Problem 31
Question
If \(z=f(x, y)\) is a function with continuous first partial derivatives in a region \(R\), then a flow \(\mathbf{V}(x, y)=(P(x, y), Q(x, y))\) in \(R\) may be defined by letting \(P(x, y)=-\frac{\partial f}{\partial y}(x, y)\) and \(Q(x, y)=\frac{\partial f}{\partial x}(x, y) .\) Show that if \(\mathbf{X}(t)=(x(t), y(t))\) is a solution of the plane autonomous system $$ \begin{aligned} &x^{\prime}=P(x, y) \\ &y^{\prime}=Q(x, y) \end{aligned} $$ then \(f(x(t), y(t))=c\) for some constant \(c .\) Thus a solution curve lies on the level curves of \(f\). [Hint: Use the Chain Rule to compute \(\left.\frac{d}{d t} f(x(t), y(t)) .\right]\)
Step-by-Step Solution
Verified Answer
The solution curve lies on the level curves of \(f\) as \(f(x(t), y(t)) = c\) for some constant \(c\).
1Step 1: Write down the given autonomous system equations
The given system of differential equations is:\[\begin{aligned}&x^{\prime} = P(x, y), \&y^{\prime} = Q(x, y)\end{aligned}\]
2Step 2: Use the chain rule to differentiate f
The function \(f(x(t), y(t))\) is a composition of functions with respect to \(t\). Use the chain rule to find \(\frac{d}{dt} f(x(t), y(t))\):\[\frac{d}{dt} f(x(t), y(t)) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}\]
3Step 3: Substitute expressions for partial derivatives
Replace \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) with \(x'\) and \(y'\) respectively:\[\frac{d}{dt} f(x(t), y(t)) = \frac{\partial f}{\partial x} P(x, y) + \frac{\partial f}{\partial y} Q(x, y)\]
4Step 4: Substitute the values of P(x,y) and Q(x, y)
Using the definitions \(P(x, y) = -\frac{\partial f}{\partial y}\) and \(Q(x, y) = \frac{\partial f}{\partial x}\), substitute back into the equation:\[\frac{d}{dt} f(x(t), y(t)) = \frac{\partial f}{\partial x} \left(-\frac{\partial f}{\partial y} \right) + \frac{\partial f}{\partial y} \frac{\partial f}{\partial x}\]
5Step 5: Simplify the expression
Simplify the expression:\[\frac{d}{dt} f(x(t), y(t)) = - \frac{\partial f}{\partial x}\frac{\partial f}{\partial y} + \frac{\partial f}{\partial y}\frac{\partial f}{\partial x}\]Now observe that the two terms on the right-hand side are equal in magnitude but opposite in sign, thus:\[\frac{d}{dt} f(x(t), y(t)) = 0\]
6Step 6: Conclude the function's behavior
Since \(\frac{d}{dt} f(x(t), y(t)) = 0\), the function \(f(x(t), y(t))\) is constant with respect to \(t\). This implies \(f(x(t), y(t)) = c\) for some constant \(c\). Thus, each solution curve \(\mathbf{X}(t)\) lies on a level curve of \(f\).
Key Concepts
Partial DerivativesChain RuleAutonomous SystemsLevel Curves
Partial Derivatives
Understanding partial derivatives is crucial when dealing with functions of multiple variables. A partial derivative of a function of several variables is its derivative with respect to one of those variables, while holding the others constant.
In the context of this exercise, if we have a function \( f(x,y) \), its partial derivatives are denoted by \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). These represent the rate of change of the function \( f \) with respect to \( x \) and \( y \) respectively, considering all other variables as constants during differentiation.
For example, \( \frac{\partial f}{\partial x} \) describes how \( f \) changes as \( x \) changes, while \( y \) remains fixed.Partial derivatives are fundamental in distinguishing the variations in multivariable systems, and they are essential when constructing flows and analyzing the behavior of functions over different parameters.
In the context of this exercise, if we have a function \( f(x,y) \), its partial derivatives are denoted by \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). These represent the rate of change of the function \( f \) with respect to \( x \) and \( y \) respectively, considering all other variables as constants during differentiation.
For example, \( \frac{\partial f}{\partial x} \) describes how \( f \) changes as \( x \) changes, while \( y \) remains fixed.Partial derivatives are fundamental in distinguishing the variations in multivariable systems, and they are essential when constructing flows and analyzing the behavior of functions over different parameters.
Chain Rule
The chain rule is a fundamental theorem in calculus, instrumental for differentiating composite functions.
When a function is a composition of other functions, the chain rule allows us to find its derivative by multiplying the derivatives of the inner and outer functions. In mathematical terms, if \( z = f(x, y) \) and \( x \) and \( y \) are functions of \( t \), then the chain rule gives:
When a function is a composition of other functions, the chain rule allows us to find its derivative by multiplying the derivatives of the inner and outer functions. In mathematical terms, if \( z = f(x, y) \) and \( x \) and \( y \) are functions of \( t \), then the chain rule gives:
- \( \frac{d}{dt} f(x(t), y(t)) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \)
Autonomous Systems
Autonomous systems are systems of differential equations where the time variable \( t \) does not explicitly appear, making the system time-independent.
This means that the equations' structure does not change over time, allowing for analysis of the system's behavior purely based on the states within it. For the given autonomous system:
Autonomous systems can reveal the stability and behavior of the system by analyzing the equilibrium points and trajectories, or flows, in the phase plane. In our exercise, these flows are defined by the partial derivatives of the function \( f(x,y) \). Understanding autonomous systems can help explore the long-term behavior or attractors within the system, crucial for many applications in physics and engineering.
This means that the equations' structure does not change over time, allowing for analysis of the system's behavior purely based on the states within it. For the given autonomous system:
- \( x' = P(x, y) \)
- \( y' = Q(x, y) \)
Autonomous systems can reveal the stability and behavior of the system by analyzing the equilibrium points and trajectories, or flows, in the phase plane. In our exercise, these flows are defined by the partial derivatives of the function \( f(x,y) \). Understanding autonomous systems can help explore the long-term behavior or attractors within the system, crucial for many applications in physics and engineering.
Level Curves
Level curves, or contour lines, are curves along which a function of two variables is constant.
In the exercise context, level curves of \( f \) are described by \( f(x, y) = c \), where \( c \) is a constant. These curves visually illustrate the "shape" of the function \( f \) in terms of its constant values.
When we demonstrate that \( f(x(t), y(t)) = c \) for a solution curve \( \mathbf{X}(t) \), we are proving that the solution remains on the same level curve as time progresses.Level curves are essential in understanding how functions behave in multi-dimensional spaces because they provide insight into the relationship between variables and the consistency of outcomes.
They help us understand the topology and gradient of the function, and are extremely useful in fields such as meteorology for weather maps, in geography for elevation plots, and in economics for utility functions.
In the exercise context, level curves of \( f \) are described by \( f(x, y) = c \), where \( c \) is a constant. These curves visually illustrate the "shape" of the function \( f \) in terms of its constant values.
When we demonstrate that \( f(x(t), y(t)) = c \) for a solution curve \( \mathbf{X}(t) \), we are proving that the solution remains on the same level curve as time progresses.Level curves are essential in understanding how functions behave in multi-dimensional spaces because they provide insight into the relationship between variables and the consistency of outcomes.
They help us understand the topology and gradient of the function, and are extremely useful in fields such as meteorology for weather maps, in geography for elevation plots, and in economics for utility functions.
Other exercises in this chapter
Problem 29
(a) Show that the plane autonomous system $$ \begin{aligned} &x^{\prime}=-x+y-x^{3} \\ &y^{\prime}=-x-y+y^{2} \end{aligned} $$ has two critical points by sketch
View solution Problem 30
If a plane autonomous system has a periodic solution, then there must be at least one critical point inside the curve generated by the solution. In Problems 27-
View solution Problem 31
Use the phase-plane method to show that \((0,0)\) is a center of the nonlinear second-order differential equation \(x^{\prime \prime}+2 x^{3}=0\).
View solution Problem 32
Use the phase-plane method to show that the solution to the nonlinear second- order differential equation \(x^{\prime \prime}+2 x-x^{2}=0\) that satisfies \(x(0
View solution