The temperature in a rod of unit length in which there is heat transfer from its right boundary into a surrounding medium kept at a constant temperature zero is determined from

Solve for u(x, t).

SOLUTION

Proceeding exactly as we did in Section 13.3, with u(x, t) = X(x)T(t) and –λ as the separation constant, we find the separated ODEs and boundary conditions to be, respectively,

X″ + λX = 0(1)

T′ + kλT = 0(2)

X(0) = 0andX′(1) = –hX(1).(3)

Equation (1) along with the homogeneous boundary conditions (3) comprise the regular Sturm–Liouville problem:

(4)

Except for the presence of the symbol h, the BVP in (4) is essentially the problem solved in Example 2 of Section 12.5. As in that example, (4) possesses nontrivial solutions only in the case λ = α² > 0, α > 0. The general solution of the DE in (4) is X(x) = c₁ cos αx + c₂ sin αx. The first boundary condition in (4) immediately gives c₁ = 0. Applying the second boundary condition in (4) to X(x) = c₂ sin αx yields

(5)

Because the graphs of y = tan x and y = –x/h, h > 0, have an infinite number of points of intersection for x > 0 (Figure 12.5.1 illustrates the case h = 1), the last equation in (5) has an infinite number of roots. Of course, these roots depend on the value of h. If the consecutive positive roots are denoted α_n, n = 1, 2, 3, … , then the eigenvalues of the problem are λ_n = , and the corresponding eigenfunctions are X(x) = c₂ sin α_n x, n = 1, 2, 3, … . The solution of the first-order DE (2) is T(t) = c₃ and so

Now at t = 0, u(x, 0) = 1, 0 < x < 1, so that

(6)

The series in (6) is not a Fourier sine series; rather, it is an expansion of u(x, 0) = 1 in terms of the orthogonal functions arising from the Sturm–Liouville problem (4). It follows that the set of eigenfunctions {sinα_n x}, n = 1, 2, 3, … , where the αs are defined by tan α = –α/h is orthogonal with respect to the weight function p(x) = 1 on the interval [0, 1]. With f(x) = 1 and ø_n(x) = sin α_nx, it follows from (8) of Section 12.1 that the coefficients A_n in (6) are

(7)

To evaluate the square norm of each of the eigenfunctions we use a trigonometric identity:

(8)

With the aid of the double angle formula sin 2α_n = 2 sin α_n cos α_n and the first equation in (5) in the form α_n cos α_n = –h sin α_n, we can simplify (8) to

Also

Consequently (7) becomes

Finally, a solution of the boundary-value problem is

≡

The twist angle θ(x, t) of a torsionally vibrating shaft of unit length is determined from

See FIGURE 13.7.1. The boundary condition at x = 1 is called a free-end condition. Solve for θ(x, t).

SOLUTION

Proceeding as in Section 13.4 with θ(x, t) = X(x)T(t) and using –λ once again as the separation constant, the separated equations and boundary conditions are

(9)

(10)

(11)

Equation (9) together with the homogeneous boundary conditions in (11),

(12)

is a regular Sturm–Liouville problem. You are encouraged to verify that for λ = 0 and for λ = –α², α > 0, the only solution of (12) is X = 0. For λ = α² > 0, α > 0, the boundary conditions X(0) = 0 and X′(1) = 0 applied to the general solution X(x) = c₁ cos αx + c₂ sin αx give, in turn, c₁ = 0 and c₂ cos α = 0. Since cos α is zero only when α is an odd integer multiple of π/2 we write α_n = (2n – 1)π/2. The eigenvalues of (12) are λ_n = = (2n – 1)²π²/4, and the corresponding eigenfunctions are X(x) = c₂ sin α_n x = c₂ sinπx, n = 1, 2, 3, … .

Since the rod is released from rest, the initial condition θ_t(x, 0) = 0 translates into X(x) T′(0) = 0 or T′(0) = 0. When applied to the general solution T(t) = c₃ cos aα_nt + c₄ sin aα_nt of the second-order DE (10), T′(0) = 0 implies c₄ = 0 leaving T(t) = c₃ cos aα_nt = c₃ cos aπt. Therefore,

In order to satisfy the remaining initial condition we form the superposition of the θ_n,

(13)

When t = 0, we must have, for 0 < x < 1,

(14)

As in Example 1, the set of eigenfunctions , n = 1, 2, 3, … , is orthogonal with respect to the weight function p(x) = 1 on the interval [0, 1]. Even though the trigonometric series in (14) looks more like a Fourier series than (6), it is not a Fourier sine series because the argument of the sine function is not an integer multiple of πx/L (here L = 1). The series is again an orthogonal series expansion or generalized Fourier series. Hence from (8) of Section 12.1 the coefficients A_n in (14) are given by

Carrying out the two integrations, we arrive at

The twist angle is then

.≡