Arms Race. A simplified mathematical model for an arms race between two countries whose expenditures for defense are expressed by the variables x(t) and y(t) is given by the linear system

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{a}\mathbf{;} \mathbf{x}\left(\mathbf{0}\right)\mathbf{=}\mathbf{1}\mathbf{,} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{+}\mathbf{b}\mathbf{;} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{4}\mathbf{,} \end{gathered}$$

Where a and b are constants that measure the trust (or distrust) each country has for the other. Determine whether there is going to be disarmament (x and y approach 0 as t increases), a stabilized arms race (x and y approach a constant as$\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$ ), or a runaway arms race (x and y approach$\mathbf{+}\mathbf{\infty }$ as $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$).

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system

$$\begin{gathered} {\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,} \end{gathered}$$

Where${\mathbf{L}}_{\mathbf{1}}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}\mathbf{,}$and${\mathbf{L}}_{\mathbf{4}}$are polynomials in $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$:

Given that:

$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{2}\mathbf{y}\mathbf{-}\mathbf{x}\mathbf{+}\mathbf{a}$…… (1)

$\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{+}\mathbf{b}$…… (2)

Rewrite the system in operator form:

$\left(\mathbf{D}\mathbf{+}1\right)\left[\mathbf{x}\right]\mathbf{-}2\mathbf{y}\mathbf{=}\mathbf{a}$ …… (3)

$\mathbf{-}4\mathbf{x}\mathbf{+}\left(\mathbf{D}\mathbf{+}3\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{b}$…… (4)

Multiply 4 on equation (3) and multiply$\mathbf{D}\mathbf{+}1$on equation (4). Then, add them together to get.

$$\begin{gathered} \mathbf{4}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{-}\mathbf{4}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\mathbf{x}\mathbf{+}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b} \\ \left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3}\mathbf{-}\mathbf{8}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{-}\mathbf{5}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{-}\mathbf{5}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b} \dots \left(5\right) \end{gathered}$$

Since the auxiliary equation to the corresponding homogeneous equation is ${\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{r}\mathbf{-}\mathbf{5}\mathbf{=}\mathbf{0}$.

Then,

$$\begin{gathered} \mathbf{r}\mathbf{=}\frac{\mathbf{-}4\pm \sqrt{{\left(4\right)}^{\mathbf{2}}\mathbf{+}4\mathbf{\times }5}}{\mathbf{2}} \\ \mathbf{=}\frac{\mathbf{-}4\pm \sqrt{16\mathbf{+}20}}{\mathbf{2}} \\ \mathbf{=}\frac{\mathbf{-}4\pm \sqrt{36}}{\mathbf{2}} \\ \mathbf{=}\frac{2\left(-2\pm 3\right)}{\mathbf{2}} \\ \mathbf{=}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{5} \end{gathered}$$

So, the roots are r =1 and r = -5 .

Then, the general solution of y is ${\mathbf{y}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}$…… (6)

Let us assume that,${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\mathbf{C}$ …… (7)

Substitute the equation (7) in equation (5).

$$\begin{gathered} \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{-}\mathbf{5}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{-}\mathbf{5}\right)\left[\mathbf{C}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b} \\ \mathbf{-}\mathbf{5}\mathbf{C}\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b} \\ \mathbf{C}\mathbf{=}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}} \end{gathered}$$

Substitute the value of C in equation (7).

$$\begin{gathered} \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{y}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{=}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}} \end{gathered}$$

So, the general solution is$\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}}$ …… (8)

Substitute the equation (8) in equation (4).

$$\begin{gathered} \mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{b} \\ \mathbf{-}\mathbf{4}\mathbf{x}\mathbf{=}\mathbf{b}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right] \\ \mathbf{=}\mathbf{b}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}}\right] \\ \mathbf{=}\mathbf{b}\mathbf{-}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{5}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\mathbf{3}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{-}\mathbf{3}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{+}\frac{\mathbf{12}\mathbf{a}\mathbf{+}\mathbf{3}\mathbf{b}}{\mathbf{5}} \\ \mathbf{=}\mathbf{-}\mathbf{4}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{+}\frac{\mathbf{12}\mathbf{a}\mathbf{+}\mathbf{8}\mathbf{b}}{\mathbf{5}} \\ \mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{4}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{+}\frac{\mathbf{12}\mathbf{a}\mathbf{+}\mathbf{8}\mathbf{b}}{\mathbf{5}}}{\mathbf{-}\mathbf{4}} \\ \mathbf{=}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{3}\mathbf{a}\mathbf{+}\mathbf{2}\mathbf{b}}{\mathbf{5}} \end{gathered}$$

So, $\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{3}\mathbf{a}\mathbf{+}\mathbf{2}\mathbf{b}}{\mathbf{5}}$…… (9)

To find:$\underset{t\to \infty }{lim}x$ and $\underset{t\to \infty }{lim}y$.

Implement the limits on equations (8) and (9).

role="math" localid="1664011682889" $$\begin{gathered} \underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\left[{\mathbf{Ae}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{3}\mathbf{a}\mathbf{+}\mathbf{2}\mathbf{b}}{\mathbf{5}}\right] \\ \mathbf{=}\mathbf{-}\frac{\mathbf{3}\mathbf{a}\mathbf{+}\mathbf{2}\mathbf{b}}{\mathbf{5}} \end{gathered}$$

role="math" localid="1664011701407" $$\begin{gathered} \underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\left[{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}}\right] \\ \mathbf{=}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}} \end{gathered}$$

Hence, the limits of the functions are constant. And the given solutions are a case of the stabilized arms race.