在本节中,我们通过分析系统来考虑葡萄糖的有限时间控制(2). 首先,我们建立了两个变量比较系统我,研究比较系统和系统之间的关系(2)并得到一些基本性质。然后我们给出了两个比较系统的上下界。系统的条件(2)得到了一个解。
3.1初步
从系统的第一和第三方程(2),我们可以
$$\开始{对齐}[b]G(T)={}&G_{0}\exp\biggl[-\int_{0{0}^{T}\biggl(\sigma_{2}+ac+\frac{am{I}(s)}{n+{I}}\bigr)\,ds\biggr]\\&+(G_{mathrm{in}}+b)\int_}^{T}\bigl[\exp\biggl(-\int_{u}^{T}\biggl(\sigma_{2}+ac+\frac{am{I}(s)}{n+{I}}\bigr)\,ds\biggr)\biggr]\,du\\={}&G_{0}\exp\bigl[-(\sigma_{2{+ac)T\bigr]\exp\biggl[-\int_}^{T} \frac{am{I}}{n+{I}(s)}\,ds\biggr]\\&+(G_{mathrm{in}}+b)\exp\bigl[-(\sigma_{2}+ac)T\bigr]\\&\times\int_{0}^{T}\biggl\{exp\bigl[(\siga_{2]+ac frac{am{I}}{n+{I}}\,ds\biggr]\biggr\}\,du。\结束{对齐}$$
(3)
此外,根据系统的第二和第四方程(2),我们制定了以下两个比较系统:
$$\left\{\textstyle\begin{array}{l@{quad}l}\frac{dI{1}(t)}{dt}=-d_{i}i{1{(t对$$
(4)
和
$$\left\{\textstyle\begin{array}{l@{quad}l}\frac{dI{2}(t)}{dt}=\sigma{1}-d_{i}i{2{(t{数组}\displaystyle\right$$
(5)
哪里\(k=1,2,点,p-1,点=T).
显然,我们有\(I_{1}(t)\leq I(t)\ leq I_{2}(t)\),\(在[0,t]\中)然后,通过集成,我们得到
$$I{1}(t)=I{0}\exp(-d_{I}t)+\frac{\sigma\exp$$
(6)
和
$$I_{2}(t)=\biggl(I_{0}-\frac{\sigma{1}}{d_{i}}\biggr)\exp(-d{i}t)+\frac{\sigma\exp$$
(7)
哪里\(t\in(k\tau,(k+1)\tau]\子集[0,t]\),\(k=0,1,\ldots,p-1).
通过计算[4]我们知道,因为\(k\tau^{+}\leqb{1}\leq b{2}\leg(k+1)\tau),
$$\int_{b_{1}}^{b_2}}\frac{am{I{1}(t)}{n+{I_1}}(t)},dt=\ln\biggl(\frac}n+{I{1}{}(b_2})}{n+{I{1}}$$
(8)
同样,我们可以从系统中获得(5),用于\(k\tau^{+}\leqb{1}\leq b{2}\leg(k+1)\tau),
$$开始{对齐}[b]\int_{b{1}}^{b{2}}\frac{am{I{2}{(t)}{n+{I{2]}(t gl(\frac{d\ln(n+{I{2}}(t))}{dt}\biggr)\,dt\\={}&\frac{am}{d_{I}}\int_{b{1}}^{b{2}}\frac}\sigma{1}{n+{I{2}{}{b{2}}\biggl(\frac{d\ln(n+{I_{2}}(t))}{dt}\bbiggr)^{-\frac{am}{d_{I}}}}。\结束{对齐}$$
(9)
根据(三)和(8),替换\(I_{1}(t)\)对于\(I(t)\),我们有
$$\开始{对齐}[b]G(T)\leq{}&G{0}\exp\bigl(-(\sigma{2}+ac)T\bigr)\exp\biggl(-\int_{0}^{T}\frac{amI{1}+ac)T\bigr)\int_{0}^{T}\exp\bigl((\sigma_{2}+ac+ac)T\bigr)\bigl(\frac{n+I_{1}(T)}{n+I_{{n+I_{1}(u)}\biggr)^{\frac{am}{d_{I)}}\\&\times\biggl\{\frac{G{0}}{(n+I{0})^{\frac{am}{d_{I}}}+(G{\mathrm{in}}+b)\int_{0}^{T}\exp\bigl,du\biggr\}。\结束{对齐}$$
(10)
类似地,替换\(I_{2}(t)\)到(三)对于\(I(t)\),我们有
$$\开始{对齐}[b]G(T)\geq{}&G{0}\exp\bigl(-(\sigma{2}+ac)T\bigr)\exp\biggl(-\int_{0}^{T}\frac{amI{2}+ac)T\bigr)\int_{0}^{T}\exp\bigl((\sigma_{2}+ac+ac)T\bigr)\exp\bigl(-\frac{am}{d_{i}}\int_{0}^{T}\frac{\sigma{1}}{n+i_{2}(s)}\,ds\biggr)\bigl(\frac{n+i_{2}(T)}{n+i_{0}}\biggr)^{\frac{am}}+ac)T\bigr)\\&&\times\int_{0}^{T}\exp\bigl((\sigma\{2}+ac)u\bigr)\exp\bigl(-\frac{am}{d_{i}}\int_{u}^{T}\ frac{\sigma\ 1}}}{n+i_{2}(s)}\,ds\biggr)\biggl(\frac{n+I{2}(T)}{n+I{2}(u)}\bigger)^{\frac}am}{d_{I}}}\,du。\结束{对齐}$$
(11)
根据(6),我们可以
$$\开始{对齐}和I{1}(0)=I{0},\qquad I{1{(\tau)=I_{0}\exp(-d_{I}\tau d_{I}T),\\&\开始{aligned}[b]I{1}(T)&\geqI{0}\exp(-d_{I}T)+\frac{\sigma[\exp_{0}-\压裂{\sigma}{1-\exp(-d{i}\tau)}\biggr)\exp$$
(12)
和
$$\开始{对齐}[b]I{1}(t)&=I{0}\exp(-d_{I}t)+\frac{\sigma\exp_{i} t吨)\exp(d_{i}\tau)_{0}-\压裂{\sigma}{1-\exp(-d{i}\tau)}\biggr)\exp$$
(13)
哪里\(k=0,1,2,\ldot,p-1,kτ<t(k+1)τ).
根据(7),我们可以
$$\开始{对齐}和I{2}(0)=I{0},\qquad I{2{(\tau)=I_{0}\exp(-d_{I}\tau}&I{0}\exp\bigl(-d_{I}(T-\tau)\bigr)+\frac{\sigma{1}}{d_{I}}\bigl[1-\exp\bigl(-d_{I{(T-\T au)\ bigr}{1-\exp(-d_{i}\tau)},\end{aligned}\\&\begin{aligned}i{2}(T)&=\biggl(i_{0}-\frac{\sigma{1}}{d_{i}}+\ frac{\ sigma[\exp(d_{i}(p-1)\tau)-1]}{1-\exp i}}\bigl(1-\exp(-d_{i}T)\bigr),\end{aligned}\\&\begin{aligned}[b]i{2}(T)\geq{}&\biggl(i_{0}-\frac{\sigma{1}}{d_{i}}\biggr)\exp(-d_{i}t)+\frac}\sigma{1}{d_{i}}+\frac{\simma(\exp[-d_{i}\tau)-\exp t)+\frac{\sigma{1}}{d{i}}\bigl(1-\exp(-d_{i} t吨)\bigr)\\&+\frac{\sigma}{1-\exp(-dd{i}\tau)}\bigl[\exp(-dd{i}\tau)-\exp(-d_{i} t吨)\bigr],\end{对齐}\end}对齐}$$
(14)
和
$$\开始{对齐}[b]I_{2}(t)&\leq\biggl(I_{0}-\frac{\sigma{1}}{d_{i}}\biggr)\exp(-d_{i}t)+\frac}\sigma{1}{d_{i}{+\frac{\simma(1-\exp_{0}-\压裂{\sigma{1}}{d_{i}}-\frac{\simma}{1-\exp(-d{i}\tau)}\biggr)\exp。\结束{对齐}$$
(15)
3.2比较系统的上下限
在下面,我们给出了比较系统的上下界(4)和(5). 这些结果将用于讨论系统解的存在性(2)初边值问题。
提议1
如果
\(I_{0}\geq\frac{\sigma}{1-\exp(-d_{I}\tau)}\),然后是系统(4)满足
\(I{1}(t)\geqI{1{(t)\triangleqI{0}\exp(-d_{I}t)+\frac{\sigma[\exp.如果
\(I_{0}<\frac{\sigma}{1-\exp(-d_{I}\tau)}\),然后是系统(4)满足
\(I_{1}(t)\geq I_{1}(\tau)\triangleq I_{0}\exp(-d_{I}\tau)\).
证明
为了方便起见,我们表示\(q=\exp(-d{i}\tau))然后我们就知道了\(0<q<1).来自(6),我们很容易得到\(I_{1}(t)\)是单调递减的\((kτ,(k+1)τ]\),\(k=0,1,2,\ldot,p-1)此外,
$$I_{1}\bigl((k+1)\tau\bigr)=我_{0}个^{k+1}+\frac{\sigmaq(1-q^{k})}{1-q}$$
和
$$\开始{对齐}[b]I_{1}\bigl[(k+1)\tau\bigr]-I_{1{[k\tau]&=I_{0}个^{k+1}+\frac{\sigmaq(1-q^{k})}{1-q}-I_{0}个^{k}-\frac{\sigmaq(1-q^{k-1})}{1-q}\\&=I_{0}个^{k} (q-1)+\σq^{k}=q^{k}\bigl[\σ-(1-q)I_{0}\bigr]。\结束{对齐}$$
(16)
如果\(I_{0}\geq\frac{\sigma}{1-\exp(-d_{I}\tau)}\),\(I_{1}(t)\)是单调递减的\((kτ,(k+1)τ]\)和\({I_{1}((k+1)τ),\(k=0,1,2,\ldot,p-1)是单调递减序列,那么\(I_{1}(t)\geqI_{1}(p\tau)=I_1}(t)=I_{0}\exp(-d_{I}t)+\frac{\sigma[\exp.
相反,如果\(I_{0}<\frac{\sigma}{1-\exp(-d_{I}\tau)}\),\(I_{1}(t)\)是单调递减的\((kτ,(k+1)τ]\)和\({I_{1}((k+1)τ),\(k=0,1,2,\ldot,p-1)是单调递增序列,那么\(I_{1}(t)\geq I_{1}(\tau)=I_{0}\exp(-d的。这就完成了证明。□
提议2
如果
\(一)_{0}-\压裂{\sigma{1}}{d_{i}}>\frac{\simma}{1-\exp(-d{i}\tau)}\),然后是系统(5)满足
$$I_{0}>I_{2}(t)\geqI_2}(t)\triangleqI_1}(d)+frac{\sigma_{1}}{d_{I}}\bigl(1-\exp(-d_{I}t)\bigr)$$
如果
\(I_{0}\geq\frac{\sigma{1}}{d_{I}}\)
和
\(一)_{0}-\压裂{\sigma{1}}{d{i}}\leq\frac{\simma}{1-\exp(-d{i}\tau)}\),然后是系统(5)满足
$$I{2}(t)\geqI{2{(\tau)\triangleqI{0}\exp(-d_{I}\tau$$
和
$$\开始{对齐}[b]I{2}(t)\leq{}&I{2neneneep \bigl[(p-1)\tau^{+}\bigr]\\={}&I{2}\bigl[(t-\tau)^{+{2}\ bigr]\triangleq I{0}\exp\bigl(-d_{I}(t-\tao)\bigr)\\&+\frac{\sigma{1}{d_{I}\ bigl[1-\exp\bigl(-d_{I}(t-\tau)\biger)\bigr]+\frac{\sigma[1-\exp。\结束{对齐}$$
证明
发件人(7),我们知道如果\(I_{0}\geq\frac{\sigma{1}}{d_{I}}\),然后\(I_{2}(t)\)是单调递减的\((kτ,(k+1)τ]\),\(k=0,1,2,\ldot,p-1)此外,
$$I_{2}\bigl((k+1)\tau\bigr)=I_{0}个^{k+1}+\frac{\sigma{1}}{d_{i}\bigl(1-q^{k+1}\biger)+\frac{\simma q(1-q_{k})}{1-q}$$
和
$$I_{2}\bigl((k+1)\tau^{+}\bigr)=我_{0}个^{k+1}+\frac{\sigma\{1}}{d_{i}}\bigl(1-q^{k+1}\bigr)+\frac{\sigma(1-q^{k+1})}{1-q}$$
通过计算,我们可以很容易地得到
$$\开始{对齐}[b]I_{2}\bigl[(k+1)\tau\bigr]-I_{2{[k\tau]&=I_{0}个^{k} (q-1)+\frac{\sigma{1}}{d_{i}}q^{k}(1-q)+\sigma q^{k}\\&=q^{k}\biggl[i{0}(q-1 d_{i}}-i{0}\biggr)+\sigma\biggr]\end{aligned}$$
(17)
和
$$I{2}\bigl[(k+1)\tau^{+}\bigr]-I{2}\ bigl[k\tau^}+}\bigr]=q^{k}\biggl[(1-q)\biggl(\frac{\sigma{1}}{d_{I}}-I{0}\bighr)+\sigma\biggr]$$
(18)
所以如果\(一)_{0}-\压裂{\sigma{1}}{d_{i}}>\frac{\simma}{1-\exp(-d{i}\tau)}\),然后\(I_{2}(t)\)是单调递减的\((kτ,(k+1)τ],{I_{2}((k+1)τ)和\({I_{2}((k+1)\tau^{+})\}\),\(k=0,1,2,\ldot,p-1),是单调递减序列,因此我们得到\(I_{2}(t)和\(I_{2}(t)\leqI_{2}(0^{+})=I_{0}\)对于\(在[0,t]\中).
如果\(I_{0}\geq\frac{\sigma{1}}{d_{I}}\)和\(一)_{0}-\压裂{\sigma{1}}{d{i}}\leq\frac{\simma}{1-\exp(-d{i}\tau)}\),然后\(I_{2}(t)\)是单调递减的\((kτ,(k+1)τ]\),\(I_{2}((k+1)\tau)和\({I_{2}((k+1)\tau^{+})\}\),\(k=0,1,2,\ldot,p-1),是单调递增序列,因此我们得到\(I_{2}(t)\geq I_{2](\tau)\)和\(I_{2}(t)\leq I_{2]((p-1)\tau^{+})=I_{2}[(t-\tau)^{+{]\)对于\(在[0,t]\中)。这就完成了证明。□
3.3系统解的存在性(2)
现在我们讨论系统解的存在性(2)用于有限时间血糖控制。
为了方便起见,我们表示
$$\beart{collected}f_{i}(t_{1},t_{2})=\biggl(\frac{n+i_{i}(t_{2})}{n+i_{i}(t_{1})}\biggr)^{\frac{am}{d_{i}}}},\ quad i=1,2,\ qquad G_{s}=\frac{G_{\mathrm{in}+b}{\sigma_{2}+ac},\\d_{1}=\sigma_{2}+ac+\frac{am\sigma\{1}}{d_{i}(n+i_{2}(t))},\quad d_{2}=\sigma\{2}+ac+\frac{am\sigma\{1}}{d_{i}(n+i_{2}(\tau)))}。\结束{聚集}$$
定理3.1
假设
\(一)_{0}-\压裂{\sigma{1}}{d_{i}}>\frac{\simma}{1-\exp(-d{i}\tau)}\)
以下两个条件成立:
$$G_{0}f_{1}(0,T)-G_{s}\leq(G_{U} -G_{s} )\exp\bigl[(\sigma_{2}+ac)T\bigr]$$
(19)
和
$$G_{0}\exp(-D_{1} 吨)+\裂缝{1}{D_{1}}(G_{\mathrm{in}}+b)\bigl(1-\exp(-D_{1} 吨)\大)\geq G_{L}f_{2}(T,0)$$
(20)
然后是系统(2)具有满足初边值问题的解.
证明
根据提议(1)和不平等(10),我们可以
$$开始{对齐}和开始{校准}[b]和\int_{0}^{T}\exp\bigl((\sigma{2}+ac)u\bigr)\bigle(n+I{1}(u)\biger)^{-\frac{am}{d_{I}},du\\&\quad\leq\int_0}{1}(T)\biger)^{-\frac{am}{d_{I}}\,du\\&\quad=\bigl(n+I{1}(T)\ biger)T\bigr)-1\bigr],\end{对齐}\end}对齐}$$
(21)
$$开始{对齐}和开始{校准}[b]G(t)\leq{}和\exp\bigl(-(\sigma{2}+ac)t\bigr)\bigle(n+I_{1}(t)\biger)^{frac{am}{d_{I}}}\\&\times\biggl\{frac}G{0}}{(n+1{0})+(G_{\mathrm{in}}+b)\bigl(n+I_{1}(t)\bigr)^{-\frac{am}{d_{I}}\frac}1}{\sigma{2}+ac}\bigl[\exp\ bigl\exp\bigl(-(\sigma{2}+ac)T\bigr)f_{1}(0,T)+G_{s}\bigl[1-\exp\bigl。\end{aligned}\end{alinged}$$
(22)
根据提议(2)和不平等(11),我们有
$$开始{聚集}开始{对齐}\exp\biggl(-\frac{am}{d_{i}}\int_{0}^{T}\frac}\sigma{1}}{n+i{2}(s)}\,ds\biggr)和\geq\exp\biggl}(T)}\,ds\biggr)\\&=\exp\biggl(-\frac{am\sigma_{1} 吨}{d_{i}(n+i{2}(T))}\biggr_{1} 吨}{d_{i}(n+i{2}(T))}\biggr)\exp\biggl(\frac{am\sigma_{1} u个}{d_{i}(n+i_{2}(T))}\biggr),\\f_{2{(u,T)\geqf_{2](0,T),\end{聚集}$$
我们可以得到
$$开始{对齐}[b]&\int_{0}^{T}\exp\bigl((\sigma{2}+ac)u\bigr(0,T)\exp\biggl(-\frac{am\sigma_{1} 吨}{d_{i}(n+i{2}(T))}\biggr)\int_{0}^{T}\exp(d_{1} u个)\,du\\&\quad=\frac{f_{2}(0,T)}{D_{1}}\exp\biggl(-\frac{am\sigma_{1} 吨}{d_{i}(n+i{2}(T))}\biggr)\bigl[\exp(d_{1} 吨)-1\bigr]\end{对齐}$$
(23)
然后根据(11)
$$\开始{对齐}[b]G(T)\geq{}&G_{0}\exp\bigl(-(\sigma{2}+ac)T\bigr)f_{2}(0,T)\exp\biggl(-\frac{am\sigma_{1} 吨}{d_{i}(n+i{2}(T))}\biggr_{1} 吨}{d_{i}(n+i{2}(T))}\biggr)\bigl[\exp(d_{1} 吨)-1\bigr]\\={}&f_{2}(0,T)\exp(-D_{1}T_{1} 吨)-1\bigr]\biggr\}\geq G_{L}。\结束{对齐}$$
(24)
这就完成了证明。□
定理3.2
假设
\(I_{0}\geq\frac{\sigma{1}}{d_{I}}\),\(I_{0}\geq\frac{\sigma}{1-\exp(-d_{I}\tau)}\),\(一)_{0}-\压裂{\sigma{1}}{d{i}}\leq\frac{\simma}{1-\exp(-d{i}\tau)}\),不平等(19)以下条件成立:
$$G_{0}f_{2}(0,T)+\frac{1}{D_{2{}(G_{\mathrm{in}}+b)\exp(-am\tau)\bigl[\exp_{2} T型)-1\bigr]\geq G_{L}\exp(D_{2} T型), $$
(25)
然后是系统(2)具有满足初边值问题的解.
证明
根据提议(2)和不平等(11),我们有
$$开始{对齐}和开始{校准}\exp\biggl(-\frac{am}{d_{i}}\int_{0}^{T}\frac}\sigma{1}}{n+i{2}(s)}\,ds\biggr)和\geq\exp\biggl 2}(\tau)}\,ds\biggr)\\&=\exp\biggl(-\frac{am\sigma_{1} 吨}{d_{i}(n+i{2}(\tau))}\biggr_{1} 吨}{d_{i}(n+i{2}(\tau))}\biggr)\exp\biggl(\frac{am\sigma_{1} u个}{d_{i}(n+i{2}(\tau))}\biggr),\\&\开始{aligned}和\frac{n+i}2}_{0}-\压裂{\sigma{1}}{d_{i}}_{0}-\压裂{\sigma{1}}{d_{i}})\exp(-d_{i}(T-\tau))+\frac{\simma(1-\exp),\结束{对齐}$$
然后我们得到
$$\开始{对齐}[b]&\int_{0}^{T}\exp\bigl((\sigma{2}+ac)u\bigr)\exp\biggl(-\frac{am}{d_{i}}\int__{u}^{T}\frac}\sigma{1}}{n+i{2}(s)}\,ds\biggr)f_2}(u,T)\,du\\&\quad\geq\exp(-am \tau)\exp\biggl(-\frac{am\sigma_{1} 吨}{d_{i}(n+i_{2}(\tau))}\biggr)\int_{0}^{T}\exp(d_{2} u个)\,du\\&\quad=\frac{\exp(-am\tau)}{D_{2}}\exp\biggl(-\frac{am\sigma_{1} 吨}{d_{i}(n+i_{2}(\tau))}\biggr)\bigl[\exp(d_{2} T型)-1\bigr]。\结束{对齐}$$
(26)
根据(11)
$$\开始{对齐}[b]G(T)\geq{}&G_{0}\exp\bigl(-(\sigma{2}+ac)T\bigr)f_{2}(0,T)\exp\biggl(-\frac{am\sigma_{1} 吨}{d_{i}(n+i_{2}(\tau))}\biggr)\\&+(G_{_{1} 吨}{d_{i}(n+i_{2}(\tau))}\biggr)\bigl[\exp(d_{2} T型)-1\bigr]\\={}&\exp(-D_{2} T型)\biggl \{G_{0}f_{2}(0,T)\&+\frac{(G_{\mathrm{in}}+b)\exp(-am\tau)}{D_{2{}}\bigl[\exp_{2} T型)-1\bigr]\biggr\}\geq G_{L}。\结束{对齐}$$
(27)
通过对定理的类似讨论3.1,我们还可以\(G(T)\leq G_{U}\)。这就完成了证明。□
定理3.3
假设
\(I_{0}\geq\frac{\sigma{1}}{d_{I}}\),\(I_{0}\leq\frac{\sigma}{1-\exp(-d_{I}\tau)}\),\(一)_{0}-\压裂{\sigma{1}}{d{i}}\leq\frac{\simma}{1-\exp(-d{i}\tau)}\),不平等(25)以下条件成立:
$$G_{0}f_{1}(0,T)-G_{s}f_1}(\tau,T)\leq\bigl(G_{U} -G_{s} f_{1}(\tau,T)\bigr)\exp\bigl((\sigma_{2}+ac)T\biger)$$
(28)
然后是系统(2)具有满足初边值问题的解.
证明
因为\(I_{0}\geq\frac{\sigma{1}}{d_{I}}\)和\(I_{0}\leq\frac{\sigma}{1-\exp(-d_{I}\tau)}\),我们有\(一)_{0}-\压裂{\sigma{1}}{d{i}}\leq\frac{\simma}{1-\exp(-d{i}\tau)}\).
根据提议(1)和不平等(10),我们可以
$$开始{对齐}和开始{校准}[b]和\int_{0}^{T}\exp\bigl((\sigma{2}+ac)u\bigr)\bigle(n+I{1}(u)\biger)^{-\frac{am}{d_{I}},du\\&\quad\leq\int_0}{1}(\tau)\bigr)^{-\frac{am}{d_{I}}\,du\\&\quad=\bigl(n+I{1}(\tau)\biger)^{\\frac{am}{d_{I}{}}\frac}1}{\sigma{2}+ac}\bigl[\exp\bigl[(\sigma{2}+ac)T\bigr)-1\bigr],\end{对齐}\end}对齐}$$
(29)
$$开始{对齐}和开始{校准}[b]G(t)\leq{}和\exp\bigl(-(\sigma{2}+ac)t\bigr)\bigle(n+I_{1}(t)\biger)^{frac{am}{d_{I}}}\\&\times\biggl\{frac}G{0}}{(n+1{0})+(G{\mathrm{in}}+b)\bigl\exp\bigl(-(\sigma{2}+ac)T\bigr)f_{1}(0,T)+G_{s}f_{1}(\tau,T)\bigl[1-\exp\bigle(-。\end{aligned}\end{alinged}$$
(30)
通过对定理的类似讨论3.2,我们还可以\(G(T)\geq G_{L}\)。这就完成了证明。□