Question

The Laplace transform was applied to the initial value problem \(\mathbf{y}^{\prime}=A \mathbf{y}, \mathbf{y}(0)=\mathbf{y}_{0}\), where \(\mathbf{y}(t)=\left[\begin{array}{l}y_{1}(t) \\ y_{2}(t)\end{array}\right], A\) is a \((2 \times 2)\) constant matrix, and \(\mathbf{y}_{0}=\left[\begin{array}{l}y_{1,0} \\ y_{2,0}\end{array}\right]\). The following transform domain solution was obtained: $$ \mathcal{L}\\{\mathbf{y}(t)\\}=\mathbf{Y}(s)=\frac{1}{s^{2}-9 s+18}\left[\begin{array}{cc} s-2 & -1 \\ 4 & s-7 \end{array}\right]\left[\begin{array}{l} y_{1,0} \\ y_{2,0} \end{array}\right] $$ (a) What are the eigenvalues of the coefficient matrix \(A\) ? (b) What is the coefficient matrix \(A\) ?

Short answer

The eigenvalues of matrix A are λ1 = 3 and λ2 = 6.

Step-by-step solution

Determine Eigenvalues of Matrix A

To find the eigenvalues of the matrix A, we look at the denominator of the transform domain solution. The characteristic equation is given by the determinant of \((A - \lambda I)\), where \(\lambda\) is an eigenvalue and \(I\) is the identity matrix. In the Laplace transform of a matrix, the characteristic equation becomes the denominator of the transform. In this case, the denominator is \((s^2 - 9s + 18)\) which can be factored as \((s-3)(s-6)\). Therefore, the eigenvalues of matrix A are \(\lambda_1 = 3\) and \(\lambda_2 = 6\).

Determine Matrix A

We know that the Laplace transform of \(\mathbf{y}'(t)= A\mathbf{y}\) is given by \(s\mathbf{Y}(s) - \mathbf{y}_0 = A\mathbf{Y}(s)\). Lets multiply both sides by the inverse of the matrix of the coefficients of the transformed solution: $$ \left[\begin{array}{cc} \frac{1}{s-3} & 0 \\\ 0 & \frac{1}{s-6} \end{array}\right]\left[\begin{array}{cc} s-2 & -1 \\\ 4 & s-7 \end{array}\right] = \left[\begin{array}{cc} 1 & \frac{s-2}{s-3} \\\ \frac{4}{s-6} & 1 \end{array}\right] $$ Now, define \(A = \left[\begin{array}{cc}a & b \\\ c & d\end{array}\right]\). We have: $$ A \left[\begin{array}{cc} 1 & \frac{s-2}{s-3} \\\ \frac{4}{s-6} & 1 \end{array}\right] = s\left[\begin{array}{cc} 1 & \frac{s-2}{s-3} \\\ \frac{4}{s-6} & 1 \end{array}\right] - \left[\begin{array}{cc}1 & 0 \\\ 0 & 1\end{array}\right] $$ Comparing the coefficients of the same powers of \(s\) on both sides, we can find individual elements of the matrix A: - For the first element (top-left corner): \(a = 1 \implies a = 1\) - For the second element (top-right corner): \(b = \frac{-2}{s-3} \implies b = -1\) - For the third element (bottom-left corner): \(c = \frac{4}{s-6} \implies c = 4\) - For the fourth element (bottom-right corner): \(d = 1 \implies d = 2\) Finally, matrix A is given by: $$ A = \left[\begin{array}{cc} 1 & -1 \\\ 4 & 2 \end{array}\right] $$

Key concepts

Eigenvalues of a Matrix

In linear algebra, eigenvalues are a set of scalars associated with a given square matrix, representing the factors by which the eigenvectors of the matrix are scaled during a transformation. Eigenvalues can reveal much about the nature of a matrix, including its stability and how it stretches or compresses vectors.

When we encounter a system of linear differential equations, these eigenvalues determine the system's behavior over time. The matrix in question often represents the coefficients of the system or transformation being studied. To find the eigenvalues, we solve the characteristic equation, which is typically a polynomial called the characteristic polynomial.

The characteristic polynomial is derived from the determinant of the matrix when an unknown scalar, typically denoted by \( \lambda \), is subtracted from each of the diagonal elements, forming a matrix \( A - \lambda I \). The determinant of this matrix equals zero, and solving this equation gives us the eigenvalues of matrix \( A \).

Transform Domain Solution

The transform domain solution is a powerful technique used to solve differential equations, which involves converting the original problem into a more manageable form using transforms, such as the Laplace Transform. Formally, this transforms a function of time, \( f(t) \), into a function of complex frequency, \( F(s) \).

In the context of linear differential equations, the Laplace Transform can convert a time-domain differential equation into an algebraic equation in the transform domain. This simplification allows us to solve for the transformed variable, \( Y(s) \), without directly dealing with the differential equation. Once we have \( Y(s) \), we can then apply the inverse Laplace Transform to revert back to the time domain and find the solution to the original problem, \( y(t) \).

This approach is particularly handy when dealing with systems described by matrix differential equations, allowing the equations to be decoupled and solved more easily.

Characteristic Equation

The characteristic equation is the equation obtained by setting the determinant of \( A - \lambda I \) equal to zero, where \( A \) is the matrix in question and \( \lambda \) represents an eigenvalue. This equation is fundamental in determining the eigenvalues of the matrix.

For a \( 2 \times 2 \) matrix, the characteristic equation is a quadratic equation and is relatively simple to solve. However, as matrices increase in size, finding the characteristic equation and its roots can become more complex. Various computational tools are available to aid in finding eigenvalues, especially for larger matrices.

In our case, using the given transform domain solution, we were able to deduce the characteristic equation \( s^2 - 9s + 18 \), which is key in finding the eigenvalues of the coefficient matrix. Solving this results in the values that profoundly affect the system's dynamics.

Matrix Differential Equations

Matrix differential equations involve a differential equation in which the variable to be found is a matrix. These are common in systems of linear differential equations, especially when dealing with multiple interrelated quantities that can be modeled simultaneously.

For instance, in our initial value problem \( \mathbf{y}^\prime=A \mathbf{y} \), the solution involves finding a matrix function \( \mathbf{y}(t) \) that satisfies the differential equation for a given initial condition, \( \mathbf{y}(0) \). The Laplace Transform is a valuable tool for solving such matrix differential equations because it reduces the problem from the time domain to the algebraic realm.

Once the Laplace Transform is applied, the transformed system can be manipulated algebraically to solve for the matrix in the transform domain. This solution can then be translated back into the time domain, providing the needed information about the original system's behavior over time.