# The Unapologetic Mathematician

## Control on the Divergence of Solutions

Now we can establish some control on how nearby solutions to the differential equation

$\displaystyle v'(t)=F(v(t))$

diverge. That is, as time goes by, how can the solutions move apart from each other?

Let $x$ and $y$ be two solutions satisfying initial conditions $x(t_0)=x_0$ and $y(t_0)=y_0$, respectively. The existence and uniqueness theorems we’ve just proven show that $x$ and $y$ are uniquely determined by this choice in some interval, and we’ll pick a $t_1$ so they’re both defined on the closed interval $[t_0,t_1]$. Now for every $t$ in this interval we have

$\displaystyle\lVert y(t)-x(t)\rVert\leq\lVert y_0-x_0\rVert e^{K(t-t_0)}$

Where $K$ is a Lipschitz constant for $F$ in the region we’re concerned with. That is, the separation between the solutions $x(t)$ and $y(t)$ can increase no faster than exponentially.

So, let’s define $d(t)=\lVert y(t)-x(t)\rVert$ to be this distance. Converting to integral equations, it’s clear that

$\displaystyle y(t)-x(t)=y_0-x_0+\int\limits_{t_0}^t\left(F(y(s))-F(x(s))\right)\,ds$

and thus

\displaystyle\begin{aligned}d(t)&\leq\lVert y(t_0)-x(t_0)\rVert+\int\limits_{t_0}^t\left\lVert F(y(s))-F(x(s))\right\rVert\,ds\\&\leq\lVert y(t_0)-x(t_0)\rVert+\int\limits_{t_0}^tK\lVert y(s)-x(s)\rVert\,ds\\&=d(t_0)+\int\limits_{t_0}^tKd(s)\,ds\end{aligned}

Now Gronwall’s inequality tells us that $d(t)\leq d(t_0)e^{K(t-t_0)}$, which is exactly the inequality we asserted above.

May 13, 2011 - Posted by | Analysis, Differential Equations

## 1 Comment »

1. […] now our result from last time tells us that these solutions can diverge no faster than exponentially. Thus we conclude […]

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