(a) Find the surface passing through the two lines z = x = 0 and z-1 = x-y = 0, and satisfying the partial differential equation…

(a) Surface through two lines satisfying the PDE

The equation is (∂/∂x − 2∂/∂y)² z = 0, since ∂²z/∂x² − 4∂²z/∂x∂y + 4∂²z/∂y² = (D − 2D′)² z.

Put u = y + 2x and v = x. Then ∂/∂x − 2∂/∂y becomes ∂/∂v, so the equation is ∂²z/∂v² = 0. Integrating twice, z = F(u) + vG(u) = F(y + 2x) + xG(y + 2x).

Line z = x = 0: its points are (0, y, 0). Putting x = 0, z = 0 gives F(y) = 0 for all y, so F ≡ 0 and z = xG(y + 2x).

Line z − 1 = x − y = 0: its points are (t, t, 1). Then 1 = tG(3t), so G(3t) = 1/t for t ≠ 0, that is G(u) = 3/u.

Hence z = 3x/(y + 2x), or equivalently the surface (2x + y)z = 3x.

Check: the PDE is satisfied wherever y + 2x ≠ 0 (direct differentiation gives zero). In the form (2x + y)z = 3x, every point of z = x = 0 satisfies it, and every point (t, t, 1) satisfies 3t = 3t, including t = 0. So the quadric surface (2x + y)z = 3x contains both complete lines, even though z is not a single-valued function of (x, y) at the origin, where the two lines have different heights 0 and 1.

(b) Gauss-Seidel solution of the linear system

The system is diagonally dominant (7 > 3, 8 > 3, 9 > 3), so the method converges. Rearranging, x₁ = (11 + x₂ − 2x₃)/7, x₂ = (9 − 2x₁ + x₃)/8, x₃ = (7 − x₁ + 2x₂)/9, using the newest values at each step. Starting from x₁ = x₂ = x₃ = 0:

Iteration 1: x₁ = 1.571429, x₂ = 0.732143, x₃ = 0.765873 Iteration 2: x₁ = 1.457200, x₂ = 0.856434, x₃ = 0.806185 Iteration 3: x₁ = 1.463438, x₂ = 0.859914, x₃ = 0.806266 Iteration 4: x₁ = 1.463912, x₂ = 0.859805, x₃ = 0.806189 Iteration 5: x₁ = 1.463918, x₂ = 0.859794, x₃ = 0.806186 Iteration 6: x₁ = 1.463918, x₂ = 0.859794, x₃ = 0.806186

The iterates have stabilised. Check: the exact solution is x₁ = 142/97, x₂ = 417/485, x₃ = 391/485.

To 4 significant figures, x₁ = 1.464, x₂ = 0.8598, x₃ = 0.8062.

(c) Removing the cross term in the Lagrangian

The transformation is a rotation, so ẋ² + ẏ² = q̇₁² + q̇₂² and the kinetic term is ½m(q̇₁² + q̇₂²).

For the potential part, xy = ½(q₁² − q₂²) sin2θ + q₁q₂ cos2θ, and the terms in x² and y² also contribute to q₁q₂. Collecting, the coefficient of q₁q₂ in L is (m/2)(w₁² − w₂²) sin2θ + k cos2θ.

For no q₁q₂ term this must vanish: tan2θ = −2k/[m(w₁² − w₂²)] for w₁² ≠ w₂². If w₁² = w₂² and k ≠ 0, then cos2θ = 0, so θ = π/4 (mod π/2).

Lagrange's equations: the potential is V = ½(λ₁q₁² + λ₂q₂²), where λ₁ and λ₂ are the eigenvalues of the matrix [m w₁², −k; −k, m w₂²]. With R = √[m²(w₁² − w₂²)² + 4k²], λ± = ½[m(w₁² + w₂²) ± R].

Then d/dt(∂L/∂q̇ᵢ) = ∂L/∂qᵢ gives m q̈₁ + ½[m(w₁² + w₂²) + R] q₁ = 0, m q̈₂ + ½[m(w₁² + w₂²) − R] q₂ = 0, for the branch of θ in which q₁ carries the larger eigenvalue. For the other branch the two equations are interchanged. These are uncoupled normal-mode equations, with frequencies depending only on m, w₁, w₂, k, not on θ.