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Q38E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 251

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

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
$\frac{dx}{dt} = 2 y - x + a; x (0) = 1, \frac{dy}{dt} = 4 x - 3 y + b; y (0) = 4,$
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 $t \to + \infty$ ), or a runaway arms race (x and y approach $+ \infty$ as $t \to + \infty$ ).

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

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: General form

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system

$L_{1} [x] + L_{2} [y] = f_{1}, L_{3} [x] + L_{4} [y] = f_{2},$

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

(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 $t \to + \infty$ 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,

If $r_{1}, r_{2} \in R$ , then $r_{1} \cdot r_{2} ⩾ 0, r_{1} + r_{2} ⩽ 0$ ,
If $r_{1}, r_{2} = α \pm βi$ , $β \neq 0$ , then $α = \frac{r_{1} + r_{2}}{2} ⩽ 0$ .

Step 2: Evaluate the given equation

Given that:

$\frac{dx}{dt} = 2 y - x + a$ …… (1)

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

Rewrite the system in operator form:

$(D + 1) [x] - 2 y = a$ …… (3)

$- 4 x + (D + 3) [y] = b$ …… (4)

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

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

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

Then,

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

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

Step 3: Solve the equations

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

Let us assume that, $y_{p} (t) = C$ …… (7)

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

$(D^{2} + 4 D - 5) [y] = 4 a + b (D^{2} + 4 D - 5) [C] = 4 a + b - 5 C = 4 a + b C = - \frac{4 a + b}{5}$

Substitute the value of C in equation (7).

$y (t) = y_{h} (t) + y_{p} (t) = {Ae}^{t} + {Be}^{- 5 t} - \frac{4 a + b}{5}$

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

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

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

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

Step 4: limit method

To find: $\underset{t \to \infty}{l i m} x$ and $\underset{t \to \infty}{l i m} y$ .

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

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

role="math" localid="1664011701407" $\lim_{t \to \infty} y (t) = \lim_{t \to \infty} [{Ae}^{t} + {Be}^{- 5 t} - \frac{4 a + b}{5}] = - \frac{4 a + b}{5}$

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

Most popular questions for Math Textbooks

A Problem of Current Interest. The motion of an ironbar attracted by the magnetic field produced by a parallel current wire and restrained by springs (see Figure 5.17) is governed by the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{x}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ = - x + }}\frac{{\bf{1}}}{{{\bf{\lambda - x}}}}\) for \({\bf{ - }}{{\bf{x}}_{\bf{o}}}{\bf{ < x < \lambda }}\)where the constants \({{\bf{x}}_{\bf{o}}}\) and \({\bf{\lambda }}\) are, respectively, the distances from the bar to the wall and to the wire when thebar is at equilibrium (rest) with the current off.

Setting\({\bf{v = }}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}\), convert the second-order equation to an equivalent first-order system.
Solve the related phase plane differential equation for the system in part (a) and thereby show that its solutions are given by\({\bf{v = \pm }}\sqrt {{\bf{C - }}{{\bf{x}}^{\bf{2}}}{\bf{ - 2ln(\lambda - x)}}} \), where C is a constant.
Show that if \({\bf{\lambda < 2}}\) there are no critical points in the xy-phase plane, whereas if \({\bf{\lambda > 2}}\) there are two critical points. For the latter case, determine these critical points.
Physically, the case \({\bf{\lambda < 2}}\)corresponds to a current so high that the magnetic attraction completely overpowers the spring. To gain insight into this, use software to plot the phase plane diagrams for the system when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\).
From your phase plane diagrams in part (d), describe the possible motions of the bar when \({\bf{\lambda = 1}}\) and when\({\bf{\lambda = 3}}\), under various initial conditions.

In Problems 15–18, find all critical points for the given system. Then use a software package to sketch the direction field in the phase plane and from this description the stability of the critical points (i.e., compare with Figure 5.12).

In Problems 25 – 28, use the elimination method to find a general solution for the given system of three equations in the three unknown functions x(t), y(t), and z(t).

$x' = x + 2 y + z, y' = 6 x - y, z' = - x - 2 y - z$

In Problems 19–24, convert the given second-order equation into a first-order system by setting v=y’. Then find all the critical points in the yv-plane. Finally, sketch (by hand or software) the direction fields, and describe the stability of the critical points (i.e., compare with Figure 5.12).

$y'' (t) + y (t) - y (t)^{3} = 0$

In Problems 3–6, find the critical point set for the given system.

$\frac{dx}{dt} = x - y, \frac{dy}{dt} = x^{2} + y^{2} - 1$

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