(r² - a²) / [r(b² - a²)].
The shear stress is τ_rθ = μ(dv/dr - v/r) = -2μB/r², so the torque per unit length on the rotating inner cylinder in case (i) is 4πμΩa²b²/(b² - a²), the principle of the Couette viscometer. As b → ∞, case (i) reduces to the free vortex Ωa²/r.
How this answer will be evaluated
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Two tasks: deduce by separation of variables that the two-dimensional Laplace equation in polar form has solutions (Ar^n + Br^-n)e^(±inθ); then determine V in the disc 0 ≤ r ≤ a subject to finiteness at r = 0 and V = Σ c_n cos(nθ) on r = a. 15 marks
Must cover
- Separation of variables V = R(r)Θ(θ), giving (r²R'' + rR')/R = -Θ''/Θ = n²
- Θ'' + n²Θ = 0, so Θ = e^(±inθ), and n an integer from 2π-periodicity
- Euler-Cauchy equation r²R'' + rR' - n²R = 0 with R = Ar^n + Br^-n; for n = 0, R = A + B ln r
- Finiteness at r = 0 gives B = 0 (also for the ln r term)
- V(r, θ) = Σ c_n (r/a)^n cos(nθ), with the constant term c_0
Loses marks
- Keeping the r^-n terms in the interior problem, which blow up at r = 0.
- Forgetting the n = 0 case (A + B ln r) so that c_0 is lost.
- Not justifying that n is an integer (single-valuedness of V).
- Writing Σ c_n r^n cos(nθ) without dividing by a^n, or including sine terms.
Earns more
- Method marks for correct separation of variables and the radial equation even if the boundary fit is wrong.
- Matching coefficients on r = a (A_n a^n = c_n) with the cosine-only data justified by orthogonality of the Fourier modes.
Extra mark
- Uniqueness of the solution by the maximum principle or Fourier completeness
- (b) For dy/dx = 1 + y/x, y(1) = 1, h = 1: (i) apply the classical fourth-order Runge-Kutta method (one step) to approximate y(2); (ii) apply Euler's method to obtain y(2) and then y(3). 15 marks
Must cover
- RK4 formula y₁ = y₀ + (1/6)(k₁ + 2k₂ + 2k₃ + k₄)
- k₁ = 2, k₂ = 2.3333, k₃ = 2.4444, k₄ = 2.7222, so y(2) = 3.3796
- Euler: y(2) = 3 and y(3) = 5.5 (second step from (2, 3))
- Comparison with the exact solution y = x(1 + ln x): y(2) = 3.3863, y(3) = 6.2958
Loses marks
- Evaluating k₂ and k₃ at x₀ + h instead of x₀ + h/2.
- Using wrong RK4 weights such as (1, 3, 3, 1)/8 instead of (1, 2, 2, 1)/6.
- Starting the second Euler step from the wrong point or omitting y(3).
Earns more
- Correct evaluation points x₀ + h/2 = 1.5 for k₂ and k₃, and x₀ + h = 2 for k₄, even if arithmetic slips occur later.
- Error comparison: RK4 error about 0.007 against Euler error about 0.386 at x = 2.
Extra mark
- Noting that with h = 1 the Euler error grows to about 0.80 at x = 3
- (c) For steady viscous incompressible flow between concentric cylinders of radii a and b (b > a) with no body forces, formulate the governing equation for the circumferential velocity and solve it for (i) inner cylinder rotating with Ω, outer at rest; (ii) outer rotating with Ω, inner at rest. 20 marks
Must cover
- Circular Couette flow: u_θ = v(r) only; governing equation d²v/dr² + (1/r)dv/dr - v/r² = 0
- General solution v = Ar + B/r
- Case (i) v(a) = Ωa, v(b) = 0: v = Ωa²(b² - r²)/[r(b² - a²)]
- Case (ii) v(a) = 0, v(b) = Ωb: v = Ωb²(r² - a²)/[r(b² - a²)]
Loses marks
- Dropping the -v/r² term from the reduced equation.
- Applying the boundary condition as v(a) = Ω instead of Ωa.
- Interchanging the two cases so that the a² and b² factors are swapped.
Earns more
- No-slip boundary conditions written as wall speeds (Ωa, Ωb), not as Ω.
- Physical discussion: rigid-rotation part Ar plus free-vortex part B/r; shear stress τ_rθ = -2μB/r² and torque per unit length 4πμΩa²b²/(b² - a²) for the inner-rotating case; radial momentum dp/dr = ρv²/r.
Extra mark
- Limit b → ∞ in case (i) giving the free vortex Ωa²/r
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