Do, or do not. There is no try, Yoda.
The problem is not the problem. The problem is your attitude about the problem, Captain Jack Sparrow
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., $\frac{dy}{dx} = 3x +5y$):
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:
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:
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.
A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:
This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.
Understanding the behavior of linear systems of differential equations is essential in many disciplines like engineering, physics, and applied mathematics. One powerful tool for analyzing such systems, without having to solve them explicitly, is the Trace-Determinant (T-D) Plane, which helps us predict the nature of solutions without solving the system explicitly.
Consider a two-linear system of differential equations: $(\begin{smallmatrix}x’\\ y’\end{smallmatrix}) = (\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix}) =[\text{This can be written in matrix form:}](\begin{smallmatrix}x’\\ y’\end{smallmatrix}) = A(\begin{smallmatrix}x\\ y\end{smallmatrix})$ where $A = (\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})$ is a 2 x 2 matrix.
To understand how the system behaves over time, we need to find its eigenvalues λ. These are special numbers that give us critical information about the system’s dynamics. They are solutions to the characteristic equation det(A -λI), where I is the identity matrix.
Finding the eigenvalues allows us to:
A phase plane is a graphical representation of the system's solutions. Solutions to the system are plotted as curves in the phase plane called trajectories or orbits. Each point on a trajectory represents the state of the system at a particular time. Following the trajectory shows how the state of the system evolves over time.
The characteristic polynomial is obtained from the determinant: det(A -λI) = 0.
For our 2 x 2 matrix, this becomes: $det(\begin{smallmatrix}a-λ & b\\ c & d-λ\end{smallmatrix}) = 0 ↭ (a-λ)(d-λ)-bc = 0↭[\text{Simplifying}] λ^2 -(a+d)λ + (ad-bc) = 0$
We define two important quantities:
Using these, the characteristic polynomial simplifies to: det(A -λI) = λ2 -Tλ+ D.
Theorem. If a 2 x 2 matrix A has eigenvalues λ1 and λ2, then the trace of A is λ1 + λ2 and det(A) = λ1·λ2.
Instead of directly computing eigenvalues, we can use the trace and understand the nature of ths solutions. Thus, we should be able to determine the phase portrait of a system x’ = AX by simply examining the trace and determinant of A.
Since the eigenvalues of A are given by: $λ = \frac{T ± \sqrt{T^2-4D}}{2}$ we can see that the sign of the discriminant Δ = T2 -4D determine the nature of the eigenvalues of A:
The Trace-Determinant Plane is a coordinate plane where each point represents a system with specific trace T and determinant D.
The parabola D = $\frac{T^2}{4}$ divides the plane into regions corresponding to different types of eigenvalues. The region corresponds to Δ = 0, i.e., repeated eigenvalues.
When we talk about changing coordinates in the context of linear systems, we’re essentially performing a similarity transformation or conjugation on the system’s matrix. This involves an invertible matrix P that transforms the original matrix A into a new matrix A′: A’ = P-1AP.
Despite this change, certain properties of the matrix remain unchanged, specifically the trace and determinant.
Theorem. The trace and determinant of a 2 x 2 matrix are invariant under a change of coordinates. Therefore, for any 2 x 2 matrix A and any invertible 2 x 2 matrix P, the following hold: det(P-1AP) = det(A) and tr(P-1AP) = tr(A).
The determinant is a measure of how the matrix scales area (in 2D). When we change coordinates, we’re just looking at the same transformation from a different perspective, so the scaling factor (determinant) remains the same. The trace is the sum of the diagonal elements of a matrix. It is related to the sum of its eigenvalues. Since eigenvalues are intrinsic to the matrix’s transformation properties, they are indeed unaffected by the coordinate system, so the trace remains unchanged.
Corollary. The discriminant T2 -4D (where T is the trace and D is the determinant of A) is invariant under a change of coordinates.
Due to the invariance under coordinate transformation, we only need to consider a few standard matrices to represent all possible behaviors of 2×2 linear systems x’ = Ax. These matrices are:
Why These Forms?
When A is diagonalizable, the general solution is: $x(t) = c_1e^{λ_1t}v_1 + c_2e^{λ_2t}v_2$ where:
Nodes:
Stable Node: Both eigenvalues are real and negative. Trajectories approach the equilibrium.
Unstable Node: Both eigenvalues are real and positive. Trajectories move away from the equilibrium.
Improper Node: Repeated eigenvalues with a full set of eigenvectors.
Saddle Point:
Eigenvalues are real with opposite signs. Solutions approach the equilibrium along one eigenvector (stable direction) and depart along another (unstable direction).
Spirals:
Eigenvalues are complex with non-zero real part. Stable Spiral: Real part is negative; solutions spiral inwards, into the equilibrium. Unstable Spiral: Real part is positive; solutions spiral outwards, away from the equilibrium.
Centers:
Eigenvalues are purely imaginary (no real part). Solutions are closed orbits (ellipses, circles); the system exhibits neutral stability.
Consider the linear system: $x’ = Ax, x’ = (\begin{smallmatrix}α & β\\ -β & α\end{smallmatrix})x$, where A = $(\begin{smallmatrix}α & β\\ -β & α\end{smallmatrix})$.
To analyze the system, we find the eigenvalues of the matrix A. The eigenvalues are solutions to the characteristic equation: det(A -λI) = 0 ↭ $det(\begin{smallmatrix}α-λ & β\\ -β & α-λ\end{smallmatrix}) = (α -λ)^2 + β^2 ⇒ (α -λ) = ±iβ$. Thus, the eigenvalues are complex conjugates: λ = α ± iβ.
Theorem: Given a real n x n matrix A with a complex eigenvalue λ = α + iβ and corresponding complex eigenvector $\vec{v}$, the general real solution to the system of differential equations x′ = Ax is: $x(t) = e^{αt}[C_1· Re(e^{iβt}\vec{v}) + C_2Im(e^{iβt}\vec{v})]$.
$\vec{v} = \vec{v_r} + i\vec{v_i}$ where $\vec{v_r}$ and $\vec{v_i}$ are real vectors representing the real and imaginary parts of v.
$e^{iβt}\vec{v} = (cos(βt) + isin(βt))(\vec{v_r}+i\vec{v_i}) = [cos(βt)\vec{v_r} -sin(βt)\vec{v_i}] + i[cos(βt)\vec{v_i} + sin(βt)\vec{v_r}]$
Two real solutions: $x_r(t) = cos(βt)\vec{v_r} -sin(βt)\vec{v_i}, x_i(t) = cos(βt)\vec{v_i} + sin(βt)\vec{v_r}$
Compute $(A−λI)\vec{v}=0↭ (\begin{smallmatrix}α-α-iβ & β\\ -β & α-α-iβ\end{smallmatrix})\vec{v} = 0 ↭ (\begin{smallmatrix}-iβ & β\\ -β & -iβ\end{smallmatrix})\vec{v} = 0$
Set up the equations:
$\begin{cases} -iβa_1 + βa_2 = 0 \\ -βa_i -iβa_2 = 0 \end{cases}$
$-iβa_1 + βa_2 = 0 ⇒[\text{Choose }a_1 = 1] -iβ + βa_2 = 0 ↭ i = a_2 ⇒\vec{v} = (\begin{smallmatrix}1\\ i\end{smallmatrix}) = (\begin{smallmatrix}1\\ 0\end{smallmatrix}) + (\begin{smallmatrix}0\\ 1\end{smallmatrix})i$ where $\vec{v_r}=(\begin{smallmatrix}1\\ 0\end{smallmatrix}), \vec{v_i} =(\begin{smallmatrix}0\\ 1\end{smallmatrix})$
Two real solutions: $x_r(t) = cos(βt)\vec{v_r} -sin(βt)\vec{v_i} = cos(βt)(\begin{smallmatrix}1\\ 0\end{smallmatrix}) -sin(βt)(\begin{smallmatrix}0\\ 1\end{smallmatrix}) = (\begin{smallmatrix}cos(βt)\\ -sin(βt)\end{smallmatrix}), x_i(t) = cos(βt)\vec{v_i} + sin(βt)\vec{v_r} = cos(βt)(\begin{smallmatrix}0\\ 1\end{smallmatrix}) + sin(βt)(\begin{smallmatrix}1\\ 0\end{smallmatrix}) = (\begin{smallmatrix}sin(βt)\\ cos(βt)\end{smallmatrix})$
The general solution to this system is $x(t) = e^{αt}[C_1· Re(e^{iβt}\vec{v}) + C_2Im(e^{iβt}\vec{v})] = e^{αt}[C_1(\begin{smallmatrix}cos(βt)\\ -sin(βt)\end{smallmatrix})+ C_2(\begin{smallmatrix}sin(βt)\\ cos(βt)\end{smallmatrix})]$ where c1 and c2 are constant determined by initial conditions.
Breaking Down the solution:
The eigenvalues of A are related to the trace and determinant of A. The eigenvalues of A are given by $λ = \frac{T±\sqrt{T^2-4D}}{2}$.
Trace: Sum of the eigenvalues. T = λ1 + λ2 = (α+iβ) + (α−iβ) = 2α.
Determinant (D): Product of the eigenvalues. D = λ1 · λ2 = (α+iβ)(α−iβ) = α2 + β2.
Discriminant: T2 -4D = (2α)2 -4(α2 + β2) = -4β2. Since β2≥ 0, the discriminant is negative ⇒ The eigenvalues are complex conjugates, the system exhibits oscillatory behavior.
Real Part of Eigenvalues (α) is directly related to the trace (T = 2α).
Classification based on Trace:
In the study of 2×2 linear systems of differential equations, we often classify the behavior of solutions based on the eigenvalues of the coefficient matrix. One crucial scenario is when the system has two distinct real eigenvalues. We start with the general form of a linear system: x’ = Ax,
We start with the general form of a linear system: x’ = Ax, where A is a 2×2 constant coefficient matrix, and x = $(\begin{smallmatrix}x(t)\\ y(t)\end{smallmatrix})$ is a vector-valued function of time t.
Consider a system with a coefficient matrix A that has two distinct real eigenvalues λ and μ. In the simplest case, where A is already diagonal, it can be written as A = $(\begin{smallmatrix}λ & 0\\ 0 & μ\end{smallmatrix})$. A is a 2 x 2 diagonal matrix with distinct eigenvalues λ and μ.
The general solution provided assumes that the system is decoupled, meaning that the equations for x and y do not depend on each other. For a system with distinct real eigenvalues λ and μ, the general solution is: $\vec{x}(t) = c_1e^{λt}\vec{v_1} + c_2e^{μt}\vec{v_2}$ where $\vec{v_1}$ and $\vec{v_2}$ are eigenvectors corresponding to λ and μ, respectively.
In this special case, the equation becomes:
$\begin{cases} x′= λx \\ y’ = μy \end{cases}$.
This allows us to write the solutions independently:
$\begin{cases} x(t) = C_1e^{λt} \\ y(t) = C_2e^{μt} \end{cases}$
x’ = λx ↭ $\frac{dx}{dt} = λx ↭ \frac{dx}{x} = λdt ↭ \int \frac{dx}{x} = λdt ↭ ln|x| = λt + C’↭ x(t) = C_1e^{λt}$. Analogously, $x(t) = C_2e^{μt}$
x0 and y0 are the initial conditions at t = 0. $x(t) = C_1e^{λ·0} = x_0 ⇒ C_1 = x_0$. Similarly, $y_0 = C_2$
For distinct real eigenvalues λ and μ (assuming neither is zero), the general solution simplifies to:
$\begin{cases} x(t) = x_0e^{λt} \\ y(t) = y_0e^{μt} \end{cases}$. where x0 and y0 are the initial conditions. This form clearly shows how each solution component evolves independently over time.
There are three cases to consider:
The origin is unstable, with trajectories approaching along the stable manifold and departing along the unstable manifold.
Relationship Between Trace, Determinant, and Eigenvalues
Our two eigenvalues are given by $λ = \frac{T±\sqrt{T^2-4D}}{2}$, Δ = T2-4D > 0 (Real and distinct eigenvalues), T = λ + μ, D = λ·μ. Let’s examine different scenarios by analyzing the signs of T and D:
This situation arises when the discriminant of the characteristic equation is zero, i.e., T2 -4D = 0.
For a 2 x 2 matrix A, the characteristic equation has a repeated real eigenvalue $λ = \frac{T±\sqrt{T^2-4D}}{2} = \frac{T}{2}$.
$\begin{cases} x’ = λx \\ y’ = λy \end{cases}$. The general solution is: $(\begin{smallmatrix}x(t)\\ y(t)\end{smallmatrix}) = c_1e^{λt}\vec{v_1} + c_2e^{λt}\vec{v_1}$ where $\vec{v_1}$ and $\vec{v_2}$ are independent eigenvectors.
In the special case of a single repeated eigenvalue and a diagonal matrix A, $\vec{v_1}$ and $\vec{v_2}$ can be chosen as the standard basis vectors: $(\begin{smallmatrix}x(t)\\ y(t)\end{smallmatrix}) = c_1e^{λt}(\begin{smallmatrix}1\\ 0\end{smallmatrix}) + c_2e^{λt}(\begin{smallmatrix}0\\ 1\end{smallmatrix})$
If λ > 0, solutions grow exponentially. Trajectories move away from the origin along straight lines. Unstable Node (Source).
If λ < 0, solutions decay exponentially. Trajectories move toward the origin along straight lines. Stable Node (Sink).
If λ > 0, solutions grow exponentially, with an additional linear term t. Trajectories exhibit a parabolic curve moving away from the origin. Unstable Improper Node.
If λ < 0, solutions decay exponentially, influenced by the linear term. Stable Improper Node.
T = 0 ⇒ $λ = \frac{T±\sqrt{T^2-4D}}{2} = \frac{±\sqrt{-4D}}{d}= ±i\frac{\sqrt{4D}}{2}=±i\sqrt{D}$. Centers correspond to systems with purely imaginary eigenvalues.
When analyzing centers in the context of linear systems, we consider the case where the eigenvalues are purely imaginary. This occurs when the real part of the eigenvalues is zero, and the eigenvalues are of the form λ=±iω, where ω is a real number ↭ For Δ to be negative, D must be positive D > 0.
Purely imaginary eigenvalues occur when the real part is zero. The real part of the eigenvalues is $\frac{T}{2}$.
T = 0, D > 0. Purely imaginary eigenvalues ⇒ No exponential growth in the solutions. The system neither grows nor decays over time. Solutions involve oscillatory terms (sin and cos functions). Trajectories are closed orbits (curves: circles or ellipses) around the equilibrium point. Systems with purely imaginary eigenvalues exhibit center behavior. The origin neither attracts nor repels trajectories.