Answers or Hints



Yüklə 393,22 Kb.
səhifə1/2
tarix13.09.2023
ölçüsü393,22 Kb.
#143033
  1   2
Документ Microsoft Word (2)



Answers or Hints
11.1. The problem y = y2/3, y(0) = 0 has solutions y1(x) = x3/27, y2(x) ≡ 0 in [0,∞). For these functions equalities hold everywhere in Theorem

    1. However, .

    2. Let y1(x) = y0, y2(x) = 1 + ǫ, ǫ > 0 and use Theorem 11.1 and

Corollary 11.2.

    1. Let y1(x) = 1 + x, y2(x) = 1/(1 − x) and use Corollary 11.2.

    2. Define f(x,y) = (f1(x,y) + f2(x,y))/2 so that f(x,y2) and y1(x0) < y2(x0). Now use Theorem 11.1.



Lecture 12
Continuous Dependence on Initial Conditions
The initial value problem (7.1) describes a model of a physical problem in which often some parameters such as lengths, masses, temperatures, etc., are involved. The values of these parameters can be measured only up to certain degree of accuracy. Thus, in (7.1) the initial condition (x0,y0) as well as the function f(x,y) may be subject to some errors either by necessity or for convenience. Hence, it is important to know how the solution of (7.1) changes when (x0,y0) and f(x,y) are slightly altered. We shall answer this question quantitatively in the following theorem.
Theorem 12.1. Let the following conditions be satisfied:

  1. f(x,y) is continuous and bounded by M in a domain D containing the points (x0,y0) and (x1,y1).

  2. f(x,y) satisfies a uniform Lipschitz condition (7.3) in D.

  3. g(x,y) is continuous and bounded by M1 in D.

  4. y(x) and z(x), the solutions of the initial value problems (7.1) and

z = f(x,z) + g(x,z), z(x1) = y1,
respectively, exist in an interval J containing x0 and x1. Then for all x J, the following inequality holds:

Proof. From Theorem 7.1 for all x J it follows that

R.P. Agarwal and D. O’Regan, An Introduction to Ordinary Differential Equations, 84 doi: 10.1007/978-0-387-71276-5_12, © Springer Science + Business Media, LLC 2008
and hence, we find
(12.2)
Now taking absolute values on both sides of (12.2) and using the hypotheses, we get
Inequality (12.3) is exactly the same as that considered in Corollary
7.6 with c0 = |y0 y1| + (M + M1)|x1 x0|, c1 = M1, c2 = L and u(x) = |y(x) − z(x)|, and hence the inequality (12.1) follows.
From the inequality (12.1) it is apparent that the difference between the solutions y(x) and z(x) in the interval J is small provided the changes in the initial point (x0,y0) as well as in the function f(x,y) do not exceed prescribed amounts. Thus, the statement “if the function f(x,y) and the initial point (x0,y0) vary continuously, then the solutions of (7.1) vary continuously” holds. It is also clear that the solution z(x) of the initial value problem z = f(x,z) + g(x,z), z(x1) = y1 need not be unique in J.

Yüklə 393,22 Kb.

Dostları ilə paylaş:
  1   2




Verilənlər bazası müəlliflik hüququ ilə müdafiə olunur ©azkurs.org 2024
rəhbərliyinə müraciət

gir | qeydiyyatdan keç
    Ana səhifə


yükləyin