Question

We consider systems of second order linear equations. Such systems arise, for instance, when Newton's laws are used to model the motion of coupled spring- mass systems, such as those in Exercises 31-32. In each of Exercises \(25-30\), let \(A=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right] .\) Note that the eigenpairs of \(A\) are \(\lambda_{1}=3, \mathbf{x}_{1}=\left[\begin{array}{l}1 \\ 1\end{array}\right]\) and \(\lambda_{2}=1, \mathbf{x}_{2}=\left[\begin{array}{r}1 \\\ -1\end{array}\right] .\) (a) Let \(T=\left[\mathbf{x}_{1}, \mathbf{x}_{2}\right]\) denote the matrix of eigenvectors that diagonalizes \(A\). Make the change of variable \(\mathbf{z}(t)=T^{-1} \mathbf{y}(t)\), and reformulate the given problem as a set of uncoupled second order linear problems. (b) Solve the uncoupled problem for \(\mathbf{z}(t)\), and then form \(\mathbf{y}(t)=T \mathbf{z}(t)\) to solve the original problem.\(\mathbf{y}^{\prime \prime}+\mathbf{y}^{\prime}+A \mathbf{y}=\mathbf{0}\)

Short answer

Answer: The general solution of the given second-order linear system of differential equations is: \(\mathbf{y}(t) =\begin{bmatrix} (c_1+c_3)e^{-\frac{1}{2}t}\cos(\frac{\sqrt{11}}{2}t) + (c_2+c_4)e^{-\frac{1}{2}t}\sin(\frac{\sqrt{11}}{2}t) \\ (c_1-c_3)e^{-\frac{1}{2}t}\cos(\frac{\sqrt{3}}{2}t) + (c_2-c_4)e^{-\frac{1}{2}t}\sin(\frac{\sqrt{3}}{2}t) \end{bmatrix}\), where \(c_1\), \(c_2\), \(c_3\), and \(c_4\) are arbitrary constants determined by the initial conditions.

Step-by-step solution

Find the matrix of eigenvectors T and its inverse T^{-1}

We are given two eigenpairs: \(\lambda_1=3\) with eigenvector \(\mathbf{x}_1=\begin{bmatrix}1\\1\end{bmatrix}\) and \(\lambda_2=1\) with eigenvector \(\mathbf{x}_2=\begin{bmatrix}1\\-1\end{bmatrix}\). We form the matrix T by placing these eigenvectors as columns in the matrix: \(T=\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}\). Now we can compute the inverse of T: \(T^{-1} = \frac{1}{(1)(-1)-(1)(1)} \begin{bmatrix}-1 & -1\\-1 & 1\end{bmatrix} = \begin{bmatrix}-\frac{1}{2} & -\frac{1}{2}\\ -\frac{1}{2} & \frac{1}{2}\end{bmatrix}\).

Perform the change of variables \(z(t) = T^{-1}y(t)\)

We are given the second-order linear system: \(\mathbf{y}^{\prime \prime} + \mathbf{y}^{\prime} + A\mathbf{y}=0\). By making the change of variables \(\mathbf{z}(t)=T^{-1} \mathbf{y}(t)\), we can rewrite the system as: \(\mathbf{z}^{\prime \prime} + \mathbf{z}^{\prime} + \Lambda \mathbf{z} = 0\) where \(\Lambda = T^{-1}AT = \begin{bmatrix}3 & 0\\0 & 1\end{bmatrix}\). Now, we have the uncoupled second-order linear problems: \(\begin{cases} z_1^{\prime \prime} + z_1^{\prime} + 3z_1 = 0 \\ z_2^{\prime \prime} + z_2^{\prime} + 1z_2 = 0 \end{cases}\).

Solve the uncoupled problems for z(t)

To solve these uncoupled problems, we can use the characteristic equation \(r^2 + r + \lambda_i = 0\) for each component (i = 1, 2) of \(\mathbf{z}(t)\). For \(i = 1\), the characteristic equation is \(r^2 + r + 3 = 0\). This equation has complex roots \(r = -\frac{1}{2} \pm \frac{\sqrt{11}i}{2}\), so the general solution is: \(z_1(t) = c_1 e^{-\frac{1}{2}t}\cos(\frac{\sqrt{11}}{2}t) + c_2 e^{-\frac{1}{2}t}\sin(\frac{\sqrt{11}}{2}t)\). Similarly, for \(i = 2\), the characteristic equation is \(r^2 + r + 1 = 0\). It has complex roots \(r = -\frac{1}{2} \pm \frac{\sqrt{3}i}{2}\). The general solution for this component is: \(z_2(t) = c_3 e^{-\frac{1}{2}t}\cos(\frac{\sqrt{3}}{2}t) + c_4 e^{-\frac{1}{2}t}\sin(\frac{\sqrt{3}}{2}t)\). The complete solution for \(\mathbf{z}(t)\) is then: \(\mathbf{z}(t) =\begin{bmatrix} z_1(t) \\ z_2(t) \end{bmatrix} = \begin{bmatrix} c_1e^{-\frac{1}{2}t}\cos(\frac{\sqrt{11}}{2}t) + c_2 e^{-\frac{1}{2}t}\sin(\frac{\sqrt{11}}{2}t) \\ c_3 e^{-\frac{1}{2}t} \cos(\frac{\sqrt{3}}{2}t) + c_4 e^{-\frac{1}{2}t}\sin(\frac{\sqrt{3}}{2}t) \end{bmatrix}\).

Transform the solution back to y(t)

To find the solution to the original problem, we need to compute \(\mathbf{y}(t) = T\mathbf{z}(t)\). By multiplying T with the z(t) solution, we get: \(\mathbf{y}(t) =\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}\begin{bmatrix}c_1 e^{-\frac{1}{2}t}\cos(\frac{\sqrt{11}}{2}t) + c_2 e^{-\frac{1}{2}t}\sin(\frac{\sqrt{11}}{2}t) \\ c_3 e^{-\frac{1}{2}t}\cos(\frac{\sqrt{3}}{2}t) + c_4 e^{-\frac{1}{2}t}\sin(\frac{\sqrt{3}}{2}t)\end{bmatrix}\). Expanding this out, our final solution for the original system is: \(\mathbf{y}(t) =\begin{bmatrix} (c_1+c_3)e^{-\frac{1}{2}t}\cos(\frac{\sqrt{11}}{2}t) + (c_2+c_4)e^{-\frac{1}{2}t}\sin(\frac{\sqrt{11}}{2}t) \\ (c_1-c_3)e^{-\frac{1}{2}t}\cos(\frac{\sqrt{3}}{2}t) + (c_2-c_4)e^{-\frac{1}{2}t}\sin(\frac{\sqrt{3}}{2}t) \end{bmatrix}\) which is the general solution of the given second-order linear system of differential equations.

Key concepts

Eigenvectors

Eigenvectors are special vectors associated with a matrix that point in a particular direction. These vectors display a unique property when multiplied by their matrix, as the result is the same vector scaled by a certain factor. This scaling factor is known as the eigenvalue. For the matrix \( A = \begin{bmatrix}2 & 1 \ 1 & 2\end{bmatrix} \), we have two distinct eigenpairs:

• Eigenvalue \( \lambda_1 = 3 \) with eigenvector \( \mathbf{x}_1 = \begin{bmatrix} 1 \ 1 \end{bmatrix} \)
• Eigenvalue \( \lambda_2 = 1 \) with eigenvector \( \mathbf{x}_2 = \begin{bmatrix} 1 \ -1 \end{bmatrix} \)

The eigenvectors are essential in transforming a system into a simpler form. We arrange them into a matrix \( T = \begin{bmatrix} 1 & 1 \ 1 & -1 \end{bmatrix} \) used for diagonalization, making complex calculations more straightforward.

Diagonalization

Diagonalization is a method used to simplify matrices, making it easier to solve systems of equations. It involves transforming a matrix into a diagonal form with the use of its eigenvectors. For the matrix \( A \), matrix \( T \) contains its eigenvectors as columns. By transforming \( A \) using matrix \( T \), the system is `diagonalized` as follows:

• Compute \( \Lambda = T^{-1} A T \).
• Mathematically, \( \Lambda = \begin{bmatrix} 3 & 0 \ 0 & 1 \end{bmatrix} \).

Here, \( \Lambda \) is a diagonal matrix, where each entry corresponds to an eigenvalue of \( A \). This transformation simplifies differential equation systems, uncoupling them into independent equations that are easier to handle.

Second Order Linear Equations

Second-order linear equations are differential equations that contain the second derivative of the unknown function. These equations take the form:\[ y''(t) + p(t) y'(t) + q(t) y(t) = f(t)\]In our exercise, \( \mathbf{y}'' + \mathbf{y}' + A \mathbf{y} = 0 \), represents a system of second-order equations. They often model physical systems, like a mass-spring system, describing how the components interact over time. To solve these equations, they are expressed in terms of transformed variables. Using \( \mathbf{z}(t) = T^{-1} \mathbf{y}(t) \) shifts us into a `decoupled` realm, where each equation in the system stands alone:

• \( z_1'' + z_1' + 3z_1 = 0 \)
• \( z_2'' + z_2' + z_2 = 0 \)

This approach separates the system's interactions, allowing for simpler individual solutions.

Characteristic Equation

The characteristic equation is a polynomial equation derived from a differential equation, significant for finding the solution of linear systems. Each second-order differential equation can be rewritten to form a characteristic equation based on the system's eigenvalues. For a second-order equation, the characteristic form is:\[ r^2 + pr + \,\lambda = 0 \]This equation determines the nature of solutions—whether they are real or complex. In our problem, we solve for \( \mathbf{z}(t) \) using this approach:

• For \( z_1(t): \) \( r^2 + r + 3 = 0 \) leading to complex roots.
• For \( z_2(t): \) \( r^2 + r + 1 = 0 \) also yielding complex roots.

The solutions, involving exponentials and trigonometric functions due to complex roots, reflect oscillatory behavior common in dynamic systems. This method is crucial in calculating the general solution for each component of the system.