Why do they say that complex exponentials are eigenfunctions of LTI systems, when there are still transient responses?

Question

Let $$\dot{x} = Ax+Bu$$ $$y = Cx + Du$$ be a linear ODE with $x(0)=0$. Here, I am assume $A$ is invertible.

As you can see, the relation $$H:u(.) \mapsto y(.),$$ where $(u(.),y(.))$ is a solution to the ODE above, forms an LTI system. If I set $u(t) = e^{st}$, for some $s \in \mathbb{C} - \text{spec}(A)$, then the output is $$y(t) = -Ce^{At}(s I - A)^{-1}B) + (C(s I - A)^{-1} B + D)e^{st}.$$

Now, my understanding of eigenfunctions is that if $f(.) \not = 0$ satisfies $Hf = \lambda f$, for some $\lambda \in \mathbb{C}$, then $f$ is an eigenfunction. This is clearly not the case for $u(t) = e^{st}$. Could someone clear up my mis-understanding?

Answer 1

Note that you only get a transient if you switch the system (or the input) on at a certain finite point in time. If the input $x(t)=e^{s_0t}$ has existed forever, the output is given by the convolution integral

\begin{align*} y(t)&=\int_{-\infty}^{\infty}h(\tau)e^{s_0(t-\tau)}d\tau\\&=e^{s_0t}\int_{-\infty}^{\infty}h(\tau)e^{-s_0\tau}d\tau\\&=e^{s_0t}H(s_0)\tag{1} \end{align*}

where $h(t)$ is the system's impulse response, and $H(s)=\mathcal{L}\{h(t)\}$ is the transfer function. Of course, in $(1)$ we've assumed that $s_0$ lies in the region of convergence of $H(s)$. From $(1)$ we can see that $e^{s_0t}$ is an eigenfunction with eigenvalue $H(s_0)$.