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A) It is possible for a wave equation to have as a solution a finite-duration function? Any closed-form example? (please share the specific wave equation with its finite-duration solution, showing how it is a solution - I want to know also How to work with a compact-supported function in more than one dimension).

B) I am specially interested in the classic electromagnetic wave equation $\nabla \vec{E}=\frac{1}{c^2}\frac{\partial^2}{\partial t^2} \vec{E}$, Could it admit compacted-supported solutions?

C) If the classic electromagnetic wave equation can´t sustained finite-duration solutions, Are there any non-linear versions that have compact-supported solutions?

I am specially interested in figure out if finite-duration functions that starts and or ends at a value different from zero could be a solution or not (that is why I am asking for a general finite-duration function). If not possible, also to know why It can´t, and what restrictions have to fulfill a finite-duration function to be an answer to a wave equation. Thinking in a laser pointer, I believe is reasonable to think that the solution function could have at least an ending point different to zero that jumps to zero, since they abruptly goes off, but I don´t know if it could be modeled by the wave equation.

I already know that there exist non-linear versions where Soliton Waves happen, which are highly localized waves, but the function that describes them is vanishing-at-infinity and not a proper finite-duration/compact-supported function (I believe Solitons waves are proportional to the square of a hyperbolic secant function).

Beforehand thanks you very much.

PS: compact-supported means here that there exists and starting time $t_0$ and a ending time $t_F$ such that the function is $f(t) = 0, \forall t<t_0$ and $f(t) = 0, \forall t>t_F$, so is of finite duration. If $f(t)$ is continuous and compact-supported, then also is bounded $\|f(t)\|_\infty < \infty$.

Joako
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    Sure, let $f(x)$ be your favorite compactly supported smooth function. Then $f(x\pm ct)$ satisfies the wave equation. – Ninad Munshi Dec 14 '21 at 23:41
  • @NinadMunshi Could you please choose one and show it that effectively fulfill the wave equation? I am really confused about it since displacements of the edges could: (1) not coincide on both sides of the equality, (2) if the value at the edges are non-zero some problems could rise on the derivatives, and (3) here I think is shown that no finite-duration function could stand the superposition principle... I am really lost with these finite-duration functions :( – Joako Dec 15 '21 at 00:41
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    It is a well known fact that every solution to the 1D wave equation is $f(x-ct)+g(x+ct)$ where $f,g$ are $C^2$ functions on $\Bbb{R}$ only, and no other restrictions. I don't have to show anything. a compactly supported smooth function is already $C^2$. – Ninad Munshi Dec 15 '21 at 00:56
  • @NinadMunshi mmm maybe we are using different assumptions... you are requiring that the solution is $C^2(\mathbb{R})$, but is nor hard to show that any finite-duration function $f(t)$ with $supp(f) = [t_0,t_F]$ that have $f(t_0)\neq 0$ and/or $f(t_F)\neq 0$ is not differentiable at the edges of the support $\partial t = {t_0,,t_F}$, so if I am right they are neither $\in C^2$... since I am asking for general finite-duration functions, I think you will see now is not so trivial the question if they can or not be the solution of wave equations (linear kind at least I think they are not). – Joako Dec 15 '21 at 01:16
  • If the solutions are compactly supported and smooth, they must also be continuous at every derivative........if you can talk about solitons I'm sure this line of reasoning is not outside of your grasp either. – Ninad Munshi Dec 15 '21 at 01:18
  • @NinadMunshi maybe you are right, but again, I am asking for general kind of finite-duration functions, not necessarily smooth, and so far, the only differential equation I found for a bump function is $g'(t)=2g(2t+1)-2g(2t-1)$ which solution is a complicated function (taken from here)... this is why I am trying to figure out how work two dimensional compact-supported functions under partial differential equations (I don´t know how to solve it). Hope you can share an non-zero starting function showing how it fulfill the wave equation. – Joako Dec 15 '21 at 02:27
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    Apologies, I misunderstood. By existence and uniqueness of the wave equation, a finite duration wave would share the same boundary conditions as the zero solution, so they cannot both be solutions to the wave equation. By finite duration I thought you meant at any specific location, not everywhere at once. – Ninad Munshi Dec 15 '21 at 03:27
  • @NinadMunshi Thanks, I see now is not possible for the classic form of the wave equation. The existence and uniqueness condition are also hold for non-linear versions of the wave equation? (as the model used in propagation of light on fiber optics).. or if the function is non-linear then it is possible to have traveling compact-supported solutions? – Joako Dec 15 '21 at 05:11
  • It seems like you are confusing two different issues. It is 100% possible to have traveling compactly supported solutions, because we don't think of the solutions of the wave equation occurring in $\Bbb{R}^4$, but rather in $\Bbb{R}^{3+1}$, and we have a series of "time slices" of $\Bbb{R}^3$. Compactly supported refers to the support in the spatial components only for any given timeslice. The machinery of foliations describes this phenomenon in rigorous detail for a general manifold, coordinate system, and foliation, but it is not necessary in this simple orthonormal, flat coordinate system. – Ninad Munshi Dec 15 '21 at 06:03
  • @NinadMunshi Thanks. I don´t have knowledge about manifolds or advanced analysis of differential eq. (nor differential geometry neither topology), but let me try to follow your answer: in the context of differential equations, the time variable is treated different from the spatial variables, where is allowed from the spatial ones to be compacted-supported but not in the time variable? If I fix every other variable dependency for each variable, selecting $x$ imply solving $\partial^2 f/\partial x^2 = c_1$, selecting $t$ imply solving $\partial^2 f/\partial t^2 = c_2$, Is not kind of the same? – Joako Dec 15 '21 at 13:01

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You are confusing compactly supported and finite duration - these do not mean the same thing in the context of PDEs that distinguish between time and spatial variables. Not many people would reasonably assume compactly supported in such a context would refer to the temporal variable. As discussed in the comments a globally finite duration solution violates existence and uniqueness. However, consider the following function $f:\Bbb{R}^3\to\Bbb{R}$

$$f(x,y,z) = \begin{cases}\exp\left[\frac{-1}{R^2-x^2-y^2-z^2}\right] & x^2+y^2+z^2 < R^2 \\ 0 & x^2+y^2+z^2 \geq R^2\end{cases}$$

Then for $k\in\Bbb{R}^3$ with $|k|=1$, we have that

$$E_i(x,y,z,t) = f(k_xx-ct,k_yy-ct, k_zz-ct)$$

satisfies the wave equation and in particular is compactly supported spatially for all times (this is a bubble of radius $R$ traveling in the $k$ direction). Below is an animation of the equivalent expression in 2D instead of 3D travelling in the $45^\circ$ direction

enter image description here

Ninad Munshi
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    Beautiful example. I would never have thought of this. – K.defaoite Dec 15 '21 at 09:59
  • thanks, for the detailed answer. I have learned through question here in SE what means that a function is compacted-supported, and at least for one-variable functions nobody tells that being of finite-duration is different from being compact-supported, so please explain where is my misconception: Why is different for the existence & uniqueness condition in time to have zeros except in a closed finite interval on the space variables? It just another variable from the perspective of the second derivatives which have to fulfill "similarly restricted" borders conditions, or It is not? – Joako Dec 15 '21 at 12:46
  • Is because the function treats time variable as a "parametrization" over the space variable? Are the "true" variables from the function point of view $\hat{x}(x,t) = x−c k_x t$, $\hat{y}(y,t) = y−c k_y t$, and $\hat{z}(z,t) = x−c k_z t$, so I have to be talking about them when speaking of the support of the function? (like being of compact-support means that $\hat{x} \neq 0$ only in some compact closed interval $[\hat{x}_0,,\hat{x}_F]$, as example) – Joako Dec 15 '21 at 13:14
  • I believe that finite-duration functions are going to be defined only through non-linear differential equations, so no finite-duration function could be the solution of the standard wave equation... this because of what I see in this paper ... hope you can comment about the veracity of this. – Joako Dec 25 '21 at 22:08
  • I have found this paper where the author explain that no scalar linear differential equation could have finite-duration solutions... but I don't know if it can be extended to ecs. of more variables, neither if it stills stand for the case $\mathbb{R}^{3+1}$ ... hope you can review it and comment... I think is quite interesting. – Joako Jan 13 '22 at 03:32
  • @NinadMunshi I have been doing a lot of research trying to understand what is going on (also made many question in the tag [tag:finite-duration], and now I believe I am understanding your answer better, specially your comment about "confusing compact-support with finite-duration". The only thing I don't get yet is if your field $E_i(t,x,y,z)$ is solving $\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)E=0$, Is that right? (I tried to do it by myself but I get lost with the derivatives at the "traveling" boundaries of the moving circle or radius $R$) – Joako May 02 '22 at 03:17
  • Now I have understood that there is a difference between the spatial coordinates and the time coordinate (don't knowing why yet), but so far, I can prove that the classical wave equation don't stand finite duration solutions because don't have a singular point in time required to break uniqueness of solutions... is really interesting that this is required by the time variable but not for the spatial coordinates to be compact-supported. I use your example and also a modification in this question where I show this difference. Hope you can see it – Joako Jun 02 '22 at 22:59
  • @NinadMunshi I cited this answer on this question, trying to figure out if a similar solution could arise from the Laplace's equation $\nabla^2 u = 0$... hope you can also visit it. – Joako Jul 04 '22 at 22:05
  • I have tried the example in $\mathbb{R^{1+1}}$ with $R=1$ and $c=1$ and it solves $E_{xx}=E_{tt}$,... – Joako Jul 06 '22 at 21:18
  • ... but with same assumptions if I try it in $\mathbb{R^{2+1}}$ with $k_x=k_y=1/\sqrt{2}$ so $|k|=1$, and it doesn't work for solve $E_{xx}+E_{yy}=E_{tt}$.... Are you sure it solves the vectorial cases of the wave eqn.? – Joako Jul 06 '22 at 21:20
  • @Joako To answer as many of your points in order as I can: $$$$ (1) Finite duration solutions may be possible for linear PDEs with infinite speed of propagation, but I am still unsure how that would play a role in their uniqueness/existence conditions. $$$$ (2) The reason spatial and temporal coordinates are treated different is because of the minus sign between the differential terms for each that treat each other differently. This affects things such as positive definitivity for the operators or even direction of motion for the variables themselves. – Ninad Munshi Jul 25 '22 at 20:39
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    (3) Vectorially you have to moderate the $f$'s so you get the correct number of copies after the derivative to cancel out. In this case I made an error - the $k$'s should fall to the $x,y,z$, not to the $t$. I have edited the original answer accordingly. – Ninad Munshi Jul 25 '22 at 20:45
  • @NinadMunshi I have tried also using the new formulation but it still don't work for two spatial coordinates: I am doing something wrong? – Joako Mar 29 '24 at 00:59
  • @Joako why are you factoring it? – Ninad Munshi Mar 29 '24 at 03:50
  • @NinadMunshi just because I found that Wolfram-Alpha shows faster that the function is solving the wave equation when factoring compared when it is not – Joako Mar 29 '24 at 04:19