Classify critical point of linear system

Question

For this linear system:

$\dfrac{dx}{dt}=x+y-2$

$\dfrac{dy}{dt}=x-y-4$

I've found the critical point to be $(1,1)$ but now I want to classify it. How do I do it?

Answer 1

$$\frac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} = \dfrac{x-y-4}{x+y-2}$$

At $(1, 1),\; \dfrac{dy}{dx}\neq 0$. Indeed, at $(1, 1)$, $\dfrac{dy}{dx}$ is not defined.

$\dfrac{dy}{dx} = 0$ when $x - y - 4 = 0$.

But this doesn't tell us too much: only that $x- y = 4$. (Two variables and one equation. Infinitely many solutions.)

On the other hand, if you find the Jacobian and evaluate the eigenvalues, you'll find that your critical point is $(3, -1)$, and is a saddle point.

Answer 2

I believe you will get an specific answer to your question but regardless this, it is very important to know some more general-theory which can be applied in more general cases.

You can use the eigenvalue-method to investigate the critical point.

Assume you have the following:

$$\begin{bmatrix} x'\\ y' \end{bmatrix} = \begin{bmatrix} a & b\\ c & d \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}$$ With constant-coefficient matrix $A$. Then the $\lambda_1, \lambda_2$ which are the eigenvalues of the matrix $A$ are solutions of the characteristic equation: $$(a - \lambda)(d-\lambda)-bc = 0$$ Then: If the eigenvalues of $A$ are:

Real, unequal, same sign $\rightarrow$ Improper node

Real, unequal, opposite sign $\rightarrow$ Saddle point

Real and equal $\rightarrow$ Proper or Improper node

Complex conjugate $\rightarrow$ Spiral point

Pure imaginary $\rightarrow$ Center