Given a control family $F:=\{f_1,\dotsc,f_n\}$, and $\phi_f^\tau(x)$ is the flowmap of the dynamical system $$ \begin{cases} z'(t)=f(z),\\ z(0)=x, \end{cases} $$ at end time point $\tau$.

Suppose $a_i>0$: can $\phi_{\sum_{i=1}^n a_i f_i(x)}^\tau$ be approximated by $\phi_{f_{\theta(t)}}^{\tau'}$?


  • $f_{\theta(t)} \in F$ for any $t\in[0,\tau']$, and
  • $f_{\theta(t)}$ is piecewise constant for $t$: for example $f_{\theta(t)}$ can be denoted as $$ f_{\theta(t)}(x)=\begin{cases} f_1(x), &t\in[0,t_1], \\ f_2(x), & t\in (t_1,t_2]. \end{cases} $$

Is there any theorem or reference to ensure this? The conclusion seems obvious.


1 Answer 1


$\renewcommand{\a}{\mathbf a}$The conclusion is indeed very intuitive. However, the proof of it is rather tedious, along the lines of the proof of the Picard–Lindelöf theorem.

Accordingly, assume the following standard conditions:

For each $j\in[n]:=\{1,\dots,n\}$, let $f_j$ be $L$-Lipshitz for some positive real $L$. Take any real $x$. Suppose that for some positive real $M$ and $b$ and for all $j\in[n]$ we have $|f_j|\le M$ on $[x-b,x+b]$. Take any $T\in(0,\min(1/L,b/M))$.

By time rescaling, without loss of generality (wlog) \begin{equation*} \sum_{j\in[n]}a_j=1. \tag{1}\label{1} \end{equation*}

Then the initial value problem \begin{equation*} z'(t)=f_\a(z(t)),\quad z(0)=x \end{equation*} has a unique solution on the interval $[0,T]$, where $f_\a:=\sum_{j\in[n]}a_j f_j$. This problem can be rewritten in integral form, as follows:
\begin{equation*} z(t)=x+\int_0^t f_\a(z(s))\,ds \tag{2}\label{2} \end{equation*} for $t\in[0,T]$.

For a natural $K$, suppose that \begin{equation*} 0=t_0<\cdots<t_K=T. \end{equation*} For $k\in[K]$ and $j\in\{0\}\cup[n]$, let \begin{equation*} h_k:=t_k-t_{k-1},\quad t_{k,j}:=t_{k-1}+h_k\sum_{r=1}^j a_r, \end{equation*} so that $t_{k,0}=t_{k-1}$ and $t_{k,n}=t_n$, in view of \eqref{1}.

The role of $f_{\theta(t)}(x)$ will be played by \begin{equation*} g_t(x):=g_{\a,t}(x):=g_\a(t,x):=\sum_{k\in[K]}\sum_{j\in[n]} f_j(x)\,1(t_{k,j-1}\le t<t_{k,j}); \end{equation*} so, $g_t=f_j$ on every interval of the form $[t_{k,j-1},t_{k,j})$.

Consider the initial value problem \begin{equation*} y'(t)=g_t(y(t)),\quad y(0)=x, \end{equation*} where $y'(t)$ is the right derivative of $y$ at $t$. This problem can be rewritten in integral form, as follows: for each $(k,j)\in[K]\times[n]$ and all $t\in[t_{k,j-1},t_{k,j}]$ \begin{equation*} y(t)=y(t_{k,j-1})+\int_{t_{k,j-1}}^t f_j(y(s))\,ds. \tag{3}\label{3} \end{equation*}

The intuition: So, within each (say small) time interval $[t_{k-1},t_k)$ of length $h_k$, we let the rate of change of $y$ be $f_j(y)$ for the fraction $a_j$ of the length $h_k$ of the time interval $[t_{k-1},t_k)$. Since $y$ will vary little within the small time interval $[t_{k-1},t_k)$, the effect of this switching between the $f_j$'s will be almost the same as the effect of using the average rate $f_\a=\sum_{j\in[n]}a_jf_j$.

Now the formalities: Reasoning as in the proof of the Picard–Lindelöf theorem, we see that problem \eqref{3} has a unique solution on the interval $[0,T]$.

Moreover, for all $(k,j)\in[K]\times[n]$ and all $t\in[t_{k,j-1},t_{k,j}]$ \begin{equation*} |y(t)-y(t_{k,j-1})|\le Mh_k a_j\le Mh_k, \tag{3.5}\label{3.5} \end{equation*} so that \begin{equation*} |f_j(y(t))-f_j(y(t_{k,j-1}))|\le LMh_k, \end{equation*} so that \begin{equation*} |f_j(y(t_{k,j-1}))-f_j(y(t_{k-1}))|\le nLMh_k \tag{4}\label{4} \end{equation*} and \begin{equation*} |y(t)-y(t_{k,j-1})-f_j(y(t_{k,j-1}))(t-t_{k,j-1})| \\ =\Big|\int_{t_{k,j-1}}^t (f_j(y(s))-f_j(y(t_{k,j-1})))\,ds\Big| \le LMh_k^2, \end{equation*} so that \begin{equation*} \Big|y(t_k)-y(t_{k-1})-\sum_{j\in[n]}f_j(y(t_{k,j-1}))h_k a_j\Big| \le nLMh_k^2, \end{equation*} so that, in view of \eqref{4}, \begin{equation*} \Big|y(t_k)-y(t_{k-1})-\sum_{j\in[n]}f_j(y(t_{k-1}))h_k a_j\Big| \le nLMh_k^2+nLMh_k^2, \end{equation*} that is, \begin{equation*} \Big|y(t_k)-y(t_{k-1})-f_\a(y(t_{k-1}))h_k\Big| \le 2nLMh_k^2 \end{equation*} for all $k\in[K]$.

Similarly, but a bit more simply, we get \begin{equation*} \Big|z(t_k)-z(t_{k-1})-f_\a(z(t_{k-1}))h_k\Big| \le 2LMh_k^2. \end{equation*} for all $k\in[K]$.

So, for all $k\in[K]$, letting $h_k:=h:=T/K$,
$D_k:=|y(t_k)-z(t_k)|$, and $E_k:=D_k+c$ for $c:=4nMh$, we get \begin{equation} |D_k-D_{k-1}|\le hLD_{k-1}+4nLMh^2, \end{equation} \begin{equation} D_k\le (1+hL)D_{k-1}+4nLMh^2, \end{equation} \begin{equation} E_k\le (1+hL)E_{k-1}, \end{equation} \begin{equation} E_k\le (1+hL)^K E_0=(1+hL)^K c\le e^{LT}c=e^{LT}4nMh, \end{equation} \begin{equation} D_k\le E_k\le Ch, \end{equation} where $C:=e^{LT}4nM$.

Also, by \eqref{3.5}, for all $k\in[K]$ and all $t\in[t_{k-1},t_k]$ we have $|y(t)-y(t_{k-1})|\le Mh$; similarly, $|z(t)-z(t_{k-1})|\le Mh$.

Thus, $|y-z|\le(C+2M)h=(C+2M)T/K$ on $[0,T]$. Choosing now $K$ to be large enough, we make $|y-z|$ however small uniformly on $[0,T]$. $\quad\Box$

  • $\begingroup$ The proof seems to be fine, but is it too strong to assume $T\in (0, min(1/𝐿,𝑏/𝑀)) $? So that it does not meet the conditions of this proposition (the proposition does not limit the range of T values) $\endgroup$ Commented Oct 30, 2023 at 11:45
  • 1
    $\begingroup$ @liangDuan : As was said in the answer, this condition on $T$ is quite standard. In fact, it is the same as the condition at the very end of the proof of the Picard–Lindelöf theorem (where they use $a$ instead of my $T$). You cannot do without such a condition. E.g., the solution $y=\frac1{1-x}$ of the initial initial value problem $y'=y^2$, $y(0)=1$ exists only for $x<1$. $\endgroup$ Commented Oct 30, 2023 at 16:31
  • $\begingroup$ @liangDuan : Do you have a response to my comment? $\endgroup$ Commented Oct 31, 2023 at 21:54
  • $\begingroup$ Thanks for your answer, recently I found that the splitting method can solve the problem immediately. But your method is also very inspiring to me! $\endgroup$ Commented Nov 8, 2023 at 6:01
  • $\begingroup$ @liangDuan : Thank you for your appreciation. Do you have a reference to "that the splitting method can solve the problem immediately"? $\endgroup$ Commented Nov 8, 2023 at 14:16

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