Continuations of Solutions to First Order ODEs

Table of Contents

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:

(1)

\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:

(2)

\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:

(3)

\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:

(4)

\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:

(5)

\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:

(6)

\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:

(7)

\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. $\blacksquare$

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Continuations of Solutions to First Order ODEs