Proof of a theorem about oscillation

Question

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?

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Cross-posted at math.se here.

Answer 1

The difference between $x$ and $y$ satisfies the equation $$\frac{\text d}{\text dt}(x(t)-y(t))=A(x-y)+R_2(x).$$

This can be solved, formally, by variation of parameters, to give $$x(t)-y(t)=\int_0^t e^{A(t-\tau)}R_2(x(\tau))\text d\tau.$$ Their separation is then bounded by $$\begin{align}|x(t)-y(t)| \leq\int_0^t | e^{A(t-\tau)}R_2(x(\tau))|\text d\tau \leq\int_0^T e^{|A|(t-\tau)}|R_2(x(\tau))|\text d\tau.\end{align}$$

Now, $R_2(x)=O(x^2)$, which means that there exists a constant $M$ and a distance $X$ such that $|R_2(x)|\leq Mx^2$ whenever $|x|\leq X$. To be able to use this, we need to ensure that $|x(t)|$ will be no greater than some $\delta$-controllable constant for all times $t\in[0,T]$, and our tool to achieve this is making the initial condition $x(0)=x_0$ small enough. This is essentially the statement that the solution is continuous with respect to the initial condition (i.e. around the solution $x(t)\equiv0$ for $x(0)=0$).

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This continuity is a standard fact in differential equations but I'll sketch a proof here. Any solution $x(t)$ must be differentiable, hence continuous, and hence bounded. This means the condition $|R_2(x(t))|\leq M_{x_0}x(t)^2$ will hold for all $t\in[0,T]$, for some suitable constant $M_{x_0}$. The right-hand side of the differential equation for $x(t)$ is therefore Lipschitz continuous: $$|Ax(t)+R_2(x(t))|\leq L|x(t)|\tag1$$ for some $L>0$. (Alternatively, you might impose this to begin with.) Setting $u(t)=|x(t)|^2$, you have $$ \frac{\text du}{\text dt} \leq\left|\frac{\text du}{\text dt}\right| = \left|2x(t)\frac{\text dx}{\text dt}\right| \leq 2L |x(t)|^2=2L u(t). $$ By Gronwall's inequality this implies that $|u(t)|\leq|u(0)|e^{2L t}$, which means that $|x(t)|\leq |x_0| e^{LT}$ for all $t\in[0,T]$.

So, in a nutshell: if you can assume or prove that (1) holds with $L$ independent of $x_0$, then your solution $|x(t)|$ will be bounded by $|x_0| e^{LT}$ for all $t\in[0,T]$. Additionally, this means that if you choose $|x_0|<\delta\leq Xe^{-LT}$, you can bind $|x(t)|<X$ for all $t\in[0,T]$.

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Let's now return to the difference $|x(t)-y(t)|$. We can now apply the smallness bound on $R_2$, which gives us $$\begin{align} |x(t)-y(t)| &\leq\int_0^t e^{|A|(t-\tau)}|R_2(x(\tau))|\text d\tau \leq\int_0^t e^{|A|(t-\tau)} M|x(\tau)|^2\text d\tau \\& \leq M\int_0^T e^{|A|T}|x_0|^2 e^{2LT}\text d\tau \leq \delta^2 MT e^{(2M+|A|)T} \end{align}$$ for all $t\in[0,T]$.

Thus, if you're given $\xi>0$ and $T<0$, then choosing $\delta=\min\{Xe^{-LT},\xi e^{-(2M+|A|)T}/MT\}$ you can ensure that $$|x(t)-y(t)|<\xi\delta$$ for all $t\in[0,T]$.

Finally, note that while this is phrased in one-dimensional language, it holds with trivial extensions (i.e. simply with appropriately defined norms) for the multidimensional case, and therefore for higher-than-first-order ODEs.

Answer 2

This is just an expanded version of my comment above. The proof is not yet complete.

For a given $T>0$, let $X(x_0,t)$ and $Y(y_0,t)$ be solutions of the first and second equations in your question. Then for any given $t\in[0,T]$ we have :

$$X(0,t)=Y(0,t)=0 \tag 1$$

( the only solution with initial condition $x_0=0$)

Also $$\displaystyle \frac {\partial X(x_0,t)}{\partial x_0}\bigg\vert _{x_0=0} = \displaystyle \frac {\partial Y(x_0,t)}{\partial x_0}\bigg\vert_{x_0=0}\tag2$$

To see this, first we notice from the integral form of the two differetial equations that

$$X(x_0, 0)=Y(x_0,0)=x_0\tag 3$$.

Now let $h_1 ( t) = \displaystyle\frac {\partial X(x_0,t)}{\partial x_0}\bigg\vert_{x_0=0}$ and $h_2(t)=\displaystyle \frac {\partial Y(x_0,t)}{\partial x_0}\bigg\vert_{x_0=0}$

Then we have (by taking derivative wrt $x_0$ on both sides of the two differential equations and using equation (3), and also the fact that $R_2(x)=O(x^2)$):

$\displaystyle \frac {dh_1 (t)}{dt}=Ah_1(t)$, with $h_1 (0)=1$

$\displaystyle \frac {dh_2 (t)}{dt}=Ah_2(t)$, with $h_2 (0)=1$

From these two equations we conclude that $h_1 (t)=h_2(t)$ for all $t\in [0,T]$, which is equation (2).

From equations (1) and (2) above we conclude that $X(x_0,t)-Y(x_0,t)$ is of the form

$$X(x_0,t)-Y(x_0,t) = x_0^2 h(x_0,t)\tag 4$$ for every $t\in[0,T]$

Assuming that for any fixed $t\in[0,T]$, Taylor series of $X(x_0,t)$ and $Y(x_0,t)$ exists (wrt $x_0$) in some neighborhood $V_t$ of $x_0=0$, we can say that $h(x_0,t)$ is smooth and bounded in $V_t$. However, in order to prove the theorem one needs to show that there exists an $a>0$ (depending upon $T$) such that $h(x_0,t)$ is bounded in the rectangle $[-a,a]\times[0,T]$ with some upper bound $M$ on its absolute value. I am not sure how to prove this part and would highly appreciate if someone can provide some hint.

Once the boundedness of $h$ is proved one can take $\delta<\xi/M$ and the statement of the theorem follows.