Question

Here is a continuous model of a person’s glucose regulatory system. Compare this with Exercise 7.1.70. Let $\mathit{g}\left(t\right)$and $\mathit{h}\left(t\right)$be the excess glucose and insulin concentration in a person’s blood. We are told that

$\left|\begin{array}{c}\frac{dg}{dt}=-g-0.2h\\ \frac{dh}{dt}=0.6g-0.2h\end{array}\right|$

Where time$\mathit{t}$ is measured in hours. After a heavy holiday dinner, we measure $\mathit{g}\left(0\right)\mathbf{=}\mathbf{30}$androle="math" localid="1664881851746" $\mathbf{h}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ . Find closed formulas for$\mathit{g}\left(t\right)$ and$\mathit{h}\mathit{\left(}\mathit{t}\mathit{\right)}$ . Sketch the trajectory.

Short answer

Thus the solution is and the sketch for the trajectory is obtained.

Step-by-step solution

Calculate the Eigen values of given system.

A linear dynamical system is modelled by $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$. Here,A and B are$\mathit{n}\mathbf{\times }\mathit{n}$matrices, where $\mathit{n}$is the number of components of the system.

It is given that $g\left(0\right)=30$and $h\left(0\right)=0$so the system is written as follows:

$\left[\begin{array}{c}\frac{dg}{dt}\\ \frac{dh}{dt}\end{array}\right]=\left[\begin{array}{cc}-1& -0.2\\ 0.6& -0.2\end{array}\right]\left[\begin{array}{c}g\\ h\end{array}\right]$

To find solutions solve as follows:

$\begin{array}{rcl}det\left(\begin{array}{c}\left[\begin{array}{cc}-1& -0.2\\ 0.6& -0.2\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\end{array}\right)& =& 0\\ det\left[\begin{array}{cc}-1-\lambda & -0.2\\ 0.6& -0.2-\lambda \end{array}\right]& =& 0\\ {\lambda }^2+1.2\lambda +0.32& =& 0\\ \left(\lambda +0.8\right)\left(\lambda +0.4\right)& =& 0\\ & & \end{array}$

Solve further as:

$\begin{array}{rcl}\lambda & =& -0.4,-0.8\end{array}$

Calculate the Eigen vectors.

Now, the corresponding Eigen vectors are obtained as follows:

First let for $\lambda =-0.4$

$\begin{array}{rcl}\left[\begin{array}{cc}-1-\left(-0.4\right)& -0.2\\ 0.6& -0.2-\left(-0.4\right)\end{array}\right]\left[\begin{array}{c}g\\ h\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ -0.6g-0.2h& =& 0\\ -0.2h& =& 0.6g\\ h& =& -3g\end{array}$

Thus, the eigen vector is $\begin{array}{rcl}& & \left[\begin{array}{c}1\\ -3\end{array}\right]\end{array}$

Now, let $\begin{array}{rcl}\lambda & =& -0.4\end{array}$

$\begin{array}{rcl}\left[\begin{array}{cc}-1-\left(-0.8\right)& -0.2\\ 0.6& -0.2-\left(-0.8\right)\end{array}\right]\left[\begin{array}{c}g\\ h\end{array}\right]& =& \left[\begin{array}{c}0\\ 0\end{array}\right]\\ -0.2g-0.2h& =& 0\\ 0.2h& =& 0.2g\\ h& =& -g\end{array}$

The Eigen vector is$\begin{array}{rcl}& & \left[\begin{array}{c}1\\ -1\end{array}\right]\end{array}$

Obtain the general solution.

The general solution is calculated as follows:

$\left[\begin{array}{c}g\\ h\end{array}\right]=c_1\left[\begin{array}{c}1\\ -3\end{array}\right]e^{-0.4t}+c_2\left[\begin{array}{c}1\\ -1\end{array}\right]e^{-0.8t}$

Now, from the initial conditions it gives:

$$\begin{gathered} \left[\begin{array}{c}30\\ 0\end{array}\right]=c_1\left[\begin{array}{c}1\\ -3\end{array}\right]+c_2\left[\begin{array}{c}1\\ -1\end{array}\right] \\ \left\{\begin{array}{l}{c}_1+c_2=30\\ -3c_1-c_2=0\end{array}\right. \end{gathered}$$

Solve the above two equations gives:

$$\begin{gathered} -2c_1=30 \\ {c}_1=-15,c_2=45 \end{gathered}$$

Thus, the solution is written as follows:

Determine graph of the system.

The graph of the system is below:

This graph is drawn according to the solution of the system where,$c_1=-15,c_2=45$and the eigen values $\lambda =-0.4,-0.8$.