MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Observing the behaviour of a few physicists "in nature", I had the impression that among the mathematical tools they use a lot (along with possibly much more sofisticated maths, of course), there is certainly Taylor expansion. They have a quantity (function) that they need to approximate: they expand it in Taylor series, keep the order of approximation that is useful for their purposes, and discard the irrelevant terms.

Appearently, there is little preoccupation for mathematically justifying this procedure, even if the to-be-approximated quantity is not given by an explicit form which is clearly known to be analytic. As Physics clearly gets no problems from the above mathematical subtleties, this may just mean that the distinction between analytic and smooth functions is somehow irrelevant to the basic equations of physics, or rather to the approximations of their solutions that are empirically testable.

If non-analytic smooth functions are irrelevant to Physics, why is it so?

Are there equations of physical importance in which non-analytic smooth solutions actually are important and cannot be safely considered "as if they were analytic" for the approximation purposes?

Remark: analogous questions may arise about Fourier series expansions.

One possible way the practice goes might be:

  1. Consider a (differential or otherwise) equation $P(f)=0$ usually with analytic coefficients.
  2. Expand the coefficients in Taylor series around a point in the scale of physical interest.
  3. Discard higher order terms obtaining an approximated equation with polynomial coefficients $\tilde{P}(f)=0$.
  4. Make the ansatz that the solutions $f$ of interest must be analytic.
  5. Find the coefficients of $f$ by hand or by other means.

This leaves open the question why the ansatz is mathematically justified, if the equation of interest was $P$ not $\tilde{P}$. Do analytic solutions of $\tilde{P}$ aptly approximate solutions of $P$? Edit: I understand now that these last two lines are not very well formulated. Perhaps, ignoring the $\tilde{P}$ thing, I should have just asked something like:

Given any $\epsilon>0$, does knowing the analytic solutions (i.e. knowing their coefficients, possibly up to an arbitrarily large but finite number of digits) of $P$ give all the information about all solutions of $P$ up to $\epsilon$-approximation? Are there physically well known classes of equations $P$ in which this may not happen (perhaps even up to taking very regular approximations of the coefficients/parameters of $P$ itself)?

share|cite|improve this question
Keep in mind that you can use Taylor's theorem instead of a full Taylor series to work with smooth functions which are not necessarily analytic. So long as you only care about finitely many terms of the series, this usually does the trick. – Jack Huizenga Nov 26 '12 at 17:05
In relativistic/hyperbolic problems the non-locality of analyticity is fundamentally incompatible with "finite speed of propagation," i.e. the fact that information cannot travel faster than light. – Yakov Shlapentokh-Rothman Nov 26 '12 at 17:48
@Yakov: not true. In fact the Klein-Gordon equation will evolve compactly supported data on one time slice to analytic data on all other time slices. – Nik Weaver Nov 26 '12 at 18:03
@Nik: I should have restricted my comment to classical relativity. However, I'm puzzled about your statement about the Klein-Gordon equation. It surely satisfies finite speed of propagation (one can establish this with energy estimates), see e.g. and…. This implies that compactly supported solutions stays compactly supported. Maybe I'm misunderstanding the claim. – Yakov Shlapentokh-Rothman Nov 26 '12 at 19:02
So hold on ... there's no reason why $f$ has to be analytic at future times. I have to retract my original comment, or at least weaken it to merely dispute the claim that analyticity is incompatible with the impossibility of superluminal signalling. (On the grounds that finite propagation is already incompatible with positive energy.) – Nik Weaver Nov 26 '12 at 20:03

16 Answers 16

up vote 89 down vote accepted

As a physicist "in nature" perhaps I can give a few examples that illustrate how non-analytic functions can appear in physics and counter the idea that physicists do not worry about the justification of these procedures.

Example 1 involves one of the most precise comparisons between experiment and theory known to physics, namely the g factor of the electron. The quantity g is a proportionality factor between the spin of the electron and its magnetic moment. Perturbation theory in QED gives a formula $$g-2= c_1 \alpha + c_2 \alpha^2 + c_3 \alpha^3 + \cdots $$ where the coefficients $c_i$ can be computed from i-loop Feynman diagrams and $\alpha=e^2/\hbar c \simeq 1/137$ is the fine structure constant. Including up to four loop diagrams gives an expression for $g$ which agrees to one part in $10^{8}$ with experiment. Yet it is known that that this perturbative series has zero radius of convergence. This is true quite generally in quantum field theory. Physicists do not ignore this, rather they regard it as evidence that QFT's are not defined by their perturbation series but must also include non-perturbative effects, generally of the form $e^{-c/g^2}$ with $g$ a dimensionless coupling constant. Much effort has gone into understanding these non-perturbative effects in a variety of quantum field theories. Instanton effects in non-Abelian gauge theory are an important example of non-perturbative phenomena.

Example 2 involves the Hydrogen atom in an electric field of magnitude $E$, aka the Stark effect. One can compute the shift in the energy eigenvalues of the Hydrogen atom Hamiltonian due to the applied electric field as a power series in $E$ using perturbation theory and again one finds excellent agreement with experiment. One can also prove that this series has zero radius of convergence. In fact, the Hamiltonian is not bounded from below and does not have any normalizable energy eigenstates. The physics of this situation explains what is going on. The electron can tunnel through the potential barrier and escape from being bound to the nucleus of the Hydrogen atom, but for reasonable size electric fields the lifetime of these states exceeds the age of the universe. The perturbation theory does not converge because there are no energy eigenstates to converge to, but it still provides an excellent approximation to the energy eigenstates measured experimentally because the experiments are done on a time scale which is very short compared to the lifetime of the metastable state.

So I would say that at least in these examples there is a very nice interplay between the physics and the mathematics. The lack of analyticity has a clear physical interpretation and this is something that is understood by physicists. Of course I'm sure there are other example where such approximations are made without a clear physical justification, but this just means that one should understand the physics better.

share|cite|improve this answer
“which agrees to one part in $10^8$”, you mean $10^{-8}$? – 36min Nov 28 '12 at 6:16
Yes. My understanding of the phrase "one part in $10^x$" is that it is equivalent to saying an accuracy of $10^{-x}$. For a detailed statement of the comparison between theory and experiment for g-2 see – Jeff Harvey Nov 28 '12 at 12:42
Oh, my bad. I thought it was a typo. Turns out my English is bad. – 36min Nov 29 '12 at 2:36
Not sure this a good example. For some field theories constructed rigorously the function is still analytic even though the Taylor series has zero radius of convergence. The issue is the location of the point around which the Taylor expansion is made: in the middle of the domain of analyticity (ordinary summation which requires positive radius of convergence) versus on the boundary (Borel summation). – Abdelmalek Abdesselam Apr 25 '13 at 17:44

Suppose that $P$ is a partial differential operator with constant coefficients. An old result of Petrowski shows that the following statements are equivalent.

  1. All the classical solutions $u$ of the differential equation $Pu=g$ are real analytic if $g$ is real analytic.
  2. The operator $P$ is elliptic.

If a physicist is interested in a well posed evolution equation, then, according to J. Hadamard, that differential equation cannot be elliptic. Thus by the above result of Petrowski you must allow non analytic solutions for analytic data. If moreover, the operator $P$ is hypoelliptic (e.g. $P$ is the operator $\partial_t -\Delta$) then any solution $u$ of $Pu=g$ is smooth once $g$ is smooth.

To conclude, yes, physicists do need to consider smooth, nonanalytic functions.

Addendum: J. Hadamard in his classic Lectures on Cauchy problem in linear partial differential equations discusses the role of smooth nonanalytic functions. I'll let the great master speak for himself.

"I have often maintained, against different geometers, the importance of this distinction. Some of them indeed argued that you may always consider any function as analytic [...] as they can be approximated with arbitrary precision by analytic ones. But, in my opinion, this objection would not apply, the question not being whether such an approximation would alter the data very little, but whether it would alter the solution very little."

In this paragraph he hints at the well posedness of the initial value problem. He then proceeds to show that the initial value problem for the Laplace operator $\partial_t^2+\partial_x^2$ is ill posed.

share|cite|improve this answer
Could you add references for these results of Petrowski and Hadamard? Nice answer btw. – Michael Bächtold Nov 27 '12 at 9:41
Check Petrowsky's collected works. This is one of his first results, in late 30s. As for Hadamard, here are some useful notes – Liviu Nicolaescu Nov 27 '12 at 10:25
Thanks. In thE notes you link there is an explanation of well posedness but not of the statement that elliptic problems are not well posed, or did I oversee that? – Michael Bächtold Nov 27 '12 at 11:26
Check Proposition 1, page 27 of these notes – Liviu Nicolaescu Nov 27 '12 at 14:14

It is worth noting that it is impossible to solve the initial value problem for the standard heat equation in the real analytic category. Here, there are asymptotic expansions available but no Taylor series.

ADDED: It should also be noted that the directionality of time, as exhibited in the heat and diffusion processes, is a phenomenon lives outside the real analytic category. I believe that any PDE in the real analytic category that is well-posed as an initial value problem can be solved in both positive and time directions. That's not true for the heat equation in the smooth category. So the need for going outside the real analytic category appears already in fundamental classical physics.

This can be avoided, I suppose, by working purely with discrete models, but that for some of us is a cure worse than the original "problem".

share|cite|improve this answer
Dear Deane, could you expand or add a reference please? – Michael Bächtold Nov 27 '12 at 9:20
Or is this a special case of what Liviu posted? – Michael Bächtold Nov 27 '12 at 9:27
This can be seen by writing the solution as a convolution of the heat kernel with the initial data. The dependence on time is easily seen to be non-smooth as time approaches zero. – Deane Yang Nov 27 '12 at 12:22
Liviu's answer, I believe, is complementary to mine. He points out that there are PDE's that model physical phenomena, where the initial value problem should not be well-posed but where the initial value problem is well-posed in the real analytic category. The key examples are elliptic PDE's modeling time-independent physical states. – Deane Yang Nov 27 '12 at 17:42

A strong argument is given above on the heat equation; let me be more specific. The heat equation, one of the most basic in PDE and mathematical physics, already known to Fourier, is $$ L=\frac{\partial }{\partial t}-\Delta_x,\quad t\in\mathbb R,\quad x\in\mathbb R^n, $$ has the fundamental solution $$E= H(t)(4\pi t)^{-n/2}e^{-\frac{\vert x\vert^2}{4t}}, $$ i.e. $LE=\delta(x)\otimes\delta(t)$ (here the Heaviside function $H$ is the indicatrix of $\mathbb R_+$). It is easy to see that the $C^\infty$ singular support of $E$ is reduced to $0_{\mathbb R^{1+n}}$ whereas the analytic singular support is the hyperplane $t=0$. Since the function $E$ is $C^\infty$ except at $x=0,t=0$, one can see that it is indeed a flat function at $t=0,x=x_0\not=0$, i.e. all derivatives vanish at such a point. It is thus impossible to understand one of the simplest PDE using only analytic functions.

A more refined -yet classical- fact is related to the notion of well-posedness as defined by Jacques Hadamard. Loosely speaking, a PDE problem is well-posed whenever the solution can be controlled by the data or the sources via suitable inequalities. A typical example of a well-posed problem: the Cauchy problem with respect to a spacelike hypersurface (e.g. $t=0$) for the wave equation. A typical example of an ill-posed problem: the Cauchy problem for the Laplace equation. Although the latter has uniqueness properties, the analytic solutions given for instance by the Cauchy-Kovalewski Theorem are extremely unstable: you have $$ \partial_x^2 u+\partial_y^2 u=0,\quad u=e^{\lambda(x+iy)}, u(0,y)=e^{i\lambda y}. $$ The Cauchy data at $x=0$ are bounded by 1, whatever is $\lambda >0$, whereas the solution increases exponentially with $x>0$: no control of $u$ by its Cauchy datum could be expected. However the solutions are analytic and uniquely determined by the Cauchy datum. The analytic method given by the CK theorem provides analytic solutions which are unstable. The CK theorem fails to deliver stable solutions in that case. No understanding of stability phenomena (a very interesting physical property) for PDE is possible within the class of analytic functions and one should use much larger classes of functional spaces in which inequalities of well-posedness could be proven.

I could have mentioned another effect, for instance for the Cauchy problem for the Laplace equation: take an analytic Cauchy datum $\phi_0$, then CK provides an analytic solution. Now, perturb $\phi_0$ by a smooth non-analytic function $w$ and take as a datum say $\phi_0+\epsilon w$. Then there is no solution to the Cauchy problem since the very existence of a (say continuous) solution is forcing the data to be analytic. It is not difficult to prove that by Fourier transformation: the analyticity will be forced by the fact that you have to compensate the exponential increase by some exponential decay of the data, triggering analyticity for this data.

share|cite|improve this answer
Regarding the last sentence, wouldn't it be possible at least in principle to understand stability phenomena by, for instance, just using polynomials? One would give polynomials as inputs to the equation, and look at the solution. If the process is stable, we would get good bounds on how the solutions changes as we vary the input a little. In particular we want bounds that are uniform in the polynomial degree in the input. But if the process is unstable, then all such bounds would blow up as we increase the polynomial degree. – timur Feb 5 '13 at 2:13
@timur It is certainly possible to do what you propose, but you have to keep in mind Hadamard's celebrated sentence about polynomials, in which he said essentially the following: "I do not care so much about approximating the data by polynomials, what matters is how this approximation is transferred to the solution." I will add other comments in a new edit of my answer above. – Bazin Feb 7 '13 at 20:20
Yes. I think what I proposed is a way to measure "how this approximation is transferred to the solution" in Hadamard's words. – timur Feb 7 '13 at 22:07

It may be that for many purposes asymptotics (in a precise sense) are needed, rather than exact formulas. The prototype for this is the fact that a finite Taylor expansion, with error term, correctly approximates a smooth function, whether or not the smooth function is actually analytic, whether or not the infinite Taylor series converges. A similar, more complicated, asymptotic idea was legitimized by Poincare, after he realized that certain series expansions "formally" solving problems in celestial mechanics were divergent, but their finite truncations provided good approximations.

There is a different objection one can have to analyticity, namely, the "action at a distance" aspect implied by "the identity principle", namely, that sufficient (but, yes, infinite) information about the function near a single point perfectly determines its behavior everywhere. This seems to me more delicate and more dubious than asymptotics.

share|cite|improve this answer

Part of your question has already been answered: a finite number of terms in a Taylor expansion can give a very good approximation, and we don't care whether the true solution is analytic; in any case, we are almost certainly only solving equations which themselves constitute an approximation.

But I also wanted to give an example where non-analytic smooth functions are important: in quantum field theory, 'instantons' give rise to terms proportional to $e^{-\frac{k}{g^2}}$, where $k$ is some numerical constant, and $g$ is a 'coupling constant'. Perturbation theory consists of a finite series expansion in $g$ (useful when $g \ll 1$), and so completely misses instanton effects. This is obviously of most interest when the 'perturbative' terms are all zero.

share|cite|improve this answer

In the field of mathematical general relativity, certain uniqueness results for black holes are known for real-analytic space-time but are open for smooth space-time. For example, the "No hair conjecture" was proven by Stephen Hawking for analytic space-time but is open in general. For more details, see slides by Klainerman:

(page 24 and on)

share|cite|improve this answer

In the BCS-theory of superconductivity the energy gap that seperates the ground state from the excited states is a non-analytic function of the exchange energy.

Here one doesn't use a taylor expansion of the interacting Hamiltonian around the non-interacting Hamiltonian. Instead one uses a Bogoliubov transformation.

share|cite|improve this answer

Here's an example that doesn't really fit the motivation. There is a setting where physicists consider discontinuous functions, continuous but not differentiable functions, and so on, all the way up to smooth but not analytic functions. Given that context, I wouldn't expect them to study Taylor series at the singularity.

Phase transitions are typically not smooth. For example, the energy as a function of temperature is discontinuous at freezing or boiling. More exotic are second-order phase transitions, which are continuous but not differentiable; and $n$-th order, which have several derivatives. But there are a few phase transitions best modeled as being the infinitely differentiable meeting of two real analytic functions.

I doubt that there are practical physical consequences distinguishing infinite order transitions from, maybe, tenth order ones. In particular, if the model is of a bulk substance made of smaller particles, then for sufficiently large derivatives (probably much larger than ten) the model breaks down. But the model is good in its domain and solving it yields smooth but not analytic functions.

share|cite|improve this answer

One example seems for me important and obvious, so I'm wondering, why no one has posted it so far: The time evolution of the state vector in quantum mechanics (Schrödinger's equation) is $|\psi(t)\rangle = \exp (-i \frac{H t}{\hbar}) \quad |\psi(0) \rangle$ and thus non-analytic in Plancks constant $\hbar$. This has important consequences in semiclassics ($\hbar \to 0$).

share|cite|improve this answer
@Andreas: I edited your post a little bit because I absolutely could not read the fraction in the exponent. – Willie Wong Nov 28 '12 at 9:10
@ Willie: Many thanks, Willie! – Andreas Rüdinger Nov 28 '12 at 19:20

Let me add to the several excellent answers. From a physicist' point of view, a concept like 'the value of a field at a point' does not have much meaning, or at least it can certainly not be measured. One can measure averages (norms), over several experiments, and over regions of space. Thus it makes more physical sense to work in the framework of distribution theory, and Sobolev spaces like $L^2$ or $H^1$, rather than of continuous functions. For instance, particles are represented by probability distributions, and their mass or energy is represented by suitable norms of such distributions. I would venture to say: if a result is only true for (say) $C^2$ but not rougher quantities, then it is unlikely to have much physical meaning.

share|cite|improve this answer

Yes, physicists really like Taylor series. However in many situations the Taylor series diverges. This is the case for almost all non-trivial "perturbation series" in physics. In the simplest situation, we have a function on $[0,1]$ which is infinitely differentiable at $0$, however its Taylor series at $0$ diverges. And there is an enormous literature about how to connect this series to some properties of the function.

share|cite|improve this answer

The answer to this question depends on what you mean by "need." In some sense, all physical measurements are integers, and usually small ones, so physicists don't even need the real number system -- in fact, physics (Galileo and Newton) predates the real number system (ca. 1870).

Some of the other answers have involved a lot of extremely exotic examples, but it isn't necessary to reach that far. Suppose a particle is released near the bottom of a bowl whose shape is given by a function $y=f(x)$, which has a minimum at $x=0$. Typically we would expect that the particle would oscillate around this minimum with a frequency given approximately by $\omega=\sqrt{gf''}$, where $g$ is the acceleration of gravity and the second derivative is evaluated at $x=0$. The result is approximately valid for small amplitudes, and for those small amplitudes it predicts correctly that the frequency is approximately independent of the amplitude.

But the bowl could be defined by some function like $y=|x|$, in which case the second derivative is undefined, and there is no amplitude small enough that the above analysis is a good approximation.

Does this mean that physicists "need" non-analytic functions? No. In reality, there are reasons why the infinitely sharp kink in the function $y=|x|$ can't be infinitely sharp. We can also come up with examples in which $y=g(x)$ is some analytic function very nearly equal to $f(x)=|x|$, and the oscillations will have exactly the same behavior for $g$ as for $f$. This will happen whenever the curvature at $x=0$ occurs on such a small scale that the particle's finite size prevents it from feeling any effect from the different behaviors of $f$ and $g$.

The issue here is not whether the function is analytic or non-analytic. The issue is whether a certain approximation is good or bad.

Often physicists find it convenient to use all kinds of badly behaved functions, such as Dirac delta functions. Convenience isn't the same as necessity. We could do quantum field theory using Egyptian fractions, if we had to --- it would just be extremely inconvenient.

share|cite|improve this answer
Even if it's well written, I've the feeling that maybe this answer doesn't address my question. In physics we have equations whose general form follows from a physically meaningful theory, and we feed those equations with some input (such as specific coefficient functions) and initial conditions, and the equations give us a prediction in the form of a solution. If we feed the equations with very irregular "input" to begin with, I'm not surprised if the solution will be badly behaved (or even that a solution, strictly speaking, will not mathematically exist). – Qfwfq Nov 27 '12 at 0:57
(...) In your $|x|$ example the source of irregularity is just in the input, not in the theory: indeed if you write the equation of motion for a very regular shape (not $|x|$ but, say, an approximation thereof) you will probably get a very regular solution. The intersting phenomenon is when the source of irregularity is "in the theory" (as -it seems to me- in the perturbative series examples in other answers), that is: you set up an equation with very regular inputs, and you get some non regular, yet physically meaningful, solutions. (...) – Qfwfq Nov 27 '12 at 0:58
(...) The question now is: which piece of mathematics -if any- justifies expanding such irregular solutions in Taylor series, and using it meaningfully, or just making an ansatz that all relevant solutions are analytic? Is anything physically relevant lost in the process? – Qfwfq Nov 27 '12 at 0:58
You may also be interested in John Earman, A primer on determinism, 1986; Norton, "Causation as folk science," 2003; and Korolev, "Indeterminism, asymptotic reasoning, and time irreversibility in classical physics," 2006. The Norton and Korolev papers can be found online by googling. Norton argues that causality doesn't hold in Newtonian mechanics. Korolev argues that the issue is functions that violate a Lipschitz condition. – Ben Crowell Nov 27 '12 at 2:29
The claim that «physics (Galileo and Newton) predates the real number system (ca. 1870)» is weird. What you mean is that the arithmetic formalization of the real number system came late into the picture, which is of course true, but most big names from before that would be shocked to hear from you that they were not dealing with real numbers! – Mariano Suárez-Alvarez Nov 27 '12 at 17:41

Here is a differential equation, a simple prototype of the equations of gas dynamics, which cannot be solved with analytic functions although coefficients and data are analytic: The $C^1$ solutions of Burgers' equation $u_t+uu_x=0$ are constant along characteristics $\dot x=u$. If the initial data satisfy $u(0,a)>u(0,b)$ for some $a<b$, then the characteristics through these points which will intersect, and the solution will, after finite time, cease to be even a $C^1$-solution. Trying a Taylor series ansatz is not an adequate approach to solve Cauchy problems with analytic data for Burgers' equation (or other hyperbolic conservation laws).

The lack of well-posedness of Cauchy problems in real analytic settings has already been pointed out in previous answers. It means that one should not confine one's attention to analytic solutions.

share|cite|improve this answer

A beautiful 19th century theorem of mechanics due to Joseph Bertrand says the following. Consider the motion of a particle which is driven by a central force potential $V$, that is, the potential $V$ is a function from the distance of the particle to some fixed origin and the force exerted on the particle is given by $\vec F=-\mathop{\rm grad}(V)$. If (almost) all trajectories are periodic, then either $V(r)\propto 1/r$ (celestial mechanics), or $V(r)\propto r^2$ (harmonic oscillator).

At some point of the proof one knows that all periods of all trajectories are rational multiples of a common period, and one needs to conclude that there is a common period. That part of the proof is always incorrect in the litterature I know. (In particular, Wikipedia's argument is not complete.)

It is in fact easy to construct smooth real families of periodic functions with non-constant rational period. The only way I can correct the proof assumes that the potential is real analytic.

share|cite|improve this answer
You may be interested in… – Willie Wong Apr 25 '13 at 8:51
@Willie. Indeed, thanks! – ACL Apr 25 '13 at 14:45

I eat everyday even if I do not know the process of digestion ...Hopefully (Dirac, Feynman and others) physicists have not waited for the justification of the mathematical objects they have discovered. The building of a theory both in mathematics and in physics is a long process which permanently needs refinements, so "less rigorous" formulations may occur. Of course there is in physics phenomenons not described by analytic expressions like shocks for example. In the solution of a physical problem we have first to build a model. This model may contain very worse functions like distributions. This is the mathematical model which describe (in an aproximative way) the qualitative behavior of the phenomenon. Then the next step is the approximation of this model by a numerical model which permit computations and simulations on calculators.

share|cite|improve this answer
I apologize, I thought the question was about the existence of non analytic functions nor smooth ones in physics. – user36539 Sep 25 '13 at 22:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.