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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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Now let us return to Eq. (7),
+1({) = f f^,(s)] ds.
0
Allowing to approach to on both sides, we obtain
() = lim f f[s, (s)] ds. (26)
^J 0
We would like to interchange the operations of integrating and taking the limit on the right side of Eq. (26) so as to obtain
() = f lim f[s,(s)] ds. (27)
Jo ^
In general, such an interchange is not permissible (see Problem 14, for example), but once again the fact that the sequence {(t)} converges uniformly is sufficient to allow us to take the limiting operation inside the integral sign. Next, we wish to take the limit inside the function f, which would give
() = f f[s, lim (s)] ds (28)
Jo ^
and hence
() = f f[s^(s)] ds. (29)
0
The statement that lim f [s,(s)] = f [s, lim (s)] is equivalent to the statement that f is continuous in its second variable, which is known by hypothesis. Hence Eq. (29) is valid and the function satisfies the integral equation (3). Therefore is also a solution of the initial value problem (2).
4. Are there other solutions of the integral equation (3) beside y = ()? To show
the uniqueness of the solution y = (), we can proceed much as in the example.
2.8 The Existence and Uniqueness Theorem
113
First, assume the existence of another solution y = (). It is then possible to show (see Problem 19) that the difference () ty(t) satisfies the inequality
({) f(t)l < A ms) f(s)l ds (30)
0
for 0 < t < h and a suitable positive number A. From this point the argument is identical to that given in the example, and we conclude that there is no solution of the initial value problem (2) other than the one generated by the method of successive approximations.
PROBLEMS In each of Problems 1 and 2 transform the given initial value problem into an equivalent problem with the initial point at the origin.
1. dy/dt = t2 + y2, y(1) = 2 2. dy/dt = 1 y3, y(1) = 3
In each of Problems 3 through 6 let () = 0 and use the method of successive approximations to solve the given initial value problem.
(a) Determine (t) for an arbitrary value of .
(b) Plot (t) for = 14. Observe whether the iterates appear to be converging.
(c) Express lim (t) = () in terms of elementary functions; that is, solve the given
^
initial value problem.
(d) Plot () (t)| for = 1,..., 4. For each of (), .4 (t) estimate the interval in which it is a reasonably good approximation to the actual solution.
> 3. y = 2(y + 1), y(0) = 0 > 4. y = y 1, y(0) = 0
> 5. y = y/2 + t, y(0) = 0 > 6. = y + 1 t, y(0) = 0
In each of Problems 7 and 8 let 0() = 0 and use the method of successive approximations to solve the given initial value problem.
(a) Determine (t) for an arbitrary value of .
(b) Plot (t) for = 1,..., 4. Observe whether the iterates appear to be converging.
> 7. y = ty + 1, y(0) = 0 > 8. y= t2y t, y(0) = 0
In each of Problems 9 and 10 let 0() = 0 and use the method of successive approximations to approximate the solution of the given initial value problem.
(a) Calculate (),..., 3().
(b) Plot (), ... ,3() and observe whether the iterates appear to be converging.
> 9. y = t2 + , y(0) = 0 > 10. y= 1 y3, y(0) = 0
In each of Problems 11 and 12 let () = 0 and use the method of successive approximations to approximate the solution of the given initial value problem.
(a) Calculate (), ..., 4(0, or (if necessary) Taylor approximations to these iterates. Keep terms up to order six.
(b) Plot the functions you found in part (a) and observe whether they appear to be converging.
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