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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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1. 7' + 2 7 = ---x,
2. 7' + 2 7 = ---x,
3. 7' + 2 7 = --- x,
4. 7' + 2 7 = --- x,
5. 7' + 2 7 = -1 -
In each of Problems 1 through 5 solve the given problem by means of an eigenfunction expansion. /(0) = 0, 7(1) = 0
y(0) = 0, 7(1) = 0; see Section 11.2, Problem 7.
/(0) = 0, /(1) = 0; see Section 11.2, Problem 3.
(0) = 0, (I) + /(1) = 0; see Section 11.2, Problem 11.
- 2x|, 7(0) = 0, 7(1) = 0
In each of Problems 6 through 9 determine a formal eigenfunction series expansion for the solution of the given problem. Assume that f satisfies the conditions of Theorem 11.3.1. State the values of for which the solution exists.
6. 7' + 7 =-f(x), 7(0) = , /(1) = 0
7. 7' + 7 =-f(x), 7(0) = 0, 7(1) = 0
8. 7' + 7 =-f(x), 7(0) = 0, 7(1) = 0
9. 7' + 7 =-f(x), 7(0) = 0, 7(1) + 7(1) = 0
In each of Problems 10 through 13 determine whether there is any value of the constant a for which the problem has a solution. Find the solution for each such value.
10. 7' + n27 = a + x, 7(0) = 0, 7(1) = 0
11. 7' + 4n27 = a + x, 7(0) = 0, 7(1) = 0
12. 7' + n27 = a, 7(0) = 0, 7(1) = 0
13. 7' + n27 = a cos nx, 7(0) = 0, 7(1) = 0
14. Let 1,.. ., be the normalized eigenfunctions of the differential equation (3) subject
TO
to the boundary conditions (2). If ^ cn(x) converges to f (x), where f (x) = 0 for each
n=1
x in 0 < x < 1, show that cn = 0 for each n.
Hint: Multiply by r(x)(x), integrate, and use the orthogonality property of the eigenfunctions.
15. Let L be a second order linear differential operator. Show that the solution 7 = (x) of the problem
L [7] = f (x), a17(0) + a27(0) = a, b17(1) + b27 (1) = can be written as 7 = u + v, where u = 1(x) and v = 2 (x) are solutions of the problems
L [u] = 0,
a1 u(0) + a2u!(0) = a, b1 u(1) + b2u'(1) =
and
L [v] = f(x), a1v(0) + a2v'(0) = 0, b1v(1) + b2v' (1) = 0,
respectively.
652
Chapter 11. Boundary Value Problems and Sturm-Liouville Theory
16. Show that the problem
y" + n 2 y = n 2 x, y(0) = 1, y(1) = 0
has the solution
y = cj sin nx + cos nx + x.
Also show that this solution cannot be obtained by splitting the problem as suggested in Problem 15, since neither of the two subsidiary problems can be solved in this case.
17. Consider the problem
y" + p(x)y + q (x)y = 0, y(0) = a, y(1) = b.
Let y = u + v, where v is any twice differentiable function satisfying the boundary conditions (but not necessarily the differential equation). Show that u is a solution of the problem
u" + p(x)d + q (x)u = g(x), u(0) = 0, u(1) = 0,
where g(x) = [v" + p(x)v' + q(x)v], and is known once v is chosen. Thus nonhomogeneities can be transferred from the boundary conditions to the differential equation. Find a function v for this problem.
18. Using the method of Problem 17, transform the problem
y" + 2y = 2 - 4x, y(0) = 1, y(1) + y (1) = -2
into a new problem in which the boundary conditions are homogeneous. Solve the latter problem by reference to Example 1 of the text.
In each of Problems 19 through 22 use eigenfunction expansions to find the solution of the given boundary value problem.
19. u( = uxx - x, u(0, t) = 0, ux(1, t) = 0, u(x, 0) = sin(nx/2);
see Problem 2.
20. ut = uxx + e-t, ux(0, t) = 0, ux(1, t) + u(1, t) = 0, u(x, 0) = 1 - x;
see Section 11.2, Problems 10 and 12.
21. ut = uxx + 1 -|1 - 2x|, u(0, t) = 0, u(1, t) = 0, u(x, 0) = 0;
see Problem 5.
22. ut = uxx + e~l(1 - x), u(0, t) = 0, ux(1, t) = 0, u(x, 0) = 0;
see Section 11.2, Problems 6 and 7.
23. Consider the boundary value problem
r(x)ut = [ p(x)ux]x - q (x)u + F(x),
u(0, t) = Tj, u(1, t) = T2, u(x, 0) = f (x).
(a) Let v (x) be a solution of the problem
[p(.x)v']' - q(x)v = -F(x), v(0) = Tj, v(1) = T2.
If w(x, t) = u(x, t) - v(x), find the boundary value problem satisfied by w. Note that this problem can be solved by the method of this section.
(b) Generalize the procedure of part (a) to the case where u satisfies the boundary conditions
ux(0, t) - h1u(0, t) = T1, ux(1, t) + h2u(1, t) = T2.
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