Question

Suppose the \(n \times n\) matrix \(A=A(t)\) is continuous on \((a, b)\) and \(t_{0}\) is in \((a, b) .\) For \(i=1,2, \ldots,\) \(n,\) let \(\mathbf{y}_{i}\) be the solution of the initial value problem \(\mathbf{y}_{i}^{\prime}=A(t) \mathbf{y}_{i}, \mathbf{y}_{i}\left(t_{0}\right)=\mathbf{e}_{i},\) where $$ \mathbf{e}_{1}=\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array}\right], \quad \mathbf{e}_{2}=\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array}\right], \quad \cdots \quad \mathbf{e}_{n}=\left[\begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \end{array}\right] $$ that is, the \(j\) th component of \(\mathbf{e}_{i}\) is 1 if \(j=i\), or 0 if \(j \neq i\). (a) Show that \(\left\\{\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}\right\\}\) is a fundamental set of solutions of \(\mathbf{y}^{\prime}=A(t) \mathbf{y}\) on \((a, b)\). (b) Conclude from (a) and Exercise 15 that \(\mathbf{y}^{\prime}=A(t) \mathbf{y}\) has infinitely many fundamental sets of solutions on \((a, b)\).

Short answer

Based on the given conditions, we showed that the solution set \(\left\{\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}\right\}\) is a fundamental set of solutions for the initial value problem \(\mathbf{y}' = A(t) \mathbf{y}\) on \((a, b)\). Then, by using the result of Exercise 15, we concluded that there are infinitely many fundamental sets of solutions for this problem on the interval \((a, b)\).

Step-by-step solution

Compute the Wronskian#

We will compute the Wronskian of \(\left\{\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}\right\}\), denoted as W[\(\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}\)]. The Wronskian of n functions is computed using the determinant of a matrix, therefore the Wronskian of \(\mathbf{y}_{1}, \mathbf{y}_{2}, ... , \mathbf{y}_{n}\) is given by: \[W[\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}]=\det\left(\begin{array}{ccc} y_{1,1} & y_{2,1} & \cdots & y_{n,1} \\ y_{1,2} & y_{2,2} & \cdots & y_{n,2} \\ \vdots & \vdots & \ddots & \vdots \\ y_{1,n} & y_{2,n} & \cdots & y_{n,n} \end{array}\right), \] where \(y_{i,j}\) denotes the \(j\)-th component of the vector function \(\mathbf{y}_{i}\). We also know that \(\mathbf{y}_{i}(t_0)=\mathbf{e}_{i}\).

Evaluate the Wronskian at \(t_0\)#

Now, we will evaluate the Wronskian at \(t_0\): \[W[\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}](t_0)=\det\left(\begin{array}{ccc} y_{1,1}(t_0) & y_{2,1}(t_0) & \cdots & y_{n,1}(t_0) \\ y_{1,2}(t_0) & y_{2,2}(t_0) & \cdots & y_{n,2}(t_0) \\ \vdots & \vdots & \ddots & \vdots \\ y_{1,n}(t_0) & y_{2,n}(t_0) & \cdots & y_{n,n}(t_0) \end{array}\right). \] Since \(\mathbf{y}_{i}(t_0)=\mathbf{e}_{i}\), each column represents an elementary basis vector \(\mathbf{e}_{i}\), and the determinant is the identity matrix, so: \[W[\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}](t_0)=\det(I)=1.\] Because the Wronskian is non-zero, the set of solutions \(\left\{\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}\right\}\) is linearly independent on \((a, b)\), and it is a fundamental set of solutions.

Use Exercise 15 to conclude part (b)#

In Exercise 15, it was shown that if a linear differential equation has one fundamental set of solutions, it has infinitely many such sets. Since we have found that \(\left\{\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}\right\}\) is a fundamental set of solutions for the given initial value problem, we can conclude that there are infinitely many fundamental sets of solutions for \(\mathbf{y}^{\prime}=A(t) \mathbf{y}\) on \((a, b)\).

Key concepts

Wronskian

The Wronskian is a mathematical tool used in the theory of differential equations. It helps us determine whether a set of solutions is linearly independent. In other words, it tells us if the solutions are distinct and do not overlap in a meaningful way. This can be very useful when working with differential equations, especially when you need a fundamental set of solutions.

To compute the Wronskian of a set of functions, say \((\mathbf{y}_1, \mathbf{y}_2, \ldots, \mathbf{y}_n)\), you form a square matrix by placing each function's derivatives in rows. The determinant of this matrix is the Wronskian. For example, suppose we have an n by n matrix as:

• Each row corresponds to the derivatives of the functions.
• The determinant of this matrix is the Wronskian, \[ W[\mathbf{y}_{1}, \mathbf{y}_{2}, \ldots, \mathbf{y}_{n}] = \det(\text{matrix}) \].

If the Wronskian is non-zero at some point in the interval, the functions are linearly independent over that interval.

When working with initial value problems, checking the Wronskian can confirm that you have the necessary independent solutions.

Fundamental set of solutions

A "fundamental set of solutions" in the realm of differential equations refers to a set of solutions that can be used to express any other solution in the system. This is vital when working with linear differential equations because it provides a complete understanding of all possible behaviors of the system.

Such a set is characterized by its linear independence, ensured by the Wronskian not equaling zero. In simpler terms, these solutions form a basis for the space of solutions. This means that any solution to the differential equation can be written as a combination of solutions from the fundamental set.

These distinct solutions allow us to express a general solution for differential equations, such as:

• Assume \(\mathbf{y}_1, \mathbf{y}_2, \ldots, \mathbf{y}_n\) is a valid fundamental set.
• Any solution \(\mathbf{y}\) to the differential equation can be expressed as \(c_1\mathbf{y}_1 + c_2\mathbf{y}_2 + \ldots + c_n\mathbf{y}_n\).

This fundamental concept enables the comprehensive analysis and solution of differential equations.

Initial value problem

An Initial Value Problem, often abbreviated as IVP, is a type of problem where you are provided with a differential equation and a set of initial conditions. These conditions specify the value of the solution at a particular point, usually at the beginning of the interval.

This problem type is crucial because it sets the starting point for solving differential equations, ensuring that the solutions fit specific real-world scenarios. For example, if you're solving a problem about motion, the initial conditions could tell you the starting position and speed.

• For a first-order IVP, specify \(\mathbf{y}(t_0) = \mathbf{e}\).
• This condition directs the solution toward exact behaviors specified at \(t_0\).

In the given exercise, solutions \(\mathbf{y}_i\) meet the initial conditions \(\mathbf{y}_i(t_0)=\mathbf{e}_i\). This ensures the solutions form a fundamental set by satisfying both the differential equation and the initial conditions.

Matrix functions

Matrix functions play a huge role in solving systems of linear differential equations. These functions are derived from matrices that evolve over time, and they enable us to simplify and solve complex differential equations.

In our context, matrix functions come from the matrix \(A(t)\), which is continuous and instrumentally defines the behavior of the system of differential equations. By manipulating this matrix:

• Solutions can be structured using matrix exponential functions, allowing for clarity in solving linear systems.
• These functions provide a framework for analyzing how initial conditions influence future states.

Thus, understanding matrix functions is essential for working with linear systems, as they encode transformation rules applicable over intervals within the differential equations.