(a) Solve the following initial value problem by using Laplace transform technique : d²y/dt² - 4 dy/dt + 3y(t) = f(t), y(0) =…

(a) Initial value problem by Laplace transform

Let Y(s)=L{y(t)} and F(s)=L{f(t)}. Using the Laplace transform of derivatives, L{y'} = sY - y(0) = sY - 1, L{y''} = s²Y - s y(0) - y'(0) = s²Y - s. Substituting in the differential equation gives s²Y - s - 4(sY - 1) + 3Y = F(s). Thus (s² - 4s + 3)Y = F(s) + s - 4, so Y = F(s)/((s-1)(s-3)) + (s-4)/((s-1)(s-3)). Now (s-4)/((s-1)(s-3)) = (3/2)·1/(s-1) - (1/2)·1/(s-3), hence its inverse transform is (3/2) exp(t) - (1/2) exp(3t). Also 1/((s-1)(s-3)) = (1/2)(1/(s-3) - 1/(s-1)), whose inverse transform is g(t) = (1/2)(exp(3t) - exp(t)). By the convolution theorem, L⁻¹{F(s)/((s-1)(s-3))} = ∫₀ᵗ f(u)g(t-u)du = (1/2)∫₀ᵗ (exp(3(t-u)) - exp(t-u))f(u)du. Therefore y(t) = (3/2) exp(t) - (1/2) exp(3t) + (1/2)∫₀ᵗ (exp(3(t-u)) - exp(t-u))f(u)du. This holds whenever f is piecewise continuous and of exponential order so that its Laplace transform exists.

(b) Path of a particle under a central acceleration

Use polar coordinates with the centre of force as origin. Let h = r²(dθ/dt). At an apse the radial velocity is zero, so the whole velocity is transverse. Given r₀ = √c and v = √(2λc³/3), h = r₀v = √c·√((2λ/3)c³) = c²√(2λ/3), so h² = 2λc⁴/3.

Take the central acceleration, directed towards the centre, as F(r) = λ(r⁵ - c²r). Its potential (per unit mass) is V(r) = ∫F(r)dr = λ(r⁶/6 - c²r²/2). At r₀ = √c, V(r₀) = λ(c³/6 - c³/2) = -λc³/3, and the kinetic energy is (1/2)v² = (1/2)·(2λc³/3) = λc³/3. Thus the total energy is E = 0. Hence (1/2)(dr/dt)² + h²/(2r²) + λ(r⁶/6 - c²r²/2) = 0. Substituting h², (dr/dt)² = λ/(3r²)(r⁴ - c²)(2c² - r⁴). Since dθ/dr = h/(r²(dr/dt)), we get dθ/dr = c²√2/[r√((r⁴ - c²)(2c² - r⁴))]. Put z = r⁴/c². Then dθ = (√2/4) dz/[z√((z-1)(2-z))]. Integrating from z = 1 at θ = 0 gives θ = (1/4) arccos(4/z - 3), so cos4θ = 4/z - 3, z = 4/(3 + cos4θ). Therefore r⁴ = 4c²/(3 + cos4θ). But cos⁴θ + sin⁴θ = (3 + cos4θ)/4. Thus r⁴(cos⁴θ + sin⁴θ) = c². Since x = r cosθ and y = r sinθ, x⁴ + y⁴ = c². So the path is the curve x⁴ + y⁴ = c², and the answer to the question asked is yes. (Check: u = 1/r satisfies Binet's equation h²u²(u″ + u) = F(1/u) for this curve.)

(c) Divergence identity and its verification

Let f = f₁i + f₂j + f₃k. Then ∇·(φf) = ∂(φf₁)/∂x + ∂(φf₂)/∂y + ∂(φf₃)/∂z = (φ_x f₁ + φ_y f₂ + φ_z f₃) + φ(f₁_x + f₂_y + f₃_z) = ∇φ·f + φ∇·f. This proves the identity.

Now let r = xi + yj + zk, r = |r|. Then ∇·(f(r)/r·r) = ∇·(f(r)e_r), where e_r = r/r. In spherical coordinates, for a radial vector f(r)e_r, ∇·(f(r)e_r) = (1/r²) d/dr(r²f(r)) = f'(r) + 2f(r)/r. So ∇·(f(r)/r·r) = f'(r) + 2f(r)/r.

To verify the identity, take φ = f(r)/r and vector f = r. Then ∇φ = (d/dr(f/r))e_r = (rf'(r) - f(r))/r² e_r. Hence ∇φ·r = (rf'(r) - f(r))/r²·r = f'(r) - f(r)/r. Also ∇·r = 3. Therefore ∇φ·r + φ∇·r = f'(r) - f(r)/r + (f(r)/r)·3 = f'(r) + 2f(r)/r, which agrees with the direct value. The identity is verified.