The Unapologetic Mathematician

Mathematics for the interested outsider

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 […]

    Pingback by Smooth Dependence on Initial Conditions « The Unapologetic Mathematician | May 16, 2011 | Reply

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