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Consider the autonomous third order scalar equation \(y^{\prime \prime \prime}+f\left(y^{\prime}\right)=0\), where \(f\) is a continuous function. Does this differential equation have a conservation law? If so, obtain the equation of the family of conserved quantities.

Short Answer

Expert verified
Question: Determine whether the autonomous third-order scalar differential equation \(y^{\prime \prime \prime} + f(y^{\prime}) = 0\) has a conservation law and, if so, find the family of conserved quantities. Answer: The given differential equation has a conservation law, and the family of conserved quantities can be represented by the equation \(Q(y, y', y'') = y^{\prime \prime} \phi + 2\int y'f'(y^{\prime})\phi dx - C\), where \(\phi(x)\) is an arbitrary function.

Step by step solution

01

Check if the equation is conservative

To determine if a differential equation has a conservation law, we can use Noether's theorem, which links the symmetries of a problem with its conserved quantities. In the given differential equation, the transformation \(y(x) \rightarrow y(x) + \varepsilon \phi(x)\) should remain invariant (\(\varepsilon\) represents a small parameter) if there is a conservation law. Let's differentiate this transformation with respect to \(x\): 1. \(y(x) \rightarrow y'(x) + \varepsilon \phi'(x)\) 2. \(y'*y(x) \rightarrow y'(x)y(x) + \varepsilon y'(x)\phi(x)\) 3. \(y'^2 * y(x) \rightarrow y'(x)^2 y(x) + 2\varepsilon y'(x)^2\phi(x) + \varepsilon^2 y'(x)^2\phi(x)^2\) Now, plug this transformation into the given differential equation: \(y^{\prime \prime \prime} + f(y^{\prime}) = 0 \Rightarrow y^{\prime \prime \prime} + f( y'(x)^2 y(x) + 2\varepsilon y'(x)^2\phi(x) + \varepsilon^2 y'(x)^2\phi(x)^2 ) = 0\). The equation is invariant under this transformation if the same equation remains for both \(O(\varepsilon^0)\) and \(O(\varepsilon^1)\), where \(O(\varepsilon^n)\) represents the terms with \(\varepsilon^n\) in the expansion.
02

Obtain the conserved quantity

We find the \(O(\varepsilon^0)\) and \(O(\varepsilon^1)\) terms of the transformed equation and match the terms on both sides. \(O(\varepsilon^0): y^{\prime \prime \prime} + f(y^{\prime}) = 0.\) \(O(\varepsilon^1): y^{\prime \prime} \phi' + 2y'f'(y^{\prime})\phi = 0.\) The last equation can be rewritten as: \(y^{\prime \prime} \phi' = - 2y'f'(y^{\prime})\phi.\) Integrating both sides of the last equation with respect to \(x\), we get: \(\int y^{\prime \prime} \phi' dx = - 2\int y'f'(y^{\prime})\phi dx \Rightarrow y^{\prime \prime} \phi = - 2\int y'f'(y^{\prime})\phi dx + C\) The result shows that the conserved quantity equation \(Q(y, y', y'') = y^{\prime \prime} \phi + 2\int y'f'(y^{\prime})\phi dx - C\) holds, which means the differential equation has a conservation law. In this case, the family of conserved quantities depends on the function \(\phi(x)\) that we choose.

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Key Concepts

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

Noether's Theorem
Understanding the connection between symmetries and conservation laws in physics and mathematics is greatly facilitated by Noether's Theorem. Introduced by Emmy Noether in 1915, this fundamental principle states that every differentiable symmetry of the action of a physical system has a corresponding conservation law. In the context of differential equations, action refers to an integral quantity that represents the dynamics of a system.

Consider a typical scenario in classical mechanics: spatial translation symmetry, which means that the laws of physics are the same regardless of where you are in space, corresponds to the conservation of momentum. Similarly, temporal translation symmetry, meaning the laws of physics are the same at any point in time, corresponds to the conservation of energy.

Thus, Noether's theorem is a powerful tool in seeking conserved quantities for differential equations, provided we can identify the relevant symmetries. In the step-by-step solution provided, the search for a conservation law within a third order differential equation uses the philosophy behind Noether's theorem—symmetries are indeed closely associated with conservation laws.
Third Order Scalar Differential Equation
A third order scalar differential equation is one that involves derivatives up to the third degree of a dependent variable with respect to a single independent variable. In our exercise, the independent variable could be considered as time, and the dependent variable as some physical quantity that changes with time. The equation given in the exercise, \( y^{\text{'''}} + f(y') = 0 \) is autonomous, meaning that the independent variable, let's call it 'x', does not explicitly appear in the function 'f'.

Autonomous equations are often easier to handle because their symmetry under translation in 'x' simplifies the analysis of their behavior. When working with such high-order non-linear differential equations, they can be challenging to solve explicitly. However, identifying conserved quantities simplifies the process of understanding the solutions and can even help in solving the equation indirectly.
Conserved Quantities
In the language of mathematics and physics, conserved quantities are elements that remain unchanged under various transformations as a system evolves. These are crucial for scientists and engineers because they imply underlying symmetries and invariants in complex systems.

Energy, momentum, and electrical charge are classic examples from physics, where these quantities remain conserved under isolation from external influences. In the context of our third order scalar differential equation, the conserved quantity would be a function involving \(y\), \(y'\), and \(y''\), which remains constant for any solution of the equation under certain conditions or symmetry transformations.

Identifying these conserved quantities is not just an academic exercise. Practically, they can provide insights into the stability of solutions or even be exploited to find particular solutions to otherwise difficult equations.
Symmetries in Differential Equations
Symmetries in differential equations are transformations that leave the form of the equation unchanged. The idea is closely related to symmetry in everyday life: just as a circle looks the same after being rotated, a differential equation has symmetry if there is a transformation of the variables that preserves the structure of the equation.

These symmetries can be spatial, temporal, or a more complex transformation. For instance, a simple spatial symmetry might be if a differential equation remains unchanged when you move the system from one location to another. In our step-by-step example, the invariance of the differential equation under a specific transformation connected to \(y(x)\) was tested. This invariance was a signal that a conserved quantity might exist, per Noether's theorem. It's these symmetries that allow us to link conservation laws to the differential equations governing physical systems.

Understanding symmetries is not only important for identifying conservation laws but also for simplifying equations and reducing the complexity of problems in various fields like physics, engineering, economics, and beyond.

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Most popular questions from this chapter

In each exercise, consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Since \(A\) is a constant invertible \((2 \times 2)\) matrix, \(\mathbf{y}=\mathbf{0}\) is the unique (isolated) equilibrium point. (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 the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} -1 & 1 \\ -1 & -1 \end{array}\right] \mathbf{y} $$

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 \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=-3 z_{1}+z_{2}+z_{1}^{2}+z_{2}^{2} \\ &z_{2}^{\prime}=2 z_{1}-2 z_{2}+\left(z_{1}^{2}+z_{2}^{2}\right)^{1 / 3} \end{aligned} $$

In each exercise, consider the linear system \(\mathbf{y}^{\prime}=A \mathbf{y}\). Since \(A\) is a constant invertible \((2 \times 2)\) matrix, \(\mathbf{y}=\mathbf{0}\) is the unique (isolated) equilibrium point. (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 the phase-plane origin. If the equilibrium point is a node, designate it as either a proper node or an improper node. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rr} 1 & -6 \\ 2 & -6 \end{array}\right] \mathbf{y} $$

Use the information provided to determine the unspecified constants. Consider the system $$ \begin{aligned} &x^{\prime}=\alpha x^{2}+\beta y+1 \\ &y^{\prime}=x+\gamma y+y^{2} . \end{aligned} $$ The slopes of the phase-plane trajectories passing through the points \((x, y)=(1,1)\) and \((1,-1)\) are 0 and 4 , respectively. The phase-plane trajectory passing through the point \((x, y)=(0,-1)\) has a vertical tangent.

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 \(6.4\). $$ \begin{aligned} &z_{1}^{\prime}=2 z_{1}+2 z_{2} \\ &z_{2}^{\prime}=-5 z_{1}-2 z_{2}+z_{1}^{2} \end{aligned} $$

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