Problem 17 - Differential Equations and Linear Algebra

Question

Show that if the vector differential equation \(\mathbf{x}^{\prime}=A \mathbf{x}\) has a solution of the form $$ \mathbf{x}(t)=e^{\lambda t}\left(\mathbf{v}_{2}+t \mathbf{v}_{1}+\frac{t^{2}}{2 !} \mathbf{v}_{2}\right) $$ then $$ \begin{array}{c} (A-\lambda I) \mathbf{v}_{0}=\mathbf{0}, \quad(A-\lambda I) \mathbf{v}_{1}=\mathbf{v}_{0}, \quad \text { and } \\ (A-\lambda I) \mathbf{v}_{2}=\mathbf{v}_{1} \end{array} $$

Step-by-Step Solution

Answer

We showed that if the given solution \(\mathbf{x}(t)\) satisfies the vector differential equation \(\mathbf{x}' = A\mathbf{x}\), then the following relationships hold: \[(A - \lambda I)\mathbf{v}_{0}=\mathbf{0}\] \[(A - \lambda I)\mathbf{v}_{1}=\mathbf{v}_{0}\] \[(A - \lambda I)\mathbf{v}_{2}=\mathbf{v}_{1}\] This implies that the given matrix \(A\), the eigenvalue \(\lambda\), and the associated eigenvectors \(\mathbf{v}_{0}\), \(\mathbf{v}_{1}\), and \(\mathbf{v}_{2}\) have the desired relationships.

Step 1: Defining the vector differential equation and proposed solution

We are given the vector differential equation as: \[\mathbf{x}' = A\mathbf{x}\] And the proposed solution: \[\mathbf{x}(t) = e^{\lambda t}\left(\mathbf{v}_{0}+t\mathbf{v}_{1}+\frac{t^{2}}{2!}\mathbf{v}_{2}\right)\]

Step 2: Calculate the derivative of \(\mathbf{x}(t)\)

We need to find the derivative of \(\mathbf{x}(t)\) with respect to time \(t\). Using the chain rule, we get: \[\mathbf{x}'(t) = e^{\lambda t}(\lambda\mathbf{v}_{0}+\lambda t\mathbf{v}_{1}+t\mathbf{v}_{1}+\frac{\lambda t^{2}}{2!} \mathbf{v}_{2}+\frac{t^{2}}{2!} \mathbf{v}_{2})\]

Step 3: Plug the derived solution into the given vector differential equation

Now, we need to substitute the derived \(\mathbf{x}'(t)\) and the proposed solution \(\mathbf{x}(t)\) into the given vector differential equation to see if they satisfy the equation: \[e^{\lambda t}(\lambda \mathbf{v}_{0}+\lambda t \mathbf{v}_{1}+t\mathbf{v}_{1}+\frac{\lambda t^{2}}{2!} \mathbf{v}_{2}+\frac{t^{2}}{2!} \mathbf{v}_{2}) = A e^{\lambda t}\left(\mathbf{v}_{0}+t\mathbf{v}_{1}+\frac{t^{2}}{2!}\mathbf{v}_{2}\right)\] Since this equation must hold for all values of \(t\), we can simplify it by dividing both sides by \(e^{\lambda t}\): \[\lambda \mathbf{v}_{0}+\lambda t \mathbf{v}_{1}+t\mathbf{v}_{1}+\frac{\lambda t^{2}}{2!} \mathbf{v}_{2}+\frac{t^{2}}{2!} \mathbf{v}_{2} = A\left(\mathbf{v}_{0}+t\mathbf{v}_{1}+\frac{t^{2}}{2!}\mathbf{v}_{2}\right)\]

Step 4: Compare coefficients to deduce relationships

Now, we need to compare the coefficients of the left side and right side of the simplified equation: For the constant term (\(t^0\)) coefficients: \[(A - \lambda I)\mathbf{v}_{0}=\mathbf{0}\] For the linear term (\(t^1\)) coefficients: \[(A - \lambda I)\mathbf{v}_{1}=\mathbf{v}_{0}\] For the quadratic term (\(t^2\)) coefficients: \[(A - \lambda I)\mathbf{v}_{2}=\mathbf{v}_{1}\] We have arrived at the desired relationships, proving that if the given solution satisfies the vector differential equation, then these relationships between eigenvalues, eigenvectors, and the matrix A hold true.

Key Concepts

Eigenvectors

Eigenvectors are critical components when dealing with vector differential equations. In the context of the exercise, an eigenvector is a special type of vector that, when multiplied by a matrix, does not change its direction, although its magnitude might be altered. Formally, for a given matrix \( A \), a non-zero vector \( \mathbf{v} \) is considered an eigenvector if there exists a scalar \( \lambda \), known as the eigenvalue, such that \( A \mathbf{v} = \lambda \mathbf{v} \). In this case, the scalar \( \lambda \) represents the factor by which the eigenvector is stretched or compressed.
In the exercise, several eigenvectors \( \mathbf{v}_0, \mathbf{v}_1, \mathbf{v}_2 \) are involved in the proposed solution for the vector differential equation. Each of these vectors plays a role in the overall expression and helps verify the relationships deduced in the equation, acting as part of the solution set that characterizes the behavior of the system over time. Understanding eigenvectors in this setting is crucial since it allows us to break down complex transformations characterized by matrices into simpler, more manageable components.

Eigenvalues

Eigenvalues are associated numbers that correspond to the eigenvectors of a matrix. They describe how much the eigenvectors are stretched during the linear transformation represented by the matrix. In mathematical terms, if \( \mathbf{v} \) is an eigenvector, then the eigenvalue \( \lambda \) is found such that \( A\mathbf{v} = \lambda \mathbf{v} \). This scalar \( \lambda \) is a crucial part of describing the dynamics of systems modeled by vector differential equations.
In the provided exercise, the value \( \lambda \) plays a significant role in defining the solution to the differential equation. The relationships between the eigenvalues and vectors, \( (A - \lambda I)\mathbf{v}_0 = \mathbf{0} \), \( (A - \lambda I)\mathbf{v}_1 = \mathbf{v}_0 \), and \( (A - \lambda I)\mathbf{v}_2 = \mathbf{v}_1 \), demonstrate how eigenvalues determine the scales of transformation and their effects on the eigenvectors. It's essential to grasp the concept of eigenvalues to appreciate how they interact with eigenvectors to solve such equations.

Matrix Algebra

Matrix algebra is the field of mathematics that deals with operations on matrices. It provides the backbone for solving vector differential equations as presented in this exercise. Understanding matrix operations—such as addition, multiplication, and the computation of determinants—is essential for solving complex linear systems.
The exercise uses matrix algebra to transform the given differential equation into a form that reveals deeper insights into the system's behavior. Specifically, operations involving \( A - \lambda I \) illustrate how algebraic manipulations with matrices help find solutions by aligning with known mathematic properties of eigenvectors and eigenvalues. This matrix operation effectively translates back the vector transformation into simpler algebraic terms, aiding in unveiling the connections stipulated by the problem.
Learning matrix algebra equips you with the tools necessary to address and simplify matrix-related solutions, thus enabling the resolution of differential equations that model various physical or theoretical phenomena.