Introduction to Systems and Phase Plane Analysis

Fluid Ejection. In the design of a sewage treatment plant, the following equation arises: $\mathbf{60}\mathbf{-}\mathbf{H}\mathbf{=}\mathbf{(}\mathbf{77}\mathbf{.}\mathbf{7}\mathbf{)}\mathbf{H}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{19}\mathbf{.}\mathbf{42}\mathbf{\left)}\mathbf{\right(}\mathbf{H}\mathbf{\text{'}}{\mathbf{)}}^{\mathbf{2}}\mathbf{;}\mathbf{H}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{H}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$ where H is the level of the fluid in an ejection chamber, and t is the time in seconds. Use the vectorized Runge–Kutta algorithm with h = 0.5 to approximate $\mathbf{H}\left(t\right)$ over the interval [0, 5].

Use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution to the initial value problem.

$$\begin{gathered} \mathrm{x}\text{'}=\mathrm{yz};\mathrm{x}\left(0\right)=0, \\ \mathrm{y}\text{'}=-\mathrm{xz};\mathrm{y}\left(0\right)=1, \\ \mathrm{z}\text{'}=\frac{-\mathrm{xy}}{2};\mathrm{z}\left(0\right)=1, \end{gathered}$$

At t=1.

Generalized Blasius Equation. H. Blasius, in his study of the laminar flow of a fluid, encountered an equation of the form $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{yy}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{=}\mathbf{(}\mathbf{y}\mathbf{\text{'}}{\mathbf{)}}^{\mathbf{2}}\mathbf{-}\mathbf{1}$. Use the Runge–Kutta algorithm for systems with h = 0.1 to approximate the solution that satisfies the initial conditions $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{32824}$. Sketch this solution on the interval [0, 2].

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

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

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

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{y}\mathbf{-}\mathbf{1} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{x}\mathbf{+}\mathit{y}\mathbf{+}\mathbf{5} \end{gathered}$$

In Problems 1 and 2, verify that the pair x(t), and y(t) is a solution to the given system. Sketch the trajectory of the given solution in the phase plane.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{3}{\mathbf{y}}^{\mathbf{3}}\mathbf{,}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{y}\mathbf{;} \\ \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{t}} \end{gathered}$$

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

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{x}\mathit{y}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{3}\mathit{x}\mathit{y}\mathbf{-}{\mathit{y}}^{\mathbf{2}}$

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

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

In Problems 7–9, solve the related phase plane differential equation (2), page 263, for the given system.

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{y}\mathbf{-}\mathbf{1}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{e}}^{\mathbf{x}\mathbf{+}\mathbf{y}}$

In Problems 7–9, solve the related phase plane differential equation (2), page 263, for the given system.

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{-}\mathbf{3}}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{x}\mathit{y}$

In Problems 7–9, solve the related phase plane differential equation (2), page 263, for the given system. $\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{2}\mathit{y}\mathbf{-}\mathit{x}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathit{y}$

Find all the critical points of the system

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{x}\mathit{y} \end{gathered}$$

and the solution curves for the related phase plane differential equation. Thereby proving that two trajectories lie on semicircles. What are the endpoints of the semicircles?

Find all the critical points of the system

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{x}\mathit{y}$

and the solution curves for the related phase plane differential equation. Thereby proving that two trajectories lie on semicircles. What are the endpoints of the semicircles?

In Problems 11–14, solve the related phase plane differential equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows).

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{8}\mathit{y}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{18}\mathit{x}$

In Problems 11–14, solve the related phase plane differential equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows).

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{(}\mathit{y}\mathbf{-}\mathit{x}\mathbf{)}\mathbf{(}\mathit{y}\mathbf{-}\mathbf{1}\mathbf{)} \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{(}\mathit{x}\mathbf{-}\mathit{y}\mathbf{)}\mathbf{(}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{)} \end{gathered}$$

In Problems 11–14, solve the related phase plane differential equation for the given system. Then sketch by hand several representative trajectories (with their flow arrows).

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{y}} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{x}} \end{gathered}$$

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 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).

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{5}\mathit{x}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{x}\mathbf{-}\mathbf{4}\mathit{y}$

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).

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{+}\mathbf{13}\mathit{y}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y}$

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).

$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{x}\mathbf{(}\mathbf{7}\mathbf{-}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{)}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{y}\mathbf{(}\mathbf{5}\mathbf{-}\mathit{x}\mathbf{-}\mathit{y}\mathbf{)}$

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).

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{-}\mathit{y}\mathbf{=}\mathbf{0}$

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).

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$

In Problems 19–24, convert the given second-order equation into the 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).

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathit{y}\mathbf{+}{\mathit{y}}^{\mathbf{5}}\mathbf{=}\mathbf{0}$

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).

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}{\mathit{y}}^{\mathbf{3}}\mathbf{=}\mathbf{0}$.

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).

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}\mathit{y}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{-}\mathit{y}\mathbf{(}\mathbf{t}{\mathbf{)}}^{\mathbf{4}}\mathbf{=}\mathbf{0}$

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).

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}\mathit{y}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{-}\mathit{y}\mathbf{(}\mathbf{t}{\mathbf{)}}^{\mathbf{3}}\mathbf{=}\mathbf{0}$

Using the software, sketch the direction field in the phase-plane for the system $\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{y}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{x}\mathbf{+}{\mathit{x}}^{\mathbf{3}}$From the sketch, conjecture whether the solution passing through each given point is periodic:

1. (0.25, 0.25)
2. (2, 2)
3. (1, 0)

Using the software, sketch the direction field in the phase-plane for the system $\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{y}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{x}\mathbf{-}{\mathit{x}}^{\mathbf{3}}\mathbf{.}$ From the sketch, conjecture whether all solutions of this system are bounded. Solve the related phase plane differential equation and confirm your conjecture.

Find the critical points and solve the related phase plane differential equation for the system $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{,}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{y}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{1}\mathbf{)}$.Describe (without using computer software) the asymptotic behavior of trajectories (as $\mathbf{t}\mathbf{\to }\mathbf{\infty }$) that start at

1. (3, 2)
2. (2, 1/2)
3. ( -2, 1/2)
4. (3, -2)

Using the software, sketch the direction field in the phase-plane for the system $\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{+}\mathit{y}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathbf{5}\mathit{x}\mathbf{-}\mathbf{4}\mathit{y}$. From the sketch, predict the asymptotic limit (as $\mathit{t}\mathbf{\to }\mathbf{\infty }$

 of the solution starting at (1, 1).

Figure 5.16 displays some trajectories for the system $\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{y}\mathbf{,}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\mathit{x}\mathbf{+}{\mathit{x}}^{\mathbf{2}}$What types of critical points (compare Figure 5.12 on page 267) occur at (0, 0) and (1, 0)?

A proof of Theorem 1, page 266, is outlined below. The goal is to show that\({\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) = g(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) = 0}}\). Justify each step

1. From the given hypotheses, deduce that\(\mathop {\lim }\limits_{t \to \infty } {\bf{x'(t) = f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}\) and\(\mathop {\lim }\limits_{t \to \infty } {\bf{y'(t) = g(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}\).
2. Suppose\({\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{) > 0}}\). Then, by continuity,\({\bf{x'(t) = }}\frac{{{\bf{f(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}\)for all large t (say, for\({\bf{t}} \ge {\bf{T}}\)). Deduce from this that\({\bf{x(t) > }}\frac{{{\bf{tf(}}{{\bf{x}}^{\bf{*}}}{\bf{,}}{{\bf{y}}^{\bf{*}}}{\bf{)}}}}{{\bf{2}}}{\bf{ + C}}\)fort>I, where C is some constant.
3. Conclude from part (b) that\(\mathop {\lim }\limits_{t \to \infty } {\bf{x(t)}} =  + \infty \), contradicting the fact that this limit is the finite number x*. Thus, f(x*, y*) cannot be positive.
4. Argue similarly that the supposition thatf(x*, y*) < 0 also leads to a contradiction; hence, f(x*, y*) must be zero.

In the same manner, argue that g(x*, y*) must bezero. Therefore, f(x*, y*) = g(x*, y*) = 0, and (x*, y*) isa critical point.

Phase plane analysis provides a quick derivation of the energy integral lemma of Section 4.8 (page 201). By completing the following steps, prove that solutions of equations of the special form \({\bf{y'' = f(y)}}\) satisfy\(\frac{{\bf{1}}}{{\bf{2}}}{{\bf{(y')}}^{\bf{2}}}{\bf{ - F(y) = constant}}\)

where F(y) is an antiderivative of f(y).

1. Set v = y’ and write \({\bf{y'' = f(y)}}\) as an equivalent first-order system.
2. Show that the solutions to the vy-phase plane equation for the system in part.
3. Satisfy\(\frac{{{{\bf{v}}^{\bf{2}}}}}{{\bf{2}}}{\bf{ = F(y) + K}}\). Replacing v by y’ then completes the proof.

Use the result of Problem 31 to prove that all solutions to the equation\({\bf{y'' + }}{{\bf{y}}^{\bf{3}}}{\bf{ = 0}}\)remain bounded. [Hint: Argue that \(\frac{{{{\bf{y}}^{\bf{4}}}}}{{\bf{4}}}\) is bounded above by the constant appearing in Problem 31.]

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.

1. Setting\({\bf{v = }}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}\), convert the second-order equation to an equivalent first-order system.
2. 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.
3. 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.
4. 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\({\bf{\lambda  = 3}}\).
5. 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.

Falling Object. The motion of an object moving vertically through the air is governed by the equation\(\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ =  - g - }}\frac{{\bf{g}}}{{{{\bf{V}}^{\bf{2}}}}}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}\left| {\frac{{{\bf{dy}}}}{{{\bf{dt}}}}} \right|\)where y is the upward vertical displacement and V is a constant called the terminal speed. Take \({\bf{g = 32ft/se}}{{\bf{c}}^{\bf{2}}}\)and V = 50 ft/sec. Sketch trajectories in the yv-phase plane for \( - 100 \le {\bf{y}} \le 100, - 100 \le {\bf{v}} \le 100\)starting from y = 0 and y = -75, -50, -25, 0, 25, 50, and 75 ft/sec. Interpret the trajectories physically; why is V called the terminal speed?

Sticky Friction. An alternative for the damping friction model F = -by′ discussed in Section 4.1 is the “sticky friction” model. For a mass sliding on a surface as depicted in Figure 5.18, the contact friction is more complicated than simply -by′. We observe, for example, that even if the mass is displaced slightly off the equilibrium location y = 0, it may nonetheless remain stationary due to the fact that the spring force -ky is insufficient to break the static friction’s grip. If the maximum force that the friction can exert is denoted by m, then a feasible model is given by

\({{\bf{F}}_{{\bf{friction}}}}{\bf{ = }}\left\{ \begin{array}{l}{\bf{ky,if}}\left| {{\bf{ky}}} \right|{\bf{ < }}\mu {\bf{andy' = 0}}\\\mu {\bf{sign(y),if}}\left| {{\bf{ky}}} \right| \ge {\bf{0andy' = 0}}\\ - \mu {\bf{sign(y'),ify'}} \ne 0.\end{array} \right.\)

(The function sign (s) is +1 when s 7 0, -1 when s 6 0, and 0 when s = 0.) The motion is governed by the equation (16) \({\bf{m}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{y}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ =  - ky + }}{{\bf{F}}_{{\bf{friction}}}}\)Thus, if the mass is at rest, friction balances the spring force if \(\left| {\bf{y}} \right|{\bf{ < }}\frac{\mu }{{\bf{k}}}\)but simply opposes it with intensity\(\mu \)if\(\left| {\bf{y}} \right| \ge \frac{\mu }{{\bf{k}}}\). If the mass is moving, friction opposes the velocity with the same intensity\(\mu \).

1. Taking m =\(\mu \) = k = 1, convert (16) into the firstorder system y′ = v (17)\({\bf{v' = }}\left\{ \begin{array}{l}{\bf{0,if}}\left| {\bf{y}} \right|{\bf{ < 1andv = 0}}{\bf{.}}\\{\bf{ - y + sign(y),if}}\left| {\bf{y}} \right| \ge {\bf{1andv = 0}}\\{\bf{ - y - sign(v),ifv}} \ne 0\end{array} \right.\) ,
2. Form the phase plane equation for (17) when v ≠ 0 and solve it to derive the solutions\({{\bf{v}}^{\bf{2}}}{\bf{ + (y \pm 1}}{{\bf{)}}^{\bf{2}}}{\bf{ = c}}\).where the plus sign prevails for v>0 and the minus sign for v<0.
3. Identify the trajectories in the phase plane as two families of concentric semicircles. What is the centre of the semicircles in the upper half-plane? The lower half-plane? 
4.  What are the critical points for (17)? 
5.  Sketch the trajectory in the phase plane of the mass released from rest at y = 7.5. At what value for y does the mass come to rest?

Rigid Body Nutation. Euler’s equations describe the motion of the principal-axis components of the angular velocity of a freely rotating rigid body (such as a space station), as seen by an observer rotating with the body (the astronauts, for example). This motion is called nutation. If the angular velocity components are denoted by x, y, and z, then an example of Euler’s equations is the three-dimensional autonomous system

\(\begin{array}{l}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = yz}}\\\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ =   - 2xz}}\\\frac{{{\bf{dz}}}}{{{\bf{dt}}}}{\bf{ = xy}}\end{array}\)

The trajectory of a solution x(t),y(t), z(t) to these equations is the curve generated by the points (x(t), y(t), z(t) ) in xyz-phase space as t varies over an interval I. 

(a) Show that each trajectory of this system lies on the surface of a (possibly degenerate) sphere centered at the origin (0, 0, 0).[Hint: Compute\(\frac{{\bf{d}}}{{{\bf{dt}}}}{\bf{(}}{{\bf{x}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + }}{{\bf{z}}^{\bf{2}}}{\bf{)}}\)What does this say about the magnitude of the angular velocity vector?

 (b) Find all the critical points of the system, i.e., all points\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{y}}_{\bf{o}}}{\bf{,}}{{\bf{z}}_{\bf{o}}}{\bf{)}}\) such that \({\bf{x(t) = }}{{\bf{x}}_{\bf{o}}}{\bf{,y(t) = }}{{\bf{y}}_{\bf{o}}}{\bf{,z(t) = }}{{\bf{z}}_{\bf{o}}}\) is a solution. For such solutions, the angular velocity vector remains constant in the body system. 

(c) Show that the trajectories of the system lie along the intersection of a sphere and an elliptic cylinder of the form\({{\bf{y}}^{\bf{2}}}{\bf{ + 2}}{{\bf{x}}^{\bf{2}}}{\bf{ = C}}\) for some constant C. [Hint: Consider the expression for dy/dx implied by Euler’s equations.] 

(d) Using the results of parts (b) and (c), argue that the trajectories of this system are closed curves. What does this say about the corresponding solutions?

(e) Figure 5.19 displays some typical trajectories for this system. Discuss the stability of the three critical points indicated on the positive axes.

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.

Radioisotopes and Cancer Detection. A radioisotope commonly used in the detection of breast cancer is technetium-99m. This radionuclide is attached to a chemical that upon injection into a patient accumulates at cancer sites. The isotope’s radiation is then detected and the site is located, using gamma cameras or other tomographic devices.

Technetium-99m decays radioactively in accordance with the equation\(\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ =  - ky}}\) with k = 0.1155>h. The short half-life of technetium-99m has the advantage that its radioactivity does not endanger the patient. A disadvantage is that the isotope must be manufactured in a cyclotron. Since hospitals are not equipped with cyclotrons, doses of technetium-99m have to be ordered in advance from medical suppliers.

Suppose a dosage of 5 millicuries (mCi) of technetium- 99m is to be administered to a patient. Estimate the delivery time from production at the manufacturer to arrival at the hospital treatment room to be 24 hours and calculate the amount of the radionuclide that the hospital must order, to be able to administer the proper

dosage.

Secretion of Hormones. The secretion of hormones into the blood is often a periodic activity. If a hormone is secreted on a 24-h cycle, then the rate of change of the level of the hormone in the blood may be represented by

the initial value problem\(\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = \alpha  - \beta cos}}\frac{{{\bf{\pi t}}}}{{{\bf{12}}}}{\bf{ - kx,x(0) = }}{{\bf{x}}_{\bf{o}}}\)where x(t) is the amount of the hormone in the blood at the time t, \({\bf{\alpha }}\) is the average secretion rate, \({\bf{\beta }}\)is the amount of daily variation in the secretion, and k is a positive constant reflecting the rate at which the body removes the hormone from the blood. If \({\bf{\alpha }}\)=\({\bf{\beta }}\) = 1, k = 2, and \({{\bf{x}}_{\bf{o}}}\) = 10, solve for x(t).

Suppose for a certain disease described by the SIR model it is determined that a = 0.003 and b = 0.5.

1. In the SI-phase plane, sketch the trajectory corresponding to the initial condition that one person is infected and 700 persons are susceptible.
2. From your graph in part (a), estimate the peak number of infected persons. Compare this with the theoretical prediction\({\bf{S = }}\frac{{\bf{k}}}{{\bf{a}}}\)=167 persons when the epidemic is at its peak.

Suppose for a certain disease described by the SIR model it is determined that a = 0.003 and b = 0.5.

1. In the SI-phase plane, sketch the trajectory corresponding to the initial condition that one person is infected and 700 persons are susceptible.
2. From your graph in part (a), estimate the peak number of infected persons. Compare this with the theoretical prediction\({\bf{S = }}\frac{{\bf{k}}}{{\bf{a}}}\)=167 persons when the epidemic is at its peak.

Show that the half-life of solutions to (2)—that is, the time required for the solution to decay to one-half of its value—equals\({\bf{(}}\frac{{{\bf{ln2}}}}{{\bf{k}}}{\bf{)}}\).

Complete the solution of the tumor growth model for example 3 on page 280 by finding P(t) and Q(t).

If p(t) is a Malthusian population that diminishes according to (2), then\({\bf{p(}}{{\bf{t}}_{\bf{2}}}{\bf{) - p(}}{{\bf{t}}_{\bf{1}}}{\bf{)}}\)is the number of individuals in the population whose lifetime lies between\({{\bf{t}}_{\bf{1}}}\,\,{\bf{and}}\,\,{{\bf{t}}_{\bf{2}}}\). Argue that the average lifetime of the population is given by the formula\(\frac{{\int_{\bf{0}}^\infty  {{\bf{t}}\left| {\frac{{{\bf{dp(t)}}}}{{{\bf{dt}}}}} \right|{\bf{dt}}} }}{{{\bf{p(0)}}}}\)and show that this equals.

Show that with the transition rate formula\({\bf{r}}\left( {\bf{N}} \right){\bf{  =   s}}\left( {{\bf{2N -- 1}}} \right)\), equation (22) takes the form of the equation for the logistic model (Section 3.2, equation

(14)). Solve (22) for this case.

Prove that the infected population I(t)in the SIR model does not increase if S(0) is less than or equal to \(\frac{{\bf{k}}}{{\bf{a}}}\).

An epidemic reported by the British Communicable DiseaseSurveillance Center in the British Medical Journal (March 4, 1978, p. 587) took place in a boarding school with 763 residents. The statistics for the infected population are shown in the graph in Figure 5.25.

Assuming that the average duration of the infection is 2 days, use a numerical differential equation solver (see Appendix G) to try to reproduce the data. Take S(0) = 762, I(0) = 1, R (0) = 0 as initial conditions. Experiment with reasonable estimates for the average number of contacts per day by the infected students, who were confined to bed after the infection was detected. What value of this parameter seems to fit the curve best?

Two springs and two masses are attached in a straight line on a horizontal frictionless surface as illustrated in Figure\(5.31\). The system is set in motion by holding the mass \({{\bf{m}}_{\bf{2}}}\) at its equilibrium position and pulling the mass \({{\bf{m}}_{\bf{1}}}\) to the left of its equilibrium position a distance \({\bf{1}}{\rm{ }}{\bf{m}}\)and then releasing both masses. Express Newton's law for the system and determine the equations of motion for the two masses if \({{\bf{m}}_{\bf{1}}}{\bf{ = 1 kg, }}{{\bf{m}}_{\bf{2}}}{\bf{ = 2 kg, }}{{\bf{k}}_{\bf{1}}}{\bf{ = 4 N / m,}}\)and\({{\bf{k}}_{\bf{2}}}{\bf{ = 10 / 3 N/ m}}\).