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Chapter 5: Q13RP (page 306)

Solve the related phase plane equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows) and describe the stability of the critical points (i.e., compare with Figure \({\bf{5}}{\bf{.12}}\), page \({\bf{267}}\)).

\(\begin{array}{c}{\bf{x' = 4 - 4y}}\\{\bf{y' = - 4x}}\end{array}\)

Short Answer

Expert verified

The related phase plane equation for the given system is\({{\bf{x}}^{\bf{2}}}{\bf{ - (y - 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}.\)

The critical point \(\left( {{\bf{0,1}}} \right)\) is the saddle and it is unstable.

Step by step solution

01

Finding the critical point

Let's first find the critical point. One has to solve the system\({\bf{x' = 0,y' = 0}}\), so, one has:

\(\begin{array}{c}{\bf{0 = 4 - 4y}} \Rightarrow {\bf{y = 1,}}\\{\bf{0 = - 4x}} \Rightarrow {\bf{x = 0}}\end{array}\)

So, the critical point is\(\left( {{\bf{x, y}}} \right){\bf{ = }}\left( {{\bf{0,1}}} \right)\). Now one can solve the related phase plane equation which is:

\(\begin{array}{c}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}\frac{{\frac{{{\bf{dy}}}}{{{\bf{dt}}}}}}{{\frac{{{\bf{dx}}}}{{{\bf{dt}}}}}}{\bf{ = }}\frac{{{\bf{ - 4x}}}}{{{\bf{4 - 4y}}}}\\{\bf{ = }}\frac{{{\bf{4x}}}}{{{\bf{4y - 4}}}}\\{\bf{(4y - 4)dy = 4xdx}}\end{array}\)

02

Integrate the given equation

Integrating the previous equation, one will get:

\(\begin{array}{c}\int {{\bf{(4y - 4)}}} {\bf{dy = }}\int {\bf{4}} {\bf{xdx}}\\{\bf{4}}\frac{{{{\bf{y}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ - 4y = 4}}\frac{{{{\bf{x}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ + C}}\\{{\bf{y}}^{\bf{2}}}{\bf{ - 2y = }}{{\bf{x}}^{\bf{2}}}{\bf{ + }}\frac{{\bf{C}}}{{\bf{2}}}\\{{\bf{x}}^{\bf{2}}}{\bf{ - (y - 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = - 1 - }}\frac{{\bf{C}}}{{\bf{2}}}\end{array}\)

Introducing a new constant \({\bf{c = - 1 - }}\frac{{\bf{C}}}{{\bf{2}}}\)one has that the solution is\({{\bf{x}}^{\bf{2}}}{\bf{ - (y - 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}\)which in \({\bf{xy - }}\)plane represents hyperbole with a focus on the critical point.

The critical point is the saddle and it is unstable. In the figure below are some trajectories with their flow arrows, the critical point, and one particular solution for \({\bf{c = 1}}\) (red line).

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Most popular questions from this chapter

In Problems 19 – 21, solve the given initial value problem.

dxdt=4x+y;    x0=1,dydt=-2x+y;   y0=0

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\({{\bf{d}}^{\bf{2}}}{\bf{ = (A - }}{{\bf{A}}^{\bf{*}}}{{\bf{)}}^{\bf{2}}}{\bf{ + 2A}}{{\bf{A}}^{\bf{*}}}{\bf{(1 - cos(}}\phi {\bf{ - }}{\phi ^ * }{\bf{))}}\).)

Two large tanks, each holding 100 L of liquid, are interconnected by pipes, with the liquid flowing from tank A into tank B at a rate of 3 L/min and from B into A at a rate of 1 L/min (see Figure 5.2). The liquid inside each tank is kept well stirred. A brine solution with a concentration of 0.2 kg/L of salt flows into tank A at a rate of 6 L/min. The (diluted) solution flows out of the system from tank A at 4 L/min and from tank B at 2 L/min. If, initially, tank A contains pure water and tank B contains 20 kg of salt, determine the mass of salt in each tank at a time t⩾0.

Solve the given initial value problem.

x'=y+z;x(0)=2y'=x+z;y(0)=2z'=x+y;z(0)=-1

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 rin (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.
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