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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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(17)
624
Chapter 11. Boundary Value Problems and Sturm Liouville Theory
EXAMPLE
1
The following example involves one boundary condition of the form (15) and is therefore more complicated than the problems in Section 10.1.
Find the eigenvalues and the corresponding eigenfunctions of the boundary value problem
/ + ky = 0, (18)
y(0) = 0, /(1) + y(1) = 0. (19)
One place where this problem occurs is in the heat conduction problem in a bar of unit length. The boundary condition at x = 0 corresponds to a zero temperature there. The boundary condition at x = 1 corresponds to a rate of heat flow that is proportional to the temperature there, and units are chosen so that the constant of proportionality is 1 (see Appendix A of Chapter 10).
The solution of the differential equation may have one of several forms, depending on k, so it is necessary to consider several cases. First, if k = 0, the general solution of the differential equation is
y = c1 x + c2. (20)
The two boundary conditions require that
C2 = 0, 2c1 + C2 = 0, (21)
respectively. The only solution of Eqs. (21) is c1 = c2 = 0, so the boundary value problem has no nontrivial solution in this case. Hence k = 0 is not an eigenvalue.
If k > 0, then the general solution of the differential equation (18) is
y = c1 sin v/k x + c2 cos v/kx, (22)
where yk > 0. The boundary condition at x = 0 requires that c2 = 0; from the boundary condition at x = 1 we then obtain the equation
q (sinVk + Vk cos yk) = 0.
For a nontrivial solution y we must have q = 0, and thus k must satisfy
sinVk + yk cos yk = 0. (23)
Note that if k is such that cos vk = 0, then sin yk = 0, and Eq. (23) is not satisfied. Hence we may assume that cos yk = 0; dividing Eq. (23) by cos vk, we obtain
yk = — tan yk. (24)
The solutions of Eq. (24) can be determined numerically. They can also be found approximately by sketching the graphs of f (Vk) = */k and g(y/k) = — tanyk for Vk > 0 on the same set of axes, and identifying the points of intersection of the two curves (see Figure 11.1.1). The point yk = 0 is specifically excluded from this argument because the solution (22) is valid only for yk = 0. Despite the fact that the curves intersect there, k = 0 is not an eigenvalue, as we have already shown. The first three positive solutions of Eq. (24) are ^Jkx = 2.029, ^k^ = 4.913, and ^k3 = 7.979.
As can be seen from Figure 11.1.1, the other roots are given with reasonable accuracy
11.1 The Occurrence of Two-Point Boundary Value Problems 625
by vX = (2n — 1)n/2 for n = 4, 5,..., the precision of this estimate improving as n increases. Hence the eigenvalues are
X1 = 4.116, X2 = 24.14,
2 2 (25)
X3 = 63.66, Xn = (2n — 1) n /4 for n = 4, 5,....
Finally, since C2 = 0, the eigenfunction corresponding to the eigenvalue Xn is
(x,Xn) = kn sin yXx; n = 1, 2,..., (26)
where the constant kn remains arbitrary.
Next consider X < 0. In this case it is convenient to let X = —fi so that fi > 0. Then Eq. (14) becomes
/ — = 0, (27)
and its general solution is
y = q sinh yl x + C2 cosh x, (28)
where „fi > 0. Proceeding as in the previous case, we find that i must satisfy the equation
«Ji = — tanh «Ji. (29)
From Figure 11.1.2 it is clear that the graphs of f (+/i) = -Jl and g(+Ji) = — tanhyl intersect only at the origin. Hence there are no positive values of „fi that satisfy Eq. (29), and hence the boundary value problem (18), (19) has no negative eigenvalues.
FIGURE 11.1.1 Graphical solution of \[X = — tan \pk.
626
Chapter 11. Boundary Value Problems and Sturm Liouville Theory
Finally, it is necessary to consider the possibility that k may be complex. It is possible to show by direct calculation that the problem (18), (19) has no complex eigenvalues. However, in Section 11.2 we consider in more detail a large class of problems that includes this example. One of the things we show there is that every problem in this class has only real eigenvalues. Therefore we omit the discussion of the nonexistence of complex eigenvalues here. Thus we conclude that all the eigenvalues and eigenfunctions of the problem (18), (19) are given by Eqs. (25) and (26).
PROBLEMS In each of Problems 1 through 6 state whether the given boundary value problem is homogeneous
I or nonhomogeneous.
1. / + 4y = 0, y(-1) = 0, y(1) = 0
2. [(1 + x2)/]' + 4 y = 0, y(0) = 0, y(1) = 1
3. /' + 4y = sinx, y(0) = 0, y(1) = 0
4. —y" + x2y = Xy, y (0) - y(0) = 0, y/(1) + y(1) = 0
5. —[(1 + x2)/]' = Xy + 1, y(—1) = 0, y(1) = 0
6. — y" = X(1 + x2)y, y (0) = 0, y(1) + 3 y(1) = 0
In each of Problems 7 through 10 determine the form of the eigenfunctions and the determinantal equation satisfied by the nonzero eigenvalues. Determine whether X = 0 is an eigenvalue, and
find approximate values for X: and X2, the nonzero eigenvalues of smallest absolute value.
Estimate Xn for large values of n. Assume that all eigenvalues are real.
7. y + Xy = 0, 8. /' + Xy = 0,
y(0) = 0, y(n) + y(n) = 0 y(0) = 0, y(1) + /(1) = 0
9. y' + Xy = 0, 10. y — Xy = 0,
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