Solutions to Assignments on Heat Equation and Laplace's Equation in One-Dimensional Rods, Assignments of Differential Equations

Download Solutions to Assignments on Heat Equation and Laplace's Equation in One-Dimensional Rods and more Assignments Differential Equations in PDF only on Docsity! Assignment Solutions of Partial Differential Equations Weijiu Liu Department of Mathematics University of Central Arkansas 201 Donaghey Avenue, Conway, AR 72035, USA 1 Assignment 1 1.2.3. Derive the heat equation for a rod assuming constant thermal properties with variable cross-sectional area A(x) assuming no sources. Denote by A the the cross-sectional area. Physical quantities: • Thermal energy density e(x, t) = the amount of thermal energy per unit volume. • Heat flux φ(x, t) = the amount of thermal energy flowing across boundaries per unit surface area per unit time. • Heat sources Q(x, t) = 0. • Temperature u(x, t). • Specific heat c = the heat energy that must be supplied to a unit mass of a substance to raise its temperature one unit. • Mass density ρ(x) = mass per unit volume. • Fourier’s Law: the heat flux is proportional to the temperature gradient φ = −K0∇u. (1) Conservation of heat energy: Rate of change of heat energy in time = Heat energy flowing across boundaries per unit time + Heat energy generated insider per unit time • heat energy = e(x, t)A(x)∆x. • Heat energy flowing across boundaries per unit time = φ(x, t)A(x)−φ(x+∆x, t)A(x+ ∆x). Then ∂ ∂t [e(x, t)A(x)∆x] = φ(x, t)A(x)− φ(x + ∆x, t)A(x + ∆x). Dividing it by ∆x and letting ∆x go to zero give A(x) ∂e ∂t = −A(x)∂φ ∂x − φ(x)∂A ∂x . (2) Heat energy per unit mass = c(x)u(x, t)ρA∆x. So e(x, t)A(x)∆x = c(x)u(x, t)ρA(x)∆x, and then e(x, t) = c(x)u(x, t)ρ. 2 2 Assignment 2 1.4.1. Determine the equilibrium temperature distribution for a one-dimensional rod with constant thermal properties with the following sources and boundary conditions: (a) Q = 0, u(0) = 0, u(L) = T. (f) Q = K0x 2, u(0) = T, u′(L) = 0. Solution. (a) Equilibrium satisfies u′′(x) = 0, whose general solution is u = c1 + c2x. The boundary condition u(0) = 0 implies c1 = 0 and u(L) = T implies c2 = T/L so that u = Tx/L. (f) In equilibrium, u satisfies u′′(x) = −Q/K0 = −x2, whose general solution (by integrating twice) is u = −x4/12 + c1 + c2x. The boundary condition u(0) = T yields c1 = T, while u ′(L) = 0 yields c2 = L3/3. Thus u = −x4/12 + L3x/3 + T. 1.4.11. Suppose ∂u ∂t = ∂2u ∂x2 + x, u(x, 0) = f(x), ∂u ∂x (0, t) = β, ∂u ∂x (L, t) = 7. (a) Calculate the total thermal energy in the one-dimensional rod (as a function of time). (b) From part (a), determine a value of β for which an equilibrium exists. For this value of β, determine lim t→∞ u(x, t). Solution. (a) Integrating the equation, we obtain: d dt ∫ L 0 u(x, t)dx = ∫ L 0 ( ∂2u ∂x2 + x ) dx = ∂u ∂x ∣∣L 0 + 1 2 L2 = 7− β + 1 2 L2. Integrating in t from 0 to t, we obtain the total thermal energy ∫ L 0 u(x, t)dx = ∫ L 0 f(x)dx + ( 7− β + 1 2 L2 ) t. (8) (b) In order for an equilibrium to exist, ( 7− β + 1 2 L2 ) t must be 0. So β = 7 + 1 2 L2. 5 The equilibrium satisfies φ′′(x) + x = 0. Its general solution (after integrating twice) is φ = −1 6 x3 + c1 + c2x. The boundary condition yields c2 = 7 + 1 2 L2. So φ = −1 6 x3 + c1 + ( 7 + 1 2 L2 ) x. Since lim t→∞ u(x, t) = φ(x), using (8), we obtain ∫ L 0 f(x)dx = ∫ L 0 u(x, t)dx = lim t→∞ ∫ L 0 u(x, t)dx = ∫ L 0 φ(x)dx = ∫ L 0 ( −1 6 x3 + c1 + ( 7 + 1 2 L2 ) x ) dx = − 1 24 L4 + c1L + 1 2 ( 7 + 1 2 L2 ) L2. Solving it gives c1 = ∫ L 0 f(x)dx− 7 2 L2 − 5 24 L4 L , and then φ = −1 6 x3 + ( 7 + 1 2 L2 ) x + ∫ L 0 f(x)dx− 7 2 L2 − 5 24 L4 L . 1.5.2. For conduction of thermal energy, the heat flux vector is φ = −K0∇u. If in addition the molecules move at an average velocity V, a process called convection, then φ = −K0∇u+cρuV. Derive the corresponding equation for heat flow, including both conduction and convection of thermal energy (assuming constant thermal properties with no sources). Solution. Physical quantities: • Thermal energy density e(x, t) = the amount of thermal energy per unit volume. • Heat flux φ(x, t) = the amount of thermal energy flowing across boundaries per unit surface area per unit time. 6 • Heat sources Q(x, t) = heat energy per unit volume generated per unit time. • Temperature u(x, t). • Specific heat c = the heat energy that must be supplied to a unit mass of a substance to raise its temperature one unit. • Mass density ρ(x) = mass per unit volume. Conservation of heat energy: Rate of change of heat energy in time = Heat energy flowing across boundaries per unit time + Heat energy generated insider per unit time • heat energy = ∫ R e(x, t)dV . • Heat energy flowing across boundaries per unit time = ∮ φ · ndS. • Heat energy generated insider per unit time = ∫ R Q(x, t)dV = 0. Then ∂ ∂t ∫ R e(x, t)dV = − ∮ φ · ndS. The divergence theorem give ∂ ∂t ∫ R e(x, t)dV = − ∫ R ∇ · φdV. and then ∂e ∂t = −∇ · φ. (9) Heat energy per unit volume = c(x)u(x, t)ρ. So e(x, t) = c(x)u(x, t)ρ. It then follows that cρ ∂u ∂t = −∇ · (cρuV) +∇ · (K0∇u). (10) and then the convection-diffusion equation ∂u ∂t +∇ · (uV) = k∇2u, (11) where k = K0 cρ is called the thermal diffusivity. 7 4 Assignment 4 2.4.1. (a). The solution is u(x, t) = a0 + ∞∑ n=1 an cos (nπx L ) e− kn2π2t L2 . where a0 = 1 L ∫ L L/2 dx = 1 2 , an = 2 L ∫ L L/2 cos (nπx L ) dx = 2 L · L nπ sin (nπx L ) ∣∣∣ L L/2 = − 2 nπ sin (nπ 2 ) . 2.4.1. (b). The solution is u(x, t) = a0 + ∞∑ n=1 an cos (nπx L ) e− kn2π2t L2 . where a0 = 1 L ∫ L 0 ( 6 + 4 cos ( 3πx L )) dx = 6, a3 = 2 L ∫ L 0 ( 6 + 4 cos ( 3πx L )) cos (nπx L ) dx = 4, and others are 0. 2.4.2. Solution. ∂u ∂t = k ∂2u ∂x2 , (19) ∂u dx (0, t) = 0, u(L, t) = 0, (20) u(x, 0) = f(x). (21) Look for a solution of the form of separation of variables: u(x, t) = φ(x)G(t), (22) Substitute the above expression into the equation (19), we obtain φ(x)G′(t) = kφ′′(x)G(t), and then G′(t) kG(t) = φ′′(x) φ(x) = −λ, (23) 10 where λ is constant to be determined. The boundary condition (20) yields that φu dx (0) = φ(L) = 0. We then have an eigenvalue problem −d 2φ dx2 = λφ, φ dx (0) = 0, φ(L) = 0. Auxiliary equations: m2 = −λ. • Case 1: λ < 0. Distinct real roots m1 = √−λ and m2 = − √−λ: φ(x) = c1e √−λx + c2e− √−λx. The boundary conditions imply that c1 = c2 = 0. So no non-zero solutions exist and then λ < 0 is not an eigenvalue. • Case 2: λ = 0. Repeated real roots m1 = m2 = 0: φ = c1 + c2x. The boundary conditions imply that c1 = c2 = 0. So no non-zero solutions exist and then λ < 0 is not an eigenvalue. • Case 3: λ > 0. Conjugate complex roots m1 = i √ λ and m2 = −i √ λ: φ = c1 cos( √ λx) + c2 sin( √ λx). φ′(0) = 0 implies that c2 = 0. φ(L) = 0 gives cos( √ λL) = 0. So √ λL = π 2 + nπ (n = 0, 1, 2, · · · ) and then we obtain the eigenvalues λn = ( π 2 + nπ )2 L2 , n = 0, 1, 2, · · · (24) and the corresponding eigenfunctions φn = cos (( π 2 + nπ ) x L ) , n = 0, 1, 2, · · · (25) 11 On the other hand, it follows from (23) that dG dt = −λkG, (26) which has solutions G(t) = ce−λkt = ce− (π2 +nπ) 2 kt L2 . We then derive the infinite series solution: u(x, t) = ∞∑ n=1 an cos (( π 2 + nπ ) x L ) e− (π2 +nπ) 2 kt L2 . The initial condition gives f(x) = u(x, 0) = ∞∑ n=1 an cos (( π 2 + nπ ) x L ) . (27) To determine an, we multiply (27) by cos ( (π2 +nπ)x L ) and integrate from 0 to L. We then find an = 2 L ∫ L 0 f(x) cos (( π 2 + nπ ) x L ) dx, n ≥ 1. (28) 2.5.1. (c) Solve Laplace’s equation ∂2u ∂x2 + ∂2u ∂y2 = 0, (29) ∂u ∂x (0, y) = 0, u(L, y) = g(Y ), u(x, 0) = 0, u(x,H) = 0, (30) Look for a solution of the form of separation of variables: u(x, y) = h(x)φ(y), (31) Substitute the above expression into the equation (29), we obtain φ(y)h′′(x) + φ′′(y)h(x) = 0, and then h′′(x) h(x) = −φ ′′(y) φ(y) = λ, (32) where λ is constant to be determined. The boundary condition yields that φ(0) = φ(H) = 0, h′(0) = 0. We then have an eigenvalue problem 12