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
An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:
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., $\frac{dy}{dx} = 3x +5y$):
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 (x_{0}, y_{0}) .
A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.
The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$
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 fields like engineering, physics, and applied mathematics. One powerful tool for analyzing such systems is the Trace-Determinant Plane, which helps us predict the nature of solutions without solving the system explicitly.
Consider a 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. These are special numbers that give us critical information about the system’s dynamics. They are solutions to the 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) = λ^{2} -Tλ+ D.
Theorem. If a 2 x 2 matrix A has eigenvalues λ_{1} and λ_{2}, then the trace of A is λ_{1} + λ_{2} and det(A) = λ_{1}·λ_{2}.
We can use the trace and determinant to establish the nature of a solution to a linear system. 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 discriminant Δ = T^{2} -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.
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^{-1}AP.
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^{-1}AP) = det(A) and tr(P^{-1}AP) = 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 T^{2} -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. Unstable Node: Both eigenvalues are real and positive. 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 and depart along another.
Spirals:
Eigenvalues are complex with a non-zero real part. Stable Spiral: Real part is negative; solutions spiral into the equilibrium. Unstable Spiral: Real part is positive; solutions spiral away from the equilibrium.
Centers:
Eigenvalues are purely imaginary (real part zero). Solutions are closed orbits; 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 $(A−λI)\vec{v}=0↭ (\begin{smallmatrix}α-α-iβ & β\\ -β & α-α-iβ\end{smallmatrix})\vec{v} = 0 ↭ (\begin{smallmatrix}-iβ & β\\ -β & -iβ\end{smallmatrix})\vec{v}$
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 c_{1} and c_{2} 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 = λ_{1} + λ_{2} = α + α = (α+iβ) + (α−iβ) = 2α.
Determinant (D): Product of the eigenvalues. T = λ_{1} · λ_{2} = (α+iβ)(α−iβ) = α^{2} + β^{2}.
Discriminant: T^{2} -4D = (2α)^{2} -4(α^{2} + β^{2}) = -4β^{2}. Since β^{2}≥ 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:
Consider the linear system:
$x’ = Ax,$ where A = $(\begin{smallmatrix}λ & 0\\ 0 & μ\end{smallmatrix})$. A is a 2 x 2 diagonal matrix with distinct eigenvalues λ and μ.
For distinct real eigenvalues λ and μ (assuming neither is zero), the general solution to the system is:
$\begin{cases} x(t) = x_0e^{λt} \\ y(t) = y_0e^{μt} \end{cases}$
EXPLANATION
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 ↭ $\frac{dx}{dt} = λx ↭ \frac{dx}{x} = λdt ↭ \int \frac{dx}{x} = λdt ↭ ln|x| = λt + C’↭ x = Ce^{λt}$
x_{0} and y_{0} are the initial conditions at t = 0. $x(t) = C_1e^{λ·0} = x_0 ⇒ C_1 = x_0$. Similarly, $y_0 = C_2$
Finally, the general solution to the system is: $\begin{cases} x(t) = x_0e^{λt} \\ y(t) = y_0e^{μt} \end{cases}$
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 = T^{2}-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., T^{2} -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.
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. That’s a borderline case.
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.