# Why must the field equations be differential?

In Landau–Lifshitz's Course of Theoretical Physics, Vol. 2 (‘Classical Fields Theory’), Ch. IV, § 27, there is an explanation why the field equations should be linear differential equations. It goes like this:

Every solution of the field equations gives a field that can exist in nature. According to the principle of superposition, the sum of any such fields must be a field that can exist in nature, that is, must satisfy the field equations.

As is well known, linear differential equations have just this property, that the sum of any solutions is also a solution. Consequently, the field equations must be linear differential equation.

Actually, this reasoning is not logically valid. Not only the authors forget to explain the word ‘differential’, but they also do not actually prove that the field equations must be linear. (Just in case: this observation is not due to me.) But it seems that the last issue can be easily overcome. However, it is exactly the word ‘differential’, not ‘linear’, that is bothering me.

There is a nice theorem of Peetre stating that the linear operator $D$ that acts on (the ring of) functions and does not increase supports, that is, $\mathop{\mathrm{supp}} f \supset \mathop{\mathrm{supp}} Df$, must be a differential operator. The property of preserving supports can be considered as a certain locality property. Hence, the field equations must be differential because all interactions must propagate with a finite velocity.

But there is another notion of ‘locality’ of an operator: the operator $D$ is called local if the function $Df$ in the neighbourhood $V$ can be computed with $f$ determined only on $V$ as well, i.e., $(Df)|_V$ is completely defined by $f|_V$. The locality in this sense is not equivalent to the locality in the sense of supports' preserving. (Unfortunately, I do not have an illustrative example at hand right now, so there is a possibility of mistake $M$ hiding here.)

The question is: what physical circumstances determine the (correct one) notion of locality for a given physical problem? (Assuming there is no mistake $M$.) And does my reasoning really justifies the word ‘differential’ in the context of field equations? If so, are there any references containing more accurate argument than the one presented in Landau–Lifshitz's Course?

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Hm, my view is that one shouldn't interpret any statement in a physics book as a rigorous statement that can be proved. There will always be some hidden assumptions and even if you could pin them down exactly, in the end they wouldn't matter anyway. The point here is to make the qualitative statement that being linear is connected to superposition and being local is connected to differentiability. That's all there is to it, IMHO. –  Marek Aug 13 '11 at 7:21
as to whether there is a mistake M, you can consider asking whether the two formulations of locality are equal at mathoverflow.net, as it is essentially a mathematics question. –  Willie Wong Aug 13 '11 at 16:28
Suggestion for a better title(v1): Why must field equations be differential equations? –  Qmechanic Feb 5 '12 at 1:24

All physical phenomena occur on a substrate, medium or 'space'. This space has properties which guarantee the stability of the system, such as the existence of a characteristic propagation speed of 'c'. The reaction allowed at each point in space, and at every moment, is strictly due to the local environment, its immediate neighborhood and so 'action at a distance' is not permitted, nor infinities.
This physical environment (1D) is illustrated in this image (by Hans de Vries) and found in his online book "Understanding Relativistic Quantum Field Theory". This medium allows the propagation of the classical wave equation $$\frac{\partial^{2}\psi}{\partial t^{2}}=c^{2}\frac{\partial^{2}\psi}{\partial\mathit{x}^{2}}$$

ψ is the vertical displacement in the mechanical model. ...the equation is satisfied by any arbitrary function which shifts along with a speed v (or −v). A function ”stretched” by a factor v has it’s slopes decreased by a factor v, while it’s second order derivatives are lower by a factor $v^2$ .

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This does not really answer your question why the equation should be differential. But I think that the two notions of locality you mentioned are just equivalent, if I am not mistaken.

Let us prove that the second definition impies the first one. One has to show that if a point $x$ does not belong to $supp(f)$ then $x$ does not belong to $supp(Df)$. Indeed, then there exists an open neighborhood $V$ of $x$ such that $f|_V=0$. Hence by the assumption $Df|_V=0$. Hence $x\not\in supp(Df)$ as requested.

Let us prove now the converse statement that the first definition implies the second one. Assume that $f|_V=g|_V$. Then $(f-g)|_V=0$. Hence $supp(f-g)\cap V=\emptyset$. Consequently $supp(D(f-g))\cap V=\emptyset$, i.e. $D(f-g)|_V=0$. That means that $Df|_V=Dg|_V$.

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