The One-Dimensional Heat Equation

Transcription

1 The One-Dimensional Heat Equation R. C. Trinity University Partial Differential Equations February 24, 2015

2 Introduction The heat equation Goal: Model heat (thermal energy) flow in a one-dimensional object (thin rod). Set up: Place rod along x-axis, and let u(x,t) = temperature in rod at position x, time t. Under ideal conditions (e.g. perfect insulation, no external heat sources, uniform rod material), one can show the temperature must satisfy u t = u c2 2 x 2. ( ) the one-dimensional heat equation The constant c 2 is called the thermal diffusivity of the rod.

3 Initial and Boundary Conditions We now assume the rod has finite length L and lies along the interval [0, L]. To completely determine u we must also specify: Initial conditions: The initial temperature profile u(x,0) = f(x) for 0 < x < L. Boundary conditions: Specific behavior at x 0 {0,L}: 1. Constant temperature: u(x 0,t) = T for t > Insulated end: u x (x 0,t) = 0 for t > Radiating end: u x (x 0,t) = Au(x 0,t) for t > 0.

4 Solving the Heat Equation Case 1: homogeneous Dirichlet boundary conditions We now apply separation of variables to the heat problem u t = c 2 u xx (0 < x < L, t > 0), u(0,t) = u(l,t) = 0 (t > 0), u(x,0) = f(x) (0 < x < L). We seek separated solutions of the form u(x,t) = X(x)T(t). In this case u t = XT } u xx = X XT = c 2 X T X T X = T c 2 T = k. Together with the boundary conditions we obtain the system X kx = 0, X(0) = X(L) = 0, T c 2 kt = 0.

5 Already know: up to constant multiples, the only solutions to the BVP in X are ( nπ ) 2, k = µ 2 n = L ( nπx ) X = X n = sin(µ n x) = sin, n N. L Therefore T must satisfy ( cnπ ) T c 2 kt = T 2 + T = 0 }{{ L } λ n T = λ 2 nt T = T n = b n e λ2 n t. We thus have the normal modes of the heat equation: u n (x,t) = X n (x)t n (t) = b n e λ2 n t sin(µ n x), n N.

6 Superposition and initial condition Applying the principle of superposition gives the general solution u(x,t) = u n (x,t) = n=1 b n e λ2 n t sin(µ n x). n=1 If we now impose our initial condition we find that f(x) = u(x,0) = ( nπx ) b n sin, L n=1 which is the sine series expansion of f(x). Hence b n = 2 L L 0 f(x)sin ( nπx ) dx. L

7 Remarks As before, if the sine series of f(x) is already known, solution can be built by simply including exponential factors. One can show that this is the only solution to the heat equation with the given initial condition. Because of the decaying exponential factors: The normal modes tend to zero (exponentially) as t. Overall, u(x,t) 0 (exponentially) uniformly in x as t. As c increases, u(x,t) 0 more rapidly. This agrees with intuition.

8 Example Solve the heat problem u t = 3u xx (0 < x < 2, t > 0), u(0,t) = u(2,t) = 0 (t > 0), u(x,0) = 50 (0 < x < 2). We have c = 3, L = 2 and, by exercise (with p = L = 2) f(x) = 50 = 200 ( ) 1 (2k +1)πx π 2k +1 sin. 2 k=0 c(2k +1)π 3(2k +1)π Since λ 2k+1 = =, we obtain L 2 u(x,t) = 200 ( 1 π 2 (2k +1)πx t/4 π 2k +1 e 3(2k+1)2 sin 2 k=0 ).

9 Solving the Heat Equation Case 2a: steady state solutions Definition: We say that u(x,t) is a steady state solution if u t 0 (i.e. u is time-independent). If u(x,t) = u(x) is a steady state solution to the heat equation then u t 0 c 2 u xx = u t = 0 u xx = 0 u = Ax +B. Steady state solutions can help us deal with inhomogeneous Dirichlet boundary conditions. Note that u(0,t) = T 1 u(l,t) = T 2 B = T 1 u = AL+B = T 2 L ( T2 T 1 ) x+t 1.

10 Solving the Heat Equation Case 2b: inhomogeneous Dirichlet boundary conditions Now consider the heat problem u t = c 2 u xx (0 < x < L, t > 0), u(0,t) = T 1, u(l,t) = T 2 (t > 0), u(x,0) = f(x) (0 < x < L). Step 1: Let u 1 denote the steady state solution from above: ( ) T2 T 1 u 1 = x +T 1. L Step 2: Let u 2 = u u 1. Remark: By superposition, u 2 still solves the heat equation.

11 The boundary and initial conditions satisfied by u 2 are u 2 (0,t) = u(0,t) u 1 (0) = T 1 T 1 = 0, u 2 (L,t) = u(l,t) u 1 (L) = T 2 T 2 = 0, u 2 (x,0) = f(x) u 1 (x). Step 3: Solve the heat equation with homogeneous Dirichlet boundary conditions and initial conditions above. This yields u 2. Step 4: Assemble u(x,t) = u 1 (x)+u 2 (x,t). Remark: According to our earlier work, lim t u 2 (x,t) = 0. We call u 2 (x,t) the transient portion of the solution. We have u(x,t) u 1 (x) as t, i.e. the solution tends to the steady state.

12 Example Solve the heat problem. u t = 3u xx (0 < x < 2, t > 0), u(0,t) = 100, u(2,t) = 0 (t > 0), u(x,0) = 50 (0 < x < 2). We have c = 3, L = 2, T 1 = 100, T 2 = 0 and f(x) = 50. The steady state solution is ( ) u 1 = x +100 = x. 2 The corresponding homogeneous problem for u 2 is thus u t = 3u xx (0 < x < 2, t > 0), u(0,t) = u(2,t) = 0 (t > 0), u(x,0) = 50 (100 50x) = 50(x 1) (0 < x < 2).

13 According to exercise (with p = L = 2), the sine series for 50(x 1) is 100 ( ) 1 2kπx π k sin, 2 k=1 i.e. only even modes occur. Since λ 2k = c2kπ L Hence u 2 (x,t) = 100 π k=1 u(x,t) = u 1 (x)+u 2 (x,t) = x 100 π = 3kπ, 1 k e 3k2 π 2t sin(kπx). k=1 1 k e 3k2 π 2t sin(kπx).

14 Solving the Heat Equation Case 3: homogeneous Neumann boundary conditions Let s now consider the heat problem u t = c 2 u xx u x (0,t) = u x (L,t) = 0 u(x,0) = f(x) (0 < x < L, 0 < t), (0 < t), (0 < x < L), in which we assume the ends of the rod are insulated. As before, assuming u(x,t) = X(x)T(t) yields the system X kx = 0, X (0) = X (L) = 0, T c 2 kt = 0. Note that the boundary conditions on X are not the same as in the Dirichlet condition case.

15 Solving for X Case 1: k = µ 2 > 0. We need to solve X µ 2 X = 0. The characteristic equation is r 2 µ 2 = 0 r = ±µ, which gives the general solution X = c 1 e µx +c 2 e µx. The boundary conditions tell us that 0 = X (0) = µc 1 µc 2, 0 = X (L) = µc 1 e µl µc 2 e µl, or in matrix form ( µ µ µe µl µe µl ) ( c1 c 2 ) = ( 0 0 ). Since the determinant is µ 2 (e µl e µl ) 0, we must have c 1 = c 2 = 0, and so X 0.

16 Case 2: k = 0. We need to solve X = 0. Integrating twice gives X = c 1 x +c 2. The boundary conditions give 0 = X (0) = X (L) = c 1. Taking c 2 = 1 we get the solution X = X 0 = 1. Case 3: k = µ 2 < 0. We need to solve X +µ 2 X = 0. The characteristic equation is r 2 +µ 2 = 0 r = ±iµ, which gives the general solution X = c 1 cos(µx)+c 2 sin(µx).

17 The boundary conditions yield 0 = X (0) = µc 1 sin0+µc 2 cos0 = µc 2 c 2 = 0, 0 = X (L) = µc 1 sin(µl)+µc 2 cos(µl) = µc 1 sin(µl). In order to have X 0, this shows that we need sin(µl) = 0 µl = nπ µ = µ n = nπ L (n Z). Taking c 1 = 1 we obtain Remarks: X = X n = cos(µ n x) (n N). We only need n > 0, since cosine is an even function. When n = 0 we get X 0 = cos0 = 1, which agrees with the k = 0 result.

18 Normal modes and superposition As before, for k = µ 2 n, we obtain T = T n = a n e λ2nt. We therefore have the normal modes u n (x,t) = X n (x)t n (t) = a n e λ2 n t cos(µ n x) (n N 0 ), where µ n = nπ/l and λ n = cµ n. The principle of superposition now gives the general solution u(x,t) = u 0 + u n = a 0 + n=1 a n e λ2 n t cos(µ n x) n=1 to the heat equation with (homogeneous) Neumann boundary conditions.

19 Initial conditions If we now impose our initial condition we find that f(x) = u(x,0) = a 0 + n=1 a n cos nπx L, which is simply the 2L-periodic cosine expansion of f(x). Hence a 0 = 1 L Remarks: L 0 f(x)dx, a n = 2 L L 0 f(x)cos nπx L dx, (n N). As before, if the cosine series of f(x) is already known, u(x,t) can be built by simply including exponential factors. Because of the exponential factors, lim t u(x,t) = a 0, which is the average initial temperature.

20 Example Solve the following heat problem: u t = 1 4 u xx, 0 < x < 1, 0 < t, u x (0,t) = u x (1,t) = 0, 0 < t, u(x,0) = 100x(1 x), 0 < x < 1. We have c = 1/2, L = 1 and f(x) = 100x(1 x). Therefore a 0 = x(1 x)dx = 50 3 a n = x(1 x)cosnπx dx = 200(1+( 1)n ) n 2 π 2, n 1.

21 Since λ n = cnπ/l = nπ/2, plugging everything into the general solution we get u(x,t) = π 2 n=1 (1+( 1) n ) n 2 e n2 π 2t/4 cosnπx. As in the case of Dirichlet boundary conditions, the exponential terms decay rapidly with t. We therefore have lim u(x,t) = 50 t 3.

22 Deriving the heat equation (Ideal) Assumptions: Rod is perfectly insulated with negligible thickness, i.e. heat only moves horizontally. No external heat sources or sinks. Rod material is uniform, i.e. has constant specific heat, s, and (linear) mass density, ρ. Recall that s = { amount of heat required to raise one unit of mass by one unit of temperature.

23 Consider a small segment of the rod at position x of length x. The thermal energy in this segment at time t is E(x,x + x,t) u(x,t)sρ x. Fourier s law of heat conduction states that the (rightward) heat flux at any point is K 0 u x (x,t), where K 0 is the thermal conductivity of the rod material. Remark: Fourier s law quantifies the notion that thermal energy moves from hot to cold.

24 Appealing to the law of conservation of energy, or t (u(x,t)sρ x) K 0 u x (x,t) +K }{{}}{{} 0 u x (x + x,t), }{{} heat flux through heat flux in heat flux in segment at left end at right end u t (x,t) K 0 sρ u x (x + x,t) u x (x,t). x Letting x 0 improves the approximation and leads to the one-dimensional heat equation where c 2 = K 0 sρ u t = c 2 u xx, is called the thermal diffusivity.