Question

Let \(A\) be an \(n \times n\) constant matrix. Then Theorem 10.2 .1 implies that the solutions of $$\mathbf{y}^{\prime}=A \mathbf{y}$$ are all defined on \((-\infty, \infty)\). (a) Use Theorem 10.2 .1 to show that the only solution of ( A) that can ever equal the zero vector is \(\mathbf{y} \equiv \mathbf{0}\). (b) Suppose \(\mathbf{y}_{1}\) is a solution of \((\mathrm{A})\) and \(\mathbf{y}_{2}\) is defined by \(\mathbf{y}_{2}(t)=\mathbf{y}_{1}(t-\tau),\) where \(\tau\) is an arbitrary real number. Show that \(y_{2}\) is also a solution of (A). (c) Suppose \(\mathbf{y}_{1}\) and \(\mathbf{y}_{2}\) are solutions of \((\mathrm{A})\) and there are real numbers \(t_{1}\) and \(t_{2}\) such that \(\mathbf{y}_{1}\left(t_{1}\right)=\mathbf{y}_{2}\left(t_{2}\right) .\) Show that \(\mathbf{y}_{2}(t)=\mathbf{y}_{1}(t-\tau)\) for all \(t,\) where \(\tau=t_{2}-t_{1} .\) HINT: Show that \(\mathbf{y}_{1}(t-\tau)\) and \(\mathbf{y}_{2}(t)\) are solutions of the same initial value problem for \((\mathrm{A}),\) and apply the uniqueness assertion of Theorem \(10.2 .1 .\)

Short answer

In summary, using Theorem 10.2.1, we showed the following properties of the solutions of a linear system of first-order differential equations: (a) The only solution that can equal the zero vector is the trivial solution \(\mathbf{y} \equiv \mathbf{0}\). (b) If \(\mathbf{y}_1(t)\) is a solution of the system, then \(\mathbf{y}_2(t) = \mathbf{y}_1(t - \tau)\) is also a solution for any arbitrary real number \(\tau\). (c) If \(\mathbf{y}_1(t_1) = \mathbf{y}_2(t_2)\), then \(\mathbf{y}_2(t) = \mathbf{y}_1(t - \tau)\) for all \(t\), where \(\tau = t_2 - t_1\).

Step-by-step solution

(a) Prove that the only solution equal to the zero vector is the trivial solution

Assume there exists a non-trivial solution \(\mathbf{y}(t)\) such that it equals the zero vector at some point \(t_0\), i.e., \(\mathbf{y}(t_0) = \mathbf{0}\). According to Theorem 10.2.1, since the solution is non-trivial and \(\mathbf{y}(t_0) = \mathbf{0}\), there must be another solution that differs from \(\mathbf{y}(t)\). However, this contradicts the uniqueness part of Theorem 10.2.1, which states that there can only be one unique solution for a given initial condition. Therefore, the only solution that can equal the zero vector is the trivial solution \(\mathbf{y} \equiv \mathbf{0}\).

(b) Show that \(\mathbf{y}_2(t) = \mathbf{y}_1(t - \tau)\) is also a solution

Let's differentiate \(\mathbf{y}_2(t)\) with respect to \(t\): $$\mathbf{y}_2'(t) = \frac{d}{dt}(\mathbf{y}_1(t - \tau)) = \mathbf{y}_1'(t - \tau).$$ Now, since \(\mathbf{y}_1(t)\) is a solution of the system, we have \(\mathbf{y}_1'(t) = A\mathbf{y}_1(t)\). Thus, we can write: $$\mathbf{y}_1'(t - \tau) = A\mathbf{y}_1(t - \tau) = A\mathbf{y}_2(t).$$ This shows that \(\mathbf{y}_2(t) = \mathbf{y}_1(t - \tau)\) is also a solution of the system, as it satisfies the original equation \(\mathbf{y}' = A\mathbf{y}\).

(c) Show that if \(\mathbf{y}_1(t_1) = \mathbf{y}_2(t_2)\), then \(\mathbf{y}_2(t) = \mathbf{y}_1(t - \tau)\) for all \(t\)

We are given that \(\mathbf{y}_1(t_1) = \mathbf{y}_2(t_2)\) and we want to show that \(\mathbf{y}_2(t) = \mathbf{y}_1(t - \tau)\) for all \(t\), where \(\tau = t_2 - t_1\). Let's consider the initial value problems: - For \(\mathbf{y}_1(t)\), the initial condition is \(\mathbf{y}_1(t_1) = \mathbf{y}_2(t_2)\). - For \(\mathbf{y}_1(t - \tau)\), the initial condition is \(\mathbf{y}_1(t_1 - \tau) = \mathbf{y}_1(t_1 - (t_2 - t_1)) = \mathbf{y}_1(t_1) = \mathbf{y}_2(t_2)\). We can see that both \(\mathbf{y}_1(t - \tau)\) and \(\mathbf{y}_2(t)\) are solutions of the same initial value problem for the system \(\mathbf{y}' = A\mathbf{y}\) with the same initial condition \(\mathbf{y}_1(t_1) = \mathbf{y}_2(t_2)\). According to the uniqueness assertion of Theorem 10.2.1, they must be the same function for all \(t\). Thus: $$\mathbf{y}_2(t) = \mathbf{y}_1(t - \tau)$$ for all \(t\).

Key concepts

Constant Coefficient Linear Systems

Understanding constant coefficient linear systems is crucial when studying differential equations. These systems are characterized by differential equations where coefficients are constants, meaning they do not change with respect to the independent variable, typically time. An example of such a system is given by the equation \( \mathbf{y}' = A\mathbf{y} \) where \( A \) is an \( n \times n \) matrix with constant entries.

This type of system is widely applicable in both theoretical and practical aspects, as it models many natural phenomena, such as mechanical and electrical systems. To solve constant coefficient linear systems, we often look for solutions that can be expressed as functions of \( e \) raised to a power, since the derivative of \( e^{at} \) is a simple multiple of itself, making it an ideal candidate for solutions of linear differential equations.

Uniqueness of Solutions

The uniqueness of solutions in differential equations reassures us that under certain conditions, a differential equation has exactly one solution satisfying a given initial condition. This concept is vital when predicting the behavior of a system over time. Uniqueness can be guaranteed by specific theorems, like the Theorem 10.2.1 referenced in the exercise.

For example, when we are given an initial value problem \( \mathbf{y}' = A\mathbf{y} \) with an initial condition \( \mathbf{y}(t_0) = \mathbf{y}_0 \) where \( A \) is a constant matrix and \( \mathbf{y}_0 \) is a given vector, the theorem assures that there is only one function \( \mathbf{y}(t) \) that fulfills both the differential equation and the initial condition. This is key in applications where we need to be certain that a system follows one and only one path given its starting point.

Initial Value Problems

Initial value problems (IVPs) are a type of differential equation that require solving not just for the general solution, but for a specific instance that aligns with given initial conditions. These problems are commonly written in the form \( \mathbf{y}' = A\mathbf{y} \) along with \( \mathbf{y}(t_0) = \mathbf{y}_0 \) where \( t_0 \) is the initial time and \( \mathbf{y}_0 \) is the initial state of the system.

IVPs are fundamental in modeling real-world scenarios, where typically only the starting state is known, and we wish to predict future behavior. The solution process usually involves finding the general solution to the differential equation and then using the initial conditions to find the particular solution that fits the specific scenario.

Matrix Exponentials

Matrix exponentials are an advanced mathematical tool used in solving systems of linear differential equations with constant coefficients. They are analogous to the exponential function for real numbers but are defined for matrices. The matrix exponential, denoted as \( e^{At} \), where \( A \) is a square matrix and \( t \) is a scalar, plays a key role in finding solutions to differential equations like \( \mathbf{y}' = A\mathbf{y} \) because it simplifies the process of finding solution functions.

The computation of matrix exponentials typically involves diagonalization or other techniques such as series expansion, which can get complex. However, understanding matrix exponentials is essential for deeper insights into how systems evolve over time, especially in control theory and other areas of applied mathematics.