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 \(|x(t)|\). $$ x^{\prime \prime}+4 x=0 $$

Short answer

The conserved quantity for the given autonomous equation \(x^{\prime \prime}+4x=0\) is \(\frac{1}{2}(x')^2 +2x^2 = C\). The particular conserved quantity passing through the point \((1,1)\) has the equation \(\frac{1}{2}y^2 +2x^2 = \frac{5}{2}\). Since the conserved quantity forms an ellipse-like shape in the phase plane, the solution passing through the point \((1,1)\) is bounded. The maximum value of \(|x(t)|\) is approximately \(\frac{\sqrt{5}}{2}\).

Step-by-step solution

Derive a conservation law for the given autonomous equation

Given the autonomous equation \(x^{\prime \prime}+4x=0\), we can rewrite it as: \(x^{\prime \prime}=-4x\). We will now multiply both sides by \(x'\) and integrate with respect to \(t\): \(x'x^{\prime \prime}= -4x x'\). Integrating both sides with respect to \(t\), $$\int x'x^{\prime \prime} dt = \int -4x x' dt$$ $$\frac{1}{2}(x')^2 = -2x^2 + C$$, where \(C\) is an arbitrary constant. So we have a one-parameter family of conservation laws: \(\frac{1}{2}(x')^2 +2x^2 = C\).

Rewrite the autonomous equation as a first order system

Setting \(y(t)=x^{\prime}(t)\), we have the following first-order system: $$ \begin{aligned} &x^{\prime}=f(x, y)=y \\ &y^{\prime}=g(x, y)=-4x \end{aligned} $$

Express the conserved quantities in terms of x and y

Using the conservation law found in Step 1 and the substitution \(y=x'\), we can write the conserved quantities as: $$\frac{1}{2}y^2 +2x^2 = C$$

Determine the equation of the particular conserved quantity

Given \((x, y) = (1,1)\), we can solve for the value of \(C\): $$\frac{1}{2}(1)^2 +2(1)^2 = C$$ $$C=\frac{5}{2}$$ Thus, the particular conserved quantity has the equation \(\frac{1}{2}y^2 +2x^2 = \frac{5}{2}\).

Plot the graph and find the velocity vector

After plotting the graph (using computer software if necessary) of the conserved quantity \(\frac{1}{2}y^2 +2x^2 = \frac{5}{2}\), we will find the velocity vector \(\mathbf{v}\) at the point \((1,1)\) using the functions \(f(x, y)\) and \(g(x, y)\): $$\mathbf{v} = f(x, y) \mathbf{i} + g(x, y) \mathbf{j} = y \mathbf{i} - 4x \mathbf{j} = \mathbf{i} - 4 \mathbf{j}$$ at the point \((1,1)\). The vector is tangent to the graph and indicates the direction in which the solution traverses the graph as time increases.

Determine whether the solution is bounded

Since the conserved quantity represents an ellipse-like shape in the phase plane, it indicates that the motion of the system is bound within the ellipse. Thus, the solution passing through the point \((1,1)\) is bounded.

Estimate the maximum value of |x(t)|

By analyzing the phase-plane graph of the conserved quantity, we can estimate the maximum value of \(|x(t)|\). The maximum value of \(x(t)\) will occur at the largest \(x\) value on the ellipse, which can be estimated from the graph or calculated from the equation of the ellipse: $$2x^2 = \frac{5}{2} - \frac{1}{2}y^2$$ At maximum \(|x(t)|\), \(y=0\), thus we have $$2x^2 = \frac{5}{2}$$ $$x^2 = \frac{5}{4}$$ $$|x(t)|=\frac{\sqrt{5}}{2}$$ So, the maximum value of \(|x(t)|\) is approximately \(\frac{\sqrt{5}}{2}\).

Key concepts

Understanding Conservation Laws

In the context of differential equations, conservation laws play an essential role by expressing a property that does not change as time progresses. These are closely tied to physical quantities that remain constant over time, such as energy or momentum in a closed system.

To derive a conservation law, we often look for a quantity that, when differentiated with respect to time, equals zero. This signifies that there is no net change in that quantity. For the autonomous equation given in the exercise, \(x^{\textprime \textprime} + u(x) = 0\), we find a conserved quantity through multiplication by \(x'\) and integrating. The result is a relationship involving \(x\), \(x'\), and an arbitrary constant, encapsulating a one-parameter family of conserved quantities.

These laws are crucial for understanding the behavior of systems governed by the differential equation in question. The conserved quantities allow us to draw conclusions about the system's evolution over time without necessarily solving the equation fully.

First Order System Transition

A first order system in differential equations is a system comprised of first order differential equations, meaning each equation involves only the first derivative of one or more functions. Converting a higher-order differential equation into a first order system can often simplify analysis and numerically solving the equations.

In the exercise, the second-order autonomous equation \(x^{\textprime \textprime} + 4x = 0\) is transformed into a first order system by introducing \(y(t) = x'(t)\). This creates a system where \(x'\) and \(y'\) are represented in terms of \(x\) and \(y\), conducive to analysis in the 'phase plane', which visually represents the solution's trajectory over time.

First order systems are often easier to work with because they enable the use of techniques like phase plane analysis, allowing us to map out the entire behavior of systems and better understand their dynamics in terms of stability, periodicity, and other features.

Phase Plane Analysis

The phase plane is a graphical representation of a system of first order differential equations. It's a powerful tool that provides insights into the system's behavior by plotting the system's states against each other. For instance, in a system with variables \(x\) and \(y\), the \(xy\)-plane becomes the stage for our analysis.

In our example, we express a family of conserved quantities as functions of \(x\) and \(y\), and the phase plane helps visualize these quantities as curves or geometric shapes. What we see are the trajectories representing the evolution of the system over time. Plotting the graph of a specific conserved quantity passing through a point like \( (1,1) \) gives us tangible information about how the system behaves near that point.

Phase plane analysis is of great benefit in studying dynamical systems. It allows us to determine equilibrium points, and understand the stability, and long-term behavior of solutions, especially when solving the system analytically is challenging.

Velocity Vector Intuition

In phase plane diagrams, the velocity vector is a pivotal concept that provides the direction and rate at which a solution's state point moves through the plane. It's defined at each point in the phase plane by the system of equations governing the behavior of variables.

For the autonomous equation in the exercise, the velocity vector at any point \( (x, y) \) is given by \(\mathbf{v} = f(x, y) \mathbf{i} + g(x, y) \mathbf{j}\), where \(f(x, y)\) and \(g(x, y)\) are the derivatives \(x'\) and \(y'\). At \( (1,1) \) this becomes \(\mathbf{v} = \mathbf{i} - 4 \mathbf{j}\). By placing this vector on the graph, we can immediately see how the state of the system will evolve: the direction of \(\mathbf{v}\) intuitively indicates the immediate future path of the system's state.

The velocity vector tells us not just about the direction but also about the speed with which the point moves. It offers a snapshot of the system's dynamics at a specific instant, which is particularly useful when analyzing conserved quantities in phase planes.