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Chapter 6: Problem 7

The conservation law for an autonomous second order scalar differential equation having the form \(x^{\prime \prime}+f(x)=0\) is given (where \(y\) corresponds to \(x^{\prime}\) ). Determine the differential equation. $$ y^{2}+x^{2} \cos x=C $$

Short Answer

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#tag_title# Step 2: Calculate the derivatives #tag_content#Compute the derivatives in the given equation: For \(\frac{d}{dt}(y^2)\): we have $$ \frac{d}{dt}(y^2) = 2y \cdot \frac{d}{dt}y = 2y \cdot y^{\prime}, $$ since \(y^{\prime} = x^{\prime \prime}(t)\), we get $$ \frac{d}{dt}(y^2) = 2y \cdot x^{\prime\prime}. $$ For \(\frac{d}{dt}(x^{2} \cos x)\): we have $$ \frac{d}{dt}(x^{2} \cos x) = 2x \cdot \frac{d}{dt}x \cos x + x^2 \cdot \frac{d}{dt}(\cos x), $$ apply the chain rule: $$ \frac{d}{dt}(x^{2} \cos x) = 2x \cdot x^{\prime} \cdot \cos x - x^2 \cdot x^{\prime} \cdot \sin x. $$ Since \(\frac{d}{dt}(C) = 0\), the equation after differentiating becomes: $$ 2y \cdot x^{\prime\prime} + 2x \cdot x^{\prime} \cdot \cos x - x^2 \cdot x^{\prime} \cdot \sin x = 0. $$ #tag_title# Step 3: Solve for \(x^{\prime\prime}\) #tag_content#Isolate \(x^{\prime\prime}\) in the equation: $$ x^{\prime\prime} = \frac{-2x \cdot x^{\prime} \cdot \cos x + x^2 \cdot x^{\prime} \cdot \sin x}{2y} $$ We know that \(y = x^{\prime}\), so we can simplify the expression: $$ x^{\prime\prime} = \frac{-2x \cdot x^{\prime} \cdot \cos x + x^2 \cdot x^{\prime} \cdot \sin x}{2x^{\prime}} $$ Now, cancel \(x^{\prime}\) from the numerator and the denominator: $$ x^{\prime\prime} = \frac{-2x \cdot \cos x + x^2 \cdot \sin x}{2} $$ Finally, we have the autonomous second-order scalar differential equation: $$ x^{\prime\prime} + f(x) = x^{\prime\prime} + 2x \cos x - x^2 \sin x = 0 $$ #tag_title# Answer #tag_content#The autonomous second-order scalar differential equation is given by \(x^{\prime\prime} + 2x \cos x - x^2 \sin x = 0\).

Step by step solution

01

Differentiate the conservation equation with respect to \(t\).

The given conservation equation is \(y^{2} + x^{2} \cos x = C\). To differentiate this equation concerning \(t\), differentiate each term related to \(t\) while considering \(x = x(t)\) and \(y = y(t) = x^{\prime}(t)\). Apply the chain rule when needed: $$ \frac{d}{dt}(y^2) + \frac{d}{dt}(x^{2} \cos x) = \frac{d}{dt}(C) $$

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Key Concepts

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

Second Order Differential Equation
A second order differential equation involves derivatives up to the second order of a function. In this context, we have an equation that includes the second derivative, noted as \(x''\). It resembles the physical principles where acceleration is proportional to a function of position, which is common in mechanics.

In general, a second order differential equation can be written as \(x'' + f(x) = 0\), where \(x''\) is the acceleration term, and \(f(x)\) is a function that depends only on \(x\). Studying such equations is crucial because they frequently model natural phenomena such as movement and oscillations in physics. It is the equation's autonomous nature鈥攎eaning it does not depend explicitly on the independent variable, usually time \(t\)鈥攖hat simplifies the analysis, focusing solely on the dependency on \(x\).
Autonomous System
An autonomous system refers to a system of differential equations that do not explicitly depend on the independent variable. This is a key concept in differential equations because it allows for the prediction of system behavior based purely on state, rather than external parameters like time.

In our exercise, \(x'' + f(x) = 0\) is autonomous because the function \(f(x)\) does not contain the variable \(t\) directly. The system's behavior depends solely on \(x\) and \(x'\). This property is important as it often leads to the conservation laws, like our given equation, where quantities remain constant over evolution, illustrating an inherent stability in the system. Understanding autonomous systems aids in simplifying and solving differential equations, as it often reduces the complexity of analysis by ignoring external variational factors.
Chain Rule Differentiation
The chain rule is a fundamental technique in calculus used for differentiating compositions of functions. When solving differential equations like our example, this rule helps in dealing with terms where the function of a function is involved.

For instance, if \(y = x'(t)\), which means \(y\) is dependent on \(t\) through \(x(t)\), differentiating \(y^2\) with respect to \(t\) requires applying the chain rule:
  • First, differentiate the outer function \(y^2 \) which gives \(2y\), and then multiply it by the derivative of the inner function \(y\), i.e., \(y'\).
  • Similarly, for \( x^2 \cos x \), differentiate using \(2x\cos x - x^2\sin x\).


Mastering the chain rule is crucial for handling more complex differential equations effectively, particularly in exercises involving conditions like the conservation law.
Integral Constants
In mathematics, an integral constant arises when solving differential equations. Whenever a function is integrated, a constant of integration \(C\) is added because indefinite integration can produce multiple possible solutions.

In our exercise, \(y^2 + x^2 \cos x = C\) includes such a constant, representing the conserved quantity in the system due to the conservation law. This constant remains unchanged over time, underscoring the concept of what is conserved.
  • The presence of \(C\) indicates that even though the variables \(x\) and \(y\) may change, this particular combination \(y^2 + x^2 \cos x\) remains fixed.
  • It expresses the inherent stability and conservation in the system.


Understanding integral constants is crucial as it helps in interpreting solutions and the set of possible states a system can attain over time, offering insight into its dynamical behavior.

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Most popular questions from this chapter

In each exercise, the given system is an almost linear system at each of its equilibrium points. (a) Find the (real) equilibrium points of the given system. (b) As in Example 2, find the corresponding linearized system \(\mathbf{z}^{\prime}=A \mathbf{z}\) at each equilibrium point. (c) What, if anything, can be inferred about the stability properties of the equilibrium point(s) by using Theorem \(6.4\) ? $$ \begin{aligned} &x^{\prime}=y^{2}-x \\ &y^{\prime}=x^{2}-y \end{aligned} $$

Each exercise lists a nonlinear system \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\), where \(A\) is a constant ( \(2 \times 2\) ) invertible matrix and \(\mathbf{g}(\mathbf{z})\) is a \((2 \times 1)\) vector function. In each of the exercises, \(\mathbf{z}=\mathbf{0}\) is an equilibrium point of the nonlinear system. (a) Identify \(A\) and \(\mathbf{g}(\mathbf{z})\). (b) Calculate \(\|\mathbf{g}(\mathbf{z})\|\). (c) Is \(\lim _{\mid \mathbf{z} \| \rightarrow 0}\|\mathbf{g}(\mathbf{z})\| /\|\mathbf{z}\|=0\) ? Is \(\mathbf{z}^{\prime}=A \mathbf{z}+\mathbf{g}(\mathbf{z})\) an almost linear system at \(\mathbf{z}=\mathbf{0}\) ? (d) If the system is almost linear, use Theorem \(6.4\) to choose one of the three statements: (i) \(\mathbf{z}=\mathbf{0}\) is an asymptotically stable equilibrium point. (ii) \(\mathbf{z}=\mathbf{0}\) is an unstable equilibrium point. (iii) No conclusion can be drawn by using Theorem \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=9 z_{1}-4 z_{2}+z_{2}^{2} \\ &z_{2}^{\prime}=15 z_{1}-7 z_{2} \end{aligned} $$

Let $$ A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right] $$ be a real invertible matrix, and consider the system \(\mathbf{y}^{\prime}=A \mathbf{y}\). (a) What conditions must the matrix entries \(a_{i j}\) satisfy to make the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) a center? (b) Assume that the equilibrium point at the origin is a center. Show that the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) is a Hamiltonian system. (c) Is the converse of the statement in part (b) true? In other words, if the system \(\mathbf{y}^{\prime}=A \mathbf{y}\) is a Hamiltonian system, does it necessarily follow that \(\mathbf{y}_{e}=\mathbf{0}\) is a center? Explain.

A linear system is given in each exercise. (a) Determine the eigenvalues of the coefficient matrix \(A\). (b) Use Table \(6.2\) to classify the type and stability characteristics of the equilibrium point at \(\mathbf{y}=\mathbf{0}\). (c) The given linear system is a Hamiltonian system. Derive the conservation law for this system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -2 & 1 \\ 5 & 2 \end{array}\right] \mathbf{y} $$

Consider a colony in which an infectious disease (such as the common cold) is present. The population consists of three "species" of individuals. Let \(s\) represent the susceptibles-healthy individuals capable of contracting the illness. Let \(i\) denote the infected individuals, and let \(r\) represent those who have recovered from the illness. Assume that those who have recovered from the illness are not permanently immunized but can become susceptible again. Also assume that the rate of infection is proportional to \(s i\), the product of the susceptible and infected populations. We obtain the model $$ \begin{aligned} &s^{\prime}=-\alpha s i+\gamma r \\ &i^{\prime}=\alpha s i-\beta i \\ &r^{\prime}=\beta i-\gamma r \end{aligned} $$ where \(\alpha, \beta\), and \(\gamma\) are positive constants. (a) Consider system (11). Use the fact that \(s(t)+i(t)+r(t)=N\) to obtain a reduced system of two differential equations for the two dependent variables \(s(t)\) and \(i(t)\). (b) For simplicity, set \(\alpha=\beta=\gamma=1\) and \(N=9\). Determine the equilibrium points of the reduced two-dimensional system. (c) Determine the linearized system at each of the equilibrium points found in part (b). Use Table \(6.2\) to analyze the stability characteristics of each of these linearized systems. (d) Show that the nonlinear system is an almost linear system at each of the equilibrium points found in part (b). What are the stability characteristics of the nonlinear system at these points?
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