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.\(\mathbf{y}^{\prime \prime}+A \mathbf{y}=\left[\begin{array}{l}1 \\\ 0\end{array}\right], \quad \mathbf{y}(0)=\left[\begin{array}{l}1 \\\ 0\end{array}\right], \quad \mathbf{y}^{\prime}(0)=\left[\begin{array}{l}0 \\\ 1\end{array}\right]\)

Short answer

Answer: The solution to the second-order linear system is given by $$ \mathbf{y}(t) = \begin{bmatrix} \frac{1}{3} + \frac{1}{2\sqrt{2}} \cos{(\sqrt{2}t)} - \frac{1}{2} e^t \\ -\frac{5}{6} + \frac{1}{2\sqrt{2}}\cos{(\sqrt{2}t)} + \frac{1}{2} e^t \end{bmatrix} $$

Step-by-step solution

Find the inverse of T

We already have the matrix of eigenvectors T, so we just need to compute its inverse: $$ T = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} $$ We can find the inverse of T using the formula for a 2x2 matrix inverse: $$ T^{-1} = \frac{1}{\det(T)}\begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} $$ So, $$T^{-1} = \frac{1}{2}\begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix}$$

Reformulate the problem in terms of z(t)

Using the change of variables, $$\mathbf{z}(t)=T^{-1} \mathbf{y}(t)$$, the original problem can be reformulated as: $$ \mathbf{y}^{\prime \prime}+A \mathbf{y}=\left[\begin{array}{l}1 \\0\end{array}\right] \implies T \mathbf{z}^{\prime \prime} + A T \mathbf{z} = \left[\begin{array}{l}1 \\0\end{array}\right] $$ To get the uncoupled set of second-order linear problems in terms of $$\mathbf{z}(t)$$, multiply both sides by $$T^{-1}$$: $$ \mathbf{z}^{\prime \prime} + T^{-1} A T \mathbf{z} = T^{-1} \left[\begin{array}{l}1 \\0\end{array}\right] $$ We are given that $$AT = TD\$$, where D is a diagonal matrix with eigenvalues on the diagonal. Therefore, $$ \mathbf{z}^{\prime \prime} + D \mathbf{z} = T^{-1} \left[\begin{array}{l}1 \\0\end{array}\right] $$

Solve the uncoupled problem for z(t)

The uncoupled problem now becomes: $$ \mathbf{z}^{\prime \prime} + \begin{bmatrix} 3 & 0 \\ 0 & 1 \end{bmatrix} \mathbf{z} = T^{-1} \left[\begin{array}{l}1 \\0\end{array}\right] $$ This gives us two uncoupled equations: $$ z_1^{\prime \prime}(t) + 3 z_1(t) = -\frac{1}{2}\text{ and } z_2^{\prime \prime}(t) + z_2(t) = -\frac{1}{2} $$ Using the integrating factor method, we can solve these uncoupled second-order linear equations. The solutions are: $$ z_1(t) = -\frac{1}{6} + C_1 e^{j\sqrt{2}t} + C_2 e^{-j\sqrt{2}t} $$ $$ z_2(t) = -\frac{1}{2} + C_3 e^{t} + C_4 e^{-t} $$ We need to find the constants $$C_1, C_2, C_3, C_4$$. To incorporate the initial conditions, we change them in terms of $$z(t)$$. Since $$\mathbf{z}(t) = T^{-1} \mathbf{y}(t)$$, we have: $$ \mathbf{z}(0) = T^{-1} \mathbf{y}(0) = \frac{1}{2} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} \\ -\frac{1}{2} \end{bmatrix} $$ $$ \mathbf{z}^\prime(0) = T^{-1} \mathbf{y}^\prime(0) = \frac{1}{2} \begin{bmatrix} -1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} \\ \frac{1}{2} \end{bmatrix} $$ Applying these initial conditions, we find the constants $$C_1, C_2, C_3, C_4$$: Substitute the initial position and derivative for $$z_1(t)$$ and $$z_2(t)$$: $$ z_1(0) = -\frac{1}{6} + C_1 e^{0} + C_2 e^{0} = -\frac{1}{2} \\ z_1^{'} (0) = C_1 \sqrt{2} - C_2 \sqrt{2} = -\frac{1}{2} $$ $$ z_2(0) = -\frac{1}{2} + C_3 e^{0} + C_4 e^{0} = -\frac{1}{2} \\ z_2^{'} (0) = C_3 - C_4 = \frac{1}{2} $$ By solving these equations simultaneously, we can find the constants, obtaining: $$ C_1 = \frac{1}{2\sqrt{2}}, \quad C_2 = 0, \quad C_3 = \frac{1}{2}, \quad C_4 = 0 $$ Now we have the solution for $$z(t)$$ as: $$ z_1(t) = -\frac{1}{6} + \frac{1}{2\sqrt{2}} \cos{(\sqrt{2}t)} $$ $$ z_2(t) = -\frac{1}{2} + \frac{1}{2} e^t $$

Find the original solution y(t)

With the solution for $$z(t)$$ in hand, we can find the solution for the original problem, $$\mathbf{y}(t) = T \mathbf{z}(t)$$: $$ \mathbf{y}(t)= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\begin{bmatrix} -\frac{1}{6} + \frac{1}{2\sqrt{2}} \cos{(\sqrt{2}t)} \\ -\frac{1}{2} + \frac{1}{2} e^t \end{bmatrix} = \begin{bmatrix} \frac{1}{3} + \frac{1}{2\sqrt{2}} \cos{(\sqrt{2}t)} - \frac{1}{2} e^t \\ -\frac{5}{6} + \frac{1}{2\sqrt{2}}\cos{(\sqrt{2}t)} + \frac{1}{2} e^t \end{bmatrix} $$ Thus, the solution is given by $$ \mathbf{y}(t) = \begin{bmatrix} \frac{1}{3} + \frac{1}{2\sqrt{2}} \cos{(\sqrt{2}t)} - \frac{1}{2} e^t \\ -\frac{5}{6} + \frac{1}{2\sqrt{2}}\cos{(\sqrt{2}t)} + \frac{1}{2} e^t \end{bmatrix} $$

Key concepts

Coupled Spring-Mass Systems

Understanding coupled spring-mass systems is crucial for students who study physics and engineering courses. A spring-mass system is a classic physics problem that involves a mass attached to a spring which can freely oscillate. However, when we talk about coupled spring-mass systems, we mean multiple masses connected to each other and to walls or other fixed points using springs. This creates a complex oscillatory system where the motion of each mass affects the other.

Mathematically, such systems can be described using second order linear equations. Why second order? Because the force exerted by a spring is proportional to the displacement, which translates to the acceleration of the mass (hence second order). These equations are particularly challenging because they usually need to consider the interaction between multiple oscillating elements, leading to a series of differential equations rather than a single one. The real power of understanding this concept manifests when solving real-world engineering problems like building suspension bridges or designing comfortable car suspension systems.

Eigenvalues and Eigenvectors

To simplify the complex problems like the one presented in our exercise, we utilize the concepts of eigenvalues and eigenvectors. These are fundamental in various areas of sciences and applied mathematics. In the context of linear algebra, an eigenvector of a matrix is a non-zero vector that changes at most by a scalar factor when that linear transformation is applied to it. The corresponding scalar is known as an eigenvalue.

For the example matrix A given in the exercise, the eigenvectors provide directions along which the matrix A acts by only stretching or compressing. The eigenvalues tell us how much that stretching or compressing occurs. When applied to coupled spring-mass systems, they can significantly simplify the analysis by decoupling the system. This is because, in the appropriate coordinate system (eigenvector basis), the problem can be split into independent single mass-spring scenarios, each oscillating according to its eigenvalue.

Diagonalization of Matrices

Diagonalization is a powerful technique in linear algebra where a matrix is expressed as the product of its eigenvectors and eigenvalues. If \( A \) is a given matrix and \( T \) is a matrix whose columns are the eigenvectors of \( A \) (as seen in our exercise), then \( A \) can be diagonalized if it can be written in the form \( A = TDT^{-1} \), where \( D \) is a diagonal matrix containing the eigenvalues.

This transformation is particularly helpful because operating with diagonal matrices is much simpler than with full matrices. When a matrix representing a coupled system is diagonalized, each of the interacting parts can be treated as if they are acting independently. This simplification into a diagonal matrix is at the heart of many analytical and computational methods in physics and engineering – it's like finding the 'straightest possible roads' through a normally 'curvy landscape' of the problem.

Integrating Factor Method

In differential equations, the integrating factor method is a technique to solve first-order linear equations of the form \( \frac{dy}{dx} + P(x)y = Q(x) \). Unfortunately, not all differential equations are this straightforward. In this exercise, after utilizing diagonalization, we're left with second-order differential equations to solve. However, we can apply methods akin to the integrating factor technique to these problems, too.

In our case, once we have decoupled our system into components that no longer interact with each other, thanks to eigenvalues and diagonalization, each independent equation can be handled using methods appropriate for second-order equations. One of which is similar to the integrating factor - we use known solutions of the homogenous equation combined with particular solutions of the non-homogenous equation to construct a general solution that addresses the initial conditions. This method, rich in its application across different fields, embodies the creative flair needed to solve complex problems by converting them into simpler, more manageable ones.