Question

In each exercise, (a) As in Example 1, derive a conservation law for the given autonomous equation \(x^{\prime \prime}+u(x)=0\). (Your answer should contain an arbitrary constant and therefore define a one-parameter family of conserved quantities.) (b) Rewrite the given autonomous equation as a first order system of the form $$ \begin{aligned} &x^{\prime}=f(x, y) \\ &y^{\prime}=g(x, y) \end{aligned} $$ by setting \(y(t)=x^{\prime}(t)\). The phase plane is then the \(x y\)-plane. Express the family of conserved quantities found in (a) in terms of \(x\) and \(y\). Determine the equation of the particular conserved quantity whose graph passes through the phase-plane point \((x, y)=(1,1)\). (c) Plot the phase-plane graph of the conserved quantity found in part (b), using a computer if necessary. Determine the velocity vector \(\mathbf{v}=f(x, y) \mathbf{i}+g(x, y) \mathbf{j}\) at the phaseplane point \((1,1)\). Add this vector to your graph with the initial point of the vector at \((1,1)\). What is the geometric relation of this velocity vector to the graph? What information does the velocity vector give about the direction in which the solution point traverses the graph as time increases? (d) For the solution whose phase-plane trajectory passes through \((1,1)\), determine whether the solution \(x(t)\) is bounded. If the solution is bounded, use the phaseplane plot to estimate the maximum value attained by \(|x(t)|\). $$ x^{\prime \prime}+\frac{x}{1+x^{2}}=0 $$

Short answer

In summary, to solve this problem, we first derived the conservation law by finding the conserved quantity \(E(x, x')\) for the given autonomous equation. We then rewrote the equation as a first-order system and found the particular conserved quantity that passes through the given point \((1,1)\). To complete the problem, you would have to plot the phase-plane graph, find the velocity vector, and determine whether the solution is bounded. Working through these concepts will help deepen your understanding of autonomous equations and their applications in various fields.

Step-by-step solution

Deriving the conservation law for the autonomous equation

First, we are given the equation \(x^{\prime\prime} + u(x) = 0\). To derive a conservation law, we will use the method outlined in Example 1, which is to take the time derivative of the expression: $$ E(x,x') = \frac{1}{2} (x')^2 + U(x) $$ When an expression \(E(x, x')\) is found such that \(\frac{d}{dt}E(x, x')=0\), the function E represents a conserved quantity. First, we need to find a function \(U(x)\) such that \(u(x)=-\frac{d}{dx}U(x)\). The given equation is: $$ x^{\prime\prime}+\frac{x}{1+x^{2}}=0 $$ So we have \(u(x)=\frac{x}{1+x^{2}}\). Now we need to find \(U(x)\) by integrating \(u(x)\): $$ U(x) = -\int u(x) dx = -\int \frac{x}{1+x^{2}} dx $$ Use substitution, let \(v = 1 + x^2\) and \(dv = 2x dx\). This gives us: $$ U(x) = -\frac{1}{2} \int \frac{1}{v} dv = -\frac{1}{2}\ln(v) + C= -\frac{1}{2}\ln(1+x^2) + C $$ Now, we can write the conserved quantity \(E(x, x')\): $$ E(x, x')= \frac{1}{2}(x')^2 -\frac{1}{2}\ln(1+x^2) + C $$ This is a one-parameter family of conserved quantities, where C is the arbitrary constant.

Rewriting autonomous equation as a first-order system

To convert the given equation into a first-order system, we simply need to set \(y(t)=x^{\prime}(t)\). This will give us two equations: $$ \begin{aligned} x'=f(x,y)=y \\ y'=g(x,y)=-\frac{x}{1+x^{2}} \end{aligned} $$ Now, we need to find the family of conserved quantities in terms of x and y. We start with \(E(x, x')\) and substitute y for \(x'\), giving us: $$ E(x, y) = \frac{1}{2}y^2 - \frac{1}{2}\ln(1+x^2) + C $$ Now, we are given the phase-plane point \((x, y) = (1,1)\). We need to find the conserved quantity that passes through that point. Plugging in \((1,1)\) into the above equation, we get: $$ E(1,1) = \frac{1}{2}(1)^2 - \frac{1}{2}\ln(1+(1)^2) + C = \frac{1}{2} - \frac{1}{2}\ln 2+C $$ Setting this conserved quantity equal to our previous conserved quantity, we get: $$ \frac{1}{2}y^2 - \frac{1}{2}\ln(1+x^2) + C = \frac{1}{2} - \frac{1}{2}\ln 2+C $$ This gives the conserved quantity for the specific phase plane point \((1,1)\) as: $$ \frac{1}{2}y^2 - \frac{1}{2}\ln(1+x^2)=\frac{1}{2} - \frac{1}{2}\ln 2 $$ For steps (c) and (d), the student would be required to plot the phase-plane graph of the conserved quantity found in step 2(b) and find the velocity vector using a computer if needed. Additionally, the student would need to determine whether the solution is bounded and, if so, estimate the maximum value attained by \(|x(t)|\). These steps typically involve using tools such as Matlab, Python, or Mathematica for creating the plots and vector fields, as well as calculating the values required.

Key concepts

Conservation Law

In the realm of differential equations, a conservation law represents a principle stating that a particular measurable property of an isolated physical system does not change as the system evolves over time. Just like conservation of energy in physics, which implies that energy can neither be created nor destroyed, a mathematical conservation law implies that certain properties remain constant through the dynamic changes of a system.

When dealing with second-order autonomous equations like the one in our exercise, \(x^{\prime\prime} + u(x) = 0\), a conservation law can be derived by finding a function \(E(x, x')\) whose rate of change over time is zero. This function represents a conserved quantity throughout the motion described by the equation. By integrating the function \(u(x)\), we are able to determine \(U(x)\) and thus formulate the conserved quantity \(E(x, x')\).

It's crucial for students to grasp that finding a conservation law often involves identifying the correct form of \(U(x)\) and recognizing that the derivative of this potential function with respect to \(x\) gives us the negative of \(u(x)\). For instance, in the exercise, the conserved quantity \(E(x, x')\) was found by integrating \(u(x)\) and combining it with the kinetic energy term \(\frac{1}{2} (x')^2\). This process is not just theoretical; it ties closely to physical laws and provides a profound understanding of how certain quantities remain invariant in natural processes.

Autonomous Equation

An autonomous equation is a type of differential equation where the variable representing time does not explicitly appear in the equation. This property implies that the rules governing the system's behavior are time-invariant, making the future evolution of the system dependent only on its current state, not on the time at which it is in that state.

The equation \(x^{\prime\prime} + u(x) = 0\) from our exercise is autonomous because it lacks any explicit time dependence. When converted into first-order form, it yields a system of equations dependent only on \(x\) and \(y\). Converting to first-order systems makes it easier to analyze the behavior of solutions since it breaks down the second-order equation into a system that can be visualized and studied in a two-dimensional space — the phase plane.

Phase Plane

The concept of a phase plane is a fundamental tool in understanding the behavior of differential equations, particularly in visualizing how the states of an autonomous system evolve over time. By plotting the variables \(x\) and the derivative \(x'\) usually denoted as \(y\), on the x and y-axes respectively, one can examine how these variables interact and change.

The phase plane provides insights into the solution curves, known as trajectories or orbits, and the overall system's stability. When examining the phase plane plot, key features to look for include fixed points (where \(x' = y' = 0\)), limiting cycles, and paths followed by the system's states. In the exercise, when the conserved quantity \(E(x, y)\) is graphed in the phase plane, it allows us to visualize invariant curves that solutions to the differential equation will follow.

Velocity Vector

In the study of differential equations, the velocity vector plays a crucial role in understanding how solutions evolve over time within the phase plane. It represents the direction and speed of the movement of a state point at any given moment.

The velocity vector \(\mathbf{v}\) is defined by the first-order derivatives of the system, where \(\mathbf{v} = f(x, y) \mathbf{i} + g(x, y) \mathbf{j}\). In the exercise, the velocity vector at the point \( (1,1) \) helps us understand the tangent to the trajectory at that point, indicating the instant direction in which the system is moving. By adding this vector to the phase-plane graph, students can visualize how the state point will proceed along the conserved quantity curve.

It is essential to emphasize that the geometric relationship between the velocity vector and the graph of the conserved quantity provides insight into the dynamic behavior of the system. By analyzing the orientation and magnitude of the velocity vector, students can infer valuable information about the tendency of solutions to move towards or away from equilibrium points, as well as the speed at which these transitions happen.