91影视

Given the national income of a country, which consists of consumption, investment, and government expenditures.

The government expenditure to be constant, at G₀, while the national income $Y\left(t\right)$, consumption $C\left(t\right)$, and investment $I\left(t\right)$change over time.

Accroding to a simple model

$$\begin{gathered} \left|\mathrm{Y}\left(\mathrm{t}\right)=\mathrm{C}\left(\mathrm{t}\right)+\mathrm{I}\left(\mathrm{t}\right)+{\mathrm{G}}_0 \\ \mathrm{C}(\mathrm{t}+1)=\mathrm{纬驰}\left(\mathrm{t}\right) \\ \mathrm{I}(\mathrm{t}+1)=\mathrm{伪}(\mathrm{C}\left(\mathrm{t}+1\right)-\mathrm{C}\left(\mathrm{t}\right))\right| \end{gathered}$$ $$\begin{gathered} \left(0<纬<1\right), \\ \left(伪>0\right) \end{gathered}$$

where $纬$is the marginal propensity to consume and $伪$is the acceleration coefficient.

To find equilibrium solution of $Y\left(t+1\right)=Y\left(t\right),C\left(t+1\right)=C\left(t\right)$, and $I\left(t+1\right)=I\left(t\right)$, use the given equations.

Since. $I\left(t+1\right)=伪\left(C\left(t+1\right)-C\left(t\right)\right)$

Put $C\left(t+1\right)=C\left(t\right)$, this implies

$I\left(t+1\right)=0$which further implies $I\left(t\right)=0$

Put this in first equation $Y\left(t\right)=C\left(t\right)+I\left(t\right)+G_0$

This implies $Y\left(t\right)=C\left(t\right)+G_0$

Since $C\left(t+1\right)=C\left(t\right)$implies $C\left(t\right)=纬Y\left(t\right)$

So the equation becomes

$$\begin{gathered} Y\left(t\right)=纬Y\left(t\right)+G_0 \\ Y\left(t\right)-纬Y\left(t\right)=G_0 \\ Y\left(t\right)=\frac{G_0}{1-纬} \end{gathered}$$

And$C\left(t\right)=\frac{纬G_0}{1-纬}$

Hence the solutions are:

role="math" localid="1668415432322" width="87" height="122">It=0Yt=G01-纬Ct=纬G01-纬

Given that $y\left(t\right),c\left(t\right)andi\left(t\right)$are the deviations of $Y\left(t\right),C\left(t\right)$and $I\left(t\right)$, respectively, from the equilibrium state.

$$\begin{gathered} I\left(t\right)=0 \\ Y\left(t\right)=\frac{G_0}{1-纬} \\ C\left(t\right)=\frac{纬G_0}{1-纬} \end{gathered}$$

Also

$$\begin{gathered} \left|y\left(t\right)=c\left(t\right)+i\left(t\right) \\ c\left(t+1\right)=纬y\left(t\right) \\ i\left(t+1\right)=伪\left(c\left(t+1\right)-c\left(t\right)\right)\right| \end{gathered}$$

Using $y\left(t\right)=c\left(t\right)+i\left(t\right)inc\left(t+1\right)=纬y\left(t\right)andi\left(t+1\right)=伪\left(c\left(t+1\right)-c\left(t\right)\right)$implies

$$\begin{gathered} c\left(t+1\right)=纬c\left(t\right)+纬i\left(t\right) \\ i\left(t+1\right)=伪\left(纬c\left(t\right)+纬i\left(t\right)-c\left(t\right)\right) \\ i\left(t+1\right)=\left(伪纬-伪\right)c\left(t\right)+伪纬i\left(t\right) \end{gathered}$$

Hence,

$c\left(t+1\right)=纬c\left(t\right)+纬i\left(t\right)$and $i\left(t+1\right)=\left(伪纬-伪\right)c\left(t\right)+伪纬i\left(t\right)$are of the form

$$\begin{gathered} \left|c\left(t+1\right)=pc\left(t\right)+qi\left(t\right) \\ i\left(t+1\right)=rc\left(t\right)+si\left(t\right)\right| \end{gathered}$$

Using$c\left(t+1\right)=纬c\left(t\right)+纬i\left(t\right)$and$i\left(t+1\right)=\left(伪纬-伪\right)c\left(t\right)+伪纬i\left(t\right)$

The state transition matrix is

role="math" localid="1668416881433" $A=\left[\begin{array}{cc}纬& 纬\\ 伪纬-伪& 伪纬\end{array}\right]$

Put $伪=5$and $纬=0.2$

So$A=\left[\begin{array}{cc}0.2& 0.2\\ -4& 1\end{array}\right]$

Find eigenvalues of A

$$\begin{gathered} \left|A-位I\right|=0 \\ \left|\begin{array}{cc}0.2-位& 0.2\\ -4& 1-位\end{array}\right|=0 \\ \left(0.2-位\right)\left(1-位\right)+0.8=0 \\ {位}^2-1.2位+1=0 \end{gathered}$$

So, $位=0.6卤0.8i$.

Here,

$\begin{array}{rcl}\left|0.6卤0.8i\right|& =& \sqrt{{\left(0.6\right)}^2+{\left(0.8\right)}^2}\\ & =& \sqrt{0.36+0.64}\\ & =& \sqrt{1}\\ & =& 1\end{array}$

Since, the modulus is not less than 1, hence the zero system is not stable.

Using$c\left(t+1\right)=纬c\left(t\right)+纬i\left(t\right)$and$i\left(t+1\right)=\left(伪纬-伪\right)c\left(t\right)+伪纬i\left(t\right)$

The state transition matrix is

role="math" localid="1668416926816" $A=\left[\begin{array}{cc}纬& 纬\\ 伪纬-伪& 伪纬\end{array}\right]$

Put $伪=1$.

So $A=\left[\begin{array}{cc}纬& 纬\\ 纬-1& 纬\end{array}\right]$

Here, tr$A=2纬$

Also,

$\begin{array}{rcl}A& =& {纬}^2-纬\left(纬-1\right)\\ & =& {纬}^2-{纬}^2+纬\\ & =& 纬\end{array}$

Since, the modulus of both the complex numbers is less than 1 (as $0<纬<1$), hence the zero system is stable.

Using$c\left(t+1\right)=纬c\left(t\right)+纬i\left(t\right)andi\left(t+1\right)=\left(伪纬-伪\right)c\left(t\right)+伪纬i\left(t\right)$

The state transition matrix is

$A=\left[\begin{array}{cc}纬& 纬\\ 伪纬-伪& 伪纬\end{array}\right]$

Here,

$\begin{array}{rcl}tr\left(A\right)& =& \left(伪+1\right)纬\\ & >& 0\end{array}$

Now,

$\begin{array}{rcl}\det A& =& 伪{纬}^2-伪{纬}^2+伪纬\\ & =& 伪纬\end{array}$

A system is stable if,

$\left|tr\left(A\right)\right|-1<\det A<1$

So the condition becomes:

$$\begin{gathered} 伪纬<1 \\ 伪纬+纬-1<伪纬 \\ 纬<1 \end{gathered}$$

From the figure,

the condition$伪纬<1$ is satisfied by the sectors I and IV.

Hence, zero system is stable in sector I and IV.