# ϵ-Approximate Solutions to Initial Values Problems of First Order ODEs

Definition: An $\epsilon$-Approximation Solution to the initial value problem $x' = f(t, x)$ with $x(\tau) = \xi$ on $J = (a, b)$ is a piecewise real-valued function $\phi \in C^1 (J, \mathbb{R})$ such that $| \phi'(t) - f(t, \phi(t)) | < \epsilon$ for all $t \in J$ where $\phi'(t)$ exists, and $\phi(\tau) = \xi$. |

*In other words, an $\epsilon$-approximation solution to the IVP $x' = f(t, x)$ with $x(\tau) = \xi$ on $J$ is a piecewise continuously differentiable real-valued function $\phi$ on $J$ such that $\phi'(t)$ is almost equal to $f(t, \phi(t))$ for all $t \in J$ for which $\phi'(t)$ exists ($\epsilon$-close), and where $\phi$ satisfies the initial condition, $\phi(\tau) = \xi$.*

Consider the initial value problem $x' = f(t, x)$ with $x(\tau) = \xi$. Then $f \in C(D, \mathbb{R})$ (where $D \subseteq \mathbb{R}^n$ is a nonempty, open, connected subset of $\mathbb{R}^2$. Consider the following set as well:

(1)Then $S$ is a closed rectangle, and for $a$ and $b$ chosen sufficiently small, $S \subseteq D$. We assume that $a$ and $b$ are chosen in this manner. Then $f$ is a continuous function defined on a closed and bounded subset of $\mathbb{R}^2$, so $f$ is bounded on $S$, i.e., there exists an $M > 0$ such that for all $(t, x) \in S$ we have that:

(2)Define the real number $c \in \mathbb{R}$ by:

(3)We use the numbers $M$ and $c$ in the following theorem:

Theorem 1 (Existence of $\epsilon$-Approximate Solutions): Let $x' = f(t, x)$ ($f \in C(D, \mathbb{R})$) with $x(\tau) = \xi$ be an initial value problem. Then for all $\epsilon > 0$ there exists an $\epsilon$-approximate solution to this IVP on the interval $[\tau - c, \tau + c]$. |

**Proof:**Let $\epsilon > 0$ be given and consider the interval $[\tau, \tau + c]$.

- Let $S \subseteq D$ be a rectangle centered at $(\tau, \xi)$ given by:

- And as mentioned above since $f$ is continuous on the closed and bounded set $S$, $f$ is bounded on $S$ and let $M > 0$ be such that $| f(t, x) | \leq M$ for all $(t, x) \in S$, and define:

- Now, since $f$ is continuous on the compact set $S$ we have that $f$ is uniformly continuous on $S$. So for the given $\epsilon > 0$ there exists a $\delta > 0$ such that if $|t - s| \leq \delta$ and $|x - y| \leq \delta$ then

- Let $P = \{ \tau = t_0, t_1, ..., t_m = \tau + c \}$ be a partition, i.e., $\tau = t_0 < t_1 < ... < t_m = \tau + c$ that divides the interval $[\tau, \tau + c]$ into $m$ subintervals $[t_0, t_1]$, $[t_1, t_2]$, …, $[t_{m-1}, t_m]$ such that for all $j \in \{ 1, 2, ..., m \}$:

- In other words, the partition $P$ divides $[\tau, \tau + c]$ into $m$ subintervals of length less than or equal to $\displaystyle{\min \left \{ \delta, \frac{\delta}{M} \right \}}$.

- We define the function $\phi$ as a piecewise union of line segments:

- Then $\phi$ is a piecewise continuously differentiable function on $[t_0, t_m] = [\tau, \tau + c]$, and $\phi'(t)$ is defined on all of $[t_0, t_m]$ except possibly at the points $t_1, ..., t_{m-1}$. Furthermore, for each interval $[t_{j-1}, t_j]$ where $j \in \{ 1, 2, ..., n \}$ we have that $\phi'(t) = f(t_{j-1}, \phi(t_{j-1}))$ and so:

- Since $| t - t_j | < \delta$ and $\displaystyle{| \phi(t) - \phi(t_j) | \leq M|t - t_j| \leq M \frac{\delta}{M} = \delta}$, from uniform continuity of $f$ we have that:

- Furthermore, $\phi'(t_0) = \phi'(\tau) = \xi$. So $\phi$ is an $\epsilon$-approximate solution on $[\tau, \tau + c]$ to the initial value problem $x' = f(t, x)$ with $x(\tau) = \xi$.

- A similar argument shows the existence of an $\epsilon$-approximate solution on $[\tau - c, \tau]$. $\blacksquare$