13.3 Heat Equation

INTRODUCTION

Consider a thin rod of length L with an initial temperature f(x) throughout and whose ends are held at temperature zero for all time t > 0. If the rod shown in FIGURE 13.3.1 satisfies the assumptions given on page 721, then the temperature u(x, t) in the rod is determined from the boundary-value problem

A thin finite rod of length L coincides with the x axis on the interval [0, L]. The cross section area at the beginning of the rod is indicated by an arrow and labeled u = 0. The cross section area in the middle of the rod is indicated by an arrow and labeled u(x, t). The cross section area at the end of the rod is indicated by an arrow and labeled u = 0. — FIGURE 13.3.1 Find the temperature u in a finite rod

(1)

(2)

(3)

In the discussion that follows next we show how to solve this BVP using the method of separation of variables introduced in Section 13.1.

Solution of the BVP

Using the product u(x, t) = X(x)T(t), and –λ as the separation constant, leads to

(4)

and X″ + λX = 0(5)

T′ + kλT = 0.(6)

Now the boundary conditions in (2) become

u(0, t) = X(0)T(t) = 0 and u(L, t) = X(L)T(t) = 0.

Since the last equalities must hold for all time t, we must have X(0) = 0 and X(L) = 0. These homogeneous boundary conditions together with the homogeneous ODE (5) constitute a regular Sturm–Liouville problem:

X″ + λX = 0, X(0) = 0, X(L) = 0.(7)

The solution of this BVP was discussed in detail in Example 2 of Section 3.9 and on page 701 of Section 12.5. In that example, we considered three possible cases for the parameter λ: zero, negative, and positive. The corresponding general solutions of the DEs are

(8)

(9)

(10)

Recall, when the boundary conditions X(0) = 0 and X(L) = 0 are applied to (8) and (9) these solutions yield only X(x) = 0 and so we are left with the unusable result u = 0. Applying the first boundary condition X(0) = 0 to the solution in (10) gives c₁ = 0. Therefore X(x) = c₂ sin αx. The second boundary condition X(L) = 0 now implies

X(L) = c₂ sin αL = 0.(11)

If c₂ = 0, then X = 0 so that u = 0. But (11) can be satisfied for c₂ ≠ 0 when sin αL = 0. This last equation implies that αL = nπ or α = nπ/L, where n = 1, 2, 3, … . Hence (7) possesses nontrivial solutions when λ_n = = n²π²/L², n = 1, 2, 3, … . The values λ_n and the corresponding solutions

(12)

are the eigenvalues and eigenfunctions, respectively, of the problem in (7).

The general solution of (6) is T(t) = , and so

(13)

where we have replaced the constant c₂c₃ by A_n. The products u_n(x, t) given in (13) satisfy the partial differential equation (1) as well as the boundary conditions (2) for each value of the positive integer n. However, in order for the functions in (13) to satisfy the initial condition (3), we would have to choose the coefficient A_n in such a manner that

u_n(x, 0) = f(x) = A_n sin x.(14)

In general, we would not expect condition (14) to be satisfied for an arbitrary, but reasonable, choice of f. Therefore we are forced to admit that u_n(x, t) is not a solution of the problem given in (1)–(3). Now by the superposition principle the function

(15)

must also, although formally, satisfy equation (1) and the conditions in (2). If we substitute t = 0 into (15), then

This last expression is recognized as the half-range expansion of f in a sine series. If we make the identification A_n = b_n, n = 1, 2, 3, … , it follows from (5) of Section 12.3 that

(16)

We conclude that a solution of the boundary-value problem described in (1), (2), and (3) is given by the infinite series

(17)

In the special case when the initial temperature is u(x, 0) = 100, L = π, and k = 1, you should verify that the coefficients (16) are given by

and that the series (17) is

(18)

Insulated Boundaries

Another boundary-value problem associated with the heat equation is

The boundary conditions in this problem indicate that both ends of the rod in Figure 13.3.1 are insulated. See the discussion on page 723 of Section 13.2. The major change in the solution of this boundary-value problem by separation of variables is that the analogue of (7) is now

In solving this type of Sturm–Liouville problem, you are cautioned to pay close attention to the case

Use of Computers

The solution u in (18) is a function of two variables and as such its graph is a surface in 3-space. We could use the 3D-plot application of a computer algebra system to approximate this surface by graphing partial sums S_n(x, t) over a rectangular region defined by 0 ≤ x ≤ π, 0 ≤ t ≤ T. Alternatively, with the aid of the 2D-plot application of a CAS we plot the solution u(x, t) on the x-interval [0, π] for increasing values of time t. See FIGURE 13.3.2(a). In Figure 13.3.2(b) the solution u(x, t) is graphed on the t-interval [0, 6] for increasing values of x(x = 0 is the left end and x = π/2 is the midpoint of the rod of length L = π). Both sets of graphs verify that which is apparent in (18)—namely, u(x, t) → 0 as t → ∞.

Graph (a) in an x, u coordinate plane labeled, u(x, t) graphed as a function of x for various fixed times, consists of five downward facing curves starting all from the point (0, 0) and ending at the estimated point (3, 0). From down to up the curves are as follows. The first curve is labeled t = 1.5 and peaks at an estimated point (1.5, 25). The second curve is labeled t = 1 and peaks at an estimated point (1.5, 40). The third curve is labeled t = 0.6 and peaks at an estimated point (1.5, 60). The fourth curve is labeled t = 0.35 and peaks at an estimated point (1.5, 85). The fifth curve is labeled t = 0.05 and peaks at an estimate point (0.75, 100) remains flat until the estimate point (2.25, 100) and then drops. A horizontal line at u = 100 is labeled t = 0. Graph (b) in a t, u coordinate plane labeled, (b) u(x, t) graphed as a function of t for various fixed positions consists of 5 convex curves starting all from the point (0, 100) and ending at the estimated point (6, 0). From down to up the curves are labeled as follows: x = pi over 12; x = pi over 6; pi over 4; pi over 2. At the end of the curves, above the point (6, 0) is labeled x = 0. — FIGURE 13.3.2 Graphs obtained using partial sums of (18)

13.3 Exercises Answers to selected odd-numbered problems begin on page ANS-34.

In Problems 1–6, solve the heat equation (1) subject to the given conditions.

In Problems 7 and 8, solve the heat equation (1) subject to the given conditions. [Hint: A solution of the boundary-value problem (1)–(3) need not be an infinite series.]

Suppose heat is lost from the lateral surface of a thin rod of length L into a surrounding medium at temperature zero. If the linear law of heat transfer applies, then the heat equation takes on the form

h a constant. Find the temperature u(x, t) if the initial temperature is f(x) throughout and the ends x = 0 and x = L are insulated. See FIGURE 13.3.3.

FIGURE 13.3.3 Rod in Problem 9
Solve Problem 9 if the ends x = 0 and x = L are held at temperature zero.
A thin rod (or wire) coinciding with the x-axis on the interval is bent into a circular ring by joining the two ends and See FIGURE 13.3.4. If the temperature u and the flux of heat are continuous at the point of juncture, then the temperature in the rod satisfies the boundary-value problem

See (15) in Section 12.5. Find the temperature u(x, t). [Hint: Think full Fourier series.]

FIGURE 13.3.4 Thin rod bent into a circular ring in Problem 11

Computer Lab Assignments

1. Solve the heat equation (1) subject to
2. Use the 3D-plot application of your CAS to graph the partial sum S₅(x, t) consisting of the first five nonzero terms of the solution in part (a) for 0 ≤ x ≤ 100, 0 ≤ t ≤ 200. Assume that k = 1.6352. Experiment with various three-dimensional viewing perspectives of the surface (called the ViewPoint option in Mathematica).

Discussion Problems

In Figure 13.3.2(b) we have the graphs of u(x, t) on the interval [0, 6] for x = 0, x = π/12, x = π/6, x = π/4, and x = π/2. Describe or sketch the graphs of u(x, t) on the same time interval but for the fixed values x = 3π/4, x = 5π/6, x = 11π/12, and x = π.