Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume II

Question

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 .

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: Differentiate x(t)

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

Step 3: Differentiate y(t)

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

Step 4: 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 \(x'(t) = -y(t)\).

Step 5: 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 \(y'(t) = 25x(t)\).

Step 6: 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 \(5x_0\).

Step 7: 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.

Step 8: 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.

Key Concepts

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.