Smooth Dependence on Initial Conditions

Now that we’ve got the existence and uniqueness of our solutions down, we have one more of our promised results: the smooth dependence of solutions on initial conditions. That is, if we use our existence and uniqueness theorems to construct a unique “flow” function $\psi:I\times U\to\mathbb{R}^n$ satisfying

$\displaystyle\begin{aligned}\frac{\partial}{\partial t}\psi(t,u)&=F(\psi(t,u))\\\psi(0,u)=u\end{aligned}$

by setting $\psi(t,u)=v_u(t)$ — where $v_u$ is the unique solution with initial condition $v_u(0)=u$ — then $\psi$ is continuously differentiable.

Now, we already know that $\psi$ is continuously differentiable in the time direction by definition. What we need to show is that the directional derivatives involving directions in $U$ exist and are continuous. To that end, let $a\in U$ be a base point and $h$ be a small enough displacement that $a+h\in U$ as well. Similarly, let $t_0$ be a fixed point in time and let $\Delta t$ be a small change in time

$\displaystyle\begin{aligned}\lVert\psi(t_0+\Delta t,a+h)-\psi(t,a)\rVert=&\lVert v_{a+h}(t+\Delta t)-v_a(t)\rVert\\\leq&\lVert v_{a+h}(t+\Delta t)-v_a(t+\Delta t)\rVert\\&+\lVert v_a(t+\Delta t)-v_a(t)\rVert\end{aligned}$

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

$\displaystyle\lVert v_{a+h}(t+\Delta t)-v_a(t+\Delta t)\rVert\leq\lVert h\rVert e^{K\Delta t}$

and so as $\lVert h\rVert\to0$ this term must go to zero as well. Meanwhile, the second term also goes to zero by the differentiability of $v_a$ . We can now see that the directional derivative at $(t_0,a)$ in the direction of $(\Delta t,h)$ exists.

But are these directional derivatives continuous. This turns out to be a lot more messy, but essentially doable by similar methods and a generalization of Gronwall’s inequality. For the sake of getting back to differential equations I’m going to just assert that not only do all directional derivatives exist, but they’re continuous, and thus the flow is $C^1$ .

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May 16, 2011 - Posted by John Armstrong | Analysis, Differential Equations

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