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Q1E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 281

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

Logistic Model. In Section 3.2 we discussed the logistic equation\(\frac{{{\bf{dp}}}}{{{\bf{dt}}}}{\bf{ = A}}{{\bf{p}}_{\bf{1}}}{\bf{p - A}}{{\bf{p}}^{\bf{2}}}{\bf{,p(0) = }}{{\bf{p}}_{\bf{o}}}\)and its use in modeling population growth. A more general model might involve the equation\(\frac{{{\bf{dp}}}}{{{\bf{dt}}}}{\bf{ = A}}{{\bf{p}}_{\bf{1}}}{\bf{p - A}}{{\bf{p}}^{\bf{r}}}{\bf{,p(0) = }}{{\bf{p}}_{\bf{o}}}\)where r>1. To see the effect of changing the parameter r in (25), take \({{\bf{p}}_{\bf{1}}}\)= 3, A = 1, and \({{\bf{p}}_{\bf{o}}}\)= 1. Then use a numerical scheme such as Runge–Kutta with h = 0.25 to approximate the solution to (25) on the interval\(0 \le {\bf{t}} \le 5\) for r = 1.5, 2, and 3What is the limiting population in each case? For r >1, determine a general formula for the limiting population.

The general formula for the limiting population is\({{\bf{3}}^{\frac{{\bf{1}}}{{{\bf{r - 1}}}}}}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Find the value of p

Here\({p_1} = 3\), \(A = 1\), and \({p_o} = 1\) then the equation is:

\(\frac{{dp}}{{dt}} = 3p - {p^r}\)

Suppose\(r > 1\). So,

\(\begin{array}{c}\int {\frac{1}{{3p - {p^r}}}dp} = \int {dt} \\\smallint \frac{1}{{{p^r}\left( {3{p^{1 - r}} - 1} \right)}}dp = t + c\\\int {\frac{1}{{3\left( {1 - r} \right)u}}du} = t + c\\\frac{{\ln \left| u \right|}}{{3\left( {1 - r} \right)}} = t + c\\\frac{{\ln \left| {3{p^{1 - r}} - 1} \right|}}{{3(1 - r)}} = t + c\\\ln \left| {3{p^{1 - r}} - 1} \right| = 3\left( {1 - r} \right)t + c\\{p^{1 - r}} = \frac{{C{e^{3(1 - r)t}} + 1}}{3}\\p = {\left( {\frac{{C{e^{3(1 - r)t}} + 1}}{3}} \right)^{\frac{1}{{1 - r}}}}\end{array}\)

Here \(p\left( 0 \right) = {p_o} = {\rm{ }}1\) then

\(\begin{array}{c}{\bf{1 = }}{\left( {\frac{{{\bf{C + 1}}}}{{\bf{3}}}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\\{\bf{C = 2}}\\{\bf{p = }}{\left( {\frac{{{\bf{2}}{{\bf{e}}^{{\bf{3(1 - r)t}}}}{\bf{ + 1}}}}{{\bf{3}}}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\end{array}\)

Step 2: Apply limits

\(\begin{array}{c}\mathop {{\bf{lim}}}\limits_{t \to \infty } {\bf{p(t) = }}\mathop {{\bf{lim}}}\limits_{t \to \infty } {\left( {\frac{{{\bf{2}}{{\bf{e}}^{{\bf{3(1 - r)t}}}}{\bf{ + 1}}}}{{\bf{3}}}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\\{\bf{ = }}{\left( {\frac{{2\mathop {{\bf{lim}}}\limits_{t \to \infty } {{\bf{e}}^{{\bf{3(1 - r)t}}}}{\bf{ + }}1}}{3}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\\{\bf{ = }}{\left( {\frac{{\bf{1}}}{{\bf{3}}}} \right)^{\frac{{\bf{1}}}{{{\bf{1 - r}}}}}}\\{\bf{ = }}{{\bf{3}}^{\frac{{\bf{1}}}{{{\bf{r - 1}}}}}}\end{array}\)

Step 3: Find the result by applying the Runge-Kutta method.

Apply this method in Mat Lab.

T	R=1.5	R=2	R=3
0	1	1	1
1	4.3024	2.7277	1.7254
1.5	6.1387	2.9343	1.7319
2	7.444	2.985	1.732
3	8.6129	2.999	1.7321
4	8.911	3	1.732
5	8.980	3	1.732

Therefore, the general formula for the limiting population is\({{\bf{3}}^{\frac{{\bf{1}}}{{{\bf{r - 1}}}}}}\).

Most popular questions for Math Textbooks

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 $t \to + \infty$ ).

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), z(t).

$x' = 3 x + y - z, y' = x + 2 y - z, z' = 3 x + 3 y - z$

In Problems 1–7, convert the given initial value problem into an initial value problem for a system in normal form.

$x'' + y - x' = 2 t; x (3) = 5, x' (3) = 2, y'' - x + y = - 1; y (3) = 1, y' (3) = - 1$

[hint] $x_{1} = x, x_{2} = x', x_{3} = y, x_{4} = y'$

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

$\frac{d x}{d t} = y^{2} - 3 y + 2, \frac{d y}{d t} = (x - 1) (y - 2)$

Show that the Poincare map for equation (1) is not chaoticby showing that if\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{\nu }}_{\bf{o}}}{\bf{)}}\)and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{o}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{o}}{\bf{)}}\)are two initial values that define the Poincare maps\({\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{\nu }}_{\bf{n}}}{\bf{)}}\) and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{n}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{n}}{\bf{)}}\), respectively, using the recursive formulas in (3), then one can make the distance between\({\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{,}}{{\bf{\nu }}_{\bf{n}}}{\bf{)}}\)and\({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{n}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{n}}{\bf{)}}\)small by making the distance between\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{\nu }}_{\bf{o}}}{\bf{)}}\) and \({\bf{(}}{{\bf{x}}^{\bf{*}}}_{\bf{o}}{\bf{,}}{{\bf{\nu }}^{\bf{*}}}_{\bf{o}}{\bf{)}}\)small. (Hint: Let \({\bf{(A,}}\phi {\bf{)}}\)and \({\bf{(}}{{\bf{A}}^{\bf{*}}}{\bf{,}}{\phi ^ * }{\bf{)}}\) be the polar coordinates of two points in the plane. From the law of cosines, it follows that the distance d between them is given by\({{\bf{d}}^{\bf{2}}}{\bf{ = (A - }}{{\bf{A}}^{\bf{*}}}{{\bf{)}}^{\bf{2}}}{\bf{ + 2A}}{{\bf{A}}^{\bf{*}}}{\bf{(1 - cos(}}\phi {\bf{ - }}{\phi ^ * }{\bf{))}}\).)

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