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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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[p(t) y]' = p(t)g(t). (5)
Since both pi and g are continuous, the function pg is integrable, and Eq. (3) follows from Eq. (5). Further, the integral of pg is differentiable, so y as given by Eq. (3) exists and is differentiable throughout the interval a < t < 3 .By substituting the expression for y in Eq. (3) into either Eq. (1) or Eq. (5), one can easily verify that this expression satisfies the differential equation throughout the interval a < t < 3. Finally, the initial
66
Chapter 2. First Order Differential Equations
Theorem 2.4.2
condition (2) determines the constant c uniquely, so there is only one solution of the initial value problem, thus completing the proof.
Equation (4) determines the integrating factor g(t) only up to a multiplicative factor that depends on the lower limit of integration. If we choose this lower limit to be t0, then
/x(t) = exp / p(s) ds, (6)
Jt0
and it follows that g(t0) = 1. Using the integrating factor given by Eq. (6) and choosing the lower limit of integration in Eq. (3) also to be t0, we obtain the general solution of Eq. (1) in the form
t
y =
ft )g(s) ds + c
g(t )
To satisfy the initial condition (2) we must choose c = y0. Thus the solution of the initial value problem (1), (2) is
// lAs)g(s) ds + y0 y = ------------77)--------- (7)
where ) is given by Eq. (6).
Turning now to nonlinear differential equations, we must replace Theorem 2.4.1 by a more general theorem, such as the following.
Let the functions f and d f/d y be continuous in some rectangle a < t < 3, y < y < S containing the point (t0, y0 ). Then, in some interval t0 h < t < t0 + h contained in a < t < 3, there is a unique solution y = $(t ) of the initial value problem
y= f(t> y)> y(t0) = (8)
Observe that the hypotheses in Theorem 2.4.2 reduce to those in Theorem 2.4.1 if the differential equation is linear. For then f (t, y) = p(t)y + g(t) and d f (t, y)/dy = p(t), so the continuity of f and d f/dy is equivalent to the continuity of p and g in this case. The proof of Theorem 2.4.1 was comparatively simple because it could be based on the expression (3) that gives the solution of an arbitrary linear equation. There is no corresponding expression for the solution of the differential equation (8), so the proof of Theorem 2.4.2 is much more difficult. It is discussed to some extent in Section 2.8 and in greater depth in more advanced books on differential equations.
Here we note that the conditions stated in Theorem 2.4.2 are sufficient to guarantee the existence of a unique solution of the initial value problem (8) in some interval t0 h < t < t0 + h, but they are not necessary. That is, the conclusion remains true under slightly weaker hypotheses about the function f. In fact, the existence of a solution (but not its uniqueness) can be established on the basis of the continuity of f alone.
We now consider some examples.
2.4 Differences Between Linear and Nonlinear Equations
67
EXAMPLE
1
EXAMPLE
2
Use Theorem 2.4.1 to find an interval in which the initial value problem
ty + 2 y = 4t2, (9)
y(1) = 2 (10)
has a unique solution.
Rewriting Eq. (9) in the standard form (1), we have
/ + (2/1) y = 4t, (11)
so p(t) = 2/1 and g(t) = 4t. Thus, for this equation, g is continuous for all t, while
p is continuous only for t < 0 or for t > 0. The interval t > 0 contains the initial point; consequently, Theorem 2.4.1 guarantees that the problem (9), (10) has a unique solution on the interval 0 < t < ?. In Example 4 of Section 2.1 we found the solution of this initial value problem to be
2 1
y = t2 + -, t > 0. (12)
t
Now suppose that the initial condition (10) is changed to y(-1) = 2. Then Theorem 2.4.1 asserts the existence of a unique solution for t < 0. As you can readily verify, the solution is again given by Eq. (12), but now on the interval ? < t < 0.
Apply Theorem 2.4.2 to the initial value problem
dy 3x2 + 4x + 2
y(0) = -1. (13)
dx 2( y - 1)
Note that Theorem 2.4.1 is not applicable to this problem since the differential equation is nonlinear. To apply Theorem 2.4.2, observe that
3x2 + 4x + 2 d f 3x2 + 4x + 2
f (x, y) = , (x, y) =-=.
' 2(y - 1) dyy 2(y - 1)2
Thus both these functions are continuous everywhere except on the line y = 1. Consequently, a rectangle can be drawn about the initial point (0, -1) in which both f and d f/dy are continuous. Therefore Theorem 2.4.2 guarantees that the initial value problem has a unique solution in some interval about x = 0. However, even though the rectangle can be stretched infinitely far in both the positive and negative x directions, this does not necessarily mean that the solution exists for all x. Indeed, the initial value problem (13) was solved in Example 2 of Section 2.2 and the solution exists only for x > -2.
Now suppose we change the initial condition to y(0) = 1. The initial point now lies on the line y = 1 so no rectangle can be drawn about it within which f and d f/dy are continuous. Consequently, Theorem 2.4.2 says nothing about possible solutions of this modified problem. However, if we separate the variables and integrate, as in Section 2.2, we find that
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