2022年下半年，看了很多关于时滞系统和引入时滞的网络化控制系统的论文，对稳定性证明中Lyapunov-Krasovskii泛函中二重积分项求导结果有些疑问，故在知乎上提问寻求帮助，幸运的被一个大佬回答了，这里转载一下。并给出大佬的知乎主页。顾玖大佬知乎主页

时滞系统中Lyapunov泛函中二重积分项求导时，用的是取消内层积分的积分号，然后再算外层积分嘛，这和考研时学习的二重积分求导方式为什么结果不一样，求问？
如果用把内层积分看成一个整体的形式，再求变现积分，结果只有论文里的第二项，有明显的不同，请各位大佬赐教，十分感谢！！！

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回答

回答：
我们可以将${\int }_{t-{\tau }_m}^t{\int }_s^t{\dot{x}}^T(v)R\dot{x}(v)dvds$这一项单独拎出来并考虑其对$t$的导数（这是唯一有难度的一项）。事实上，我们可以得到一个更一般的结论：即，在满足一定条件下，我们可以得到函数$$F(t)={\int }_{\alpha (t)}^{\beta (t)}[{\int }_{\phi (x,t)}^{\psi (x,t)}f(x,y)dy]dx\text{\hspace{1em}(1)}$$关于变量$t$的导数。
如果求得了$F(t)$对$t$的导数，只需要令$\alpha (t)=t-{\tau }_m$，$\beta (t)=t$，$\phi (x,t)=x$，$\psi (x,t)=t$即可得到我们想要的答案。
注：为了阅读方便，建议先跳到最后看结论，然后再回来看引理 1 和定理 1 的证明。
在给出$F(t)$对$t$的导数之前，我们先证明一个引理。

[引理1]

[引理1] 设函数$\alpha (t)$，$\beta (t)$在区间$[t_0,t_1]$上可导，函数$f(x,t)$在区域$D$上连续，其中区域$D$为$\{(x,t)|x\in [\alpha (t),\beta (t)],t\in [t_0,t_1]\}$，且$f(x,t)$关于$t$的偏导数在$D$上连续，则有$$\frac{d}{dt}{\int }_{\alpha (t)}^{\beta (t)}f(x,t)dx=f(\beta (t),t){\beta }^{\prime }(t)-f(\alpha (t),t){\alpha }^{\prime }(t)+{\int }_{\alpha (t)}^{\beta (t)}\frac{\partial }{\partial t}f(x,t)dx\text{\hspace{1em}(2)}$$证明： 记$F(t)={\int }_{\alpha (t)}^{\beta (t)}f(x,t)dx$，则有$$\begin{array}{rl}F(t+\Delta t)-F(t)& ={\int }_{\alpha (t+\Delta t)}^{\beta (t+\Delta t)}f(x,t+\Delta t)dx-{\int }_{\alpha (t)}^{\beta (t)}f(x,t)dx\\ & ={\int }_{\beta (t)}^{\beta (t+\Delta t)}f(x,t+\Delta t)dx-{\int }_{\alpha (t)}^{\alpha (t+\Delta t)}f(x,t+\Delta t)dx\\ & +{\int }_{\alpha (t)}^{\beta (t)}[f(x,t+\Delta t)-f(x,t)]dx\end{array}\text{\hspace{1em}(3)}$$根据导数定义$$\begin{array}{rl}{F}^{\prime }(t)& =\underset{\Delta t\to 0}{\lim }\frac{F(t+\Delta t)-F(t)}{\Delta t}\\ & =\underset{\Delta t\to 0}{\lim }\frac{1}{\Delta t}{\int }_{\beta (t)}^{\beta (t+\Delta t)}f(x,t+\Delta t)dx-\underset{\Delta t\to 0}{\lim }\frac{1}{\Delta t}{\int }_{\alpha (t)}^{\alpha (t+\Delta t)}f(x,t+\Delta t)dx\\ & +\underset{\Delta t\to 0}{\lim }\frac{1}{\Delta t}{\int }_{\alpha (t)}^{\beta (t)}[f(x,t+\Delta t)-f(x,t)]dx\end{array}\text{\hspace{1em}(4)}$$由积分中值定理，可知$\exists {\xi }_1\in [\beta (t),\beta (t+\Delta t)]$以及$\exists {\xi }_2\in [\alpha (t),\alpha (t+\Delta t)]$使得$$\begin{array}{rl}& \underset{\Delta t\to 0}{\lim }\frac{1}{\Delta t}{\int }_{\beta (t)}^{\beta (t+\Delta t)}f(x,t+\Delta t)dx\\ =& \underset{\Delta t\to 0}{\lim }\frac{1}{\Delta t}f({\xi }_1,t+\Delta t)(\beta (t+\Delta t)-\beta (t))\\ =& f(\beta (t),t){\beta }^{\prime }(t)\end{array}\text{\hspace{1em}(5)}$$同理可得$$\begin{array}{rl}& \underset{\Delta t\to 0}{\lim }\frac{1}{\Delta t}{\int }_{\alpha (t)}^{\alpha (t+\Delta t)}f(x,t+\Delta t)dx\\ =& \underset{\Delta t\to 0}{\lim }\frac{1}{\Delta t}f({\xi }_1,t+\Delta t)(\alpha (t+\Delta t)-\alpha (t))\\ =& f(\alpha (t),t){\alpha }^{\prime }(t)\end{array}\text{\hspace{1em}(6)}$$由$f(x,t)$具有关于$t$的偏导数，我们有$$\begin{array}{rl}& \underset{\Delta t\to 0}{\lim }\frac{1}{\Delta t}{\int }_{\alpha (t)}^{\beta (t)}[f(x,t+\Delta t)-f(x,t)]dx\\ =& {\int }_{\alpha (t)}^{\beta (t)}\underset{\Delta t\to 0}{\lim }\left[\frac{f(x,t+\Delta t)-f(x,t)}{\Delta t}\right]dx\\ =& {\int }_{\alpha (t)}^{\beta (t)}\frac{\partial }{\partial t}f(x,t)dx\end{array}\text{\hspace{1em}(7)}$$综上所述，我们可以得到(1)式，证毕$\square$
有了引理1，我们可以很容易的给出$f(t)$对$t$的导数，即下面的定理1。

[定理1]

[定理1] 设函数$\alpha (t),\beta (t)$在区间$[t_0,t_1]$上可导，函数$\phi (x,t),\psi (x,t)$在区域$D$上连续，其中区域$D$为$\{(x,t)|x\in [\alpha (t),\beta (t)],t\in [t_0,t_1]\}$，当取定$t\in [t_0,t_1]$时，$f(x,y)$在相应区域$E=\{(x,y)|x\in [\alpha (t),\beta (t)],y\in [\phi (x,t),\psi (x,t)]\}$上连续，则有$$F(t)={\int }_{\alpha (t)}^{\beta (t)}[{\int }_{\phi (x,t)}^{\psi (x,t)}f(x,y)dy]dx\text{\hspace{1em}(8)}$$在$[t_0,t_1]$上可导，且$$\begin{array}{rl}{F}^{\prime }(t)& =\frac{d}{dt}[{\int }_{\phi (x,t)}^{\psi (x,t)}f(x,y)dy]dx\\ & ={\beta }^{\prime }(t){\int }_{\phi (\beta (t),t)}^{\psi (\beta (t),t)}f(\beta (t),y)dy-{\alpha }^{\prime }{\int }_{\phi (\alpha (t),t)}^{\psi (\alpha (t),t)}f(\alpha (t),y)dy\\ & +{\int }_{\alpha (t)}^{\beta (t)}[{\psi }_t^{\prime }(x,t)f(x,\psi (x,t))-{\phi }_t^{\prime }(x,t)f(x,\phi (x,t))]dx\end{array}\text{\hspace{1em}(9)}$$证明： 记$G(x,t)={\int }_{\phi (t)}^{\psi (t)}f(x,y)dy$，则$F(t)={\int }_{\alpha (t)}^{\beta (t)}G(x,t)dx$。
由引理1得$$F^{\prime }(t)=G(\beta (t),t){\beta }^{\prime }(t)-G(\alpha (t),t){\alpha }^{\prime }(t)+{\int }_{\alpha (t)}^{\beta (t)}\frac{\partial }{\partial t}G(x,t)dx\text{\hspace{1em}(10)}$$其中，$$\frac{\partial }{\partial t}G(x,t)=\frac{d}{dt}{\int }_{\phi (x,t)}^{\psi (x,t)}f(x,y)dy=f(x,\psi (x,t)){\psi }_t^{\prime }(x,t)-f(x,\phi (x,t)){\phi }_t^{\prime }(x,t)\text{\hspace{1em}(11)}$$因此，带入得$$\begin{array}{rl}{F}^{\prime }(t)& ={\beta }^{\prime }(t){\int }_{\phi (\beta (t),t)}^{\psi (\beta (t),t)}f(\beta (t),y)dy-{\alpha }^{\prime }{\int }_{\phi (\alpha (t),t)}^{\psi (\alpha (t),t)}f(\alpha (t),y)dy\\ & +{\int }_{\alpha (t)}^{\beta (t)}[{\psi }_t^{\prime }(x,t)f(x,\psi (x,t))-{\phi }_t^{\prime }(x,t)f(x,\phi (x,t))]dx\end{array}\text{\hspace{1em}(12)}$$证毕$\square$
到此为止，${\int }_{t-{\tau }_m}^t{\int }_s^t{\dot{x}}^T(v)R\dot{x}(v)dvds$对时间$t$的导数就只是定理1的一个特例。
对于$F(t)={\int }_{\alpha (t)}^{\beta (t)}[{\int }_{\phi (x,t)}^{\psi (x,t)}f(x,y)dy]dx$，我们令$\alpha (t)=t-{\tau }_m$，$\beta (t)=t$，$\phi (t)=x$，$\psi (t)=t$，则有$$\begin{array}{rl}& \frac{d}{dt}{\int }_{t-{\tau }_m}^t{\int }_s^t{\dot{x}}^T(v)R\dot{x}(v)dvds\\ =& -{\int }_{t-{\tau }_m}^t{\dot{x}}^T(v)R\dot{x}(v)dv+{\int }_{t-{\tau }_m}^t{\dot{x}}^T(t)R\dot{x}(t)dt\\ =& -{\int }_{t-{\tau }_m}^t{\dot{x}}^T(v)R\dot{x}(v)dv+{\tau }_m{\dot{x}}^T(t)R\dot{x}(t)\end{array}\text{\hspace{1em}(13)}$$推导完毕$\square$

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