Extending solutions to Differential Equations

Consider the differential equation \[ \dot x = f(t, x), \] where the right hand side of the equation is a function $f(t,x)$ that is defined on some open subset $\cO\subset\R^2$.

Assumptions about $f$

We assume that the function $f$ and its partial derivative $f_x$ both are continuous functions on $\cO$. By a theorem from real analysis it follows that for every closed and bounded rectangle $R = \{(t, x) : c\le t\le d, a\le x \le b\}$ there exist numbers $M$ and $L$ such that \[ |f(t, x)| \le M, \text{ and } |f_x(t, x)|\le L \] for all $(t, x)\in R$. The constants $M$ and $L$ may depend on which particular rectangle $R$ was chosen. Boundedness of $f_x$ implies that $f$ satisfies the Lipschitz condition on the rectangle $R$: namely, the Mean Value Theorem implies that for any $t\in[c, d]$ and any two $x, y \in [a,b]$ there is a $c$ such that \[ \left| f(t, x) - f(t,y)\right| = \left| f_x(t, c) (x-y)\right|, \] which implies \[ \left| f(t, x) - f(t,y)\right| \le L |x-y|. \] These assumptions guarantee that Picard’s local existence and uniqueness theorems applies on any closed and bounded rectangle $R\subset \cO$.

The maximal solution

Theorem. For every $(t_0, x_0)\in \cO$ there is a maximal solution $x:[t_0, T)\to\R$ of the differential equation $\dot x = f(t, x)$ with $x(t_0)=x_0$

By definition, a solution is maximal if every other solution $\tilde x:[t_0, t_1)\to\R$ is contained in the maximal solution, i.e.

$t_1\le T$ (so the solution is not defined on a longer time interval than the maximal solution), and
$\tilde x(t) = x(t)$ for all $t\in[t_0, t_1)$ (i.e. the solution coincides with the maximal solution whenever both are defined.)

Where the maximal solution ends

Theorem. Let $x:[t_0, T)\to\R$ be the maximal solution of $\dot x=f(t, x)$, $x(t_0) = x_0$. Then one of the following holds

$T=\infty$
$T\lt\infty$, but $\lim_{t\nearrow T} x(t)$ does not exist
$T\lt \infty$, and $X = \lim_{t\nearrow T} x(t)$ exists, but $(T, X)\not\in\cO$

How to show $T=\infty$

In many cases the right hand side $f(t, x)$ in the differential equation is defined for all $(t, x)$, so the domain of $f$ is $\cO=\R^2$. In this case the third possibility above cannot occur, so for any solution one either has $T=\infty$ or $T\lt \infty$ and $x(t)$ has no limit as $t\nearrow T$. In other words: if a solution only exists up to some finite time $t=T\lt \infty$, then $x(t)$ has no limit as $t\nearrow T$.

Theorem. Let $x(t)$ be a solution of the differential equation $\dot x=f(t,x)$, and assume that on the time interval $[t_0, t_1)$ one knows that $|f(t, x(t))| \le M$ for some constant $M$. Then $\lim_{t\nearrow t_1} x(t)$ exists, and the maximal solution exists on a larger time interval $[t_0, t_1)$.

Proof

For any $t\in [t_0, t_1)$ we have \begin{equation}\label{eq-1} x(t) = x(t_0) + \int _{t_0}^t x'(s)\, ds = x(t_0) + \int _{t_0}^t f(s, x(s))\, ds. \end{equation} If $|f(s, x(s))|\le M$ then the integral \[ \int_{t_0}^{t_1} f(s, x(s))\, ds \] exists and we can prove that \[ \lim_{t\nearrow t_1} x(t) = A, \] where \begin{equation}\label{eq-2} A\stackrel{\sf def}= x(t_0) + \int _{t_0}^{t_1} f(s, x(s))\, ds \end{equation} by the following reasoning. Due to \eqref{eq-1} and \eqref{eq-2} we have \[ A- x(t) = \int_t^{t_1} f(s, x(s))\, ds \] so that \[ |A-x(t)| \le \int_t^{t_1} |f(s, x(s))|\, ds \le \int_t^{t_1} M\, ds =M(t_1-t). \] This implies that $\lim_{t\nearrow t_1} |A-x(t)| = 0$, i.e. $x(t)$ has a limit as $t\nearrow t_1$.