Let $a$, $b$ be two real positive parameters with $a>b$, and consider the following nonlinear differential equation: \begin{align} \dot{x}_{\varepsilon}(t) = a - b\sin(x_{\varepsilon}(t))+\varepsilon,\quad t\ge 0,\ x_\varepsilon(0)\in\mathbb{R}, \end{align} where $\varepsilon$ is a positive real constant. Let us define $$ \Delta(t,\varepsilon):= |{x}_{\varepsilon}(t)-{x}_{0}(t)|, $$ and note that $\Delta(t,0)\equiv 0$.

My question. Suppose that $x_\varepsilon(0)=x_0(0)$. Is $\Delta(t,\varepsilon)$ linearly bounded in $t\ge 0$ and $\varepsilon$, that is $$ \Delta(t,\varepsilon)\le K \varepsilon t, $$ where $K$ being a suitable constant? [In case of positive answer, if $x_\varepsilon(0)\ne x_0(0)$, is it possible to find an upper bound (similar to the above one) which takes into account the initial conditions?]


$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \renewcommand{\th}{\theta} \newcommand{\om}{\omega} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$

Let $x=x(a,t)$ be a solution to the diff. eq. \begin{equation*} \dot{x}(a,t) = a - b\sin x(a,t). \tag{0} \end{equation*} The variables $t$ and $x$ are separable. Indeed, this diff. eq. can be rewritten as \begin{equation*} \int\frac{dx}{a - b\sin x}=\int dt, \end{equation*} whence \begin{equation*} \frac2\om\, \tan ^{-1}\frac{b-a \tan(\frac{x}{2})}{\om}=t+c \end{equation*} and \begin{equation*} x(a,t)=f(a,t+c), \tag{1} \end{equation*} for some real constant $c$, where \begin{equation*} f(a,t):=2 \tan ^{-1}\frac{b+\om \tan (\frac{1}{2}\om t)}{a} \end{equation*} and \begin{equation*} \om:=\om(a):=\sqrt{a^2-b^2}. \end{equation*} Letting \begin{equation*} T:=2\pi/\om, \tag{2} \end{equation*} note that $f(a,t)$ is smooth in $t\in(-T/2,T/2)$, with the partial derivative \begin{equation*} f'_t(a,t)=\frac{a\om^2}{a^2-b(b \cos \om t+\om \sin \om t)}\in\Big(\frac{\om^2}{a+b},\frac{\om^2}{a-b}\Big) \tag{3} \end{equation*} in $t$. Also, $f(a,t)\to\pm\pi$ as $\pm t\uparrow T/2$.

Thus (and this is the crucial point), the function $(-T/2,T/2)\ni t\mapsto f(a,t)$ can be extended to a smooth function $\R\ni t\mapsto f(a,t)$ such that \begin{equation*} \psi(a,t):=f(a,t)-\om(a) t \end{equation*} is periodic in $t$ with period $T$ as in (2), and then (1) will define the general solution $x$ to the diff. eq. (0).

The graph of $f$ for $a=3,b=2.5$ is shown here:

enter image description here

Fix any real $x_0$. In view of (3), there is a unique real $c(a)$ such that \begin{equation*} f(a,c(a))=x_0, \tag{4} \end{equation*} and the the formula \begin{equation*} x(a,t)=f(a,t+c(a))=\om(a)t+\psi(a,t+c(a)) \end{equation*} will define the unique solution to (0) satisfying the initial condition $x(a,0)=x_0$. For each real $t$, \begin{equation*} x'_a(a,t)=\frac d{da}\,f(a,t+c(a))=\om'(a)t+\psi'_a(a,t+c(a))+\psi'_t(a,t+c(a))c'(a). \end{equation*} By (4), the implicit function theorem, and (3), \begin{equation*} |c'(a)|=|f'_a(a,c(a))/f'_t(a,c(a))|\ll|f'_a(a,c(a))|\ll\om'(a)|c(a)|+|\psi'_a(a,c(a))|; \end{equation*} the constants in $\ll$ may depend on $a$ and $b$, but not on $t$. Since $\psi(a,t)$ is smooth in $(a,t)$ and periodic in $t$, it follows from the last two displays that \begin{equation*} |x'_a(a,t)|\ll t+1. \tag{5} \end{equation*} So, for $t>1$ we have $|x'_a(a,t)|\ll t$ and hence \begin{equation*} |x(a+\ep,t)-x(a,t)|\ll\ep t. \tag{6} \end{equation*}

It remains to show that (6) holds for $t\in[0,1]$ as well. This is now easy: because of the smoothness of $f(a,t)$ in $(a,t)$ and the above bound on $|c'(a)|$, for $t\in[0,1]$ \begin{equation*} |x(a+\ep,t)-x(a,t)|\le\int_0^t|f'_t(a+\ep,s+c(a+\ep))-f'_t(a,s+c(a))|\,ds\ll \int_0^t\ep\,ds=\ep t. \end{equation*} So, (6) holds for all $t\ge0$.


First write the differential equation as an integral equation $\newcommand{\ve}{{\varepsilon}}$

$$ x_\ve(t)=x_0+(a+\ve)t-b\int_0^t \sin x_\ve(s) ds. $$

We deduce

$$ x_\ve(t)-x_0(t)= \ve t-b\int_0^t\big(\; \sin x_\ve (s)-\sin x_0(s)\;\big) ds $$

so (for $t\geq 0$)

$$ \Delta(t,\ve)\leq \ve t +b\int_0^t \Delta(s,\ve) ds. $$

Gronwall's inequality now implies

$$\Delta(t,\ve)\leq\ve t +b\ve e^t\int_0^t s e^{-s} ds = \ve t +b\ve e^t \Big(\; 1-(t+1)e^{-t}\;\Big). $$

  • $\begingroup$ Do you think this approach can give the bound $\ll\epsilon t$? $\endgroup$ – Iosif Pinelis May 4 '18 at 22:56
  • $\begingroup$ No. It is too general. Your approach seems to be the only one that takes into account the specific situation. $\endgroup$ – Liviu Nicolaescu May 5 '18 at 8:44

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