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It is possible to (continuous-time) finite-duration continuous systems to be linear?

I am trying to understand which effects are introduced in continuous-time systems described by continuous functions just for defined them as finite-duration systems, meaning with this, they are time limited.

Let define an arbitrary finite-duration function as: $$f(t) = \begin{cases} 0,\, t<t_0 \\ x(t),\, t_0\leq t \leq t_F \\ 0,\, t> t_F \end{cases}$$ with $x(t)$ and arbitrary continuous function. Note that $f(t)$ is compact-supported, and since is continuous within its domain $[t_0,\,t_F]$, it is also bounded $\|f\|_\infty < \infty$.

If I want to this function $f(t)$ to be linear, it has to fulfill that for any variables $x$ and $y$, and arbitrary constant $a\in\mathbb{R}$:

  • $f(x+y)=f(x)+f(y)$
  • $f(ax) = af(x)$

Now, I will do an analysis that I don´t know if is right, so please follow it and explain me with detail where I am making wrong assumptions:

  1. Without loss of generality I can say that $\lim_{t\to t_0^+} f(t) = \lim_{t\to t_0} x(t) = f(t)\big|_{t=t_0} = f(t_0)$ (I am actually defining it), and let $f(t_0) \neq 0$.
  2. Now, without full generality (but I think is general enough), let set the constants $a$, $t_0$, and $t_F$ such that they fulfill $0 < a < t_0 < t_F$, so the constants fulfill $0 < t_0 - a < t_0$.
  3. Now, I can use the linearity properties by adding $0$ to the argument: $$ f(t_0) = f(t_0 + a - a) = f(t_0 - a)+f(a) = 0$$ since $f(a)$ and $f(t-a)$ lives in the part where the piece-wise function $f(t)$ is defined to be zero since it has not started yet to have non-zero values, contradicting that the arbitrary function $x(t)$ could have non zero values (at least a $t_0$).

I think that the same analysis can be done for every time $t \in [t_0,\,t_F]$ by introducing other on-purpose-defined constants, meaning that the only finite-duration linear function is the zero function.

so,

A) It is true that no finite-duration continuous-time phenomena can be defined through a continuous linear system??

  1. not by a linear function.
  2. neither through homogeneous or inhomogeneous ordinary or partial linear differential equations.
  3. neither as the output of continuous-time LTI systems defined as a convolution of an input and a impulse response function (because LTI means LINEAR and Time-Invariant system).

Please, if you can, answer with counterexamples since my math knowledge is limited, and also comment is possible for each numbered point separately.

B) Does the same holds for discrete-time systems?

Added later

I believe I can generalize my analysis through this: lets define $\Delta T = t_F - t_0 > 0$ the length of the support of the finite duration function, and lets choose a constant $b \geq 0$ such that the time $t = t_0 + b$ is contained into the support of $f(t)$, so $t_0 \leq t_0 + b \leq t_F$.

Then, for any arbitrary point within the support of $f(t)$ I can state that if $f(t)$ fulfill the additive property of a linear operator $f(x+y) = f(x)+f(y)$, then by adding zero it will becomes: $$\begin{array}{r c l} f(t) & = & f(t_0+b) \\ & = & f(t_0 +b +\Delta T - \Delta T) \\ & = & f(t_0 + b + t_F - t_0)-f(\Delta T) \\ & = & f(t_F+b) - f(t_F) + f(t_0) \\ & = & f(t_0) - f(t_F) \end{array}$$ because $f(t_F + b) = 0$ since $t_F+b$ is outside the support for any $b>0$, so if the operator has the additive property, every point within the support must be equal to a constant $f(t) = f(t_0)-f(t_F)$ (or $f(t) = f(t_0)$ if $b=0$), so the only operator $f(t)$ will be a constant function, not an arbitrary function as is intended. So no compact-support/finite-duration non-constant function could be described through an additive operator....

Since derivatives are additive,

C) Is this meaning that no finite-duration function could be described with differential equations? or just means that the differential equation will be non-linear? Or maybe it will be described by differential equation in piecewise domains?

Joako
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    The first condition gives linearity on all rational grids, the second leads directly to $f(x)=xf(1)$, both are incompatible with a compact support for anything but the zero function. – Lutz Lehmann Dec 12 '21 at 23:44
  • @LutzLehmann sorry but I don´t fully understand your answer since I don´t know what a "rational grid" means (I have been learning through questions here in SE what a compact-support is - I have only the formal training in math of an electrician), and also I don´t catch what you want so said with $f(x)=xf(1)$ (I believe that the property is $f(1\cdot x) = 1\cdot f(x)$)... so please extend your explanation on how what you mention makes impossible for a linear system to describe a finite-duration function. Thanks beforehand. – Joako Dec 13 '21 at 02:54
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    If you have additivity $f(x+y)=f(x)+f(y)$, then for any fixed real number $r$ you get $f(\frac{m}{n}r)=\frac{m}{n}f(r)$. With continuity the slopes $f(r)/r$ have all to be the same. The second property also implies $f(ax)=xf(a)$. – Lutz Lehmann Dec 13 '21 at 05:50
  • @LutzLehmann but $f(ax)=xf(a)$ with variable "x" and constant "a" (so $f(a)$ is also a constant): Does it will implying that the only possible 1-D linear map is the linear function $f(x)=mx$? Is that so?.... also, I don´t directly understand How the linear properties are incompatible with the compact-support idea... Is the analysis I present right? – Joako Dec 13 '21 at 15:18
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    If $m\ne 0$, then every value, apart from at zero, is non-zero. The support is the whole of $\Bbb R$, which is not compact. – Lutz Lehmann Dec 13 '21 at 15:36
  • @LutzLehmann Thanks, I understand that if $f(t)$ is of finite duration, $f(t)$ is necessarily non-linear. I add an analysis more general, but also a question, since I don´t really understand what this fully means for operators as the derivatives. – Joako Dec 13 '21 at 15:44
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    I'm still not sure what you are trying to achieve, my previous comments were on making sure that some technical termini you used were used in the common sense, or if they would require special definitions. // Any autonomous or time-invariant ODE system $x'=F(x)$ can not have solutions with non-trivial compact support. Because outside that support, you would get initial conditions $x(t_*)=0$ with additionally $F(0)=0$, which implies, under the usual uniqueness assumptions, that the only solution is $x=0$ , the constant zero function. – Lutz Lehmann Dec 13 '21 at 15:51
  • @LutzLehmann I am trying to understand this finite-duration properties: every function I see as engineer where analytical, and I have see this last months that no non-zero analytical function could be compact supported, neither the solutions of ODE, so everything I know are just approximations, which really shock me... that is why I am trying to figure out what will affect to consider that function are time-limited, as "naive" physical systems should be (note here as example, that quantum physics Schrödinge eq. is a linear differential eq, so, with unbounded domain solutions in time and space) – Joako Dec 13 '21 at 16:00
  • @LutzLehmann (...), so I am trying to understand how finite-duration systems are model: Could be being described through non-linear differential equations? How this finite-duration equation will look alike?... I have recently asked in the signal forums if continuous time LTI systems (Linear and Time Invariant) could be finite-duration, and nobody really know that is impossible (as you have answer me now), so, since LTI system are the basis of Signals theory, maybe this obvious question for mathematicians, is of widespread ignorance on engineers community – Joako Dec 13 '21 at 16:09
  • @LutzLehmann Hope you can answer letter C as an answer to give you the points... thanks again. – Joako Dec 13 '21 at 16:11
  • @LutzLehmann as example, $y(t) = \sqrt{e^{(1-t)/t}-1}$ lives on $\mathbb{R}$ only in the domain $(0,,1]$, and has differential equation $y''/y' +y'/y+(2t+1)/t^2 = 0$... Is this s an example of a finite duration differential equation? or since it has complex values outside its $(0,,1]$ is not compact-supported?.. at least for me, is not at all trivial to figure out How finite-duration functions' non-linear differential equations will be (if they exist)? It will be restricted only into the compact support? like $f(t)$ follows $f'+f -c= 0,, |t| \leq 1$, and $f(t) = 0$ otherwise, as example – Joako Dec 13 '21 at 16:26
  • @LutzLehmann as another example, in here is said that exist a "bump function" $\in C_c^\infty$ (so is compact-supported and smooth), that fulfill the following differential equation $y'(t) =2\cdot y(2t+1)-2\cdot y(2t-1)$ for $\forall,t\in\mathbb{R}$, but unfortunately it doesn´t show the close form of its solution $y(t)$ with $\mathrm{supp}(y) = [-1,,1]$ (if there is any) – Joako Dec 13 '21 at 19:17

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I have found the following paper named "Finite time differential equations" by V. T. Haimo (1985), where continuous time differential equations with finite-duration solutions are studied, an it is stated the following:

"One notices immediately that finite time differential equations cannot be Lipschitz at the origin. As all solutions reach zero in finite time, there is non-uniqueness of solutions through zero in backwards time. This, of course, violates the uniqueness condition for solutions of Lipschitz differential equations."

Since, linear differential equations have solutions that are unique, and finite-duration solutions aren´t, finite-duration phenomena models must be non-linear to show the required behavior (non meaning this, that every non-linear dynamic system support finite-duration solutions).

The paper also show which conditions must fulfill the non-linear differential equation to support finite-duration solutions, at least for first and second order scalar ODEs.

Joako
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