|
 |
|
|
|
|
|
|
The most terrifying words in the English language are: I’m from the government and I’m here to help, Ronald Reagan
Forget the politicians. The politicians are put there to give you the idea you have freedom of choice. You don’t. You have no choice. You have owners. They own you. They own everything. They own all the important land, they own and control the corporations that’ve long since bought and paid for, the senate, the congress, the state houses, the city halls, they got the judges in their back pocket, and they own all the big media companies so they control just about all of the news and the information you get to hear. They got you by the balls. They spend billions of dollars every year lobbying to get what they want. Well, we know what they want. They want more for themselves and less for everybody else. But I’ll tell you what they don’t want. They don’t want a population of citizens capable of critical thinking. They don’t want well informed, well educated people capable of critical thinking. They’re not interested in that. That doesn’t help them, George Carlin
In advanced calculus and differential equations, solving systems of linear differential equations is a fundamental topic. Specifically, we focus on systems of the form: $\vec{x'}(t) = A\vec{x}(t)$ where:
Our goal is to find the general solution to this system using matrix exponentials. This method provides a powerful and efficient way to solve linear systems of differential equations.
A fundamental matrix, X(t), for the system $\vec{x'} = A\vec{x}$, is a matrix whose columns are linearly independent solutions to the system. In other words, if $\vec{x_1}$ and $\vec{x_2}$ are two linearly independent solutions, then we can construct the fundamental matrix as: X(t) = $(\begin{smallmatrix}\vec{x_1}(t) : \vec{x_2}(t)\end{smallmatrix}) = (\begin{smallmatrix}x_1(t) & x_2(t)\\\ y_1(t) & y_2(t)\end{smallmatrix})$ where the first colum is the vector $\vec{x_1}(t)$ and the second column is the vector $\vec{x_2}(t)$.
There is not a unique fundamental matrix because its columns can be formed by any two linearly independent solutions, and there can be many such pairs.
Once a fundamental matrix X(t) is found, we can express the general solution to the system $\vec{x'} = A\vec{x}$ as a linear combination of the fundamental solutions: $\vec{x}(t) = c_1\vec{x_1}(t)+ c_2\vec{x_2}(t)$
This can be elegantly rewritten in matrix form: $\vec{x} = X(t)\vec{c} = [\vec{x_1}\vec{x_2}](\begin{smallmatrix}c_1\\\ c_2\end{smallmatrix})$ where $\vec{c} = (\begin{smallmatrix}c_1\\\ c_2\end{smallmatrix})$ is a vector of constants determined by the system’s initial conditions.
This expression emphasizes that any solution $\vec{x}(t)$ is simply a linear combination of the fundamental solutions.
Futhermore, if C is any invertible 2 x 2 matrix ($\det(C) \neq 0$), then multiplying it by the fundamental matrix X(t) yields another valid fundamental matrix, namely $\tilde{X}(t) = X(t)C$. This shows that the general solution can be expressed with different constants or scaling factors, but the essential mathematical structure remains the same.
The matrix exponential is a powerful and versatile tool for solving systems of linear differential equations. For a system of the form: $\vec{x'} = A\vec{x}$ the solution can be directly expressed using the matrix exponential $e^{At}$.
To understand how the matrix exponential works, let’s first consider the simpler case of a scalar differential equation: x’ = ax where a is a constant. The solution is $x(t) = ce^{at}$.
The exponential function $e^{at}$ can be expressed as its Maclaurin power series expansion: $e^{at} = 1 + at + a^2\frac{t^2}{2!}+ a^3\frac{t^3}{3!}+ ···$
This series converges everywhere, and differentiating it (term by term) gives the derivative:
$\frac{de^{at}}{dt} = 0 + 1·a + a^2t + a^3\frac{t^2}{2} + ··· =[\text{Factoring a}] a(1 +at + a^2\frac{t^2}{2} + ···) = ae^{at}$.
Thus, $e^{at}$ perfectly solves the equation $x' = ax$.
By extension, for a system of the form $\vec{x}' = A\vec{x}$, the solution can be expressed using the matrix exponential of $A$, denoted $e^{At}$. It is defined exactly by the same series: $e^{At} := I_2 + At + \frac{A^2 t^2}{2!} + \frac{A^3 t^3}{3!} + \dots$ where $I_2$ is the $2 \times 2$ identity matrix, and $A^n$ represents the matrix $A$ multiplied by itself $n$ times (in other words, the matrix A to the n-th power).
The matrix exponential $e^{At}$ provides the standard fundamental matrix for the system $\vec{x'} = A\vec{x}$. The general solution to the system is given by $\vec{x}(t)=e^{At}\vec{c}$ where $\vec{c}$ is a constant vector determined by the initial conditions of the system.
For a system $\vec{x'} = A\vec{x}$, the matrix exponential $e^{At}$ plays the same role as the scalar exponential $e^{at}$ for the scalar case. Just as $e^{at}$ solve x’ = ax, the matrix exponential solves: $\frac{d}{dt}e^{At}=Ae^{At}$
First, note that the derivative of $e^{At}$ can be computed by differentiating its series term-by-term:
$\frac{d}{dt}e^{At} =[e^{At} \text{ is defined as }] \frac{d}{dt}(I_2 + At + A^2\frac{t^2}{2!} + A^3\frac{t^3}{3!} + A^4\frac{t^4}{4!} + ···) = A + A^2t + \frac{A^3t^2}{2!} + \frac{A^4t^4}{3!} ··· =[\text{Factoring A}] A(I + At + \frac{A^2t^2}{2!} + \frac{A^3t^3}{3!}+···) = Ae^{At}$
Thus, $e^{At}$ satisfies $X' = AX$, proving that it is a fundamental matrix for the system. The general solution to the system $\vec{x'} = A\vec{x}$ can be expressed as $\vec{x}(t) = e^{At}\vec{c}$ where $\vec{c}$ is a vector of constants determined by initial conditions.
Initial condition. At t = 0, the matrix exponential reduces to the identity matrix: $e^{A \cdot 0} = I_2$. Thus, $\det(e^{At}) \neq 0$ (specifically, it equals $1$ at $t=0$), ensuring that $e^{At}$ is always non-singular (invertible) and can serve as a valid fundamental matrix.
Consider the system of differential equations: $\begin{cases} x' = y \\\ y' = x \end{cases}$
We can rewrite this system in matrix form as: $\vec{x'} = A\vec{x} = (\begin{smallmatrix}0 & 1\\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}x\\\ y\end{smallmatrix})$ where $A = (\begin{smallmatrix}0 & 1\\\ 1 & 0\end{smallmatrix})$ and $\vec{x} = (\begin{smallmatrix}x\\\ y\end{smallmatrix})$
To solve the system, we will compute the matrix exponential $e^{At}$, which is defined by the power series: $e^{At} = I_2 + At + A^2\frac{t^2}{2!} + A^3\frac{t^3}{3!} + ··· $ where $I_2$ is the 2×2 identity matrix.
Computing Powers of A: $A^2 = (\begin{smallmatrix}0 & 1\\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}0 & 1\\\ 1 & 0\end{smallmatrix}) = (\begin{smallmatrix}1 & 0\\\ 0 & 1\end{smallmatrix}) = I_2, A^3 = A^2A = I_2A = A, A^4 = A^3A = AA = I_2$
Because $A^2 = I_2$, the powers alternate. Substituting these into the series: $e^{At} = I_2 + At + A^2\frac{t^2}{2!} + A^3\frac{t^3}{3!} + ··· = (\begin{smallmatrix}1 & 0\\\ 0 & 1\end{smallmatrix}) + (\begin{smallmatrix}0 & 1\\\ 1 & 0\end{smallmatrix})t + (\begin{smallmatrix}1 & 0\\\ 0 & 1\end{smallmatrix})\frac{t^2}{2!} + (\begin{smallmatrix}0 & 1\\\ 1 & 0\end{smallmatrix})\frac{t^3}{3!}+···$
Grouping the terms by their position in the matrix yields: $e^{At} = (\begin{smallmatrix}1+\frac{t^2}{2!}+\frac{t^4}{4!}+··· & t+\frac{t^3}{3!}+···\\\ t+\frac{t^3}{3!}+··· & 1+\frac{t^2}{2!}+\frac{t^4}{4!}+···\end{smallmatrix})$
Recognizing the Taylor series expansions for hyperbolic sine and cosine: $e^{At} = (\begin{smallmatrix}cosh(t) & sinh(t)\\\ sinh(t) & cosh(t)\end{smallmatrix})$.
Recall that: $\cosh(t) = 1 + \frac{t^2}{2!} + \frac{t^4}{4!} + \dots = \frac{e^t + e^{-t}}{2}$ and $\sinh(t) = t + \frac{t^3}{3!} + \dots = \frac{e^t - e^{-t}}{2}$
If we are given an Initial Value Problem (IVP) with conditions $\vec{x}(0) = \vec{x_0}$, we apply our general solution $\vec{x}(t) = e^{At}\vec{c}$.
At $t = 0$: $\vec{x}(0) = e^{A \cdot 0}\vec{c} = I_2\vec{c} = \vec{c} \implies \vec{c} = \vec{x_0}$
Therefore, the specific solution to the IVP is simply: $\boxed{\vec{x}(t) = e^{At}\vec{x_0}}$
While the power series method worked elegantly in the previous example, computing infinite matrix series is highly impractical for most matrices. In practice, we use diagonalization.
If the matrix $A$ has distinct real eigenvalues, it is diagonalizable. We can factor $A$ into $A = PDP^{-1}$, where:
Because $A = PDP^{-1}$, we get $A^n = (PDP^{-1})^n = PD^nP^{-1}$. This works because: $P^{-1}P = 1$; The middle factors collapse telescopically; Indeed, only the D’s accumulate.
And powers of a diagonal matrix are trivial: $D^n = \begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix}$. So instead of computing powers of a full matrix, we compute powers of scalars.
Now apply the power series to the diagonal form. Start with $e^{At} = \sum_{n=0}^\infty \frac{(At)^n}{n!} = \sum_{n=0}^\infty \frac{(PDP^{-1}t)^n}{n!} = \sum_{n=0}^\infty P \frac{(Dt)^n}{n!}P^{-1} = P(\sum_{n=0}^\infty \frac{(Dt)^n}{n!})P^{-1}$
But the inner sum is just $e^{Dt}$, so $\boxed{e^{At} = Pe^{Dt}P^{-1}}$
Since $D$ is diagonal, $e^{Dt}$ is trivially computed by exponentiating each diagonal entry independently: $D = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} \implies e^{Dt} = \begin{pmatrix} e^{\lambda_1 t} & 0 \\ 0 & e^{\lambda_2 t} \end{pmatrix}$
In standard scalar algebra, the exponential law states that $e^{a+b} = e^a e^b$.
However, in matrix algebra, the exponential law $e^{A+B} = e^A e^B$ does not generally hold for matrices unless matrices $A$ and $B$ commute, meaning $AB = BA$.
If $A$ and $B$ do not commute, $e^{A+B} \neq e^A e^B$. Some special cases where the exponential law does hold include:
Computing the matrix exponential $e^{At}$ is essential for solving linear system of differerential equations of the form: $\vec{x}'(t) = A \vec{x}(t)$. There are serveral methods to compute $e^{At}$:
Recall that any fundamental matrix can be written as $X(t) = e^{At}C$. If we evaluate this at $t = 0$, we get $X(0) = e^{A \cdot 0}C = I_2 C = C$. Therefore, $C = X(0)$. Substituting $C$ back into our equation yields $X(t) = e^{At}X(0)$, which rearranges algebraically to $e^{At} = X(t)X(0)^{-1}$.
This method: (i) avoids directly calculating the matrix exponential from the infinite series definition, which can be complex for larger matrices; (ii) provides a systematic approach applicable to various matrices; (iii) guarantees that $e^{At}$ satisfies both the differential equation and the initial condition at t = 0.
Consider the system of firt-order linear differential equations: $\begin{cases} x' = y \\\ y' = x \end{cases}$
This can be written in matrix form as $\vec{x'} = A\vec{x}$ where $\vec{x} =(\begin{smallmatrix} x\\\ y\end{smallmatrix})$ and $A = (\begin{smallmatrix}0 & 1\\\ 1 & 0\end{smallmatrix})$.
Our goal is to find the general solution $\vec{x}$. One method to solve such systems is to use the fundamental matrix and the matrix exponential.
Step 1: Find Eigenvalues and Eigenvectors
We start by finding the eigenvalues of A. The characteristic equation is: $\det(A -\lambda I) = 0$ ↭ $\det \begin{pmatrix} -\lambda & 1 \\ 1 & -\lambda \end{pmatrix} = 0 \implies \lambda^2 - 1 = 0 \implies \lambda_1 = 1, \quad \lambda_2 = -1$
Now, find the eigenvectors
This yields the system $-a_1 + a_2 = 0$ and $a_1 - a_2 = 0$. Choosing $a_1 = 1$ gives $a_2 = 1$. Thus, $\vec{v_1} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
This yields $a_1 + a_2 = 0$. Choosing $a_1 = 1$ gives $a_2 = -1$. Thus, $\vec{v_2} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.
Step 2: Form the General Solution and the Fundamental Matrix X(t)
The general solution to $\vec{x'} = A\vec{x}$ is a linear combination of our eigenvalue/eigenvector pairs: $\vec{x}(t) = c_1e^{t}(\begin{smallmatrix}1\\\ 1\end{smallmatrix}) + c_2e^{-t}(\begin{smallmatrix}1\\\ -1\end{smallmatrix})$
Thus, the fundamental matrix is formed by placing these independent solutions as columns into a matrix: $X(t) = (\vec{x_1}(t)~\vec{x_2}(t)) = (\begin{smallmatrix}e^{t} & e^{t}\\\ e^{t} & -e^{-t}\end{smallmatrix})$
The fundamental matrix X(t) is directly constructed from a set of linearly independent solutions.
Step 3: Compute $X(0)$ and $X(0)^{-1}$
Substitute t = 0 into the fundamental matrix: X(0) = $(\begin{smallmatrix}e^{0} & e^{-0}\\\ e^{-0} & -e^{-0}\end{smallmatrix}) = (\begin{smallmatrix}1 & 1\\\ 1 & -1\end{smallmatrix})$
Calculate the determinant to verify invertibility and find the inverse: det(X(0)) = $det(\begin{smallmatrix}1 & 1\\\ 1 & -1\end{smallmatrix}) = -1 -1 = -2 ≠ 0$. Since det(X(0)) = -2, X(0) is invertible.
$X(0)^{-1} = \frac{1}{-2}(\begin{smallmatrix}-1 & -1\\\ -1 & 1\end{smallmatrix}) = \frac{1}{2}(\begin{smallmatrix}1 & 1\\\ 1 & -1\end{smallmatrix})$
Step 4: Calculate the determinant to verify invertibility and find the inverse:
Multiply the fundamental matrix by the inverse matrix we just found: $e^{At} = X(t)X(0)^{-1} = (\begin{smallmatrix}e^{t} & e^{-t}\\\ e^{t} & -e^{-t}\end{smallmatrix})·\frac{1}{2}(\begin{smallmatrix}1 & 1\\\ 1 & -1\end{smallmatrix}) = \frac{1}{2}(\begin{smallmatrix}e^{t} + e^{-t} & e^{t} - e^{-t}\\\ e^{t} -e^{-t} & e^{t} + e^{-t}\end{smallmatrix})$ This result matches the earlier expression for $e^{At}$.
Recognizing the definitions of hyperbolic sine and cosine, this matches the standard expression: $e^{At} = \begin{pmatrix} \cosh(t) & \sinh(t) \\ \sinh(t) & \cosh(t) \end{pmatrix}$
We have successfully found the solution to the system: $\vec{x}(t) = e^{At}\vec{x_0} = (\begin{smallmatrix}cosh(t) & sinh(t)\\\ sinh(t) & cosh(t)\end{smallmatrix})\vec{x_0}$. This confirms that computing the fundamental matrix provides a rigorous, powerful pathway to solving linear systems of differential equations.
JustToThePoint Copyright © 2011 - 2026 Anawim. ALL RIGHTS RESERVED. Bilingual e-books, articles, and videos to help your child and your entire family succeed, develop a healthy lifestyle, and have a lot of fun. Social Issues, Join us.
This website uses cookies to improve your navigation experience.
By continuing, you are consenting to our
use of cookies, in accordance with our Cookies Policy and
Website Terms and Conditions of use.