# Nonhomogeneous linear dynamic system exercise

I have the following system of equations:

\begin{align*} x_{t+1} & = 3x_t + y_t \\ \\ y_{t+1} & = 2 + 5y_t. \end{align*}

I know how to solve it when I do not have a constant, but I couldn't write it in the matrical form. What is the method for this kind of system to find the long-term equilibrium of this system? And how can I comment on whether equilibrium is unique or there are multiple equilibria? Any help is appreciated in advance.

Preamble

1. This answer does not provide a step-by-step solution to the above exercise. It shows how if you can solve a linear dynamic system without the constants (a homogeneous linear dynamic system) then generally you can solve the one with the constants as well.

2. I will use the $$\triangleq$$ symbol to define things. I will also write the discrete time dynamic equations in their 'difference' form, meaning instead of $$x_{t+1} = f(x_t)$$ I will write $$\Delta x_t = g(x_t),$$ where $$\Delta x_t \triangleq x_{t+1} - x_t.$$

3. I will use a somewhat unconvential matrix notation, I will simply use capital letters for matrices, I will not bold and nonitalicize them. Similarly I will simply underline vector variables.

Matrix form

Suppose we have a dynamic system that is linear but non-homogeneous (the origin point is not an equilibrium): \begin{align*} \Delta x_t & = a_{11} x_t + a_{12} y_t + b_1 \\ \Delta y_t & = a_{21} x_t + a_{22} y_t + b_2. \end{align*} If we define the (vector) variable $$\underline{x}_t$$ for all $$t$$ time periods as \begin{align*} \underline{x}_t & \triangleq \begin{bmatrix}{x_t \\ y_t} \end{bmatrix} \end{align*} we can write the above dynamic system in matrix form: \begin{align*} \Delta \underline{x}_t & = A \underline{x}_t + \underline{b}, \end{align*}

Shifting/homogenizing the system

It is possible to 'homogenize' the above linear equation system, to redefine the variables in such a way that $$\underline{b}$$ disappears.

For simplicity's sake let us assume that $$A$$ is invertable. Let us define a new (vector) variable $$\underline{z}_t$$ for all $$t$$ time periods as \begin{align*} \underline{z}_t & \triangleq \underline{x}_t + A^{-1}\underline{b}. \end{align*} For $$t+1$$ this is \begin{align*} \underline{z}_{t+1} & = \underline{x}_{t+1} + A^{-1}\underline{b}. \end{align*} Taking the difference of the two equations above: \begin{align*} \underline{z}_{t+1} - \underline{z}_t & = (\underline{x}_{t+1} + A^{-1}\underline{b}) - (\underline{x}_t + A^{-1}\underline{b}) \\ \\ \Delta \underline{z}_t & = \Delta \underline{x}_t \end{align*} The change in $$\underline{z}_t$$ is the same as the change in $$\underline{x}_t$$; by creating $$\underline{z}$$ we merely a shifted $$\underline{x}$$. Thus if we can find equilibrium values for $$\underline{z}$$ we will also find equilibrium values for $$\underline{x}$$. Let us first get the dynamic system for $$\underline{z}$$:
\begin{align*} \Delta \underline{z}_t & = A \underline{x}_t + \underline{b} \\ \\ \Delta \underline{z}_t & = A(\underline{x}_t + A^{-1}\underline{b}) = A \underline{z}_t. \end{align*} This is a homogeneous linear dynamic system. In equilibrium there is no change in the varible, so \begin{align*} \Delta \underline{z}^* & = \underline{0} \\ \\ A \underline{z}^* & = \underline{0} \\ \\ \underline{z}^* & = A^{-1}\underline{0} = \underline{0}. \end{align*} We have found the only equilibrium. (We relied heavily on $$A$$ being invertible.) The corresponding $$\underline{x}_t$$ value is \begin{align*} \underline{x}^* & = \underline{z}^* - A^{-1}\underline{b} \\ & = 0 - A^{-1}\underline{b}. \end{align*}

Stability

We have not discussed the stability of the equilibrium in either system. The shift operation we performed above does not change distance, meaning for an any two $$\underline{x}$$ states and their corresponding $$\underline{z}$$ states, the distance is the same between the two state pairs. Specifically: $$d(\underline{x}^*,\underline{x}_t) = d(\underline{z}^*,\underline{z}_t).$$ A 2D illustration of the concept:

Since the definitions of stability (global, local, asymptotic, etc.) are all dependent on how a trajectory's distance from equilibrium evolves, and these are the same in the two systems, the equilibria's stability will be the same as well.

• Thank you so much; it is very helpful. I really appreciate it. Dec 18, 2023 at 14:23

Hi: Using the lag operator, L, for $$y_t$$, you can write $$y_{t+1}(1- 5L) = 2 \longrightarrow y_{t+1} = \frac{2}{(1-5L)}$$. But you cannot write that an an infinite series because the thing multiplying the L is greater than 1. So, this implies that there is no equilibrium for $$y_t$$. In the discrete time series literature, it's referred to as an explosive process. Think of the model for $$y_{t}$$ as an AR(1) where the lagged coefficient, 5, implies the non-stationarity of $$y_{t}$$.

Also, if you are not familiar with discrete time series, check out Box-Jenkins or Hamilton. Box Jenkins is better for an introduction.

Since, $$y_t$$ does not have an equilibrium, this implies that $$x_t$$ does not have one either.

• Hi! Is $y=-1/2$ not an equilibrium for $y$? Dec 17, 2023 at 14:30
• @Giskard: It's an interesting question. Your solution satisfies the second equation but equilibrium obviously means "does it always get to that point from any starting point". I come at it from a time-series perspective ( I'm not an economist AT ALL ) so, from an econ perspective, maybe that is an equilibrium ? I don't feel confident enough to say. In stat time-series world, one would definitely say that that model for $y_t$ is non-stationary because the long term mean doesn't exist because the process explodes to infinity. Dec 18, 2023 at 6:00