(a) Reduce the following partial differential equation to a canonical form and hence solve it: yu_xx + (x+y)u_xy + xu_yy = 0 (15…

(a) Canonical form and general solution

Comparing with Au_xx + Bu_xy + Cu_yy = 0: A = y, B = x + y, C = x, so B² − 4AC = (x − y)². The equation is hyperbolic for x ≠ y (parabolic on y = x).

Characteristic equation: y m² − (x + y)m + x = 0, m = dy/dx, i.e. (m − 1)(ym − x) = 0. m = 1 ⟹ y − x = const. m = x/y ⟹ y dy = x dx ⟹ y² − x² = const.

Put ξ = y − x, η = y² − x². Then ξ_x = −1, ξ_y = 1, η_x = −2x, η_y = 2y, and u_xx = u_ξξ + 4x u_ξη + 4x² u_ηη − 2u_η, u_xy = −u_ξξ − 2(x + y)u_ξη − 4xy u_ηη, u_yy = u_ξξ + 4y u_ξη + 4y² u_ηη + 2u_η.

Substituting, the u_ξξ and u_ηη coefficients vanish, the u_ξη coefficient is 8xy − 2(x + y)² = −2(x − y)², and the u_η coefficient is 2(x − y). With x − y = −ξ: −2ξ² u_ξη − 2ξ u_η = 0 ⟹ u_ξη + (1/ξ)u_η = 0 (canonical form, ξ ≠ 0).

Solution: ξu_ξη + u_η = ∂/∂ξ(ξu_η) = 0, so ξu_η = F(η) and u = f(η)/ξ + g(ξ). u(x, y) = f(y² − x²)/(y − x) + g(y − x), f and g arbitrary. (Direct substitution gives zero identically.)

(b) Runge–Kutta method of fourth order

f(x, y) = x + y², y(0) = 1, h = 0.1. k₁ = h f(xₙ, yₙ), k₂ = h f(xₙ + h/2, yₙ + k₁/2), k₃ = h f(xₙ + h/2, yₙ + k₂/2), k₄ = h f(xₙ + h, yₙ + k₃), yₙ₊₁ = yₙ + (k₁ + 2k₂ + 2k₃ + k₄)/6 (all values rounded to four places).

Step 1 (x₀ = 0, y₀ = 1): k₁ = 0.1(0 + 1) = 0.1000 k₂ = 0.1(0.05 + 1.0500²) = 0.1153 k₃ = 0.1(0.05 + 1.0577²) = 0.1169 k₄ = 0.1(0.1 + 1.1169²) = 0.1347 Δy = (0.1000 + 0.2306 + 0.2338 + 0.1347)/6 = 0.6991/6 = 0.1165 y(0.1) = 1.1165.

Step 2 (x₁ = 0.1, y₁ = 1.1165): k₁ = 0.1(0.1 + 1.1165²) = 0.1347 k₂ = 0.1(0.15 + 1.1839²) = 0.1552 k₃ = 0.1(0.15 + 1.1941²) = 0.1576 k₄ = 0.1(0.2 + 1.2741²) = 0.1823 Δy = (0.1347 + 0.3104 + 0.3152 + 0.1823)/6 = 0.9426/6 = 0.1571 y(0.2) = 1.2736.

A high-accuracy integration gives 1.27356, so the RK4 value is correct to four decimal places.

(c) Flow about a cylinder near a plane wall, and the force

Let z − ia = r₁e^(iθ₁), z + ia = r₂e^(iθ₂). Then w = ik[log(r₁/r₂) + i(θ₁ − θ₂)], so φ = −k(θ₁ − θ₂), ψ = k log(r₁/r₂).

Verification:

1. w is analytic except at z = ±ia, so φ and ψ are harmonic in the fluid: the flow is irrotational and incompressible.
2. On y = 0, every point is equidistant from ±ia, so r₁ = r₂ and ψ = 0: the plane y = 0 is a streamline and can be a rigid boundary.
3. Streamlines are r₁/r₂ = λ. For λ below 1, |z − ia|² = λ²|z + ia|² gives x² + y² − 2cy + a² = 0, c = a(1 + λ²)/(1 − λ²), a circle with centre (0, c) and radius R, R² = c² − a², lying in y > 0 and enclosing ia. Any such circle may be taken as a rigid cylinder.
4. dw/dz = −2ka/(z² + a²) → 0 at infinity, and the circulation round the cylinder is 2πk. The flow does not depend on time, so it is steady.

Hence w is the complex potential of a steady flow about the cylinder (centre height c, radius R, a² = c² − R²) with y = 0 as a rigid wall: a line vortex at ia with its image at −ia.

Force (Blasius theorem): X − iY = (iρ/2)∮_C (dw/dz)² dz, (dw/dz)² = 4k²a²/[(z − ia)²(z + ia)²]. Only the double pole z = ia lies inside C, with residue d/dz[4k²a²/(z + ia)²] at z = ia = −8k²a²/(2ia)³ = −ik²/a. X − iY = (iρ/2)(2πi)(−ik²/a) = iπρk²/a, so X = 0 and Y = −πρk²/a.

The fluid exerts a force πρk²/a per unit length on the cylinder, perpendicular to the wall and directed towards it, where a = √(c² − R²).