Question

Let \(\Phi(t)\) denote the fundamental matrix satisfying \(\Phi^{\prime}=A \Phi, \Phi(0)=L\) In the text we also denoted this matrix by \(\exp (A t)\), In this problem we show that \(\Phi\) does indeed have the principal algebraic properties associated with the exponential function. (a) Show that \(\Phi(t) \Phi(s)=\Phi(t+s) ;\) that is, \(\exp (\hat{\mathbf{A}} t) \exp (\mathbf{A} s)=\exp [\mathbf{A}(t+s)]\) Hint: Show that if \(s\) is fixed and \(t\) is variable, then both \(\Phi(t) \Phi(s)\) and \(\Phi(t+s)\) satisfy the initial value problem \(\mathbf{Z}^{\prime}=\mathbf{A} \mathbf{Z}, \mathbf{Z}(0)=\mathbf{\Phi}(s)\) (b) Show that \(\Phi(t) \Phi(-t)=\mathbf{I}\); that is, exp(At) \(\exp [\mathbf{A}(-t)]=\mathbf{1}\). Then show that \(\Phi(-t)=\) \(\mathbf{\Phi}^{-1}(t) .\) (c) Show that \(\mathbf{\Phi}(t-s)=\mathbf{\Phi}(t) \mathbf{\Phi}^{-1}(s)\)

Short answer

Based on the given step by step solution, the question is as follows: Show that \(\Phi(t) \Phi(s)=\Phi(t+s)\), \(\Phi(t) \Phi(-t)=\mathbf{I}\), and \(\Phi(t-s)=\Phi(t) \Phi^{-1}(s)\) for the fundamental matrix \(\Phi(t)\), given that it satisfies the initial value problem \(\mathbf{Z}'=\mathbf{A}\mathbf{Z},\ \mathbf{Z}(0)=\Phi(s)\).

Step-by-step solution

Show that both \(\Phi(t)\Phi(s)\) and \(\Phi(t+s)\) satisfy the given IVP.

Given the IVP: \(\mathbf{Z}'=\mathbf{A}\mathbf{Z},\ \mathbf{Z}(0)=\Phi(s)\). Let \(\mathbf{Z}_1(t)=\Phi(t)\Phi(s)\) and \(\mathbf{Z}_2(t)=\Phi(t+s)\). Then we compute their derivatives: \[\mathbf{Z'}_1 = \frac{d (\Phi(t) \Phi(s))}{dt} = \Phi'(t) \Phi(s) = A \Phi(t) \Phi(s)\] \[\mathbf{Z'}_2 = \frac{d (\Phi(t+s))}{dt} = \Phi'(t+s) = A \Phi(t+s)\] Now check the initial conditions: \[\mathbf{Z}_1(0) = \Phi(0) \Phi(s) = L \Phi(s) = \Phi(s)\] \[\mathbf{Z}_2(0) = \Phi(0+s) = \Phi(s)\] Both \(\mathbf{Z}_1\) and \(\mathbf{Z}_2\) satisfy the IVP.

Show that \(\mathbf{Z}_1\) and \(\mathbf{Z}_2\) are equivalent

Since both \(\mathbf{Z}_1\) and \(\mathbf{Z}_2\) satisfy the same IVP with the same initial condition, they must be equivalent. Thus, we have that \(\Phi(t)\Phi(s)=\Phi(t+s)\). (b) Show that \(\Phi(t) \Phi(-t)=\mathbf{I}\) and that \(\Phi(-t)=\Phi^{-1}(t)\).

Prove \(\Phi(t) \Phi(-t)=\mathbf{I}\)

Using the property from part (a), we have: \[\Phi(t) \Phi(-t) = \Phi(t + (-t)) = \Phi(0) = L = \mathbf{I}\]

Prove \(\Phi(-t)=\Phi^{-1}(t)\)

Since \(\Phi(t) \Phi(-t) = \mathbf{I}\), we can conclude that \(\Phi(-t) = \Phi^{-1}(t)\). (c) Show that \(\mathbf{\Phi}(t-s)=\mathbf{\Phi}(t) \mathbf{\Phi}^{-1}(s)\)

Use properties of the exponential function

Using the properties established in parts (a) and (b), we have: \[\Phi(t-s) = \Phi(t) \Phi(-(t-s))\]

Show that \(\mathbf{\Phi}(t-s)=\mathbf{\Phi}(t) \mathbf{\Phi}^{-1}(s)\)

From part (b), we know that \(\Phi(-(t-s))=\Phi^{-1}(t-s)\). Now we can rewrite the equation above as: \[\Phi(t-s) = \Phi(t) \Phi^{-1}(t-s)\]

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. When it comes to physical, chemical, biological, or financial problems, these equations often represent how a certain quantity changes over time and are fundamental in predicting behaviors within those systems.

For example, consider the temperature of a warming object, the growth of bacteria, or the rate of cash flow: These all can be modeled with differential equations. The solutions to these equations typically involve finding a function (or set of functions) that satisfy the given relationship between the variable, usually time, and its rate of change.

Initial Value Problem

An initial value problem is a specific type of differential equation that also provides an initial condition. This combination allows for the determination of a unique solution. In our exercise, \( \Phi(0)=L \) serves as the initial condition, anchoring the solution at time \(t=0\).

This is akin to knowing where a vehicle starts before calculating its position at future times, making it easier to track the vehicle's trajectory accurately. The solution to an initial value problem gives you one specific path out of many possible ones, dictated by the starting position.

Exponential Function Properties

The exponential function, often denoted as \( e^x \) or \( \exp(x) \) when involving matrices, has some remarkable properties that make it unique and very useful in differential equations:

• It is its own derivative, meaning \( \frac{d}{dx}e^x = e^x \).
• It has a multiplicative property, where \( e^{x+y} = e^x \cdot e^y \).
• \( e^0 = 1 \), which serves as an identity in multiplication.
• The function \( e^{-x} \) acts as the multiplicative inverse, so \( e^x \cdot e^{-x} = 1 \) (similar to a reciprocal).

The properties of the exponential function translate into our matrix context, and as such, you can see the exponential function playing a central role in the solutions of matrix-based differential equations.

Matrix Exponentiation

Matrix exponentiation is a process where a square matrix is raised to a power, which in our exercise, is analogous to the matrix equivalent of the exponential function in calculus. In our context, \( \exp(At) \) represents the matrix exponential, and it mimics the behavior of the scalar exponential function.

Just like the scalar case, the matrix exponential has special properties:

• It serves as the fundamental solution to linear systems of differential equations.
• It retains the property \( \exp(A(t+s)) = \exp(At) \cdot \exp(As) \), which is essential when dealing with the composition of influences over time.
• It also respects inverses: \( \exp(-At) \) behaves like \( \exp(At) \) inverse.

These properties allow us to solve complex system behaviors by leveraging the exponential function's structure and behavior in the matrix domain.