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A.1 Existence of Solutions

The approach we employ is the method of successive approximations, which was developed by the French mathematician Emile Picard (1856–1941). This method is based on the fact that the function y(x) satisfies the initial value problem in (1) on the open interval I containing $x = a$ $x = a$ if and only if it satisfies the integral equation

y (x) = b + \int_{a}^{x} f (t, y (t)) d t

$y (x) = b + \int_{a}^{x} f (t, y (t)) d t$ (4)

for all x in I. In particular, if y(x) satisfies Eq. (4), then clearly $y (a) = b,$ $y (a) = b,$ and differentiation of both sides in (4)—using the fundamental theorem of calculus—yields the differential equation $y^{′} (x) = f (x, y (x))$ $y^{′} (x) = f (x, y (x))$ .

To attempt to solve Eq. (4), we begin with the initial function

y_{0} (x) \equiv b

$y_{0} (x) \equiv b$ (5)

and then define iteratively a sequence $y_{1}, y_{2}, y_{3}, \dots$ $y_{1}, y_{2}, y_{3}, \dots$ of functions that we hope will converge to the solution. Specifically, we let

y_{1} (x) = b + \int_{a}^{x} f (t, y_{0} (t)) d t and y_{2} (x) = b + \int_{a}^{x} f (t, y_{1} (t)) d t .

$y_{1} (x) = b + \int_{a}^{x} f (t, y_{0} (t)) d t and y_{2} (x) = b + \int_{a}^{x} f (t, y_{1} (t)) d t .$ (6)

In general, $y_{n + 1}$ $y_{n + 1}$ is obtained by substitution of $y_{n}$ $y_{n}$ for y in the right-hand side in Eq. (4):

y_{n + 1} (x) = b + \int_{a}^{x} f (t, y_{n} (t)) d t .

$y_{n + 1} (x) = b + \int_{a}^{x} f (t, y_{n} (t)) d t .$ (7)

Suppose we know that each of these functions ${y_{n} (x)}_{0}^{\infty}$ ${y_{n} (x)}_{0}^{\infty}$ is defined on some open interval (the same for each n) containing $x = a$ $x = a$ and that the limit

y (x) = \lim_{n \to \infty} y_{n} (x)

$y (x) = \lim_{n \to \infty} y_{n} (x)$ (8)

exists at each point of this interval. Then it will follow that

\begin{array}{rcl} y (x) & = & \lim_{n \to \infty} y_{n + 1} (x) = \lim_{n \to \infty} [b + \int_{a}^{x} f (t, y_{n} (t)) d t] \\ = & b + \lim_{n \to \infty} \int_{a}^{x} f (t, y_{n} (t)) d t \end{array}

$\begin{array}{rcl} y (x) & = & \lim_{n \to \infty} y_{n + 1} (x) = \lim_{n \to \infty} [b + \int_{a}^{x} f (t, y_{n} (t)) d t] \\ = & b + \lim_{n \to \infty} \int_{a}^{x} f (t, y_{n} (t)) d t \end{array}$ (9)

\begin{array}{rcl} = & b + \int_{a}^{x} f (t, \lim_{n \to \infty} y_{n} (t)) d t \end{array}

$\begin{array}{rcl} = & b + \int_{a}^{x} f (t, \lim_{n \to \infty} y_{n} (t)) d t \end{array}$ (10)

and, hence, that

y (x) = b + \int_{a}^{x} f (t, y (t)) d t,

$y (x) = b + \int_{a}^{x} f (t, y (t)) d t,$

provided that we can validate the interchange of limit operations involved in passing from (9) to (10). It is therefore reasonable to expect that, under favorable conditions, the sequence ${y_{n} (x)}$ ${y_{n} (x)}$ defined iteratively in Eqs. (5) and (7) will converge to a solution y(x) of the integral equation in (4), and hence to a solution of the original initial value problem in (1).

Example 1

To apply the method of successive approximations to the initial value problem

\frac{d y}{d x} = y, y (0) = 1,

$\frac{d y}{d x} = y, y (0) = 1,$ (11)

we write Eqs. (5) and (7), thereby obtaining

y_{0} (x) \equiv 1, y_{n + 1} (x) = 1 + \int_{0}^{x} y_{n} (t) d t .

$y_{0} (x) \equiv 1, y_{n + 1} (x) = 1 + \int_{0}^{x} y_{n} (t) d t .$ (12)

The iteration formula in (12) yields

\begin{array}{rcl} y_{1} (x) & = & 1 + \int_{0}^{x} 1 d t = 1 + x, \\ y_{2} (x) & = & 1 + \int_{0}^{x} (1 + t) d t = 1 + x + \frac{1}{2} x^{2}, \\ y_{3} (x) & = & 1 + \int_{0}^{x} (1 + t + \frac{1}{2} t^{2}) d t = 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3}, \end{array}

$\begin{array}{rcl} y_{1} (x) & = & 1 + \int_{0}^{x} 1 d t = 1 + x, \\ y_{2} (x) & = & 1 + \int_{0}^{x} (1 + t) d t = 1 + x + \frac{1}{2} x^{2}, \\ y_{3} (x) & = & 1 + \int_{0}^{x} (1 + t + \frac{1}{2} t^{2}) d t = 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3}, \end{array}$

and

\begin{array}{rcl} y_{4} (x) & = & 1 + \int_{0}^{x} (1 + t + \frac{1}{2} t^{2} + \frac{1}{6} t^{3}) d t \\ = & 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{1}{24} x^{4} . \end{array}

$\begin{array}{rcl} y_{4} (x) & = & 1 + \int_{0}^{x} (1 + t + \frac{1}{2} t^{2} + \frac{1}{6} t^{3}) d t \\ = & 1 + x + \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + \frac{1}{24} x^{4} . \end{array}$

It is clear that we are generating the sequence of partial sums of a power series solution; indeed, we immediately recognize the series as that of $y (x) = e^{x} .$ $y (x) = e^{x} .$ There is no difficulty in demonstrating that the exponential function is indeed the solution of the initial value problem in (11); moreover, a diligent student can verify (by using a proof by induction on n) that $y_{n} (x),$ $y_{n} (x),$ obtained in the aforementioned manner, is indeed the nth partial sum for the Taylor series with center zero for $y (x) = e^{x}$ $y (x) = e^{x}$ .

Example 2

To apply the method of successive approximations to the initial value problem

\frac{d y}{d x} = 4 x y, y (0) = 3,

$\frac{d y}{d x} = 4 x y, y (0) = 3,$ (13)

we write Eqs. (5) and (7) as in Example 1. Now we obtain

y_{0} (x) \equiv 3, y_{n + 1} (x) = 3 + \int_{0}^{x} 4 t y_{n} (t) d t .

$y_{0} (x) \equiv 3, y_{n + 1} (x) = 3 + \int_{0}^{x} 4 t y_{n} (t) d t .$ (14)

The iteration formula in (14) yields

\begin{array}{rcl} y_{1} (x) & = & 3 + \int_{0}^{x} (4 t) (3) d t = 3 + 6 x^{2}, \\ y_{2} (x) & = & 3 + \int_{0}^{x} (4 t) (3 + 6 t^{2}) d t = 3 + 6 x^{2} + 6 x^{4}, \\ y_{3} (x) & = & 3 + \int_{0}^{x} (4 t) (3 + 6 t^{2} + 6 t^{4}) d t = 3 + 6 x^{2} + 6 x^{4} + 4 x^{6}, \end{array}

$\begin{array}{rcl} y_{1} (x) & = & 3 + \int_{0}^{x} (4 t) (3) d t = 3 + 6 x^{2}, \\ y_{2} (x) & = & 3 + \int_{0}^{x} (4 t) (3 + 6 t^{2}) d t = 3 + 6 x^{2} + 6 x^{4}, \\ y_{3} (x) & = & 3 + \int_{0}^{x} (4 t) (3 + 6 t^{2} + 6 t^{4}) d t = 3 + 6 x^{2} + 6 x^{4} + 4 x^{6}, \end{array}$

and

\begin{array}{rcl} y_{4} (x) & = & 3 + \int_{0}^{x} (4 t) (3 + 6 t^{2} + 6 t^{4} + 4 t^{6}) d t \\ = & 3 + 6 x^{2} + 6 x^{4} + 4 x^{6} + 2 x^{8} . \end{array}

$\begin{array}{rcl} y_{4} (x) & = & 3 + \int_{0}^{x} (4 t) (3 + 6 t^{2} + 6 t^{4} + 4 t^{6}) d t \\ = & 3 + 6 x^{2} + 6 x^{4} + 4 x^{6} + 2 x^{8} . \end{array}$

It is again clear that we are generating partial sums of a power series solution. It is not quite so obvious what function has such a power series representation, but the initial value problem in (13) is readily solved by separation of variables:

\begin{array}{rcl} y (x) & = & 3 \exp (2 x^{2}) = 3 \sum_{n = 0}^{\infty} \frac{{(2 x^{2})}^{n}}{n!} \\ = & 3 + 6 x^{2} + 6 x^{4} + 4 x^{6} + 2 x^{8} + \frac{4}{5} x^{10} + \dots . \end{array}

$\begin{array}{rcl} y (x) & = & 3 \exp (2 x^{2}) = 3 \sum_{n = 0}^{\infty} \frac{{(2 x^{2})}^{n}}{n!} \\ = & 3 + 6 x^{2} + 6 x^{4} + 4 x^{6} + 2 x^{8} + \frac{4}{5} x^{10} + \dots . \end{array}$

In some cases, it may be necessary to compute a much larger number of terms, either in order to identify the solution or to use a partial sum of its series with large subscript to approximate the solution accurately for x near its initial value. Fortunately, computer algebra systems such as Maple and Mathematica can perform the symbolic integrations (as opposed to numerical integrations) of the sort in Examples 1 and 2. If necessary, you could generate the first hundred terms in Example 2 in a matter of minutes.

In general, of course, we apply Picard’s method because we cannot find a solution by elementary methods. Suppose that we have produced a large number of terms of what we believe to be the correct power series expansion of the solution. We must have conditions under which the sequence ${y_{n} (x)}$ ${y_{n} (x)}$ provided by the method of successive approximations is guaranteed in advance to converge to a solution. It is just as convenient to discuss the initial value problem

\frac{d x}{d t} = f (x, t), x (a) = b

$\frac{d x}{d t} = f (x, t), x (a) = b$ (15)

for a system of m first-order equations, where

x = [\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ ⋮ \\ x_{m} \end{array}], f = [\begin{array}{c} f_{1} \\ f_{2} \\ f_{3} \\ ⋮ \\ f_{m} \end{array}], and b = [\begin{array}{c} b_{1} \\ b_{2} \\ b_{3} \\ ⋮ \\ b_{m} \end{array}] .

$x = [\begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ ⋮ \\ x_{m} \end{array}], f = [\begin{array}{c} f_{1} \\ f_{2} \\ f_{3} \\ ⋮ \\ f_{m} \end{array}], and b = [\begin{array}{c} b_{1} \\ b_{2} \\ b_{3} \\ ⋮ \\ b_{m} \end{array}] .$

It turns out that with the aid of this vector notation (which we introduced in Section 7.1), most results concerning a single [scalar] equation $x ′ = f (x, t)$ $x ′ = f (x, t)$ can be generalized readily to analogous results for a system of m first-order equations, as abbreviated in (15). Consequently, the effort of using vector notation is amply justified by the generality it provides.

The method of successive approximations for the system in (15) calls for us to compute the sequence ${x_{n} (t)}_{0}^{\infty}$ ${x_{n} (t)}_{0}^{\infty}$ of vector-valued functions of t,

x_{n} (t) = [\begin{array}{c} x_{1 n} (t) \\ x_{2 n} (t) \\ x_{3 n} (t) \\ ⋮ \\ x_{m n} (t) \end{array}],

$x_{n} (t) = [\begin{array}{c} x_{1 n} (t) \\ x_{2 n} (t) \\ x_{3 n} (t) \\ ⋮ \\ x_{m n} (t) \end{array}],$

defined iteratively by

x_{0} (a) \equiv b, x_{n + 1} (t) = b + \int_{a}^{t} f (x_{n} (s), s) d s .

$x_{0} (a) \equiv b, x_{n + 1} (t) = b + \int_{a}^{t} f (x_{n} (s), s) d s .$ (16)

Recall that vector-valued functions are integrated componentwise.

Example 3

Consider the m-dimensional initial value problem

\frac{d x}{d t} = Ax, x (0) = b

$\frac{d x}{d t} = Ax, x (0) = b$ (17)

for a homogeneous linear system with $m \times m$ $m \times m$ constant coefficient matrix A. The equations in (16) take the form

x_{0} (t) = b, x_{n + 1} = b + \int_{0}^{x} A x_{n} (s) d s .

$x_{0} (t) = b, x_{n + 1} = b + \int_{0}^{x} A x_{n} (s) d s .$ (18)

Thus

\begin{array}{rcl} x_{1} (t) & = & b + \int_{0}^{t} A b d s = b + A b t = (I + A t) b, \\ x_{2} (t) & = & b + \int_{0}^{t} A (b + A b s) d s = b + A b t + \frac{1}{2} A^{2} b t^{2} = (I + A t + \frac{1}{2} A^{2} t^{2}) b, \end{array}

$\begin{array}{rcl} x_{1} (t) & = & b + \int_{0}^{t} A b d s = b + A b t = (I + A t) b, \\ x_{2} (t) & = & b + \int_{0}^{t} A (b + A b s) d s = b + A b t + \frac{1}{2} A^{2} b t^{2} = (I + A t + \frac{1}{2} A^{2} t^{2}) b, \end{array}$

and

x_{3} (t) = b + \int_{0}^{t} A (b + A b s + \frac{1}{2} A^{2} b s^{2}) d s = (I + A t + \frac{1}{2} A^{2} t^{2} + \frac{1}{6} A^{3} t^{3}) b .

$x_{3} (t) = b + \int_{0}^{t} A (b + A b s + \frac{1}{2} A^{2} b s^{2}) d s = (I + A t + \frac{1}{2} A^{2} t^{2} + \frac{1}{6} A^{3} t^{3}) b .$

We have therefore obtained the first several partial sums of the exponential series solution

x (t) = e^{A t} b = (\sum_{n = 0}^{\infty} \frac{{(A t)}^{n}}{n!}) b

$x (t) = e^{A t} b = (\sum_{n = 0}^{\infty} \frac{{(A t)}^{n}}{n!}) b$ (19)

of (17), which was derived earlier in Section 8.1.

The key to establishing convergence in the method of successive approximations is an appropriate condition on the rate at which f(x, t) changes when x varies but t is held fixed. If R is a region in $(m + 1)$ $(m + 1)$ -dimensional (x, t)-space, then the function f(x, t) is said to be Lipschitz continuous on R if there exists a constant $k > 0$ $k > 0$ such that

| f (x_{1}, t) - f (x_{2}, t) | ≦ k | x_{1} - x_{2} |

$| f (x_{1}, t) - f (x_{2}, t) | ≦ k | x_{1} - x_{2} |$ (20)

if $(x_{1}, t)$ $(x_{1}, t)$ and $(x_{2}, t)$ $(x_{2}, t)$ are points of R. Recall that the norm of an m-dimensional point or vector x is defined to be

| x | = \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + \dots + x_{m}^{2}} .

$| x | = \sqrt{x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + \dots + x_{m}^{2}} .$ (21)

Then $| x_{1} - x_{2} |$ $| x_{1} - x_{2} |$ is simply the Euclidean distance between the points $x_{1}$ $x_{1}$ and $x_{2}$ $x_{2}$ .

Example 4

Let $f (x, t) = x^{2} \exp (- t^{2}) \sin t$ $f (x, t) = x^{2} \exp (- t^{2}) \sin t$ and let R be the strip $0 ≦ x ≦ 2$ $0 ≦ x ≦ 2$ in the xy-plane. If $(x_{1}, t)$ $(x_{1}, t)$ and $(x_{2}, t)$ $(x_{2}, t)$ are both points of R, then

| f (x_{1}, t) - f (x_{2}, t) | = | \exp (- t^{2}) \sin t | \cdot | x_{1} + x_{2} | \cdot | x_{1} - x_{2} | ≦ 4 | x_{1} - x_{2} |,

$| f (x_{1}, t) - f (x_{2}, t) | = | \exp (- t^{2}) \sin t | \cdot | x_{1} + x_{2} | \cdot | x_{1} - x_{2} | ≦ 4 | x_{1} - x_{2} |,$

because $| \exp (- t^{2}) \sin t | ≦ 1$ $| \exp (- t^{2}) \sin t | ≦ 1$ for all t and $| x_{1} + x_{2} | ≦ 4$ $| x_{1} + x_{2} | ≦ 4$ if $x_{1}$ $x_{1}$ and $x_{2}$ $x_{2}$ are both in the interval [0, 2]. Thus f satisfies the Lipschitz condition in (20) with $k = 4$ $k = 4$ and is therefore Lipschitz continuous in the strip R.

Example 5

Let $f (x, t) = t \sqrt{x}$ $f (x, t) = t \sqrt{x}$ on the rectangle R consisting of the points (x, t) in the xt-plane for which $0 ≦ x ≦ 1$ $0 ≦ x ≦ 1$ and $0 ≦ t ≦ 1.$ $0 ≦ t ≦ 1.$ Then, taking $x_{1} = x, x_{2} = 0,$ $x_{1} = x, x_{2} = 0,$ and $t = 1,$ $t = 1,$ we find that

| f (x, 1) - f (0, 1) | = \sqrt{x} = \frac{1}{\sqrt{x}} | x - 0 | .

$| f (x, 1) - f (0, 1) | = \sqrt{x} = \frac{1}{\sqrt{x}} | x - 0 | .$

Because $x^{- 1/2} \to + \infty$ $x^{- 1/2} \to + \infty$ as $x \to 0^{+},$ $x \to 0^{+},$ we see that the Lipschitz condition in (20) cannot be satisfied by any (finite) constant $k > 0.$ $k > 0.$ Thus the function f, though obviously continuous on R, is not Lipschitz continuous on R.

Suppose, however, that the function f(x, t) has a continuous partial derivative $f_{x} (x, t)$ $f_{x} (x, t)$ on the closed rectangle R in the xt-plane, and denote by k the maximum value of $| f_{x} (x, t) |$ $| f_{x} (x, t) |$ on R. Then the mean-value theorem of differential calculus yields

| f (x_{1}, t) - f (x_{2}, t) | = | f_{x} (\bar{x}, t) \cdot (x_{1} - x_{2}) |

$| f (x_{1}, t) - f (x_{2}, t) | = | f_{x} (\bar{x}, t) \cdot (x_{1} - x_{2}) |$

for some $\bar{x}$ $\bar{x}$ in $(x_{1}, x_{2}),$ $(x_{1}, x_{2}),$ so it follows that

| f (x_{1}, t) - f (x_{2}, t) | ≦ k | x_{1} - x_{2} |

$| f (x_{1}, t) - f (x_{2}, t) | ≦ k | x_{1} - x_{2} |$

because $| f_{x} (\bar{x}, t) | ≦ k .$ $| f_{x} (\bar{x}, t) | ≦ k .$ Thus a continuously differentiable function f(x, t) defined on a closed rectangle is Lipschitz continuous there. More generally, the multivariable mean value theorem of advanced calculus can be used similarly to prove that a vector-valued function f(x, t) with continuously differentiable component functions on a closed rectangular region R in (x, t)-space is Lipschitz continuous on R.

Example 6

The function $f (x, t) = x^{2}$ $f (x, t) = x^{2}$ is Lipschitz continuous on any closed [bounded] region in the xt-plane. But consider this function on the infinite strip R consisting of the points (x, t) for which $0 ≦ t ≦ 1$ $0 ≦ t ≦ 1$ and x is arbitrary. Then

| f (x_{1}, t) - f (x_{2}, t) | = | x_{1}^{2} - x_{2}^{2} | = | x_{1} + x_{2} | \cdot | x_{1} - x_{2} | .

$| f (x_{1}, t) - f (x_{2}, t) | = | x_{1}^{2} - x_{2}^{2} | = | x_{1} + x_{2} | \cdot | x_{1} - x_{2} | .$

Because $| x_{1} + x_{2} |$ $| x_{1} + x_{2} |$ can be made arbitrarily large, it follows that f is not Lipschitz continuous on the infinite strip R.

If I is an interval on the t-axis, then the set of all points (x, t) with t in I is an infinite strip or slab in $(m + 1)$ $(m + 1)$ -space (as indicated in Fig. A.1). Example 6 shows that Lipschitz continuity of f(x, t) on such an infinite slab is a very strong condition. Nevertheless, the existence of a solution of the initial value problem

\frac{d x}{d t} = f (x, t), x (a) = b

$\frac{d x}{d t} = f (x, t), x (a) = b$ (15)

under the hypothesis of Lipschitz continuity of f in such a slab is of considerable importance.

FIGURE A.1.

An infinite slab in $(m + 1)$ $(m + 1)$ -space.

Proof

We want to show that the sequence ${x_{n} (t)}_{0}^{\infty}$ ${x_{n} (t)}_{0}^{\infty}$ of successive approximations determined iteratively by

x_{0} (a) = b, x_{n + 1} (t) = b + \int_{0}^{t} f (x_{n} (s), s) d s

$x_{0} (a) = b, x_{n + 1} (t) = b + \int_{0}^{t} f (x_{n} (s), s) d s$ (16)

converges to a solution x(t) of (15). We see that each of these functions in turn is continuous on I, as each is an (indefinite) integral of a continuous function.

We may assume that $a = 0,$ $a = 0,$ because the transformation $t \to t + a$ $t \to t + a$ converts (15) into an equivalent problem with initial point $t = 0.$ $t = 0.$ Also, we will consider only the portion $t ≧ 0$ $t ≧ 0$ of the interval I; the details for the case $t ≦ 0$ $t ≦ 0$ are very similar.

The main part of the proof consists in showing that if [0, T] is a closed (and bounded) interval contained in I, then the sequence ${x_{n} (t)}$ ${x_{n} (t)}$ converges uniformly on [0, T] to a limit function x(t). This means that, given $ϵ > 0,$ $ϵ > 0,$ there exists an integer N such that

| x_{n} (t) - x (t) | < ϵ

$| x_{n} (t) - x (t) | < ϵ$ (22)

for all $n ≧ N$ $n ≧ N$ and all t in [0, T]. For ordinary (perhaps nonuniform) convergence the integer N, for which (22) holds for all $n ≧ N,$ $n ≧ N,$ might depend on t, with no single value of N working for all t in I. Once this uniform convergence of the sequence ${x_{n} (t)}$ ${x_{n} (t)}$ has been established, the following conclusions will follow from standard theorems of advanced calculus [see pages 620–622 of A. E. Taylor and W. R. Mann, Advanced Calculus, 3rd ed. (New York: John Wiley, 1983)].

The limit function x(t) is continuous on [0, T].
If N is so chosen that the inequality in (22) holds for $n ≧ N,$ $n ≧ N,$ then the Lipschitz continuity of f implies that

$| f (x_{n} (t), t) - f (x (t), t) | ≦ k | x_{n} (t) - x (t) | < k ϵ$ $| f (x_{n} (t), t) - f (x (t), t) | ≦ k | x_{n} (t) - x (t) | < k ϵ$

for all t in [0, T] and $n ≧ N,$ $n ≧ N,$ so it follows that the sequence ${f (x_{n} (t), t)}_{0}^{\infty}$ ${f (x_{n} (t), t)}_{0}^{\infty}$ converges uniformly to f(x(t), t) on [0, T].
But a uniformly convergent sequence or series can be integrated termwise, so it follows that, on taking limits in the iterative formula in (16),

$\begin{array}{rcl} x (t) & = & \lim_{n \to \infty} x_{n + 1} (t) = b + \lim_{n \to \infty} \int_{0}^{t} f (x_{n} (s), s) d s \\ = & b + \int_{0}^{t} \lim_{n \to \infty} f (x_{n} (s), s) d s; \end{array}$ $\begin{array}{rcl} x (t) & = & \lim_{n \to \infty} x_{n + 1} (t) = b + \lim_{n \to \infty} \int_{0}^{t} f (x_{n} (s), s) d s \\ = & b + \int_{0}^{t} \lim_{n \to \infty} f (x_{n} (s), s) d s; \end{array}$

thus

$x (t) = b + \int_{0}^{t} f (x (s), s) d s .$ $x (t) = b + \int_{0}^{t} f (x (s), s) d s .$ (23)
Because the function x(t) is continuous on [0, T], the integral equation in (23) [analogous to the 1-dimensional case in (4)] implies that $x ′ (t) = f (x (t), t)$ $x ′ (t) = f (x (t), t)$ on [0, T]. If this is true on every closed subinterval of the open interval I, however, then it is true on the entire interval I as well.

It therefore remains only to prove that the sequence ${x_{n} (t)}_{0}^{\infty}$ ${x_{n} (t)}_{0}^{\infty}$ converges uniformly on the closed interval [0, T]. Let M be the maximum value of $| f (b, t) |$ $| f (b, t) |$ for t in [0, T]. Then

| x_{1} (t) - x_{0} (t) | = | \int_{0}^{t} f (x_{0} (s), s) d s | ≦ \int_{0}^{t} | f (b, s) | d s ≦ M t .

$| x_{1} (t) - x_{0} (t) | = | \int_{0}^{t} f (x_{0} (s), s) d s | ≦ \int_{0}^{t} | f (b, s) | d s ≦ M t .$ (24)

Next,

| x_{2} (t) - x_{1} (t) | = | \int_{0}^{t} [f (x_{1} (s), s) - f (x_{0} (s), s)] d s | ≦ k \int_{0}^{t} | x_{1} (s) - x_{0} (s) | d s,

$| x_{2} (t) - x_{1} (t) | = | \int_{0}^{t} [f (x_{1} (s), s) - f (x_{0} (s), s)] d s | ≦ k \int_{0}^{t} | x_{1} (s) - x_{0} (s) | d s,$

and hence

| x_{2} (t) - x_{1} (t) | ≦ k \int_{0}^{t} M s d s = \frac{1}{2} k M t^{2} .

$| x_{2} (t) - x_{1} (t) | ≦ k \int_{0}^{t} M s d s = \frac{1}{2} k M t^{2} .$ (25)

We now proceed by induction. Assume that

| x_{n} (t) - x_{n - 1} (t) | ≦ \frac{M}{k} \cdot \frac{{(k t)}^{n}}{n!} .

$| x_{n} (t) - x_{n - 1} (t) | ≦ \frac{M}{k} \cdot \frac{{(k t)}^{n}}{n!} .$ (26)

It then follows that

\begin{array}{rcl} | x_{n + 1} (t) - x_{n} (t) | & = & | \int_{0}^{t} f (x_{n} (s), s) - f (x_{n - 1} (s), s)] d s | \\ ≦ & k \int_{0}^{t} | x_{n} (s) - x_{n - 1} (s) | d s; \end{array}

$\begin{array}{rcl} | x_{n + 1} (t) - x_{n} (t) | & = & | \int_{0}^{t} f (x_{n} (s), s) - f (x_{n - 1} (s), s)] d s | \\ ≦ & k \int_{0}^{t} | x_{n} (s) - x_{n - 1} (s) | d s; \end{array}$

consequently,

| x_{n + 1} (t) - x_{n} (t) | ≦ k \int_{0}^{t} \frac{M}{k} \cdot \frac{{(k s)}^{n}}{n!} d s .

$| x_{n + 1} (t) - x_{n} (t) | ≦ k \int_{0}^{t} \frac{M}{k} \cdot \frac{{(k s)}^{n}}{n!} d s .$

It follows, upon evaluation of this integral, that

| x_{n + 1} (t) - x_{n} (t) | ≦ \frac{M}{k} \cdot \frac{{(k t)}^{n + 1}}{(n + 1)!} .

$| x_{n + 1} (t) - x_{n} (t) | ≦ \frac{M}{k} \cdot \frac{{(k t)}^{n + 1}}{(n + 1)!} .$

Thus (26) holds on [0, T] for all $n ≧ 1$ $n ≧ 1$ .

Hence the terms of the infinite series

x_{0} (t) + \sum_{n = 1}^{\infty} [x_{n} (t) - x_{n - 1} (t)]

$x_{0} (t) + \sum_{n = 1}^{\infty} [x_{n} (t) - x_{n - 1} (t)]$ (27)

are dominated (in magnitude on the interval [0, T]) by the terms of the convergent series

\sum_{n = 1}^{\infty} \frac{M}{k} \cdot \frac{{(k T)}^{n + 1}}{(n + 1)!} = \frac{M}{k} (e^{k T} - 1),

$\sum_{n = 1}^{\infty} \frac{M}{k} \cdot \frac{{(k T)}^{n + 1}}{(n + 1)!} = \frac{M}{k} (e^{k T} - 1),$ (28)

which is a series of positive constants. It therefore follows (from the Weierstrass M-test on pages 618–619 of Taylor and Mann) that the series in (27) converges uniformly on [0, T]. But the sequence of partial sums of this series is, however, simply our original sequence ${x_{n} (t)}_{0}^{\infty}$ ${x_{n} (t)}_{0}^{\infty}$ of successive approximations, so the proof of Theorem 1 is finally complete.

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Table of Contents for A.1 Existence of Solutions

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Table of Contents for
A.1 Existence of Solutions