# Preface

This section presents some results about existence and uniqueness of initial value problems for systems of differential equations as well as for single higher order differential equations because the latter can be transfered to the former.

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# Existence and Uniqueness

Our main interest in this section is the initial value problem for the system of differential equations:

$\dot{\bf x} ( t ) \equiv \frac{{\text d}{\bf x}}{{\text d}t} = {\bf f} \left( t, {\bf x} ( t) \right) , \qquad {\bf x} ( t_0 ) = {\bf x}_0 ,$
where x(t) is n-vector function defined on some interval including the initial point. There are known many statements that guarantee existence and uniqueness of a solution to this initial value problem. They all include some modification of Lipschitz continuity of the slope vector function f(x,t).

Theorem: : Suppose f is a vector-valued function of n+1 variables (t, y1, ... , yn) defined for (t,y) on a set Ω of the form

$\left\vert t - t_0 \right\vert \le a, \qquad \| {\bf y} - {\bf y}_0 \| \le b \qquad (a,b > 0),$
or of the form
$\left\vert t - t_0 \right\vert \le a, \qquad \| {\bf y} \| < \infty \qquad (a> 0).$
If ∂f/ ∂yk (k = 1,2, ... , n) exists, is continuous on Ω, and there is a constant K > 0 such that
$\left\| \frac{\partial {\bf f}}{\partial y_k} \, \left( x, {\bf y} \right) \right\| \le \mbox{K} \qquad (k=1,2,\ldots n),$
for all (t,y) in Ω, then f satisfies a Lipschitz condition $$\| {\bf f}(t, {\bf y}) - {\bf f}(t, {\bf z}) \| \le \mbox{K} \,\| {\bf y} - {\bf z} \|$$ on Ω with Lipschitz constant K.    ▣

Theorem: : Suppose f is a continuous vector-valued function of n+1 variables (t, y1, ... , yn) defined for (t,y) on a set Ω of the form

$\left\vert t - t_0 \right\vert \le a, \qquad \| {\bf y} - {\bf y}_0 \| \le b \qquad (a,b > 0).$
Then the initial value problem
${\bf y}' (t) = {\bf f} (t, {\bf y}) , \qquad {\bf y}(x_0 ) = {\bf y}_0 ,$
has a solution.

If, in addition, f satisfies a Lipschitz condition and bounded $$\| {\bf f}(t, {\bf y}) \| \le M$$ on Ω, then the initial avlue problem has a unique solution on the interval $$|t - t_0 | \le \min \left\{ a, b/M \right\} .$$    ▣

The idea to proof this theorem is to apply a fixed point theorem. The main obstakle in its application is the derivative operator, which is unbounded one. To get rid of it, we apply the inverse operator $$\texttt{D}^{-1} f(t) = f(t_0 ) + \int_{t_0}^t f(s)\,{\text d}s$$ to both sides of the differential equation. Since the inverse operator $$\texttt{D}^{-1}$$ is a bounded operator expressed through a definite integral, we can reduce the given initial value problem to the integral equation. Therefore, the first step in establishing this statement is to rewrite the given initial value problem in equivalent form as Volterra integral equation:

${\bf x} ( t ) ={\bf x}_0 + \int_{t_0}^t {\bf f} ({\bf x} (s) ,s ) \, {\text d}s .$

The equivalence follows from the Fundamental Theorem of Calculus. It suffices to find a continuous function $${\bf x}(t)$$ that satisfies the integral equation within the interval $$t_0 -h < t < t_0 +h ,$$ for some small value $$h$$ since the right hand-side (integral) will be continuously differentiable in $$t .$$

The proof of existence and uniqueness theorem for the above initial value problem is based on technique known as Picard iteration to construct the required solution:

${\bf x}_{n+1} ( t ) ={\bf x}_0 + \int_{t_0}^t {\bf f} ({\bf x}_n (s) ,s ) \,{\text d}s , \qquad n=0,1,2,\ldots .$ The initial approximation is chosen to be the initial value (constant): $${\bf x}_0 (t) \equiv {\bf x}_0 .$$ It can be shown that this iterative sequence converges uniformly for $$t \in [t_0 -h, t_0 +h ]$$ when h is small enough.

Theorem: : Suppose $$a_0 (x), a_1 (x) , \ldots , a_{n-1} (x)$$ and f(x) are continuous real-valued functions on an interval $$(a,b) \ni x_0 .$$ Then for any choice of (initial conditions) $$y_0 (x), y_1 (x) , \ldots , y_{n-1} (x) ,$$ there exists a unique solution on the whole interval (a,b) to the initial value problem

$y^{(n)} (x) + a_{n-1} (x)\, y^{(n-1)} (x) + \cdots + a_0 (x)\, y(x) = f(x) , \qquad y^{(i)} (x_0 ) = y_i , \quad i=0,1, \ldots , n-1;$
for all x ∈ (a,b).    ▣

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