For an open set $U \subseteq \mathbb{R}^4$, if $f:U \to \mathbb{R}$ is a "good" (for example, smooth) function, is there a solution to the following equation?

$$\left( \Delta - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\chi(x, y, z, t)=f(x, y, z, t)$$


I want to transform Maxwell's equations

$$\operatorname{rot}E(x,t)+\frac{\partial B(x, t)}{\partial t}=0$$ $$\operatorname{div}B(x,t)=0$$ $$\operatorname{rot}H(x,t)-\frac{\partial D(x,t)}{\partial t}=i(x,t)$$ $$\operatorname{div}D(x,t)=\rho(x,t)$$

into the following form with the electrical potential $\phi$ and the vector potential $A$:

$$B(x,t)=\operatorname{rot}A_L(x,t)$$ $$E(x,t)=-\frac{\partial A_L(x,t)}{\partial t} -\operatorname{grad}\phi_L(x,t)$$ $$\square A_L(x,t)=-\mu_0i(x,t)$$ $$\square \phi_L(x,t) = -\frac{1}{\epsilon_0}\rho(x,t)$$ $$\operatorname{div}A_L(x, t)+\frac{1}{c^2}\frac{\partial\phi_L(x,t)}{\partial t}=0$$

In order to do this, we need the existence of a solution to the equation

$$\square\chi = -\left(\operatorname{div}A_0 + \frac{1}{c^2}\frac{\partial \phi_0}{\partial t}\right)$$

where $A_0$ and $\phi_0$ is a special solution to the following equations:

$$B(x,t)=\operatorname{rot}A(x,t)$$ $$E(x,t)=-\frac{\partial A(x,t)}{\partial t} -\operatorname{grad}\phi(x,t)$$ $$\operatorname{grad}\left( \operatorname{div}A(x,t)+\frac{1}{c^2}\frac{\partial \phi(x,t)}{\partial t}\right) + \left( \frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\Delta\right)A(x,t)=\mu_0 i(x,t)$$ $$-\operatorname{div}\left(\frac{\partial A(x,t)}{\partial t}\right) - \Delta \phi(x,t)=\frac{\rho(x,t)}{\epsilon_0}$$

If it exists, $A_L$ and $\phi_L$ are defined as follows:

$$A_L = A_0 + \operatorname{grad}\chi$$ $$\phi_L = \phi_0 - \frac{\partial}{\partial t}\chi$$


The answer to the question if whether a solution $\chi$ to the the following equation exists $$ -\frac{1}{c^{2}}\square=\left(\Delta - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\chi=f\;\text{ in }\;\Bbb R^4\equiv \Bbb R^3\times \Bbb R \label{w}\tag{W} $$ under mild smoothness requirements on the datum $f$ is yes: I explain below why it is so in a constructive way, by actually constructing an explicit solution in two steps:

  1. Construction of a fundamental solution: what is needed is a slightly modified fundamental solution of the D'Alembert operator, precisely the solution of the following equation: $$ \square \mathscr{E}(x,t)=-c^2\delta(x,t)\label{da}\tag{DA} $$ where $\delta(x,t)\equiv \delta(x)\times\delta(t)$ is the usual tensor product of Dirac measures respectively on the spatial and on the time domain. Once $\mathscr{E}(x,t)$ has been determined, we can find, provided certain compatiility conditions on $f$ are fulfilled (see below), a distributional solution $\chi(x,t)$ to the posed problem by convolution $$ \chi(x,t)=\mathscr{E}\ast f(x,t)\label{s}\tag{S} $$ The minimal requirements on $f$ is that the convolution product at the right term of \eqref{s} should exists as a distribution.

  2. The regularity problem: prove that, provided $f$ is a "good" (for example $C^2$ smooth) function, the distribution $\chi$ in \eqref{s} is a "good" function in the same way.

Calculation of the modified fundamental solution for the D'Alembert operator in $\Bbb R^{3+1}$

We construct $\mathscr{E}$ as a distribution of slow growth (i.e. $\mathscr{E}\in \mathscr{S}^\prime$, see for example [1] §8.1-§8.2, pp. 113-116 or [2], §5.1-§5.2, pp. 74-78) by applying to PDE \eqref{da} the Fourier transform $\mathscr{F}_{x\to\xi}$ respect to the spatial variable $x$. By proceeding this way, \eqref{da} is transformed into the following ODE: $$ \frac{\partial^2 \hat{\mathscr{E}}(\xi,t)}{\partial t^2} + c^2|\xi|^2\hat{\mathscr{E}}(\xi,t)=-c^2\delta(t)\label{1}\tag{1} $$ Consider its equivalent standard form $$ \frac{\partial^2 \hat{\mathscr{E}}_p(\xi,t)}{\partial t^2} + c^2|\xi|^2\hat{\mathscr{E}}_p(\xi,t)=\delta(t)\label{1'}\tag{1'} $$ which has the same solutions, just multiplied by the constant $-c^2$: by solving it (see here, [1] §10.5, p. 147 or [2], §4.9, example 4.9.6 pp. 77-74 and §15.4, example 15.4.4) we get the following distribution $$ \hat{\mathscr{E}}_p(\xi,t)= H(t)\frac{\sin c|\xi|t}{c|\xi|}\iff\hat{\mathscr{E}}(\xi,t)= -cH(t)\frac{\sin c|\xi|t}{|\xi|}\label{2}\tag{2} $$ where $H(t)$ is the usual Heaviside function. Then, taking the inverse Fourier transform $\mathscr{F}_{\xi\to x}^{-1}\big(\hat{\mathscr{E}}\big)$ we get the sought for solution of \eqref{da} (see [1] §9.8, p. 135 and §10.7, p. 149) $$ \mathscr{E}(x,t)=-\frac{H(t)}{4\pi t}\delta_{S_{ct}}(x)=-c\frac{H(t)}{2\pi }\delta\big(c^2t^2-|x|^2\big)\label{3}\tag{3} $$ where

  • $S_{ct}=\{x\in\Bbb R^3 | |x|^2=x_1^2+x_2^2+x_3^2=c^2t^2\}$ is the spherical light wave surface,
  • $\delta_{S_{ct}}(x)$ is the Dirac measure supported on $S_{ct}$, otherwise called single layer measure.

Now, given any distribution $f\in\mathscr{D}(\Bbb R^{3+1})$ for which the convolution with $\mathscr{E}$ exists (for example any distribution of compact support) using \eqref{3} in formula \eqref{s} gives a generalized solution of \eqref{w}.

Construction of a regular solution

Instead of recurring to the standard (and complex) methods of regularity theory we will try a trickier way by looking carefully at the structure of \eqref{3} and on how this distribution acts on the space of infinitely smooth rapidly decreasing functions: precisely, given $\varphi\in\mathscr{S}$ we have that $$ \begin{split} \langle\mathscr{E},\varphi\rangle&=-\frac{1}{4\pi}\int\limits_{0}^{+\infty}\langle\delta_{S_{ct}},\varphi\rangle\frac{\mathrm{d}t}{t}\\ &=-\frac{1}{4\pi}\int\limits_{0}^{+\infty}\frac{1}{t}\int\limits_{S_{ct}}\varphi(x,t)\,\mathrm{d}\sigma_x\mathrm{d}t \end{split}\label{4}\tag{4} $$ From \eqref{4} we see that $\mathscr{E}$ acts on $\varphi\in\mathscr{S}$ as a spherical mean respect to the spatial $x\in \Bbb R^3$ variable and as a weighted time integral mean with weight function $t\mapsto {1\over t}\in L^1_\mathrm{loc}$ respect to the time variable $t\in\Bbb R_+$.
This implies that \eqref{4} is meaningful also for functions which are not in $\mathscr{S}$ nor are infinitely smooth. Precisely, provided that

  • $\varphi(\cdot,t)\in L^1_\mathrm{loc}(\Bbb R^3)$ for almost all $t\in\Bbb R_+$, without any growth condition at infinity and
  • $\varphi(x,\cdot)\in L^1_\mathrm{loc}(\Bbb R)$ with $|\varphi(x,t)|=O(t^{-\varepsilon})$ as $t\to\infty$ a.e. on $\Bbb R^3$ with $0<c\le\varepsilon$.

equation \eqref{4} is meaningful. Then, by putting $$ \varphi(y,\tau)=f(x-y,t-\tau) $$ and by using \eqref{4} jointly with the definition of convolution between a distribution and a function, i.e. $$ \mathscr{E}\ast\varphi (x,t) \triangleq \langle \mathscr{E}, \varphi(x-y,t-\tau)\rangle $$ we get the sought for solution $$ \chi(x,t)=\mathscr{E}\ast f(x,t)=-\frac{1}{4\pi}\int\limits_{0}^{+\infty}\frac{1}{\tau}\int\limits_{S_{c\tau}}f(x-y,t-\tau)\,\mathrm{d}\sigma_y\mathrm{d}\tau \label{S}\tag{WS} $$


  • The hypothesis $n=3$, i.e. the fact that we are working in a $3$D space, is essential for defining the structure of \eqref{3}. The inverse transform of $\hat{\mathscr{E}}$ in \eqref{2} has not the same structure on every $\Bbb R^n$: in monographs on hyperbolic PDEs, this concept is also stated by saying that Huygens's principle does not hold in even spatial dimension.
  • The regularity of the solution we have obtained is very weak: in particularly we do not know the smoothness of $\chi$ for a given smoothness of $f$. Deeper methods are required for the investigation of this problems.

[1] V. S. Vladimirov (1971)[1967], Equations of mathematical physics, Translated from the Russian original (1967) by Audrey Littlewood. Edited by Alan Jeffrey, (English), Pure and Applied Mathematics, Vol. 3, New York: Marcel Dekker, Inc., pp. vi+418, MR0268497, Zbl 0207.09101.

[2] V. S. Vladimirov (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, Vol. 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR2012831, Zbl 1078.46029.


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