(a) Reduce the partial differential equation ∂²z/∂y² - ∂²z/∂x∂y + ∂z/∂x - ∂z/∂y(1+1/x) + z/x = 0 to canonical form. (15…

(a) Canonical form of the partial differential equation

Compare the equation with A zₓₓ + B zₓᵧ + C zᵧᵧ + lower-order terms = 0. Here A = 0, B = −1, C = 1, so B² − 4AC = 1 > 0 and the equation is hyperbolic.

The characteristic equation A dy² − B dx dy + C dx² = 0 becomes dx dy + dx² = 0, that is dx(dx + dy) = 0. The characteristics are x = constant and x + y = constant.

Take ξ = x, η = x + y and Z(ξ, η) = z(x, y). By the chain rule, zₓ = Z_ξ + Z_η, zᵧ = Z_η, zᵧᵧ = Z_ηη, zₓᵧ = Z_ξη + Z_ηη.

Substituting in the equation: Z_ηη − (Z_ξη + Z_ηη) + (Z_ξ + Z_η) − Z_η(1 + 1/ξ) + Z/ξ = 0, which simplifies to −Z_ξη + Z_ξ − Z_η/ξ + Z/ξ = 0.

Hence the canonical form is Z_ξη − Z_ξ + (1/ξ)(Z_η − Z) = 0, with ξ = x ≠ 0.

It can also be written as (∂/∂ξ + 1/ξ)(Z_η − Z) = 0, which makes the equation easy to integrate.

(b) Root of log₁₀(2x + 1) − x² + 3 = 0 by Regula-Falsi

Let f(x) = log₁₀(2x + 1) − x² + 3. Then f(0) = 3 > 0 and f(3) = log₁₀7 − 6 = −5.154902 < 0, so a root lies in [0, 3].

Regula-Falsi formula for a bracket [a, b]: x_new = (a f(b) − b f(a))/(f(b) − f(a)), and the sub-interval where f changes sign is kept. Starting with a = 0, b = 3, f(a) > 0 at every stage, so b = 3 stays fixed and a moves up:

x₁ = 1.103631, f = 2.288134 x₂ = 1.686611, f = 0.796143 x₃ = 1.862319, f = 0.206135 x₄ = 1.906064, f = 0.049258 x₅ = 1.916418, f = 0.011544 x₆ = 1.918839, f = 0.002693 x₇ = 1.919404, f = 0.000628 x₈ = 1.919535, f = 0.000146 x₉ = 1.919566, f = 0.000034 x₁₀ = 1.919573, f = 0.000008 x₁₁ = 1.919575 (1.9195747), f = 0.0000019 x₁₂ = 1.919575 (1.9195751), f = 0.0000004

Successive values x₁₁ and x₁₂ agree to six decimal places. Hence the root, correct to 6 decimal places, is x = 1.919575.

The one-sided approach (b stays at 3) is typical of Regula-Falsi and explains the slow, linear convergence.

(c) Navier-Stokes conditions and the pressure distribution

Take the flow as steady and incompressible, with constant density ρ and viscosity μ, velocity (u, v, w) = (c(x² − y²), −2cxy, 0), where c is a constant.

Continuity: ∂u/∂x + ∂v/∂y + ∂w/∂z = 2cx − 2cx + 0 = 0, so mass is conserved for every c.

Viscous terms: ∇²u = 2c − 2c = 0 and ∇²v = 0 + 0 = 0, so the viscous terms vanish identically, and viscosity drops out.

Momentum equations, with D/Dt = u∂/∂x + v∂/∂y for the steady flow: u uₓ + v uᵧ = c(x² − y²)(2cx) + (−2cxy)(−2cy) = 2c²x(x² + y²), u vₓ + v vᵧ = c(x² − y²)(−2cy) + (−2cxy)(−2cx) = 2c²y(x² + y²), Dw/Dt = 0.

The equations ρ(Dv/Dt) = −∇p + μ∇²v + ρB then become pₓ = −2ρc²x(x² + y²), pᵧ = −2ρc²y(x² + y²), p_z = −ρg (since B_z = −g and Dw/Dt = 0).

These are compatible, since pₓᵧ = pᵧₓ = −4ρc²xy, so a pressure field exists. The velocity field is therefore a solution of the Navier-Stokes equations for any constant c (the flow is irrotational: ∂v/∂x − ∂u/∂y = 0), provided it is steady and incompressible.

Integrating pₓ gives p = −ρc²(x⁴/2 + x²y²) + φ(y, z). Comparing with pᵧ gives ∂φ/∂y = −2ρc²y³, so φ = −ρc²y⁴/2 + ψ(z), and p_z = −ρg gives ψ = −ρgz + C. Hence p = −ρgz − (ρc²/2)(x² + y²)² + C, where C is an arbitrary constant. This is Bernoulli's relation p + ρgz + ½ρ|v|² = constant, with |v|² = c²(x² + y²)².