Problem 52 - Advanced Engineering Mathematics

Question

(a) The given system can be written as $$x_{1}^{\prime \prime}=-\frac{k_{1}+k_{2}}{m_{1}} x_{1}+\frac{k_{2}}{m_{1}} x_{2}, \quad x_{2}^{\prime \prime}=\frac{k_{2}}{m_{2}} x_{1}-\frac{k_{2}}{m_{2}} x_{2}$$ In terms of matrices this is \(\mathbf{X}^{\prime \prime}=\mathbf{A X}\) where $$\mathbf{X}=\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) \text { and } \mathbf{A}=\left(\begin{array}{cc} -\frac{k_{1}+k_{2}}{m_{1}} & \frac{k_{2}}{m_{1}} \\ \frac{k_{2}}{m_{2}} & -\frac{k_{2}}{m_{2}} \end{array}\right)$$ (b) If \(\mathbf{X}=\mathbf{K} e^{\omega t}\) then \(\mathbf{X}^{\prime \prime}=\omega^{2} \mathbf{K} e^{\omega t}\) and \(\mathbf{A X}=\mathbf{A K} e^{\omega t}\) so that \(\mathbf{X}^{\prime \prime}=\mathbf{A} \mathbf{X}\) becomes \(\omega^{2} \mathbf{K} e^{\omega t}=\mathbf{A} \mathbf{K} e^{\omega t}\) or \(\left(\mathbf{A}-\omega^{2} \mathbf{I}\right) \mathbf{K}=0 .\) Now let \(\omega^{2}=\lambda\) (c) When \(m_{1}=1, m_{2}=1, k_{1}=3,\) and \(k_{2}=2\) we obtain \(\mathbf{A}=\left(\begin{array}{rr}-5 & 2 \\ 2 & -2\end{array}\right) .\) The eigenvalues and corresponding eigenvectors of \(\mathbf{A}\) are \(\lambda_{1}=-1, \lambda_{2}=-6, \mathbf{K}_{1}=\left(\begin{array}{l}1 \\\ 2\end{array}\right), \mathbf{K}_{2}=\left(\begin{array}{r}-2 \\\ 1\end{array}\right) .\) since \(\omega_{1}=i, \omega_{2}=-i, \omega_{3}=\sqrt{6} i,\) and \(\omega_{4}=-\sqrt{6} i\) a solution is $$\mathbf{X}=c_{1}\left(\begin{array}{l} 1 \\ 2 \end{array}\right) e^{i t}+c_{2}\left(\begin{array}{l} 1 \\ 2 \end{array}\right) e^{-i t}+c_{3}\left(\begin{array}{r} -2 \\ 1 \end{array}\right) e^{\sqrt{6} i t}+c_{4}\left(\begin{array}{r} -2 \\ 1 \end{array}\right) e^{-\sqrt{6} i t}$$ (d) Using \(e^{i t}=\cos t+i \sin t\) and \(e^{\sqrt{6} i t}=\cos \sqrt{6} t+i \sin \sqrt{6} t\) the preceding solution can be rewritten as $$\begin{aligned} \mathbf{X}=c_{1}\left(\begin{array}{c} 1 \\ 2 \end{array}\right)(\cos t+i \sin t)+c_{2}\left(\begin{array}{c} 1 \\ 2 \end{array}\right)(\cos t-i \sin t) \\ +c_{3}\left(\begin{array}{c} -2 \\ 1 \end{array}\right)(\cos \sqrt{6} t+i \sin \sqrt{6} t)+c_{4}\left(\begin{array}{c} -2 \\ 1 \end{array}\right)(\cos \sqrt{6} t+i \sin \sqrt{6} t) \\ =\left(c_{1}+c_{2}\right)\left(\begin{array}{c} 1 \\ 2 \end{array}\right) \cos t+i\left(c_{1}-c_{2}\right)\left(\begin{array}{c} 1 \\ 2 \end{array}\right) \sin t \\ +\left(c_{3}+c_{4}\right)\left(\begin{array}{c} -2 \\ 1 \end{array}\right) \cos \sqrt{6} t+i\left(c_{3}-c_{4}\right)\left(\begin{array}{c} -2 \\ 1 \end{array}\right) \sin \sqrt{6} t \\ =b_{1}\left(\begin{array}{c} 1 \\ 2 \end{array}\right) \cos t+b_{2}\left(\begin{array}{c} 1 \\ 2 \end{array}\right) \sin t+b_{3}\left(\begin{array}{c} -2 \\ 1 \end{array}\right) \cos \sqrt{6} t+b_{4}\left(\begin{array}{c} -2 \\ 1 \end{array}\right) \sin \sqrt{6} t \end{aligned}$$ where \(b_{1}=c_{1}+c_{2}, b_{2}=i\left(c_{1}-c_{2}\right), b_{3}=c_{3}+c_{4},\) and \(b_{4}=i\left(c_{3}-c_{4}\right)\)

Step-by-Step Solution

Answer

The solution is expressed in terms of cosine and sine functions using Euler's formula.

Step 1: Identify the Matrix Representation

We start by identifying the matrix representation of the given system of equations. The system is given as:\[ x_{1}'' = -\frac{k_{1} + k_{2}}{m_{1}} x_{1} + \frac{k_{2}}{m_{1}} x_{2}, \; x_{2}'' = \frac{k_{2}}{m_{2}} x_{1} - \frac{k_{2}}{m_{2}} x_{2} \]This can be written in matrix form as \( \mathbf{X}'' = \mathbf{A X} \), where \( \mathbf{X} = \begin{pmatrix} x_{1} \ x_{2} \end{pmatrix} \) and the matrix \( \mathbf{A} \) is \[ \mathbf{A} = \begin{pmatrix} -\frac{k_{1} + k_{2}}{m_{1}} & \frac{k_{2}}{m_{1}} \ \frac{k_{2}}{m_{2}} & -\frac{k_{2}}{m_{2}} \end{pmatrix} \]

Step 2: Propose the Solution Form and Substitute

Assume the solution to the matrix equation is of the form \( \mathbf{X} = \mathbf{K} e^{\omega t} \). Then the second derivative is \( \mathbf{X}'' = \omega^{2} \mathbf{K} e^{\omega t} \).Substitute this back into the matrix equation:\[ \omega^{2} \mathbf{K} e^{\omega t} = \mathbf{A} \mathbf{K} e^{\omega t} \]This simplifies to:\[ (\mathbf{A} - \omega^2 \mathbf{I}) \mathbf{K} = 0 \]

Step 3: Solve for Eigenvalues and Eigenvectors

Set \( \omega^2 = \lambda \) to find the eigenvalues and eigenvectors of matrix \( \mathbf{A} \).For given values \( m_{1} = 1, m_{2} = 1, k_{1} = 3, k_{2} = 2 \), the matrix \( \mathbf{A} \) becomes:\[ \mathbf{A} = \begin{pmatrix} -5 & 2 \ 2 & -2 \end{pmatrix} \]The eigenvalues of \( \mathbf{A} \) are \( \lambda_{1} = -1 \) and \( \lambda_{2} = -6 \), with corresponding eigenvectors \( \mathbf{K}_{1} = \begin{pmatrix} 1 \ 2 \end{pmatrix} \) and \( \mathbf{K}_{2} = \begin{pmatrix} -2 \ 1 \end{pmatrix} \).

Step 4: Construct the General Solution

Using the identified eigenvalues, construct the general solution:\[ \mathbf{X} = c_{1} \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{i t} + c_{2} \begin{pmatrix} 1 \ 2 \end{pmatrix} e^{-i t} + c_{3} \begin{pmatrix} -2 \ 1 \end{pmatrix} e^{\sqrt{6} i t} + c_{4} \begin{pmatrix} -2 \ 1 \end{pmatrix} e^{-\sqrt{6} i t} \]

Step 5: Apply Euler's Formula

Replace complex exponentials with trigonometric expressions using Euler's formula: \( e^{i t} = \cos t + i \sin t \) and \( e^{\sqrt{6} i t} = \cos(\sqrt{6} t) + i \sin(\sqrt{6} t) \).Transform the solution into a trigonometric identity:\[ \mathbf{X} = b_{1} \begin{pmatrix} 1 \ 2 \end{pmatrix} \cos t + b_{2} \begin{pmatrix} 1 \ 2 \end{pmatrix} \sin t + b_{3} \begin{pmatrix} -2 \ 1 \end{pmatrix} \cos(\sqrt{6} t) + b_{4} \begin{pmatrix} -2 \ 1 \end{pmatrix} \sin(\sqrt{6} t) \] where coefficients \( b_{1}, b_{2}, b_{3}, b_{4} \) are derived from initial conditions.

Key Concepts

Eigenvalues and Eigenvectors

Eigenvalues and eigenvectors are fundamental in understanding the behavior of linear transformations, especially in systems represented by matrices. Let's break these down further to grasp their significance. An eigenvalue, denoted by \( \lambda \), represents a scalar that scales an eigenvector after a matrix transformation, while the direction of the eigenvector remains unchanged. To find eigenvalues for a matrix \( \mathbf{A} \), we solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \mathbf{I} \) is the identity matrix. This equation yields several values for \( \lambda \), known as the eigenvalues of the matrix.

Once eigenvalues are determined, we proceed to find the corresponding eigenvectors. These vectors are the non-zero vectors \( \mathbf{K} \) that satisfy \( (\mathbf{A} - \lambda \mathbf{I}) \mathbf{K} = 0 \). In our example, with the specific matrix given\[\mathbf{A} = \begin{pmatrix} -5 & 2 \ 2 & -2 \end{pmatrix}\] this approach uncovers the eigenvalues \( \lambda_1 = -1 \) and \( \lambda_2 = -6 \), with their respective eigenvectors \( \mathbf{K}_1 = \begin{pmatrix} 1 \ 2 \end{pmatrix} \) and \( \mathbf{K}_2 = \begin{pmatrix} -2 \ 1 \end{pmatrix} \). Each eigenvalue and eigenvector pair provides a distinct solution component to differential equations using matrix \( \mathbf{A} \).

The eigenvectors tell us essential directions and modes of operation within the system, providing insight into stability and long-term behavior.

Matrix Representation

Matrix representation is a mathematical structure that encodes systems of linear equations. By organizing constants and coefficients into a matrix, it allows for elegance in expression and solution. Consider the given second-order differential equation system in the problem. Initially, the system presents a series of complicated expressions, but by using a matrix representation \( \mathbf{X}'' = \mathbf{A} \mathbf{X} \), the system becomes easier to analyze.

Let's explore how it works:

• \( \mathbf{X} \) represents the vector of dependent variables, \( \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \)
• \( \mathbf{A} \) embodies interaction coefficients between these variables, \( \begin{pmatrix} -\frac{k_{1} + k_{2}}{m_{1}} & \frac{k_{2}}{m_{1}} \ \frac{k_{2}}{m_{2}} & -\frac{k_{2}}{m_{2}} \end{pmatrix} \)
• \( \mathbf{X}'' \) is the second derivative vector of the variables.

Simplifying systems into matrices unveils powerful computational methods like row reduction, determinant calculations, and solving linear systems algebraically. This matrix compactly summarizes interactions between the variables. In physics, for example, it paves the way for understanding oscillations and coupling between masses in mechanical systems. Filling in specific constants reveals how parameters like mass (\( m_1 \), \( m_2 \)) and spring constants (\( k_1 \), \( k_2 \)) specifically determine the system's behavior.

Adopting matrix representation cuts through complexity with precision and makes transformational solutions methodical.

Complex Exponentials

Complex exponentials are a key tool in solving differential equations involving oscillatory behavior. By leveraging Euler's formula \( e^{i \theta} = \cos \theta + i \sin \theta \), they convert back-and-forth oscillations into manageable expressions. In the given differential equation problem, proposed solutions took the form \( \mathbf{X} = \mathbf{K} e^{\omega t} \), where \( \omega \) could be a complex number.

Let's examine the role of complex exponentials:

• They transform sinusoidal functions into exponential form, encapsulating oscillation properties.
• Provide a way to express solutions that naturally occur in coupled oscillatory systems.
• Handle systems with damping or external driving forces smoothly.

In our solution, we observe resultant terms like \( e^{i t} \), which upon expansion using Euler's formula becomes \( \cos t + i \sin t \). This transformation simplifies both the consumption and practical application in equations of motion.

By transforming and evaluating solutions in terms of trigonometric functions, complex exponentials aid in easily perceiving underlying oscillations, their frequency, and amplitude. The application gives sustenance for both analyzing wave properties and employing in mechanical and electric circuits. Euler's formula thus acts as a bridge, linking the exponential form to intuitive sine and cosine forms used in physical interpretation.