Question

Convert the linear scalar equation $$ P_{0}(t) y^{(n)}+P_{1}(t) y^{(n-1)}+\cdots+P_{n}(t) y(t)=F(t) $$ into an equivalent \(n \times n\) system $$ \mathbf{y}^{\prime}=A(t) \mathbf{y}+\mathbf{f}(t) $$ and show that \(A\) and \(\mathbf{f}\) are continuous on an interval \((a, b)\) if and only if \((\mathrm{A})\) is normal on \((a, b)\).

Short answer

Question: Show that the matrix A(t) and the vector f(t) are continuous on an interval (a, b) if and only if condition (A) is normal on that interval. Answer: A(t) and f(t) are continuous on an interval (a, b) if and only if condition (A) is normal on that interval, because the continuity of A(t) and f(t) depends on the continuity of the functions P_i(t) and F(t) that define the system. If A(t) and f(t) are continuous, then all the P_i(t) terms in the A(t) matrix and the F(t) term in the f(t) vector are continuous, making condition (A) normal. Conversely, if condition (A) is normal, all functions P_i(t) and F(t) are continuous in that interval, resulting in the continuity of A(t) and f(t).

Step-by-step solution

Rewrite given equation as a first-order system

To convert the given linear scalar equation of the form $$ P_{0}(t) y^{(n)}+P_{1}(t) y^{(n-1)}+\cdots+P_{n}(t) y(t)=F(t), $$ we will first define the following state variables: $$ y_1(t) = y(t), \\ y_2(t) = y'(t), \\ y_3(t) = y''(t), \\ \vdots \\ y_n(t) = y^{(n)}(t). $$ Now, we can write the system of first-order differential equations using these state variables as follows: $$ \begin{aligned} y_1'(t) &= y_2(t), \\ y_2'(t) &= y_3(t), \\ y_3'(t) &= y_4(t), \\ &\ \ \vdots \\ y_{n-1}'(t) &= y_n(t), \\ y_n'(t) &= \frac{1}{P_0(t)} \left(F(t) - P_n(t)y_1(t) - \cdots - P_1(t)y_{n-1}(t) \right). \end{aligned} $$ This system of linear first-order differential equations can be represented in the matrix form as: $$ \mathbf{y}'(t) = A(t) \mathbf{y}(t) + \mathbf{f}(t), $$ where $$ \mathbf{y}(t) = \begin{pmatrix} y_1(t) \\ y_2(t) \\ y_3(t) \\ \vdots \\ y_n(t) \end{pmatrix}, $$ $$ A(t) = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots &\vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ -\frac{P_n(t)}{P_0(t)} & -\frac{P_{n-1}(t)}{P_0(t)} & \cdots & -\frac{P_1(t)}{P_0(t)} & 0 \end{pmatrix}, $$ and $$ \mathbf{f}(t) = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ \frac{F(t)}{P_0(t)} \end{pmatrix}. $$

Show A(t) and f(t) are continuous if and only if condition (A) is normal

We have already found the equivalent system of first-order differential equations for the given scalar equation and represented it in matrix form. Now, let's analyze the continuity of A(t) and f(t) in the interval (a, b). For A(t) and f(t) to be continuous, the functions P_i(t) and F(t) must also be continuous in that interval. If A(t) and f(t) are continuous, then all the P_i(t) terms in the A(t) matrix and the F(t) term in the f(t) vector are continuous, since they are part of the continuous functions that define the system. Therefore, if A(t) and f(t) are continuous on an interval (a, b), condition (A) holds for that interval, being normal. Conversely, if condition (A) is normal on the interval (a, b), all functions P_i(t) and F(t) are continuous in that interval. In this case, A(t) and f(t) can be built using these continuous functions, making A(t) and f(t) continuous as well. So, we conclude that A(t) and f(t) are continuous on an interval (a, b) if and only if condition (A) is normal on that interval.

Key concepts

First-Order Differential Equation Systems

Understanding first-order differential equation systems is crucial for breaking down more complex equations into simpler parts that can be analyzed and solved. This concept revolves around transforming higher-order differential equations into a set of first-order equations.

For instance, consider a higher-order equation such as \( P_{0}(t) y^{(n)}+P_{1}(t) y^{(n-1)}+\cdots+P_{n}(t) y(t)=F(t) \). To reframe this as a first-order differential equation system, we introduce new variables to represent the derivatives of \(y\), up to the \(n-1\)th derivative. Each of these variables becomes a component of the state vector \(\mathbf{y}\).

The key advantage of this method lies in the reduction of complexity; by converting to a first-order system, we can leverage matrix algebra and vector calculus to understand the behavior of the original higher-order equation. Moreover, this approach facilitates the application of numerical methods for solving differential equations, which are generally geared towards first-order systems.

Matrix Representation of Differential Equations

Matrix representation offers a compact and powerful way to represent differential equation systems. In the context of our problem, we utilized a matrix \(A(t)\) to represent the coefficients of the new system, and a vector \(\mathbf{f}(t)\) for the inputs to the system.

The structure of \(A(t)\) reflects the relationships between the derivatives of \(y\), while \(\mathbf{f}(t)\) incorporates the forcing function \(F(t)\) into our system equation, with all the other terms being zero, emphasizing its direct impact only on the \(n\)th equation.

Numerical Stability and Interpretation

In analyzing these systems, the matrix representation provides insights into the numerical stability and the dynamic behavior of the system, predicting how the system evolves over time. The sparsity pattern of the transition matrix \(A(t)\)—primarily zeroes except for specific entries—shows how each state variable directly influences only the next one in line, providing a clear depiction of the system dynamics.

Continuity of Differential Equation Solutions

When we talk about the continuity of differential equation solutions, we're referring to whether solutions can be described smoothly within a given interval. This property is crucial for ensuring that the solutions are not just theoretically correct, but also practically meaningful.

In our exercise, for the matrix \(A(t)\) and vector \(\mathbf{f}(t)\) to be continuous, the scalar functions from which they are derived, \(P_i(t)\) and \(F(t)\), must also be continuous. Continuity of \(A(t)\) and \(\mathbf{f}(t)\) means the system's description does not change abruptly, which is essential for the existence and uniqueness of solutions to the differential equation.

Practical Implications

From a practical standpoint, the continuity requirement ensures that numerical solutions to the differential equations do not exhibit unexpected 'jumps' or 'breaks.' This forms the basis for algorithms used in computational applications to predict system behavior reliably. It's not just a mathematical nicety—it significantly influences how we study physical systems and predict their behavior under various conditions.