(a) Solve the partial differential equation (∂)/(∂ y)((∂ φ)/(∂ x) + φ) + 2x^2y((∂ φ)/(∂ x) + φ) = 0 by transforming it to the canonical form. (15 marks)

(b) Using Newton's forward difference formula for interpolation, estimate the value of f(2·5) fr

Question

(a) Solve the partial differential equation (∂)/(∂ y)((∂ φ)/(∂ x) + φ) + 2x^2y((∂ φ)/(∂ x) + φ) = 0 by transforming it to the canonical form. (15 marks)

(b) Using Newton's forward difference formula for interpolation, estimate the value of f(2·5) from the following data: x : 1 2 3 4 5 6, f(x) : 0 1 8 27 64 125. (15 marks)

(c) Suppose an infinite liquid contains two parallel, equal and opposite rectilinear vortices at a distance 2a. Show that the streamlines relative to the vortex are given by the equation log (x² + (y-a)²)/(x² + (y+a)²) + y/a = C, where C is a constant, the origin is the middle point of the join, and the line joining the vortices is the axis of y. (20 marks)

UPSC Answer Check · Accepted Answer

(a) Write the equation as
φ_xy + φ_y + 2x²y φ_x + 2x²y φ = 0.
For the second-order part, A = 0, B = 1, C = 0, so B² − 4AC = 1 > 0. Hence it is hyperbolic. The characteristic equation is
A(dy)² − B dx dy + C(dx)² = 0, i.e. −dx dy = 0.
Thus the characteristics are x = constant and y = constant, so take ξ = x, η = y. The canonical form is
φ_ξη + φ_η + 2ξ²η φ_ξ + 2ξ²η φ = 0,
or
(φ_ξ + φ)_η + 2ξ²η(φ_ξ + φ) = 0.

Set v = φ_ξ + φ. Then
v_η + 2ξ²η v = 0.
The integrating factor in η is
exp(∫2ξ²η dη) = exp(ξ²η²).
Therefore
(v exp(ξ²η²))_η = 0,
so
v exp(ξ²η²) = F(ξ),
where F is arbitrary. Hence
φ_ξ + φ = F(ξ) exp(−ξ²η²).
Multiply by exp(ξ):
(φ exp(ξ))_ξ = F(ξ) exp(ξ − ξ²η²).
Integrating with respect to ξ,
φ exp(ξ) = ∫^ξ F(s) exp(s − s²η²) ds + G(η).
Thus the general solution is
φ(x,y) = exp(−x)[∫^x F(s) exp(s − s²y²) ds + G(y)],
where F and G are arbitrary functions.
Check: φ_x + φ = F(x) exp(−x²y²), so the original equation is satisfied.

(b) Use Newton’s forward difference formula:
f(x₀ + ph) = f₀ + pΔf₀ + p(p−1)/2! Δ²f₀ + p(p−1)(p−2)/3! Δ³f₀ + …
Take x₀ = 2, h = 1, x = 2·5, so
p = (2·5 − 2)/1 = 0·5.

At x₀ = 2:
f₀ = f(2) = 1
Δf₀ = f(3) − f(2) = 8 − 1 = 7
Δ²f₀ = f(4) − 2f(3) + f(2) = 27 − 16 + 1 = 12
Δ³f₀ = f(5) − 3f(4) + 3f(3) − f(2) = 64 − 81 + 24 − 1 = 6
Δ⁴f₀ = 0.

Therefore
f(2·5) = 1 + 0·5×7 + [(0·5)(−0·5)/2]×12 + [(0·5)(−0·5)(−1·5)/6]×6
= 1 + 3·5 − 1·5 + 0·375
= 3·375.
So
f(2·5) = 27/8 = 3·375.

(c) Let the vortices be at A(0,a) and B(0,−a). Let the circulation at A be Γ and at B be −Γ. For a rectilinear vortex of strength k, the stream function at distance r is
ψ = −(k/2π) ln r.
Hence the total stream function for the two vortices is
ψ = −(Γ/2π) ln r₁ + (Γ/2π) ln r₂
= (Γ/2π) ln(r₂/r₁),
where
r₁² = x² + (y − a)²,
r₂² = x² + (y + a)².

The vortex at A is convected by the field of B. The speed induced at distance 2a is
U = Γ/(2π·2a) = Γ/(4πa).
With the chosen signs, the pair moves along the positive x-direction.

In the frame moving with the vortices, subtract the uniform stream U in the x-direction. The relative stream function is
ψ_rel = ψ − Uy
= (Γ/2π) ln(r₂/r₁) − (Γ/(4πa))y.
For a streamline relative to the vortex, ψ_rel = constant. Thus
(Γ/2π) ln(r₂/r₁) − (Γ/(4πa))y = constant.
Divide by Γ/(2π):
ln(r₂/r₁) − y/(2a) = constant.
Multiply by −2 and absorb the constant:
2 ln(r₁/r₂) + y/a = C.
Equivalently,
ln(r₁²/r₂²) + y/a = C.
Substituting r₁² and r₂²,
log([x² + (y − a)²]/[x² + (y + a)²]) + y/a = C,
where C is a constant. This holds for all points except the vortex singularities r₁ = 0 and r₂ = 0.

(a) Solve the partial differential equation (∂)/(∂ y)((∂ φ)/(∂ x) + φ) + 2x^2y((∂ φ)/(∂ x) + φ) = 0 by transforming it to the…

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