2D Homogeneous Linear Systems with Constant Coefficients

$$\frac{d\vec{x}}{dt}=A\vec{x}\text{where }\vec{x}(t)=\left[\begin{array}{c}{x}_1(t)\\ x_2(t)\end{array}\right],A\text{ is a }2\times 2\text{ real constant matrix}$$

Distinct Real & Nonzero Eigenvalues

Loosey goosey explanation

Recall $u^{\prime }=\lambda u⟹u=c{e}^{\lambda t}$ “Think of matrices as a generalization of the number system” Well it turns out that ${\vec{x}}^{\prime }=A\vec{x}⟹\vec{x}=\vec{c}{e}^{At}$, where $\dim A=2$ A is diagonalizable $⟺\vec{x}=c_1{e}^{{\lambda }_{1}t}{\vec{x}}_1+c_2{e}^{{\lambda }_{2}t}{\vec{x}}_2$ where ${\lambda }_n$ are A’s eigenvalues, and ${\vec{v}}_n$ are $A$‘s eigenvectors. See Eigenvectors, Eigenvalues, and Eigenspaces.

Find the eigenvalues of $A$ by solving the Characteristic Polynomial We assume that $\lambda ={\lambda }_1,{\lambda }_2\in \mathbb{R}\ne 0{\lambda }_1\ne {\lambda }_2$

$$\begin{array}{r}\det (A-\lambda I)=0\end{array}$$

We find ${\lambda }_1$‘s eigenvector ${\vec{u}}_1$ by solving $(A-{\lambda }_{1}I)\vec{x}=0$ We find ${\lambda }_2$‘s eigenvector ${\vec{u}}_1$ by solving $(A-{\lambda }_{2}I)\vec{x}=0$ The general solution for $\vec{x}$ is

$$\vec{x}(t)=C_1\exp ({\lambda }_{1}t){\vec{u}}_1+C_2\exp ({\lambda }_{2}t){\vec{u}}_2\text{where }{C}_1,C_2\text{ are free parameters}$$

If an initial condition is given, then there is a unique solution.

Stability & Phase Portrait

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![[Pasted image 20250905173724.png|${\lambda }_1,{\lambda }_2<0⟹$Stable Nodal Sink, Asymptotically Stable]]

![[Pasted image 20250905173930.png|${\lambda }_1,{\lambda }_2>0⟹$ Nodal Source, Unstable]]

![[Pasted image 20250905180026.png|${\lambda }_1>0{\lambda }_2<0⟹$ Saddle, Unstable]]

A Zero Eigenvalue

This is a specific case of the previous section. This is the same as the previous section Distinct Real & Nonzero Eigenvalues, except We assume $\lambda ={\lambda }_1,{\lambda }_2,\text{WLOG }{\lambda }_1=0,{\lambda }_2\ne {\lambda }_1$, which implies

$$\vec{x}=C_1{\vec{u}}_1+C_2\exp ({\lambda }_{2}t){\vec{u}}_2$$

This is a simplification of the general formula from Distinct Real & Nonzero Eigenvalues On the eigenspace of the zero eigenvalue ${\lambda }_1$, the solutions are time independent; i.e., the eigenspace of the zero eigenvalue is a line of equilibrium. The other eigenvalue determines the behavior of the phase space outside of the equilibrium line.

Stability & Phase Portrait

Both cases have ${\lambda }_1=0$

Col

![[Pasted image 20250912160920.png|${\lambda }_2<0$, Attractive line of equilibirum, strictly stable|225]]

![[Pasted image 20250912160948.png|${\lambda }_2>0$, Repulsive line of equilibrium, unstable|200]]

Repeated Real Eigenvalues

This is the same as the previous sections except We assume $\lambda ={\lambda }_1,{\lambda }_2\in \mathbb{R},{\lambda }_1={\lambda }_2$

Easy Case

where $A={\lambda }_{1}I$

$$\vec{x}=\exp ({\lambda }_{1}t)\vec{C}$$

This is a simplification of the general formula from Distinct Real & Nonzero Eigenvalues, where $\vec{C}$ is formed by $C_1$ and $C_2$, and we chose the Standard Basis Vectors for our eigenvectors, because any linearly independent pair of vectors in ${\mathbb{R}}^2$ would be satisfactory.

Hard Case

where $A\ne {\lambda }_{1}I$

$$\vec{x}=C_1\exp ({\lambda }_{1}t)\vec{u}+C_2\exp ({\lambda }_{1}t)\left(\vec{v}+t\vec{u}\right)$$

We get the first term for this general solution from the typical method of finding eigenvectors, but because we have multiplicity of 2, our result is only one eigenspace, i.e. a partial solution. In order to get a full solution, we have to find a generalized eigenvector as well (the eigenvector for the second term). We get this eigenvector $\vec{v}$ by solving $(A-{\lambda }_{1}I)\vec{x}=\vec{u}$, where $\vec{u}$ was the eigenvector from the first term.

Stability and Phase Portrait

Complex Eigenvalues

This is the same as the previous sections except We assume $\lambda ={\lambda }_1,{\lambda }_2\in \mathbb{C},{\lambda }_1=\alpha +\beta i,{\lambda }_2=\alpha -\beta i,\beta \ne 0$

$$\text{Complex-valued: }\vec{x}=C_1\exp ({\lambda }_{1}t){\vec{u}}_1+C_2\exp ({\lambda }_{2}t){\vec{u}}_2$$ $$\text{Real-valued: }\vec{x}=C_1\mathrm{Re}(\exp ({\lambda }_{1}t){\vec{u}}_1)+C_2\mathrm{Im}(\exp ({\lambda }_{1}t){\vec{u}}_1)=C_1\exp (\alpha t)\left(\cos (\beta t)\vec{a}-\sin (\beta t)\vec{b}\right)+C_2\exp (\alpha t)\left(\sin (\beta t)\vec{a}+\cos (\beta t)\vec{b}\right)$$

where $\exp ({\lambda }_{1}t){\vec{u}}_1$ is a complex solution to the system where $\alpha =\mathrm{Re}{\lambda }_1$ gives the growth/decay rate where $\beta =\mathrm{Im}{\lambda }_1$ is the frequency of the oscillation of the system

When finding eigenvectors, for $A\in {\mathbb{C}}^{2\times 2}$ and $\det A\ne 0$, ignore the second row, because it can be eliminated by definition. See Misc. When you get $\lambda =\alpha \pm \beta i$, you can use either eigenvalue-eigenvector pair for the general equation. They will both give you the same solution, as the sign change will get absorbed in the $C$s

Stability and Phase Portrait

Apply matrix $A$ once, to get the initial velocity, to determine the direction of rotation $\exp \alpha t$ scales the whole solution