Continuations of Solutions to First Order ODEs

# Continuations of Solutions to First Order ODEs

Definition: Let $\phi$ be a solution to the differential equation $x' = f(t, x)$ on an interval $J = (a, b)$. A Continuation of $\phi$ is an extension $\phi_0$ to an interval $J_0$ with $J \subseteq J_0$ which is also a solution to $x' = f(t, x)$. If no such continuation exists, $\phi$ is said to be Noncontinuable. |

Theorem 1: Let $\phi$ be a solution to the IVP $x' = f(t, x)$ with $x(\tau) = \xi$ on the interval $J = (a, b)$ with $f \in C(D, \mathbb{R})$ and let $f$ be bounded on $D$. Define $\displaystyle{\lim_{t \to a^+} \phi(t) = \phi(a^+)}$ and $\displaystyle{\lim_{t \to b^-} \phi(t) = \phi(b^-)}$. Then:a) $\phi(a^+)$ and $\phi(b^-)$ both exist.b) If $(a, \phi(a+)) \in D$ then $\phi$ can be continued to the left.c) If $(b, \phi(b^-)) \in D$ then $\phi$ can be continued to the right. |

**Proof of a)**Since $f$ is bounded on $D$ there exists an $M > 0$ such that $| f(t, x) | \leq M$ for all $(t, x) \in D$. Take $\tau \in J$ and set $\phi (\tau) = \xi$. Let $t, u$ be such that $a < t < u < b$. Then the solution $\phi$ of $x' = f(t, x)$ satisfies the IVP $x' = f(t, x)$ with $x(\tau) = \xi$ and so:

\begin{align} \quad \phi (u) = \xi + \int_{\tau}^{u} f(s, \phi(s)) \: ds \quad \mathrm{and} \quad \phi (t) = \xi + \int_{\tau}^{t} f(s, \phi(s)) \: ds \end{align}

- Therefore:

\begin{align} \quad | \phi (u) - \phi (t) | &= \biggr \lvert \left ( \xi + \int_{\tau}^{u} f(s, \phi(s)) \: ds \right ) - \left ( \xi + \int_{\tau}^{t} f(s, \phi(s)) \: ds \right ) \biggr \rvert \\ &= \biggr \lvert \int_{t}^{u} f(s, \phi(s)) \: ds \biggr \rvert \\ & \leq \int_{t}^{u} | f(s, \phi(s)) | \: ds \\ & \leq \int_{t}^{u} M \: ds \\ & \leq M(u - t) \end{align}

- Hence:

\begin{align} \quad | \phi (u) - \phi(t) | \leq M |u - t| \end{align}

- Let $(t_m)_{m=1}^{\infty}$ be a sequence of numbers in $(a, \tau)$ such that $(t_m)_{m=1}^{\infty}$ converges to $a$. Then $|t_m - t_n|$ can be made arbitrarily small which means from the inequality above that $| \phi(t_m) - \phi(t_n) |$ can be made arbitrarily small by taking $m, n \geq N$ for some $N \in \mathbb{N}$. So $(\phi(t_m))_{m=1}^{\infty}$ is a Cauchy sequence, so $(\phi(t_m))_{m=1}^{\infty}$ converges to $\displaystyle{\phi(a^+) = \lim_{t \to a^+} \phi(t)}$, so $\phi(a^+)$ exists.

- Now let $(t_m)_{m=1}^{\infty}$ be a sequence of numbers in $(\tau, b)$ such that $(t_m)_{m=1}^{\infty}$ converges to $b$. Then $|t_m - t_n|$ can be made arbitrarily small which means from the inequality above that $| \phi(t_m) - \phi(t_n) |$ can be made arbitrarily small by taking $m, n \geq N$ for some $N \in \mathbb{N}$. So $(\phi(t_m))_{m=1}^{\infty}$ is a Cauchy sequence, so $(\phi(t_m))_{m=1}^{\infty}$ converges to $\displaystyle{\phi(b^-) = \lim_{t \to b^-} \phi(t)}$, so $\phi(b^-)$ exists.

**Proof of b)**Suppose t hat $(a, \phi(a^+)) \in D$. By The Local Existence Theorem for Solutions to Initial Value Problems of First Order ODEs, there exists a solution $\phi_0$ to the IVP $x' = f(t, x)$ with $x(a) = \phi(a^+)$ defined on $[a - c, a]$ for some $c > 0$.

- On $a < t$, we define $\phi_0(t) = \phi(t)$. Then consider the function extension:

\begin{align} \quad \phi_0(t) = \left\{\begin{matrix} \phi(a^+) + \int_{a}^t f(s, \phi_0(s)) \: ds & \mathrm{if} \: t \in (a - c, a] \\ \xi + \int_{\tau}^t f(s, \phi_0(s)) \: ds & \mathrm{if} \: t \in (a, b) \end{matrix}\right. \end{align}

- Then $\phi_0$ satisfies the IVP for $t \in (a, b)$, and for $t \in (a - c, a]$ we have that:

\begin{align} \quad \phi_0(t) &= \phi(a^+) + \int_a^t f(s, \phi_0(s)) \: ds \\ &= \lim_{t \to a^+} \left ( \xi + \int_{\tau}^{t} f(s, \phi_0(s)) \: ds \right ) + \int_a^t f(s, \phi_0(s)) \: ds \\ &= \xi + \int_{\tau}^{a} f(s, \phi_0(s)) \:ds + \int_a^t f(s, \phi_(s)) \: ds \\ &= \xi + \int_{\tau}^{t} f(s, \phi_0(s)) \: ds \end{align}

- So $\phi_0$ is a solution to the IVP $x' = f(t, x)$ with $x(\tau) = \xi$ on the interval $(a - c, b)$, so $\phi_0$ is a continuation of the solution $\phi$ to the left. $\blacksquare$

**Proof of c)**Suppose that $(b, \phi(b^-)) \in D$. By The Local Existence Theorem for Solutions to Initial Value Problems of First Order ODEs, there exists a solution $\phi_0$ to the IVP $x' = f(t, x)$ with $x(b) = \phi(b^-)$ defined on $[b, b + c]$ for some $c > 0$.

- On $t < b$, we define $\phi_0(t) = \phi(t)$. Then consider the function extension:

\begin{align} \quad \phi_0(t) = \left\{\begin{matrix} \xi + \int_{\tau}^{t} f(s, \phi_0(s)) \: ds & \: \mathrm{if} \: t \in (a, b)\\ \phi(b^-) + \int_{b}^{t} f(s, \phi_0(s)) \: ds & \mathrm{if} t \in [b, b+c) & \end{matrix}\right. \end{align}

- Then $\phi_0$ satisfies the IVP for $t \in (a, b)$, and for $t \in [b, b + c)$ we have that:

\begin{align} \quad \phi_0(t) &= \phi(b^-) + \int_b^t f(s, \phi_0(s)) \: ds \\ &= \lim_{t \to b^-} \left ( \xi + \int_{\tau}^{t} f(s, \phi_0(s)) \: ds \right ) + \int_b^t f(s, \phi_0(s)) \: ds \\ &= \xi + \int_{\tau}^{b} f(s, \phi_0(s)) \: ds + \int_b^t f(s, \phi_0(s)) \: ds \\ &= \xi + \int_{\tau}^{t} f(s, \phi_0(s)) \: ds \end{align}

- So $\phi_0$ is a solution to the IVP $x' = f(t, x)$ with $x(\tau) = \xi$ on the interval $(a, b + c)$, so $\phi_0$ is a continuation of the solution $\phi$ to the right. $\blacksquare$