Answers or Hints

Answers or Hints
11.1. The problem y ^′ = y^2/3, 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.

3. 
   Let y₁(x) = 1 + x, y₂(x) = 1/(1 − x) and use Corollary 11.2.
   
4. 
   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.