# It is possible for a (electromagnetic) wave equation to have as a solution a finite-duration/compact-supported function? Any closed-form examples?

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 $$f(t) = 0, \forall t and $$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$$.

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

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

• Beautiful example. I would never have thought of this. Dec 15, 2021 at 9:59
• 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? Dec 15, 2021 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) Dec 15, 2021 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. Dec 25, 2021 at 22:08
• (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. Jul 25, 2022 at 20:45