The system of differential equations is in steady state

Question

We have a system of non-homogeneous differential equations $$X'=AX+B$$

What does it mean that the system is in steady state??

$X$ is the vector $\begin{pmatrix} x_1(t) \\ x_2(t) \\ x_3(t)\end{pmatrix}$, $A$ is a $3 \times 3$ matrix and $B$ is a $3 \times 1$ matrix.

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Answer 1

Consider the solution to the univariate differential equation $$x'(t)=-ax(t)$$ for $a>0$, which is $x(t)=x_0e^{-at}$. As $t$ grows large, $x(t) \to 0$. Then we call $0$ the steady state of $x(t)$. This is a sort of fixed point: if you begin at the steady state (if it exists), then your solution does not change in time, i.e. $x'(t)=0$. (For this example, that would mean $x_0=0$)

(It's essentially the same for higher spatial dimensions.)

Answer 2

the non homogenous equation $$x' = Ax + b \tag 1$$ sometimes has a constant solution $x = x_\infty.$ by the very definition, it must satisfy $$Ax_\infty + b = 0 \tag 2 $$ the equation $(2)$ may or may not have a solution. if all the eigenvalues of the matrix $A$ has negative real part and $b$ does not grow badly, then the solutions of $(1)$ approach $x_\infty$ as $t\to \infty.$

if system is ever in steady state it means $$x(t) = x_\infty , \, \forall t. $$

Answer 3

You will ease the task by using an auxiliary variable that absorbs the steady-state.

$$X'(t)=AX(t)+B=A(X(t)+A^{-1}B),$$ so that with $Y(t)=X(t)+A^{-1}B$, $$Y'(t)=AY(t),$$ and the steady-state condition is just $Y(\infty)=0$, i.e. $X(\infty)=-A^{-1}B$.

Now the solution of the homogenous equation is

$$Y(t)=e^{At}Y_0=Pe^{Dt}P^{-1}Y_0,$$ where $P$ is a matrix of Eigenvectors of $A$ and $D$ a diagonal matrix of Eigenvalues.

As in this case the Eigenvalues are real negative, the solution can be expressed as the sum of three negative exponentials, $$Y(t)=e^{d_0t}U+e^{d_1t}V+e^{d_2t}W,$$

and it can be approximated for large $t$ as the term with the slowest decay*,

$$Y(t)\approx e^{d_2t}U.$$

The system "reaches" the steady state when, say only $1\%$ of the transitory signal remains, i.e. when

$$d_2t\approx\ln(0.01).$$

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*Unless the initial conditions are such that this term is missing or very small.