Question

Prove that \(\varphi(t)\) is a solution of the equation $$ \dot{x}=A x $$ with the Initial condition \(\varphi(0)=x_{0}\), where \(A: R^{n} \rightarrow R^{n}\) is the linear operator \(\left(\equiv \boldsymbol{R}\right.\)-endomorphism) defined by the relation \(A \boldsymbol{x}=\left.\frac{d}{d t}\right|_{t=0}\left(g^{\mathrm{t}} \boldsymbol{x}\right)\) for all \(x \in \boldsymbol{R}^{\mathrm{n}} .\) Hint. Cf. 5 4, Sect. \(4 .\) Equation (1) is called linear. Thus to describe all one-parameter groups of linear transformations it suffices to study the solutions of the linear equations \((1)\) We shall see below that the correspondence between one-parameter groups \(\left\\{g^{t}\right\\}\) of linear transformations and the linear equations (1) is one-to-one and ontos each operator \(A: R^{n} \rightarrow R^{n}\) defines a one-parameter group \(\left\\{g^{\prime}\right\\} .\)

Short answer

Question: Show that the function \(\varphi(t) = g^{t} \boldsymbol{x}\) is a solution of the linear equation \(\dot{x} = Ax\) with the initial condition \(\varphi(0) = x_{0}\). Answer: To show that \(\varphi(t)\) is a solution, we first rewrote the given equation using the relation \(A x=\left.\frac{d}{d t}\right|_{t=0}\left(g^{\mathrm{t}} \boldsymbol{x}\right)\) and the initial condition \(\varphi(0) = x_{0}\). Then, we found the one-to-one correspondence between one-parameter groups of linear transformations and linear equations by differentiating both sides of the equation and evaluating at \(t=0\). Finally, we integrated both sides of the equation to find the solution and showed that \(\varphi(t) = g^{t} \boldsymbol{x}\) satisfies both the linear equation and the initial condition, making it a solution of the given equation.

Step-by-step solution

Define the given equation

Rewrite the given equation with the linear operator \(A: R^{n} \rightarrow R^{n}\) as: $$ \dot{x} = Ax $$

Using the hint, rewrite the equation with the relation

Rewrite the equation using the relation \(A x=\left.\frac{d}{d t}\right|_{t=0}\left(g^{\mathrm{t}} \boldsymbol{x}\right)\): $$ \dot{x} = \left.\frac{d}{d t}\right|_{t=0}\left(g^{\mathrm{t}} \boldsymbol{x}\right) $$

Write down the initial condition

The given initial condition is: $$ \varphi(0) = x_{0} $$

Determine the one-to-one correspondence between one-parameter groups and linear equations

To show the one-to-one correspondence, we can first differentiate both sides of the equation and then evaluate the result at \(t=0\): $$ \frac{d}{d t} \left(g^{\mathrm{t}} \boldsymbol{x}\right) = A \left(g^{\mathrm{t}} \boldsymbol{x}\right) \\ \left.\frac{d}{d t}\right|_{t=0}\left(g^{\mathrm{t}} \boldsymbol{x}\right) = A \boldsymbol{x} $$

Establish the initial condition

Now, we can use the initial condition \(\varphi(0) = x_{0}\) to find the value of \(g^{t}\) at \(t=0\): $$ g^{0} \boldsymbol{x} = \left.\frac{d}{d t}\right|_{t=0}\left(g^{\mathrm{t}} \boldsymbol{x}\right) = A \boldsymbol{x} $$

Integrate both sides

Integrate both sides of the equation to find the solution of the linear equation: $$ \int \dot{x} dt = \int A \boldsymbol{x} dt $$

Show that \(\varphi(t)\) is a solution

Now, we can show that \(\varphi(t)\) is a solution of the equation with the initial condition. Since $$ \frac{d}{d t} \left(g^{\mathrm{t}} \boldsymbol{x}\right) = A \left(g^{\mathrm{t}} \boldsymbol{x}\right) \\ $$ At \(t=0\), we have $$ g^{0} \boldsymbol{x} = \left.\frac{d}{d t}\right|_{t=0}\left(g^{\mathrm{t}} \boldsymbol{x}\right)= A \boldsymbol{x} \\ $$ Thus, the function \(\varphi(t) = g^{t} \boldsymbol{x}\) satisfies both the linear equation and the initial condition, so it is a solution of the given equation.

Key concepts

Linear Differential Equations

In mathematics, particularly in calculus and differential equations, a linear differential equation is an equation involving a function and its derivatives. It's characterized by the linearity of the unknown function and its derivative terms. Specifically, for a linear differential equation, the terms involving the dependent variable and all its derivatives are of the first degree. This means no powers or products of these terms are present.

For instance, the given equation \( \dot{x} = Ax \) is a linear differential equation. Here, \( A \) is a constant matrix acting on the vector \( x \), making it linear. In general, the simplicity and predictability of linear differential equations make them easier to solve compared to their nonlinear counterparts. Solving such equations can provide insights into various phenomena in physics, engineering, and other fields. They are foundational in modeling systems where the principle of superposition applies, meaning solutions can be linearly summed to form new solutions.

Initial Value Problem

An initial value problem (IVP) is a type of problem in differential equations where you're given an equation along with specific initial conditions. These conditions provide starting values for the dependent variable at a given point. The goal is to find a solution that satisfies both the equation and these initial conditions.

For example, in the exercise, we have \( \varphi(0) = x_{0} \) as the initial condition. This means we are tasked with finding a function \( \varphi(t) \) such that it satisfies \( \dot{x} = Ax \) and at \( t = 0 \), \( \varphi(t) = x_0 \). This initial condition is crucial because it offers a 'starting point' for our solution, ensuring it fits the specific scenario we're investigating.

• Initial conditions help determine a particular solution among potential general solutions to a differential equation.
• They are essential for modeling real-world situations like predicting population growth or determining the motion of projectiles.

Linear Operators

Linear operators are functions from one vector space to another that respect the operations of vector addition and scalar multiplication. This means applying the operator to a sum of vectors or a scalar multiple of a vector yields the same result as if you applied the operator separately and combined the results. In mathematical notation, an operator \( A \) is linear if for all vectors \( u \) and \( v \) and any scalar \( c \):

• \( A(u + v) = A(u) + A(v) \)
• \( A(cu) = cA(u) \)

In the context of the exercise, the operator \( A: R^n \rightarrow R^n \) is defined such that \( A \boldsymbol{x} = \left.\frac{d}{d t}\right|_{t=0}\left(g^{\mathrm{t}} \boldsymbol{x}\right) \). This formalizes how the transformation operates over the vector space \( R^n \). Being linear, \( A \) simplifies the analysis and solution of the differential equation, since it allows for predictable behavior of solutions under vector combinations.

One-Parameter Groups

In mathematical analysis and the study of differential equations, a one-parameter group refers to a continuous group of transformations that depend on a single parameter, typically denoted as \( t \).

These groups form an essential part of the theory of continuous symmetries and dynamical systems. A one-parameter group \( \{g^t\} \) can be thought of as a family of transformations or functions depending continuously on the parameter \( t \). Each \( g^t \) performs a specific transformation.

• The parameter \( t \) typically represents time, suggesting how a system evolves over time.
• In the exercise, \( \{g^t\} \) represents the flow of the system dictated by the linear equation \( \dot{x}=Ax \).

Importantly, the exercise connects the concept of one-parameter groups with linear operators, showing a unique correspondence between them. Understanding this correspondence is crucial for analyzing how solutions to linear differential equations behave over time, and it lays the foundation for more advanced topics in dynamical systems and group theory.