Causality, Linearity, and Time Invariance for Systems Described by Linear Constant Coefficient Differential Equations

Question

I am currently using Signals and Systems by Alan Oppenheim as a reference to learn about LTI systems.

Before introducing systems represented by linear constant coefficient differential equations, the author first considers general systems of the form $x(t) \rightarrow y(t)$. When discussing linearity of systems, he introduces the "zero in-zero out" property of linear systems (not necessarily time invariant or causal), that is, given a linear system, if the input $x(t)$ is zero for all $t$, then the output $y(t)$ must be zero for all $t$. In addition, a linear system is causal if and only if $y(t) = 0$ for all $t < t_0$ if $x(t) = 0$ for all $t < t_0$, a condition known as initial rest, from which I concluded that initial rest for linear systems implies causality but not time invariance.

Now given a linear constant coefficient differential equation $$Ly(t) = x(t),$$ where $L$ is a linear differential operator with constant coefficients, the author emphasizes that although the differential equation is linear, the system that it represents may or may not be linear depending on the initial conditions. I find this confusing because given two input output pairs $x_1(t) \rightarrow y_1(t)$ and $x_2(t) \rightarrow y_2(t)$, we have $L(ay_1(t) + by_2(t)) = ax_1(t) + bx_2(t)$ for any constants $a, b \in \mathbb C$ by virtue of the linearity of the operator $L$. It seems that $ax_1(t) + bx_2(t) \rightarrow ay_1(t) + by_2(t)$ regardless of the initial conditions, why is the system not necessarily linear in that case? What am I missing here?

In addition, the author states that for the system to be linear, the initial conditions have to be zero, and for it to be causal and time invariant in addition to being linear, its initial conditions must obey initial rest. But isn't initial rest in general a condition required for causality, not time invariance? Besides, doesn't this imply that a differential equation can only represent a causal system if the input is zero before some time? Given an input that is nonzero for all time, how do we impose causality via initial conditions that correspond to initial rest? I hope my question makes sense. Thank you in advance.

Answer 1

Note that the system is not uniquely characterized by the linear differential operator. You need auxiliary conditions, and dependent on these conditions, the system may be linear, causal and time-invariant or not. So if you just use the operator $L$ to show that the system is linear then you're actually not showing anything because $L$ is a linear operator by definition.

The output of a system described by a linear constant coefficient differential equation can be split up into two contributions: the zero-state response (ZSR) and the zero-input response (ZIR). The ZSR is the response of the system with zero initial conditions, and, consequently, the ZSR is fully determined by the input signal. The ZIR response is the response of the system caused by non-zero initial conditions, and it is independent of the input signal. Since non-zero initial conditions cause an output term that does not depend on the input signal, scaling or shifting the input signal will not affect that term. Consequently, a system with non-zero initial conditions can neither be linear nor time-invariant, because if it were, the output should scale and shift according to the scaling and shifting of the input signal.

Causality simply means that the system's response at a given time does not depend on the future of the input signal. Clearly, causality is a property of the system, so it cannot depend on the input signal. A causal linear time-invariant (LTI) system is characterized by an impulse response $h(t)$ that satisfies $h(t)=0$ for $t<0$, and it is irrelevant whether the input signal is zero before a certain time or not.

Also take a look at this related answer.