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 $\|f(t)\|_\infty < \infty$.

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    $\begingroup$ Sure, let $f(x)$ be your favorite compactly supported smooth function. Then $f(x\pm ct)$ satisfies the wave equation. $\endgroup$ Dec 14, 2021 at 23:41
  • $\begingroup$ @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 :( $\endgroup$
    – Joako
    Dec 15, 2021 at 0:41
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    $\begingroup$ 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 $C^2$. $\endgroup$ Dec 15, 2021 at 0:56
  • $\begingroup$ @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). $\endgroup$
    – Joako
    Dec 15, 2021 at 1:16
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    $\begingroup$ 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. $\endgroup$ Dec 15, 2021 at 3:27

1 Answer 1


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

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    $\begingroup$ Beautiful example. I would never have thought of this. $\endgroup$
    – K.defaoite
    Dec 15, 2021 at 9:59
  • $\begingroup$ 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? $\endgroup$
    – Joako
    Dec 15, 2021 at 12:46
  • $\begingroup$ 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) $\endgroup$
    – Joako
    Dec 15, 2021 at 13:14
  • $\begingroup$ 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. $\endgroup$
    – Joako
    Dec 25, 2021 at 22:08
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    $\begingroup$ (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. $\endgroup$ Jul 25, 2022 at 20:45

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