There is a theorem in page 100 of Arnold's Mathematical Methods of Classical Mechanics, which says that:

If $\cfrac{dx}{dt} = f(x) = Ax + R_2(x)$, where $A = \cfrac{\partial f}{\partial x}|_{x = 0}$, $R_2(x) = O(x^2)$, and $\cfrac{dy}{dt} = Ay$, $y(0) = x(0)$, then for any $\vphantom{\cfrac12} T>0$ and for any $\xi > 0$ there exists $\delta > 0$ such that if $|x(0)| < \delta$, then $\vphantom{\cfrac12}|x(t) - y(t)| < \xi \delta$ for all $t$ in the interval $0 < t < T$.

How can I prove this theorem rigorously?

Cross-posted at physics.se here.

  • 1
    $\begingroup$ This follows from continuity of solution of initial value problems with respect to parameters. For example, for the equation with parameters $x'=Ax+\lambda R_2(x)$, with parameter $\lambda\in\mathbb R$. When $\lambda=0$ you get the linear one, with $\lambda=1$ the original non-linear one (and somework, depending on what statement of this theorem you have :-) ) $\endgroup$ – Mariano Suárez-Álvarez Dec 8 '13 at 2:53
  • $\begingroup$ Thanks! Can you explain it further? $\endgroup$ – Eden Harder Dec 9 '13 at 14:31

Lets assume that $R$ is smooth. Therefore if $|f|,|g| < M \le 1$, then $$ |R(f) - R(g)| \le C_1 M |f-g| .$$

Also define $$ C_2 = \sup_{0 \le t \le T} \|e^{tA}\| .$$

Define $$ B(\epsilon) = \{f(t) \in C([0,T]) : \|f\|_\infty \le \epsilon\} $$ $$ G_{x_0}:B(\epsilon) \to C([0,T]) $$ $$ G_{x_0}f(t) = e^{tA} x_0 + \int_0^t e^{(t-s)A} R(f(s)) \, ds $$ Then $$ \|G_{x_0}(f-g)\|_\infty \le T C_2 C_1\epsilon \|f-g\|_\infty $$ Hence there exists $\delta_1>0$ such that if $\epsilon<\delta_1$, then $G_{x_0}$ is a contraction mapping. Also $$ \|G_{x_0}(f)\|_\infty \le C_2 |x_0| + T C_2 C_1\epsilon^2 $$ Hence there exists a $\delta_2$ such that if $\epsilon, |x_0| < \delta_2$, then the range of $G_{x_0}$ is contained in $B(\epsilon)$.

Therefore if $\epsilon, |x_0| < \min\{\delta_1,\delta_2\}$, then $G$ has a fixed point. That will be the function $x(t)$.

Let $z = x-y$. Then $$ \frac{dz}{dt} = Az + R(x) .$$ $$ z(t) = \int_0^t e^{(t-s)A} R(x(s)) \, ds $$ Therefore $$ \|z\|_\infty \le C_2 C_1 \epsilon^2 .$$ Choose $\delta = \min\{\delta_1,\delta_2,\xi/(C_1 C_2)\}$.

  • $\begingroup$ Thanks very much! How can you make sure that $x(t) \in A(\delta)$? If can not, then $x(t)$ is not the fixed point of $T$. $\endgroup$ – Eden Harder Dec 8 '13 at 3:56
  • $\begingroup$ I didn't explain that argument very well. But I edited the document to say that a similar argument will tell you this. $\endgroup$ – Stephen Montgomery-Smith Dec 8 '13 at 4:13
  • $\begingroup$ Thanks! Can you explain $Tsup_{0≤t≤T}∥e^{tA}∥(∥R(f)−R(g)∥_∞)≤Kδ∥f−g∥_∞$ further more? I can not figure out where δ comes from. $\endgroup$ – Eden Harder Dec 8 '13 at 5:53
  • $\begingroup$ If it were the case that $R(x) = x^2$, then we would have $R(f) - R(g) = (f+g)(f-g)$. I was saying that if $R$ is smooth, then we can do something similar, were the $f+g$ is replaced by something from a mean value theorem. $\endgroup$ – Stephen Montgomery-Smith Dec 8 '13 at 15:48
  • $\begingroup$ Ok~ your proof assumes that $||x||_\infty \leq \delta$ at the last step while the theorem only under the condition that $|x(0)| \leq \delta$. $\endgroup$ – Eden Harder Dec 9 '13 at 0:10

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