Any theorem shows that flowmap $\phi_{\sum_{i=1}^n a_i f_i(x)}^\tau$ can be approximated by $\phi_{f_{\theta(t)}(x)}^{\tau'}$?

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 $$\phi_{f_{\theta(t)}}^{\tau'}$$?

Here

• $$f_{\theta(t)} \in F$$ 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.

$$\renewcommand{\a}{\mathbf a}$$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 $$\begin{equation*} 0=t_0<\cdots For $$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 $$\begin{equation*} g_t(x):=g_{\a,t}(x):=g_\a(t,x):=\sum_{k\in[K]}\sum_{j\in[n]} f_j(x)\,1(t_{k,j-1}\le t so, $$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 $$$$|D_k-D_{k-1}|\le hLD_{k-1}+4nLMh^2,$$$$ $$$$D_k\le (1+hL)D_{k-1}+4nLMh^2,$$$$ $$$$E_k\le (1+hL)E_{k-1},$$$$ $$$$E_k\le (1+hL)^K E_0=(1+hL)^K c\le e^{LT}c=e^{LT}4nMh,$$$$ $$$$D_k\le E_k\le Ch,$$$$ where $$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 $$[0,T]$$. $$\quad\Box$$

• 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) Commented Oct 30, 2023 at 11:45
• @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 $x<1$. Commented Oct 30, 2023 at 16:31
• @liangDuan : Do you have a response to my comment? Commented Oct 31, 2023 at 21:54
• 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！ Commented Nov 8, 2023 at 6:01
• @liangDuan : Thank you for your appreciation. Do you have a reference to "that the splitting method can solve the problem immediately"? Commented Nov 8, 2023 at 14:16