Finding equilibrium points of differential equation

Question

Given the system $$ x'=xe^{y-3}$$ $$\space \space \space \space \space \space \space \space \space \space \space \space \space \space \space y'=2\sin(x)+3-y$$

Find the equilibrium points and decide if they are stable.

I know that the equilibrium points of the system are the solutions $X$ such that $X'=0$, which means $x'=0=y'$. By this condition, I get the points $X=(x,y) \in \{\{(0,3),(k\pi,3)\},k \in \mathbb Z\}$.

I don't know how to decide if these points are stable, I would appreciate suggestions to do that part and if someone could tell me if what I've done to find the equilibrium points is correct.

Answer 1

I like $\dot x$ etc. better than $x'$ etc., so I'll use the "dot" notation for time derivatives. This being said, the given equations become

$\dot x = xe^{y - 3} \tag{1}$

and

$\dot y = 2\sin x + 3 - y. \tag{2}$

At an equilibrium point, we have $\dot x = \dot y = 0$, so

$xe^{y - 3} = 0 \tag{3}$

and

$2\sin x + 3 - y = 0. \tag{4}$

Since $e^{y - 3} \ne 0$ for all $y \in \Bbb R$, (3) yields $x = 0$; then by (4), $y = 3$; there are no other equilibria. $(x, y) = (3, k\pi)$, $k \in \Bbb Z$, is not an equilibrium as direct substitution of these values into (1)-(4) will reveal.

The Jacobian $J(x, y)$ of the system (1)-(2) at the point $(x, y)$ is given by the matrix

$J(x, y) = \begin{bmatrix} \dfrac{\partial {\dot x}}{\partial x} & \dfrac{\partial {\dot x}}{\partial y} \\ \dfrac{\partial {\dot y}}{\partial x} & \dfrac{\partial {\dot y}}{\partial y} \end{bmatrix} = \begin{bmatrix} e^{y - 3} & xe^{y - 3} \\ 2\cos x & -1 \end{bmatrix}; \tag{5}$

at $(0, 3)$ we have

$J(0, 3) = \begin{bmatrix} 1 & 0 \\ 2 & -1 \end{bmatrix}. \tag{6}$

It is easy to see that the characteristic polynomial of the matrix $J(0, 3)$ is $\lambda^2 -1$; the eigenvalues are thus $\pm 1$, as may also easily be seen by inspection of the triangular matrix $J(0, 3)$. Since $J(0, 3)$ has one positive and one negative eigenvalue, the point $(0, 3)$ is a saddle; hence, it is an unstable equilibrium.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 2

I don't know where you're getting the $(3,k\pi)$. You want $x e^{y-3}=0$, which says $x=0$, and $2\sin(x) + 3 - y = 0$, which when $x=0$ says $y=3$.

For stability, you want to find the Jacobian matrix and evaluate it at the equilibrium point, then see what are the eigenvalues.