Complex Eigenvalues in Systems of Differential Equations

Recall

An expert is a person who has made all the mistakes that can be made in a very narrow field, Niels Bohr

Mistakes are a great educator when one is honest enough to admit them and willing to learn from them, Aleksandr Solzhenitsynv

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

• Dependent variables: Variables that depend on one or more other variables (y).
• Independent variables: Variables upon which the dependent variables depend (x).
• Derivatives: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

First-Order Linear Ordinary Differential Equations (ODEs)

Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:

• y = y(x) is the unknown function of the independent variable x.
• y′ = $\frac{dy}{dx}$ is the derivative of y with respect to x.
• a(x), b(x), and c(x) are known functions of x.

These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.

If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

• The function f(x, y) (the right-hand side of the ODE) in the equation y’ = f(x, y) is continuous in some neighborhood around a point (x₀, y₀) and
• Its partial derivative with respect to y, denoted as $\frac{∂f}{∂y}$, is also continuous near (x₀, y₀).

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it satisfies the initial condition y(x₀) = y₀.

This theorem ensures that under these conditions, the solution exists and is unique near x = x₀.

General Form of a Linear First-Order System

A linear first-order system has the form:

$\begin{cases} x’ = a(t)x + b(t)y + r_1(t) \\ y’ = c(t)x + d(t)y + r_2(t) \end{cases}$ where:

• a(t), b(t), c(t), and d(t) are functions of time t (they can be constants in simpler cases).
• r₁(t) and r₂(t) represent external forcing functions or inputs that depend on t.
• x(t₀) = x₀ and y(t₀) = y₀ are the initial conditions.

Complex Eigenvalues in Systems of Differential Equations

In the study of linear systems of differential equations, we often encounter systems of the form: x′(t) = Ax(t) where:

• x(t) is a vector-valued of function of t.
• A is a constant n × n matrix with real entries.
• x’(t) denotes the derivative of x(t) with respect to time.

Solving such systems involves finding the eigenvalues and eigenvectors of the matrix A. When A has distinct real eigenvalues, λ₁ and λ₂ with corresponding eigenvectors v₁ and v₂, the general solution is straightforward: $x = C_1e^{λ_1t}v_1 + C_2e^{λ_2t}v_2$.

However, when the eigenvalues are complex λ = a ± ib with a corresponding complex eigenvector v, additional steps are necessary to obtain real-valued solutions, which are essential for real-world applications.

Steps to Find the Real-Valued Solution

Suppose A has a pair of complex conjugate eigenvalues, λ = a ± ib with a corresponding complex eigenvector v. Since the system x′(t) = Ax(t) is real, we seek real-valued solutions. We can achieve this by utilizing Euler’s formula and separating the complex solution into its real and imaginary parts.

Theorem: Given a real n x n matrix A with a complex eigenvalue λ = a + ib and corresponding complex eigenvector $\vec{v}$, :the general real solution to the system of differential equations x′ = Ax is: $x(t) = e^{at}[C_1· \mathbb{R}(e^{ibt}\vec{v}) + C_2\mathbb{ℑ}(e^{ibt}\vec{v})]$ where:

• $\mathbb{R}=$Re and $\mathbb{ℑ}$=Im denote the real and imaginary parts, respectively.
• C:₁ and C₂ are real constants determined by initial conditions.

1. Finding Eigenvalues and Eigenvectors. Solve the characteristic equation det(A− λI) = 0 to find the eigenvalues λ. If λ = a +ib is a complex eigenvalue, its complex conjugate λ = a −ib is also an eigenvalue of A because A has real entries. Find a corresponding complex eigenvector v by solving (A −λI)v=0.
2. Constructing the Complex Solution. The complex solution to the system is x_complex(t) = e^λtv = e^at(e^ibtv).
3. Separating into Real and Imaginary Parts. Use Euler’s Formula : e^ibt = cos(bt) + isin(bt). Express v in terms of its real and imaginary parts: v = v_real + iv_imaginary. Multiply e^ibt and v to obtain the real and imaginary components.

   x_complex(t) = e^λtv = e^at(e^ibtv) = e^at(cos(bt) + isin(bt))(v_real + iv_imaginary).

4. Forming Real-Valued Solutions. The real part x_real(t) = $e^{at}[cos(bt)v_{real}-sin(bt)v_{imaginary}]$ is a real-valued solution. The imaginary part x_imaginary(t) = $e^{at}[sin(bt)v_{real}+cos(bt)v_{imaginary}]$ is also a real-valued solution. Both x_real(t) and x_imaginary(t) satisfy the differential equation.
5. General Solution. The general real solution is a linear combination of the real and imaginary parts: x(t) = C₁x_real(t) + C₂x_imaginary(t).

Give the general solution to the system:

$\begin{cases} x’ = 2x + -5y \\ y’ = x -2y \end{cases}$

Our goal is to solve this system and find expressions for x(t) and y(t).

We can express this system as a matrix equation: x’ = Ax where x = $\left( \begin{smallmatrix}x\\ y\end{smallmatrix} \right)$ and the matrix A is $\left( \begin{smallmatrix}2 & -5\\ 1 & -2\end{smallmatrix} \right)$

The first step in solving this system is to find the eigenvalues of matrix A. The characteristic equation is given by: |A - λI| = $\left| \begin{smallmatrix}2-λ & -5\\ 1 & -2-λ\end{smallmatrix} \right| = (2-λ)(-2-λ)+5 = -4-2λ +2λ +λ^2 + 5 = λ^2 + 1 = 0 ⇒[\text{Solving for λ}] λ = ± i$. The eigenvalues are complex conjugates.

Now we solve for the eigenvector to λ = i:

$(A−λI)v=0 ↭ \left( \begin{smallmatrix}2-i & -5\\ 1 & -2-i\end{smallmatrix} \right)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

This leads to the following system of equations:

$\begin{cases} (2-i)a_1 + -5a_2 = 0 \\ a_1 + (-2-i)a_2 = 0 \end{cases}$

From the second equation (ii), $a_1 = (2+i)a_2 ⇒[\text{Replace this in (i)}] (2-i)(2+i)a_2 -5a_2 = 0 ↭ (4 -i^2)a_2 -5a_2 = 0 ↭ 5a_2 -5a_2 = 0,$ so any a_2 is a solution. Therefore, for simplicity, let’s choose $a_2 = 1, a_1 = (2 +i)$. Eigenvector Corresponding to λ=i: $v = (\begin{smallmatrix}2+i\\1\end{smallmatrix})$

The complex solution to the system is given by: $x(t)= e^{λt}v = e^{it}(\begin{smallmatrix}2+i\\1\end{smallmatrix}) =[\text{Using Euler’s formula}] (cos(t)+isin(t))(\begin{smallmatrix}2+i\\1\end{smallmatrix}) = (\begin{smallmatrix}(cos(t)+isin(t))(2+i)\\cos(t)+isin(t)\end{smallmatrix}) = \text{Simplifying, we get} (\begin{smallmatrix}(2cos(t)-sin(t))+i(cos(t)+2sin(t))\\cos(t)+isin(t)\end{smallmatrix}) $

Combining the real and imaginary parts, we get two real-valued solutions:

x_real(t) = $(\begin{smallmatrix}2cos(t)-sin(t)\\cos(t)\end{smallmatrix})$

x_imaginary(t) = $(\begin{smallmatrix}cos(t)+2sin(t)\\sin(t)\end{smallmatrix})$

Thus, the general solution to the system is a linear combination of the real and imaginary parts:

x(t) = c₁x_real(t) + c₂x_imaginary(t) = $c_1(\begin{smallmatrix}2cos(t)-sin(t)\\cos(t)\end{smallmatrix})+c_2(\begin{smallmatrix}cos(t)+2sin(t)\\sin(t)\end{smallmatrix})$

Intuition:

• The eigenvalues are purely imaginary (λ = ±i), indicating that the system exhibits oscillatory behavior without exponential growth or decay.
• The solutions involve sine and cosine functions, reflecting periodic oscillations.
• c₁ and c₂ are constants determined by the initial conditions of the system.

Modelling a love affair

In this problem, we model the dynamics of a romantic relationship between Susan and George using a system of differential equations. Let x(t) represents Susan’s love for George, and y(t) represents George’s love for Susan. Their love dynamics (interaction of their feelings) are modeled by the following system of differential equations:

$\begin{cases} x’ = x + 2y \\ y’ = -x -y \end{cases}$

Our goal is to solve this system and find expressions for x(t) and y(t).

Step 1: Representing the System in Matrix Form

We can express this system as a matrix equation: x’ = Ax where x = $\left( \begin{smallmatrix}x\\ y\end{smallmatrix} \right)$ and the matrix A is $\left( \begin{smallmatrix}1 & +2\\ -1 & -1\end{smallmatrix} \right)$

Thus, the system becomes: $\left( \begin{smallmatrix}x’\\ y’\end{smallmatrix} \right) = \left( \begin{smallmatrix}1 & +2\\ -1 & -1\end{smallmatrix} \right)\left( \begin{smallmatrix}x\\ y\end{smallmatrix} \right)$

Step 2: Finding the Eigenvalues of Matrix A

The second step in solving this system is to find the eigenvalues of the coefficient matrix A. The characteristic equation is given by: |A - λI| = $\left| \begin{smallmatrix}1-λ & +2\\ -1 & -1-λ\end{smallmatrix} \right| = (1-λ)(-1-λ)+2 = -1+λ -λ +λ^2 + 2 = λ^2 + 1 = 0 ⇒[\text{Solving for λ}] λ = ± i$. The eigenvalues are complex conjugates.

• Step 3: Finding the Eigenvectors for λ = i

We solve the system: $(A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

which is equivalent to solving:

$\left( \begin{smallmatrix}1-i & +2\\ -1 & -1-i\end{smallmatrix} \right)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$.

This leads to the following system of equations:

$\begin{cases} (1-i)a_1 + 2a_2 = 0 \\ -a_1 + (-1-i)a_2 = 0 \end{cases}$

From the first equation: $a_1 = \frac{-2a_2}{1-i}$. Substitute in the second equation $-\frac{-2a_2}{1-i} + (-1-i)a_2 = 0 ↭ \frac{2a_2(1+i)}{(1-i)(1+i)}+ (-1-i)a_2 = 0 ↭ \frac{2a_2(1+i)}{2}+ (-1-i)a_2 = 0 ↭ a_2(1+i)+ (-1-i)a_2 = 0 ↭ a_2(1+i-1-i) = 0↭ a_2·0 = 0$. Since this equation is always true, a₂ can be any non-zero value. Let’s choose a₂​ = 1 for simplicity. Then, (ii) $a_1 = (-1-i)a_2 = -1-i$

The solution is: $(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = (\begin{smallmatrix}-1-i\\ 1 \end{smallmatrix})$

Step 4: Constructing the Complex Solution

The complex solution to the system is given by: $(\begin{smallmatrix}-1-i\\ 1 \end{smallmatrix})e^{it} = [(\begin{smallmatrix}-1\\ 1 \end{smallmatrix})+i(\begin{smallmatrix}-1\\ 0 \end{smallmatrix})](cos(t) + isin(t))$

Step 5: Separating into Real and Imaginary Parts

Combining the real and imaginary parts, we get two real-valued solutions:

x_real(t) = $(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (\begin{smallmatrix}-1\\ 1 \end{smallmatrix})cos(t)-(\begin{smallmatrix}-1\\ 0 \end{smallmatrix})sin(t) = (\begin{smallmatrix}-cos(t)+sin(t)\\ cos(t) \end{smallmatrix})$

x_imaginary(t) = $(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (\begin{smallmatrix}-1\\ 1 \end{smallmatrix})sin(t)+(\begin{smallmatrix}-1\\\ 0 \end{smallmatrix})cos(t) = (\begin{smallmatrix}-cos(t)-sin(t)\\\ sin(t) \end{smallmatrix})$

Step 6: Forming the General Real-Valued Solution

Thus, the general solution to the system is a linear combination of these:

x(t) = c₁x_real(t) + c₂x_imaginary(t) = $c_1(\begin{smallmatrix}-cos(t)+sin(t)\\ cos(t) \end{smallmatrix})+c_2(\begin{smallmatrix}-cos(t)-sin(t)\\\ sin(t) \end{smallmatrix})$

$x(t) = (\begin{smallmatrix}-c_1cos(t)+c_1sin(t)-c_2cos(t)-c_2sin(t)\\ c_1cos(t)+c_2sin(t)\end{smallmatrix}) = (\begin{smallmatrix}-(c_1+c_2)cos(t)+(c_1-c_2)sin(t)\\ c_1cos(t)+c_2sin(t)\end{smallmatrix})$

Step 7. Interpretation of the Solution

• The functions cos(t) and sin(t) indicate oscillatory behavior in the dynamics of Susan’s and George’s feelings, which could represent cycles of attraction and estrangement in their relationship. Yeah!, love is a rollercoaster of ups and downs.
• The constants c₁ and c₂ are determined by the initial conditions of the system, i.e., the initial levels of affection between Susan and George.

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