A second order matrix differential equation

Here's an analytic solution of the problem. The matrix $B(t)$ will be determined as a function of $t$ .

Similar to the scalar case, it can be shown that the general solution of the given differential equation is

$B(t) = \cos(A t) C_1 + \sin(A t) C_2 \hspace{12pt} (1)$

where the square matrices $C_1$ and $C_2$ depend on the initial conditions.

To determine the matrix sinusoid $\sin(A t)$ , we use the following procedure which stems from consequences of the Cayley-Hamilton theorem. According to this procedure, we can express a given function of a matrix as the following expansion:

$\sin(A t) = \alpha_0 I + \alpha_1 (A t) + \alpha_2 (A t)^2 + \cdots + \alpha_{n-1} (A t)^{n-1} \hspace{12pt} (2)$

where $n$ is the size of the square matrix $A$ . In our case, $n = 2$ , so

$\sin(A t) = \alpha_0 I + \alpha_1 (A t) \hspace{12pt} (3)$

To determine the constants $\alpha_0$ and $\alpha_1$ , we use the fact that the eigenvalues of the matrix $(A t)$ satisfy the same equation. So if $\lambda$ is an eigenvalue of $A$ , then

$\sin( \lambda t ) = \alpha_0 + \alpha_1 (\lambda t) \hspace{12pt} (4)$

Now it is a straight forward excercise to determine the eigenvalues of the given matrix $A$ . They are $\lambda_1 = 1$ and $\lambda_2 = 5$ . Plugging these two values in equation (4), we obtain a linear equation in the two unknowns $\alpha_0$ and $\alpha_1$ , which upon solving yields,

$\alpha_0 = \frac{5}{4} \sin(t) - \frac{1}{4} \sin(5 t) \hspace{12pt} (5a)$

and,

$\alpha_1 = \dfrac{ \sin(5t) - \sin(t) } { 4 t} \hspace{12 pt} (5b)$

Plugging these expressions in equation (3), gives us the desired matrix $\sin(A t)$ , namely

$\sin(A t) = \begin{bmatrix} -\frac{1}{4} \sin t + \frac{5}{4} \sin 5 t && - \frac{1}{4} \sin t + \frac{1}{4} \sin 5 t \\ \frac{5}{4} \sin t - \frac{5}{4} \sin 5 t && \frac{5}{4} \sin t - \frac{1}{4} \sin 5 t \end{bmatrix} \hspace{12pt} (6)$

And, in exactly the same manner, we determine the matrix $\cos(A t)$ to be,

$\cos(A t) = \begin{bmatrix} -\frac{1}{4} \cos t + \frac{5}{4} \cos 5 t && - \frac{1}{4} \cos t + \frac{1}{4} \cos 5 t \\ \frac{5}{4} \cos t - \frac{5}{4} \cos 5 t && \frac{5}{4} \cos t - \frac{1}{4} \cos 5 t \end{bmatrix} \hspace{12pt} (7)$

Now, we need to determine the matrices $C_1$ and $C_2$ . Plugging $t = 0$ in equation (1), yields,

$B(0) = C_1 \hspace{12pt} (8a)$

And differentiating equation (1), and plugging in $t = 0$ yields,

$\dfrac{d B}{dt} (0) = A C_2 \hspace{12pt} (8b)$

Since $A$ is invertible, we can determine $C_2$ . What remains is to plug in the above-obtained expressions in equation (1), and the result is,

$B(t) = \begin{bmatrix} -\frac{3}{4} \cos t + \frac{7}{4} \cos 5 t - \frac{1}{2} \sin t + \frac{1}{2} \sin 5 t && - \cos t + 2 \cos 5 t - \frac{5}{4} \sin t + \frac{1}{4} \sin 5 t \\ \frac{15}{4} \cos t - \frac{7}{4} \cos 5 t + \frac{5}{2} \sin t - \frac{1}{2} \sin 5 t && 5 \cos t - 2 \cos 5 t + \frac{25}{4} \sin t - \frac{1}{4} \sin 5 t \end{bmatrix} \hspace{12pt} (9)$

Finally, we can evaluate $B_{11}(2)$ ,

$B_{11}(2) = -\frac{3}{4} \cos 2 + \frac{7}{4} \cos 10 - \frac{1}{2} \sin 2 + \frac{1}{2} \sin 10 =\boxed{ -1.882924}$

Here's a very elementary numerical integration approach, just to illustrate that the process works the same with matrices as with scalars.

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import math
import numpy as np

##############################################

# Constants

A = np.array([[6.0,1.0],[-5.0,0.0]])
Asq = np.dot(A,A)

# Initialize B and derivatives

B = np.array([[1.0,1.0],[2.0,3.0]])
Bd = np.array([[2.0,0.0],[0.0,5.0]])
Bdd = -np.dot(Asq,B)

# Initialize time

t = 0.0
dt = 10.0**(-6.0)

##############################################

while t <= 2.0:

    B = B + Bd * dt      # Euler integration for B and first derivative
    Bd = Bd + Bdd * dt

    Bdd = -np.dot(Asq,B)

    t = t + dt

##############################################

print B

# Result

#[[-1.88296442 -2.53466354]
 #[ 2.45311577  5.41656049]]

Consider the given differential equation:

Let us introduce a change of variables as follows:

$X_1 = B$ $X_2 = \frac{dB}{dt}$

By doing so, the second order differential equation can be transformed into a system of first order equations. Note that $X_1$ and $X_2$ each are square matrices of size 2.

$\frac{dX_1}{dt} = X_2$ $\frac{dX_2}{dt} = -A^2X_1$

This can be combined in a matrix form as such:

$\begin{bmatrix}\dot{X_1}\\\dot{X_2}\\\end{bmatrix} = \begin{bmatrix}0_{2\times2} & I_2\\-A^2 & 0_{2\times2}\end{bmatrix}\begin{bmatrix}{X_1}\\{X_2}\\\end{bmatrix}$

We Denote: $K = \begin{bmatrix}0_{2\times2} & I_2\\-A^2 & 0_{2\times2}\end{bmatrix}$

and:

$X = \begin{bmatrix}{X_1}\\{X_2}\\\end{bmatrix}$

Here $0_{2\times2}$ is a matrix of 2 rows and 2 columns with all entries as zero. $I_2$ is the identity matrix of size 2. The subscripts are dropped in further computations for convenience. The matrix $X$ has 4 rows and 4 columns.

The system of differential equations becomes:

$\dot{X} = KX$

Where the initial condition is:

$X_o = \begin{bmatrix}B(0)\\\frac{dB}{dt}(0)\\\end{bmatrix}$

The solution to this equation is of the form: $X = e^{Kt}X_o$

Here $e^{Kt}$ is the matrix exponential since K is a square matrix of size 4.

Now, it is just a matter of substituting values and obtaining the required answer. I used a computer for the last step since computing a matrix exponential is not so trivial except in some special circumstances.

The required answer is: -1.883