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Found in: Page 251

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# 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$\frac{\mathbf{dx}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{2}}{\mathbf{y}}{\mathbf{-}}{\mathbf{x}}{\mathbf{+}}{\mathbf{a}}{\mathbf{;}}{ }{ }{ }{ }{\mathbf{x}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{1}}{\mathbf{,}}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{dy}}{\mathbf{dt}}{\mathbf{=}}{\mathbf{4}}{\mathbf{x}}{\mathbf{-}}{\mathbf{3}}{\mathbf{y}}{\mathbf{+}}{\mathbf{b}}{\mathbf{;}}{ }{ }{ }{\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{4}}{\mathbf{,}}$ 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 $\mathbf{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$).

Therefore, the given solutions are a case of the stabilized arms race.

See the step by step solution

## Step 1: General form

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system

${\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,}\phantom{\rule{0ex}{0ex}}{\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,}$

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

(a) Make sure that the system is written in operator form.

(b) Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.

(c) (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), and y(t) give the desired general solution.

(d) Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants- twice as many as needed.]

(e) Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t) and y(t) in terms of the remaining constants.

Vieta’s formulas for finding roots:

For function y(t) to be bounded when $\mathbf{t}\to \mathbf{+}\infty$ we need for both roots of the auxiliary equation to be non-positive if they are reals and, if they are complex, then the real part has to be non-positive. In other words,

1. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\in \mathbf{R}$, then ${\mathbf{r}}_{\mathbf{1}}·{\mathbf{r}}_{\mathbf{2}}⩾\mathbf{0}\mathbf{,}{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}⩽\mathbf{0}$,
2. If ${\mathbf{r}}_{\mathbf{1}}\mathbf{,}{\mathbf{r}}_{\mathbf{2}}\mathbf{=}\mathbf{\alpha }±\mathbf{\beta i}$,$\mathbf{\beta }\ne 0$ , then $\mathbf{\alpha }\mathbf{=}\frac{{\mathbf{r}}_{\mathbf{1}}\mathbf{+}{\mathbf{r}}_{\mathbf{2}}}{\mathbf{2}}⩽0$.

## Step 2: Evaluate the given equation

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.

$\mathbf{4}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{-}\mathbf{4}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\mathbf{x}\mathbf{+}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}\phantom{\rule{0ex}{0ex}}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{-}\mathbf{8}\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}\phantom{\rule{0ex}{0ex}}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{+}\mathbf{3}\mathbf{-}\mathbf{8}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}\phantom{\rule{0ex}{0ex}}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{-}\mathbf{5}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}\phantom{\rule{0ex}{0ex}}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{-}\mathbf{5}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b} \dots \left(5\right)$

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,

$\mathbf{r}\mathbf{=}\frac{\mathbf{-}4±\sqrt{{\left(4\right)}^{\mathbf{2}}\mathbf{+}4\mathbf{×}5}}{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\frac{\mathbf{-}4±\sqrt{16\mathbf{+}20}}{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\frac{\mathbf{-}4±\sqrt{36}}{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\frac{2\left(-2±3\right)}{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{-}\mathbf{5}$

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

## Step 3: Solve the equations

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).

$\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{-}\mathbf{5}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}\phantom{\rule{0ex}{0ex}}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{D}\mathbf{-}\mathbf{5}\right)\left[\mathbf{C}\right]\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}\phantom{\rule{0ex}{0ex}}\mathbf{-}\mathbf{5}\mathbf{C}\mathbf{=}\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}\phantom{\rule{0ex}{0ex}}\mathbf{C}\mathbf{=}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}}$

Substitute the value of C in equation (7).

$\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{y}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\phantom{\rule{0ex}{0ex}}\mathbf{=}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}}$

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).

$\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{+}\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\mathbf{=}\mathbf{b}\phantom{\rule{0ex}{0ex}}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{=}\mathbf{b}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[\mathbf{y}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{b}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{3}\right)\left[{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{b}\mathbf{-}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{5}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\mathbf{3}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{-}\mathbf{3}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{+}\frac{\mathbf{12}\mathbf{a}\mathbf{+}\mathbf{3}\mathbf{b}}{\mathbf{5}}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}\mathbf{4}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{+}\frac{\mathbf{12}\mathbf{a}\mathbf{+}\mathbf{8}\mathbf{b}}{\mathbf{5}}\phantom{\rule{0ex}{0ex}}\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{4}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{+}\frac{\mathbf{12}\mathbf{a}\mathbf{+}\mathbf{8}\mathbf{b}}{\mathbf{5}}}{\mathbf{-}\mathbf{4}}\phantom{\rule{0ex}{0ex}}\mathbf{=}{\mathbf{Ae}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{3}\mathbf{a}\mathbf{+}\mathbf{2}\mathbf{b}}{\mathbf{5}}\phantom{\rule{0ex}{0ex}}$

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)

## Step 4: limit method

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" $\underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}\underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\left[{\mathbf{Ae}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{3}\mathbf{a}\mathbf{+}\mathbf{2}\mathbf{b}}{\mathbf{5}}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}\frac{\mathbf{3}\mathbf{a}\mathbf{+}\mathbf{2}\mathbf{b}}{\mathbf{5}}$

role="math" localid="1664011701407" $\underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\underset{\mathbf{t}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\left[{\mathbf{Ae}}^{\mathbf{t}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{5}\mathbf{t}}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}\frac{\mathbf{4}\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{5}}$

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