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Exercise \(18.2 .3 \quad\) a. Show that for any number, \(x_{0},\) the solution to $$ \begin{array}{c} x(0)=x_{0} \quad x^{\prime}=-y \\ y(0)=0 \quad y^{\prime}=25 x \\ \text { is } x(t)=x_{0} \cos 5 t, \quad y(t)=5 x_{0} \sin 5 t \end{array} $$ b. Draw the graph in the \(x, y-\) plane of the solutions for \(x_{0}=1\) and \(x_{0}=0.5\). c. Show that the origin (0,0) is a stable equilibrium point of Equations 18.28 . d. Show that the origin (0,0) not an asymptotically stable equilibrium of Equations 18.28 .

Short Answer

Expert verified
The solution satisfies the differential equations; the origin is stable but not asymptotically stable.

Step by step solution

01

Understand the given differential equations

The differential equations provided are \(x'(t) = -y(t)\) and \(y'(t) = 25x(t)\). Initial conditions are \(x(0) = x_0\) and \(y(0) = 0\). We aim to verify that \(x(t) = x_0 \cos(5t)\) and \(y(t) = 5x_0 \sin(5t)\) satisfy these equations.
02

Differentiate x(t)

Differentiate \(x(t) = x_0 \cos(5t)\) with respect to \(t\):\[x'(t) = -5x_0 \sin(5t)\].
03

Differentiate y(t)

Differentiate \(y(t) = 5x_0 \sin(5t)\) with respect to \(t\):\[y'(t) = 25x_0 \cos(5t)\].
04

Verify x'(t) = -y(t)

Plug \(y(t) = 5x_0 \sin(5t)\) into \(x'(t) = -y(t)\):\[x'(t) = -5x_0 \sin(5t) = -y(t)\]. This confirms \(x'(t) = -y(t)\).
05

Verify y'(t) = 25x(t)

Plug \(x(t) = x_0 \cos(5t)\) into \(y'(t) = 25x(t)\):\[y'(t) = 25x_0 \cos(5t) = 25x(t)\]. This confirms \(y'(t) = 25x(t)\).
06

Draw the phase portrait

Plot the paths described by \(x(t) = x_0 \cos(5t)\) and \(y(t) = 5x_0 \sin(5t)\) when \(x_0 = 1\) and \(x_0 = 0.5\). The paths trace ellipses centered at the origin, with principal radii corresponding to \(x_0\) and \(5x_0\).
07

Define stability at the origin

The solution remains bounded, tracing elliptical paths around the origin, indicating stability. For small perturbations at \((0,0)\), the trajectories return near the origin but not decay towards it.
08

Analyze asymptotic stability

Although the trajectories remain close to the origin, they do not approach zero as \(t \to \infty\). Hence, the origin is not asymptotically stable because disturbances do not diminish over time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase Portraits
In the study of differential equations, phase portraits provide a visual way to understand the dynamics of a system. They are plots that show the trajectory of solutions in the phase plane, usually depicted as the \(x, y\) plane for a two-dimensional system.

For this exercise, the differential equations transform into a set of parametric equations:\(x(t) = x_0 \cos{(5t)}\) and \(y(t) = 5x_0 \sin{(5t)}\). These represent ellipses centered at the origin with varying radii depending on \(x_0\).

When plotting the phase portraits for \(x_0 = 1\) and \(x_0 = 0.5\), you will notice distinct ellipses. The ellipse for \(x_0 = 1\) will be larger, indicating a greater amplitude of oscillation compared to \(x_0 = 0.5\). These paths circumnavigate the origin continuously, illustrating the system's periodic nature.

Phase portraits are crucial for visualizing how solutions evolve over time, providing insights into the long-term behavior of the system.
Stability Analysis
Stability analysis assesses how a system behaves in response to small perturbations. In this context, it specifically examines how the system's solutions react as time progresses.

The equilibrium point at the origin \( (0,0) \) is considered stable if solutions starting close to it remain nearby for all future times. As seen in this exercise, trajectories of the system maintain bounded paths along elliptical trajectories around the origin. This bounded behavior ensures that the system is confirmed to be stable because solutions don't diverge uncontrollably from the origin.

However, stability here does not imply convergence to the equilibrium point, but rather a consistent distance from it.

This is a critical distinction, as not all stable systems are necessarily asymptotically stable, which will be explored in more detail shortly.
Equilibrium Points
Equilibrium points are where the system does not change over time. For the given system of differential equations, the only equilibrium point is at the origin, \((0,0)\).

At this point, both derivatives become zero: \(x' = -y\) and \(y' = 25x\) both resolve to zero when \(x = 0\) and \(y = 0\). This indicates a no-movement state where the solution is at rest.

Equilibrium points are crucial in determining the overall behavior of the system. They can be seen as fixed points that greatly influence the nature of surrounding solutions and their trajectories within the phase portrait.

Understanding their stability characteristics — whether solutions starting near them stay near or move away — is pivotal to comprehensively analyzing dynamical systems.
Oscillatory Solutions
Oscillatory solutions cover solutions that exhibit periodic behavior, like those modeled in this exercise. The system yields solutions of the form \(x(t) = x_0 \cos{(5t)}\) and \(y(t) = 5x_0 \sin{(5t)}\), indicating harmonic oscillations.

This periodic motion is characteristic of a harmonic oscillator, where solutions oscillate indefinitely over time without converging to a point. The frequency of oscillation is determined by the coefficients in the cosine and sine functions.

In physical terms, these solutions resemble the behavior of a simple harmonic oscillator, like a spring or pendulum, which moves in a repetitive cycle. It's important to note here how oscillatory solutions help predict long-term system behavior and the potential for resonance or constructive interference, especially in real-world applications.

Understanding such solutions is essential in fields ranging from physics to engineering, where controlled periodic motion is often a goal or must be guarded against.

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Most popular questions from this chapter

Anderson and May \(^{4}\) give the following model of immune effector cells (helper and cytotoxic T-cells), \(E,\) that limit viral population, \(V\), growth in a human body. $$ \begin{array}{l} d E / d t=\Lambda-\mu E+\epsilon V E \\ d V / d t=r V-\sigma V E \end{array} $$ a. \(\Lambda\) is intrinsic production rate of effector cells from bone marrow. Give similar meaning to each of the other four terms on the RHS of Equations 18.64 . b. Find the equilibrium effector cell population, \(\hat{E},\) in the absence of virus. c. Suppose an inoculum \(V_{0}\) of virus is introduced into the body with \(E=\hat{E}\). Find conditions on \(r\), \(\sigma,\) and \(\hat{E}\) in order that the viral population will increase. d. The Jacobian matrix at any \((E, V)\) is $$ J(E, V)=\left[\begin{array}{rc} -\mu+\epsilon V & \epsilon E \\ -\sigma V & r-\sigma V \end{array}\right] $$ Show that $$ J(\hat{E}, 0)=\left[\begin{array}{rr} -\mu & \sigma \frac{\Lambda}{\mu} \\ 0 & r-\sigma \frac{\Lambda}{\mu} \end{array}\right] $$ e. The characteristic values of the upper diagonal matrix \(J(\hat{E}, 0)\) are the diagonal entries, \(-\mu\) and \(r-\sigma \frac{\Lambda}{\mu} .\) What happens to a small introduction of virus into a healthy individual if both characteristic values are negative? f. In order that the viral population to expand it is necessary that \(r-\sigma \frac{\Lambda}{\mu}>0 .\) What is the role of \(r\) in the model? g. If that condition is met and the viral population increases, show that there will be an equilibrium state, $$ E^{*}=r / \sigma, \quad V^{*}=\frac{\mu r-\Lambda \sigma}{\epsilon r} $$ It is clear that in order for \(V^{*}\) to be positive, we must have (again) $$ \mu r-\Lambda \sigma>0 \quad \text { so that } \quad r>\frac{\Lambda \sigma}{\mu} $$ h. Anderson and May report this system to be asymptotically stable, at \(\left(E^{*}, V^{*}\right),\) but only weakly so, meaning that it is subject to wide oscillations. Show that $$ J\left(E^{*}, V^{*}\right)=\left[\begin{array}{cc} -\frac{\Lambda \sigma}{r} & \epsilon \frac{r}{\sigma} \\ -\sigma \frac{\mu r-\Lambda \sigma}{\epsilon r} & 0 \end{array}\right] $$ i. The characteristic equation of a \(2 \times 2\) matrix \(M\) is is \(s^{2}-\operatorname{trace}(M) s+\operatorname{det}(M)=0\). Show that the characteristic equation of \(J\left(E^{*}, V^{*}\right)\) is $$ s^{2}+\frac{\Lambda \sigma}{r} s+\mu r-\Lambda \sigma=0 $$ j. The roots of the characteristic equation 18.65 are $$ \frac{-\frac{\Lambda \sigma}{r} \pm \sqrt{\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)}}{2} $$ Argue that the real part is negative so that the system is stable. Note: Two cases: $$ \left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)>0 \quad \text { and } \quad\left(\frac{\Lambda \sigma}{r}\right)^{2}-4(\mu r-\Lambda \sigma)<0 $$ k. Use \(\Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01\) and compute \(E^{*}, V^{*},\) and the stability at \(\left(E^{*}, V^{*}\right)\) l. Let \(E_{0}=2, V_{0}=1, \Lambda=1, \mu=0.5, \epsilon=0.02, r=0.25,\) and \(\sigma=0.01 .\) Approximate the solutions to Equations 18.64 using the trapezoid rule. Observe that the rise in viral load precedes the increase in effector cells.

Compute and graph the solutions to $$ y^{\prime \prime}+p y^{\prime}+9 y=0 \quad y(0)=0 \quad y^{\prime}(0)=1 $$ for \(p=10, p=6, p=4, p=0,\) and \(p=-6\).

Exercise 18.3 .6 Interpret the output of the MATLAB program. $$ \begin{array}{l} \mathrm{A}=[0.770 .1 ; 0.0680 .9] \\ \mathrm{T} 1=[\exp (\mathrm{A}) \operatorname{expm}(\mathrm{A})] \\ \mathrm{T} 2=[\log (\mathrm{A}) \operatorname{logm}(\mathrm{A})] \\ \mathrm{T} 3=[\exp (\log (\mathrm{A})) \operatorname{expm}(\operatorname{logm}(\mathrm{A}))] \\ \mathrm{T} 4=\left[\log \left(\mathrm{A}^{-2}\right) 2 * \log (\mathrm{a})\right] \end{array} $$ \(\mathrm{T} 5=\left[\operatorname{logm}\left(\mathrm{A}^{-2}\right) 2 * \operatorname{logm}\left(\mathrm{A}^{-2}\right)\right]\) \(\mathrm{T} 6=[\exp (2 * \mathrm{~A}) \exp (\mathrm{A}) * \exp (\mathrm{A})]\) \(\mathrm{T} 7=[\operatorname{expm}(2 * \mathrm{~A}) \operatorname{expm}(\mathrm{A}) * \operatorname{expm}(\mathrm{A})]\) $$ \text { Output } $$ $$ \begin{array}{rrrrr} \mathrm{T} 1= & 2.1598 & 1.1052 & 2.1674 & 0.2309 \\ & 1.0704 & 2.4596 & 0.1570 & 2.4676 \\ \mathrm{~T} 2= & -0.2614 & -2.3026 & -0.2666 & 0.1204 \\ & -2.6882 & -0.1054 & 0.0819 & -0.1100 \\ \mathrm{~T} 3= & 0.7700 & 0.1000 & 0.7700 & 0.1000 \\ & 0.0680 & 0.9000 & 0.0680 & 0.9000 \\ \mathrm{~T} 4= & -0.5113 & -1.7898 & -0.5227 & -4.6052 \\ & -2.1754 & -0.2024 & -5.3765 & -0.2107 \\ \mathrm{~T} 5= & -0.5331 & 0.2408 & -0.5331 & 0.2408 \\ & 0.1637 & -0.2201 & 0.1637 & -0.2201 \\ \mathrm{~T} 6= & 4.6646 & 1.2214 & 5.8475 & 5.1052 \\ & 1.1457 & 6.0496 & 4.9444 & 7.2326 \\ \mathrm{~T} 7= & 4.7341 & 1.0703 & 4.7341 & 1.0703 \\ & 0.7278 & 6.1254 & 0.7278 & 6.1254 \end{array} $$

Consider now the case that there is resistance \(r>0\) in the spring-mass equation with a harmonic forcing function, $$ m y^{\prime \prime}(t)+r y^{\prime}(t)+k y(t)=\cos \omega t $$ a. Show that in the real root case, \(\mu_{1}\) and \(\mu_{2}\) are negative. b. Show that in the repeated root case, the value of \(\mu\) is negative. c. Show that in the and complex roots case, the value of \(\mu\) is negative. d. Show that in all of the cases, $$ \lim _{t \rightarrow \infty} y_{h}(t)=0 $$ The next few exercises examine the importance of \(y_{p},\) the particular solution of Equation \(18.20 .\) We will need $$ \begin{array}{ll} y_{p}=A \cos 0.3 t+B \sin 0.3 t & \text { and one of } \\ y_{h}=C_{1} e^{\mu_{1} t}+C_{2} e^{\mu_{2} t}, & \text { real roots, } \\ y_{h}=C_{1} e^{\mu t}+C_{2} t e^{\mu t}, & \text { repeated root, and } \\ y_{h}=e^{\mu t}\left(C_{1} \cos (\omega t)+C_{2} \sin (\omega t)\right) & \text { complex roots. } \end{array} $$

Symbiotic relationships are common and persist for long periods. It is curious that there are no known or very few symbiotic relationships between mammals. There are, however, many symbiotic relationships between mammals and other organisms, Escherichia coli, for example. Analysis of the equations for symbiosis, Equations 18.63: $$ \begin{aligned} x^{\prime}(t) &=r_{x} \times x(t) \times(1-a x(t)+b y(t)) \\ y^{\prime}(t) &=r_{y} \times y(t) \times(1+c x(t)-d y(t)) \end{aligned} $$ a. Show that the equilibrium point without zeros is $$ x_{1}=\frac{b+d}{a d-b c}, \quad y_{1}=\frac{a+c}{a d-b c} $$ $$ \text { if } a d-b c \neq 0 $$ b. \(x_{1}\) and \(y_{1}\) are positive only if \(a d-b c>0 .\) This is a surprise to us. Set \(b=c=d=1\) and examine the equilibrium point for \(a>1\) and \(a<1\). c. Assume \(a d-b c>0\) so that \(x_{1}\) and \(y_{1}\) are positive. Either work it out (no!) or accept our analysis that the Jacobian at \(\left(x_{1}, y_{1}\right)\) is $$ J=\left[\begin{array}{cc} -a \frac{b+d}{a d-b c} & b \frac{b+d}{a d-b c} \\ c \frac{a+c}{a d-b c} & -d \frac{a+c}{a d-b c} \end{array}\right]=\frac{1}{a d-b c}\left[\begin{array}{cc} -a(b+d) & b(b+d) \\ c(a+c) & -d(a+c) \end{array}\right] $$ Argue that if \(a d-b c>0,\left(x_{1}, y_{1}\right)\) is an asymptotically stable equilibrium. d. With persistence you might show that the characteristic roots are not complex. It requires showing that the discrimiant $$ \begin{array}{l} (a(b+d)+d(a+c))^{2}-4(a(b+d) d(a+c)-c(a+c) d(b+d))= \\ (a(b+d)-d(a+c))^{2}+4 c(a+c) d(b+d) \geq 0 \end{array} $$

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