Paper II — Q7
(a) Find the complete integral of z(p²-q²) = x-y; p≡∂z/∂x, q≡∂z/∂y. (15 marks) (b) Find the unique polynomial of degree 2 or…
Find the complete integral of z(p²-q²) = x-y; p≡∂z/∂x, q≡∂z/∂y. 15 marks
Find the unique polynomial of degree 2 or less which fits the following data: x : 0 1 3 f(x) : 1 3 55 Also obtain the bound on the truncation error. 15 marks
Show that for an incompressible steady flow with constant viscosity, the velocity components u(y) = (U/h)y - (hy/2μ)(dp/dx)(1-y/h) v = 0 = w, with p = p(x), satisfy the equation of motion in the absence of body force. Given that U, h and dp/dx are constants. 20 marks
हिंदी में प्रश्न पढ़ें
z(p²-q²) = x-y; p≡∂z/∂x, q≡∂z/∂y का पूर्ण समाकल प्राप्त कीजिए। (15 अंक)
घात 2 या 2 से कम का वह अद्वितीय बहुपद, जो आँकड़ों x : 0 1 3 f(x) : 1 3 55 पर ठीक बैठता है, प्राप्त कीजिए। क्षण त्रुटि पर परिबंध भी प्राप्त कीजिए। (15 अंक)
दर्शाइए कि अचर विस्कांशता के एक असंपीड्य अपरिवर्ती प्रवाह के लिए वेग घटक u(y) = (U/h)y - (hy/2μ)(dp/dx)(1-y/h) v = 0 = w, p = p(x) के साथ, पिण्ड बल की अनुपस्थिति में गति के समीकरण को संतुष्ट करते हैं। यह दिया गया है कि U, h और dp/dx अचर हैं। (20 अंक)
Model answer
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(a)
Using Charpit’s method, take F = z(p² − q²) − x + y = 0. Then F_p = 2zp, F_q = −2zq, F_x = −1, F_y = 1, F_z = p² − q².
The Charpit equations are dx/(2zp) = dy/(−2zq) = dz/(2(x−y)) = −dp/(−1 + p(p²−q²)) = −dq/(1 + q(p²−q²)).
Since p² − q² = (x−y)/z, along a characteristic, d(p+q)/dλ = −(p+q)(x−y)/z, dz/dλ = 2(x−y). Thus d(p+q)/(p+q) = −dz/(2z), so (p+q)√z = a, where a is an arbitrary constant. Hence p+q = a/√z.
Also p−q = (p²−q²)/(p+q) = (x−y)/(a√z). Therefore p = (a² + x − y)/(2a√z), q = (a² − x + y)/(2a√z).
Now use dz = p dx + q dy: 2a√z dz = (a² + x − y)dx + (a² − x + y)dy = a²(dx+dy) + (x−y)(dx−dy) = a² d(x+y) + (1/2)d((x−y)²).
Integrating, (4a/3) z^(3/2) = a²(x+y) + (x−y)²/2 + b, where a and b are arbitrary constants, a ≠ 0. Complete integral: (4a/3) z^(3/2) = a²(x+y) + (x−y)²/2 + b.
(b)
Using Newton’s divided-difference interpolation, f[0] = 1, f[0,1] = (3−1)/(1−0) = 2, f[1,3] = (55−3)/(3−1) = 26, f[0,1,3] = (26−2)/(3−0) = 8.
Hence P(x) = 1 + 2x + 8x(x−1) = 1 + 2x + 8x² − 8x = 8x² − 6x + 1.
P(x) = 8x² − 6x + 1.
The truncation error for quadratic interpolation at nodes 0,1,3 is E(x) = f(x) − P(x) = f'''(ξ)/6 · x(x−1)(x−3), for some ξ between the nodes and x, provided f is three times differentiable.
On [0,3], max |x(x−1)(x−3)| = (20 + 14√7)/27, attained at x = (4 + √7)/3. Therefore, if M = max_[0,3] |f'''(t)|, then
|E(x)| ≤ M(20 + 14√7)/162 = M(10 + 7√7)/81.
Bound: |E(x)| ≤ M(10 + 7√7)/81, M = max |f'''| on the interval.
(c)
For an incompressible steady flow with constant viscosity μ and no body force, the Navier–Stokes equations are ∂u/∂x + ∂v/∂y + ∂w/∂z = 0, ρ(u∂u/∂x + v∂u/∂y + w∂u/∂z) = −∂p/∂x + μ∇²u, and similarly for v and w.
Given u(y) = (U/h)y − (h y/(2μ))(dp/dx)(1 − y/h), v = 0 = w, p = p(x).
Continuity: ∂u/∂x + ∂v/∂y + ∂w/∂z = 0 + 0 + 0 = 0. So the incompressibility condition is satisfied.
Let P = dp/dx. Then u = (U/h)y − (hP/(2μ))y + (P/(2μ))y². Differentiate: du/dy = U/h − hP/(2μ) + (P/μ)y, d²u/dy² = P/μ.
For the x-momentum equation, since u = u(y), v = w = 0, ρ(u∂u/∂x + v∂u/∂y + w∂u/∂z) = 0. Also ∂²u/∂x² = ∂²u/∂z² = 0, so −∂p/∂x + μ d²u/dy² = −P + μ(P/μ) = 0. Thus the x-momentum equation is satisfied.
For the y-momentum equation, v = 0 and p = p(x) imply −∂p/∂y + μ∇²v = 0 + 0 = 0. Similarly, the z-momentum equation is satisfied.
Hence the given velocity field and pressure satisfy the equation of motion in the absence of body force.
How this answer will be evaluated
What earns full marks: Complete, stepwise derivation with all assumptions stated and results verified.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Complete integral of the first-order PDE z(p² - q²) = x - y. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Charpit equations with F = z(p² − q²) − x + y
- First integral (p + q)√z = a; p − q = (x − y)/(a√z) ⇒ p, q expressions
- Integrate dz = p dx + q dy
- Complete integral (4a/3)z^(3/2) = a²(x + y) + (x − y)²/2 + b
Loses marks
- Skipping the derivation of the first integral
- Failing to integrate dz = p dx + q dy
- Answer without working
Earns more
- Correctly identifies the PDE as non-linear
- Explicitly states the two arbitrary constants a and b
- Verifies the result by substituting p and q back into the PDE
Extra mark
- Verifies that p and q satisfy the compatibility condition or back-substitutes the integral into the PDE
- (b) Unique polynomial of degree ≤ 2 fitting the data and the truncation error bound. 15 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Divided differences f[0,1] = 2, f[1,3] = 26, f[0,1,3] = 8
- P(x) = 8x² − 6x + 1 (matches all three data points)
- Error formula E = f‴(ξ)/6 · x(x − 1)(x − 3)
- Bound |E| ≤ M(10 + 7√7)/81 (max |x(x−1)(x−3)| = (20 + 14√7)/27)
Loses marks
- Incorrect construction of the divided difference table
- Failing to calculate the numerical error bound
- Answer without working
Earns more
- States that the interpolating polynomial is unique (degree ≤ 2 through three distinct nodes)
- Verifies the polynomial by substituting x=0, 1, 3
- Clearly defines the notation for divided differences
Extra mark
- Cross-checks P(x) using the Lagrange form
- (c) Show that the given velocity components satisfy the equation of motion. 20 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- Continuity ∂u/∂x + ∂v/∂y + ∂w/∂z = 0 satisfied
- u = (U/h)y − (hP/2μ)y + (P/2μ)y²; d²u/dy² = P/μ
- x-momentum: convective terms vanish; −P + μ(P/μ) = 0
- y- and z-momentum satisfied (p = p(x), v = w = 0)
Loses marks
- Failing to substitute the given velocity components
- Incorrect calculation of the derivatives
- Answer without working
Earns more
- Explicitly states the assumptions (steady, incompressible, constant viscosity)
- Verifies the continuity equation ∂u/∂x + ∂v/∂y + ∂w/∂z = 0
- Clearly defines the notation for the velocity components
Extra mark
- Neat figure of the flow geometry
What "Solve" is asking you to do
Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.
Structure that answers it
Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity
Where marks are lost
Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.
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