Problem 4 - Advanced Engineering Mathematics

Question

The general solution of the linear system \(\mathbf{X}^{\prime}=\mathbf{A} \mathbf{X}\) is given. (a) In each case discuss the nature of the solution in a neighborhood of \((0,0)\). (b) With the aid of a graphing utility plot the solution that satisfies \(\mathbf{X}(0)=(1,1)\) $$ \begin{aligned} &\mathbf{A}=\left(\begin{array}{rr} -1 & -4 \\ 1 & -1 \end{array}\right) \\ &\mathbf{X}(t)=e^{-t}\left[c_{1}\left(\begin{array}{c} 2 \cos 2 t \\ \sin 2 t \end{array}\right)+c_{2}\left(\begin{array}{c} -2 \sin 2 t \\ \cos 2 t \end{array}\right)\right] \end{aligned} $$

Step-by-Step Solution

Answer

The solution spirals into the origin. Initial values lead to \( c_1 = \frac{1}{2} \), \( c_2 = 1 \). Use plotting to visualize.

Step 1: Characterize the Solution

The given solution \( \mathbf{X}(t) = e^{-t} \left[ c_1 \begin{pmatrix} 2 \cos 2t \ \sin 2t \end{pmatrix} + c_2 \begin{pmatrix} -2 \sin 2t \ \cos 2t \end{pmatrix} \right] \) involves terms with exponential decay \( e^{-t} \) and oscillatory behaviors involving sine and cosine terms.

Step 2: Analyze the Eigenvalues

The matrix \( \mathbf{A} \) can be analyzed to find eigenvalues using the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Since the solution provided is exponential with oscillations, the eigenvalues must be complex conjugates: \( \lambda = -1 \pm 2i \). This indicates a spiral solution with a decay factor.

Step 3: Nature of Solution Near (0,0)

Given the eigenvalues \( \lambda = -1 \pm 2i \), the solution spirals into the origin because of the negative real part \(-1\). Thus, in the neighborhood of \((0,0)\), the trajectory spirals inward towards the origin, implying a stable focus.

Step 4: Determine Constants \( c_1 \) and \( c_2 \)

To find specific \( c_1 \) and \( c_2 \), use the initial condition \( \mathbf{X}(0) = (1,1) \). Substituting \( t = 0 \) into \( \mathbf{X}(0) \), we get: \( \mathbf{X}(0) = \left[c_1 \begin{pmatrix} 2 \ 0 \end{pmatrix} + c_2 \begin{pmatrix} 0 \ 1 \end{pmatrix} \right] = \begin{pmatrix} 2c_1 \ c_2 \end{pmatrix} \). Equating to \( (1,1) \) gives \( 2c_1 = 1 \) and \( c_2 = 1 \), which solve to \( c_1 = \frac{1}{2} \) and \( c_2 = 1 \).

Step 5: Plot the Solution

With \( c_1 = \frac{1}{2} \) and \( c_2 = 1 \), the solution becomes \( \mathbf{X}(t) = e^{-t} \left[ \frac{1}{2} \begin{pmatrix} 2 \cos 2t \ \sin 2t \end{pmatrix} + \begin{pmatrix} -2 \sin 2t \ \cos 2t \end{pmatrix} \right] \). Use a graphing utility to plot this trajectory to visualise the inward spiral towards the origin.

Key Concepts

Eigenvalues

Eigenvalues are crucial in determining the behavior of solutions to differential equations, particularly when dealing with systems of linear equations. In the exercise, the matrix \( \mathbf{A} \) is used to identify the system's eigenvalues through the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). Here, this analysis reveals complex eigenvalues \( \lambda = -1 \pm 2i \), indicating the presence of a real part and an imaginary part.
These complex conjugate eigenvalues suggest oscillatory behavior, typical of spiraling trajectories, because of the imaginary components \( 2i \). The real part \(-1\) indicates a decay over time, leading to solutions that spiral towards the origin instead of diverging. Such information is vital for predicting the system's long-term behavior.

Spiral Solutions

Spiral solutions occur when solutions to differential equations involve both oscillations and exponential decay. In the solution \( \mathbf{X}(t) = e^{-t} \left[ c_1 \begin{pmatrix} 2 \cos 2t \ \sin 2t \end{pmatrix} + c_2 \begin{pmatrix} -2 \sin 2t \ \cos 2t \end{pmatrix} \right] \), the presence of trigonometric functions like sine and cosine contributes to the oscillatory motion.
This oscillation, combined with the exponential decay factor \( e^{-t} \), means that as time progresses, the magnitude of the trajectory diminishes, pulling the trajectory inwards. This results in a spiral pattern that converges towards the origin, characterized as a stable focus due to the negative real eigenvalue.

Initial Conditions

Initial conditions are essential for defining the specific trajectory of a solution within its general framework. They uniquely determine the constants in the general solution. In our exercise, the initial condition \( \mathbf{X}(0) = (1,1) \) was utilized to find specific values for \( c_1 \) and \( c_2 \).
By substituting \( t = 0 \) into the solution, the equation \( \begin{pmatrix} 2c_1 \ c_2 \end{pmatrix} = \begin{pmatrix} 1 \ 1 \end{pmatrix} \) was formed. Solving provides \( c_1 = \frac{1}{2} \) and \( c_2 = 1 \). This process shows how initial conditions influence the solution's form and describe the initial state of the system's behavior.

Matrix Analysis

Matrix analysis is a powerful tool in understanding systems of linear differential equations. The matrix \( \mathbf{A} \), containing the coefficients of the system, plays a pivotal role in determining the dynamics of the solutions. It is through the eigenvalues and eigenvectors derived via matrix analysis that the nature of the system's solutions is discerned.
By calculating \( \mathbf{A} = \begin{pmatrix} -1 & -4 \ 1 & -1 \end{pmatrix} \), we found its complex conjugate eigenvalues \( \lambda = -1 \pm 2i \). These results allow us to classify the system’s behavior, interpret the spiral solution, and understand its approach towards the origin. Matrix analysis thus offers a systematic approach to dissect and predict the trajectories of linear differential systems.