1. In general, what is meant by non-linear system in classical mechanics? Does it always concern the differential equations one ends up with (any examples would be greatly appreciated)? If so, is it considered as non-linear because of: Higher powers of system variables? ($x^2,x^3...$), or does also any function of $x$ makes the system non-linear? like $\cos(x)$, $\ln(x)$, $e^x$ etc. I am confused.

  2. Furthermore, why is it that most non-linear systems are considered non-integrable? Is it due to this fact that such systems are usually considered to be unpredictable even classically? (because we can't have exact analytical solutions?).


4 Answers 4


(1) In general, what is meant by non-linear system in classical mechanics?

A linear system is described by a set of differential equations that are a linear combination of the dependent variable and its derivatives. Some examples of linear systems in classical mechanics:

  • A damped harmonic oscillator, $$m \frac{d^2 x(t)}{dt^2} + c \frac{d x(t)}{dt} + k x(t) = 0$$
  • The heat equation, $$\frac{\partial u(\vec x, t)}{\partial t} -\alpha \nabla^2 u(\vec x, t) = 0$$
  • The wave equation, $$\frac{\partial^2 u(\vec x, t)}{\partial t^2} -c \nabla^2 u(\vec x, t) = 0$$

Non-linear systems cannot be described by a linear set of differential equations. Some examples of non-linear systems in classical mechanics:

  • Aerodynamic drag, where the drag force is proportional to the square of velocity, $$F_d = \frac 1 2 \rho v^2 C_D A$$
  • The Navier-Stokes equations, which are notoriously non-linear, $$\rho \left( \frac{\partial v}{\partial t} + \vec v \cdot \vec \nabla v \right) = -\vec \nabla p + \vec \nabla T + \vec f$$
  • Gravitational systems, where the force is inversely proportional to the square of distance between objects, $$\vec F = -\frac {GMm}{||\vec r||^3}\vec r$$

(2) Furthermore, why is it that most non-linear systems are considered non-integrable?

That term, "non-integrable" has two very distinct meanings. One sense is that the integral cannot be expressed as a finite combination of elementary functions. The elementary functions are polynomials, rational roots, the exponential function, the logarithmic function, and the trigonometric functions. This is a rather arbitrary division. For example, the integrals $\operatorname{li}(x) = \int_0^x 1/\ln(t)\,dt$ and $\operatorname{Si}(x) = \int_0^x \sin(t)/t\,dt$ cannot be expressed in the elementary functions. These are the logarithmic and sine integrals. These "special functions" appear so often that algorithms have been devised to estimate their values. Categorizing functions as elementary versus non-elementary is a bit arbitrary.

Just because the solution to a problem can't be expressed in elementary functions doesn't mean the problem is unsolvable. It just mean it's not solvable in the elementary functions. For example, people oftentimes say the three body problem is not "solvable". That's nonsense (ignoring collision cases). In the sense of solvability in the elementary functions, even the two body problem is not "integrable". Kepler's equation, $M = E - e\sin E$, gets in the way. Just because the two body problem cannot be expressed in terms of a finite combination of elementary functions does not mean we can't solve the two body problem.

There's another sense of "integrability", which is "does the integral exist?" Going back to the n body problem, a problem exists with collisions. These collisions introduce singularities, so that one could say that the n body problem is not integrable in the case of collisions. Collisions represent one kind of singularity. Painlevé conjectured that the n body problem has collisionless singularities in when $n\ge 4$. This has been proven to be true when $n \ge 5$. Newtonian mechanics allows some configurations of gravitating point masses to be sent to infinity in finite time. This truly is an example of non-integrability.

Proving integrability (or lack thereof) in this sense is a much tougher problem than showing that a problem is (or is not) solvable in the elementary functions. There's a million dollar prize for the first person who can either prove that the Navier-Stokes equations have globally-defined, smooth solutions, or come up with a counterexample that shows that that the Navier-Stokes equations are not "integrable."

  • $\begingroup$ "linear system is described by a set of differential equations" Why should they be differential? $\endgroup$
    – anderstood
    Commented Oct 15, 2014 at 18:38
  • 2
    $\begingroup$ @anderstood - Snarky answer: $Ax=\lambda x$ is a differential equation. It's a zeroth order differential equation! Less snarky: I answered from the perspective of physics. While disciplines other than physics do deal with static linear systems (e.g., $Ax=\lambda x$) or with step-wise linear systems (e.g., a discrete Kalman filter), physics is typically concerned with differential changes over time and/or space. Moreover, Mateus Sampaio specifically asked about integrability, which implies that Mateus was using the term in the context of a set of differential equations. $\endgroup$ Commented Oct 15, 2014 at 19:15
  • $\begingroup$ I'd go for your snarky answer, as you mention gravitational force as an example of nonlinearity (in $r$) :D. I somehow disagree with the fact that physics exclusively study differential changes: what about the gravitational force, or a linear spring, in static? $\endgroup$
    – anderstood
    Commented Oct 15, 2014 at 19:27
  • $\begingroup$ Thank you for this nice answer, learned a lot. I upvoted for you. I have two questions: 1)why is Navier-Stokes notoriously non-linear? because of the term v.div(v)? 2) But if you say they're not solvable by elementary functions, how are they solvable? only by numerical methods? Sorry you have given a very nice answer, I just have these small doubts still :( $\endgroup$
    – user929304
    Commented Oct 16, 2014 at 11:59
  • $\begingroup$ @user929304 - Conceptually, what's the difference between calculating $\sin(x)$ versus some non-elementary function for which there exists a well-known algorithm? The answer is, not much. The distinction between "elementary" versus "non-elementary" is rather arbitrary. Your calculator or computer is using numerical methods to calculate $\sin(x)$, and most assuredly it is not using a Taylor approximation. $\endgroup$ Commented Oct 16, 2014 at 12:21

As requested by the OP, I gather my points in an answer.

Linear systems Linear systems are systems which are linear with respect to a physical quantity. Mathematically, their evolution can be written as a (possibly differential) equation.

Examples: A linear spring is linear in the sense that is produces a force proportional to the displacement it undergoes: $F(x)=kx$ (assuming its rest position is 0). If $x$ is doubled, the reaction force $F$ is doubled too.

If you now consider a mass at the end of the spring, and you don't neglect the inertia, you get another example which involves derivatives (linear dynamics): $$m \ddot x+kx=0$$ It's linear with respect to $t\mapsto x(t)$: any linear combination of solutions is solution.

Mathematics offer very powerful tool for such systems. In linear dynamics, for example, it is quite easy to calculate modes of a given system. Then, any motion can be decomposed as a linear combination of these modes! The computations only have to be done for modes (there are as many modes as degrees of freedom) and then you can calculate any motion with no computational cost!

Nonlinear systems Again, the system is nonlinear with respect to a physical quantity.

In continuum mechanics, you can observe :

  • material nonlinearities (if you double the pressure the material does not double the strains, e.g.); in 1D, an example could be: $$F(x)=kx^2$$ or: $$F(x)=\sqrt{|cos(x)|}$$ The latter is never used/exposed because no material obeys such a law. But if ever you find one, then you'll have a nonlinear system!

In 3D, see strain tensor vs Euler-Lagrange tensor e.g.

  • geometrical nonlinearities (if you have large displacements).
  • contact nonlinearities: if you double the position $x(t)$ of ball bouncing on the ground, you're new motion could penetrate the ground (see "unilateral constraints"), which would obviously not be a solution.

In nonlinear dynamics, modes cannot be used as in linear dynamics, because the sum of two motions is no longer a possibly motion (i.e. a solution of the system's equations).

The equations of nonlinear systems can usually not be solved analytically, unfortunately. Luckily, it is often possible to use numerical methods (such as Finite Element Methods) to solve nonlinear differential equations by approaching the exact solution, provided it exists. This is used in an extremely wide range of fields in mechanics: weather forecast, car crashes tests, modeling the combustible of rockets, cracks in concrete, thermal diffusion, acoustics, etc.

The drawback of solving equations numerically is the computational cost, which has no upper bond (the more precision you want, the smaller step you will take). And for complex problems, relevant preliminary results can take weeks...

Note: computers only solve linear problems. Numerical methods consists in making small linear step in such a way that overall the nonlinearity(ies) is(are) conserved.

  • $\begingroup$ Thanks for taking the time and writing it up as an answer finally.+1 very nice $\endgroup$
    – user929304
    Commented Oct 15, 2014 at 15:49
  • $\begingroup$ Can you explain how $F(x) = k^2 x$ is nonlinear while $F(x) = kx$ is? $\endgroup$
    – Señor O
    Commented Oct 15, 2014 at 17:07
  • $\begingroup$ @SeñorO It's of course a typo. Edited. ($k^2x$ is indeed linear with respect to $x$) $\endgroup$
    – anderstood
    Commented Oct 15, 2014 at 17:13
  • $\begingroup$ The mathematics of classical physics is based on the assumption of instantanoues change propagation ("recomputation" of dependent variables in a function of a differential equation), which was obsoleted by Einstein, and used in Verilog and reactive programming, but math has not caught up for some reason yet. $\endgroup$ Commented Nov 18, 2022 at 0:18

A linear system is one whose dynamics obeys linear differential equations, in contrast with those that are non-linear whose dynamics obeys non-linear differential equations. So if the dyanmics of the variable $x(t)$ obeys a a differential equation $$f\left(x(t),\frac{d}{dt}x(t),\dots,\frac{d^n}{dt^n}x(t),t\right)=0,$$ if $x_1(t)$ and $x_2(t)$ are differente solutions, for a linear system it's true that $ax_1(t)+bx_2(t)$, with $a$ and $b$ constants, will also be a solution, while it's not in general true for a non-linear system. For example, take the unidimensional harmonic oscilator, with potential $V(x)=\dfrac{k}{2}x^2$: $$mx''(t)+kx(t)=0,$$ it's easy to check that the linear combination of solutions is also a solution. However for the anharmonic potential $V(x)=\dfrac{\lambda}{3}x^3$, the dynamics will obey $$mx''(t)+\lambda x(t)^2=0,$$ and it's not true that the linear combinationof two solutions will in general yield another solution, since $$m(x_1(t)+x_2(t))''+\lambda(x_1(t)+x_2(t))^2=\\ (mx_1''(t)+\lambda x_1(t)^2)+(mx_2''(t)+\lambda x_2(t)^2)+2\lambda x_1(t)x_2(t)=\\ 2\lambda x_1(t)x_2(t)\neq0,$$ in general.

The problem of non-linear systems is that the theory of non-linear differential equations is much harder and lacks a lot of results that the linear theory does, so in general the systems may not be integrable given a particular differential equation and you have to study each separate case most of the times.

  • 2
    $\begingroup$ Few additional points: i) "Nonlinear" is often used without specifying with respect to what it is nonlinear. ii) In continuum mechanics, you can observe material nonlinearities (if you double the pressure the material does not double the strains, e.g.), geometrical nonlinearities (if you have big displacements) or contact nonlinearities. iii) Properties of linear systems are absolutely essential to facilitate calculations. For example in linear dynamics it is easy to find a basis of any displacement whereas in nonlinear dynamics, we don't know much! $\endgroup$
    – anderstood
    Commented Oct 15, 2014 at 14:06
  • $\begingroup$ iv) Using numerical methods to approach the exact solution of a nonlinear equation (assuming there exists a solution), smooth nonlinear problems can be linearized locally, at each calculation step. Numerical methods are essential because they allow to find approximate solutions of nonlinear equations. The drawback is that they need to be implemented, and requires computation time. $\endgroup$
    – anderstood
    Commented Oct 15, 2014 at 14:10
  • $\begingroup$ @anderstood thanks for the additional points, very relevant, it would be really awesome if you could write it up maybe as an answer, so you can possibly elaborate slightly more on each point. Thanks either way, very nice! $\endgroup$
    – user929304
    Commented Oct 15, 2014 at 15:09
  • $\begingroup$ Thanks for this again. David Hammen in his answer here, talks about two different non-integrabilities, which type do you refer to in your last paragraph? I am still a bit confused about "how one concludes that an equation is not integrable"... Thanks $\endgroup$
    – user929304
    Commented Oct 16, 2014 at 13:20
  • $\begingroup$ Dear Mateus, could you lend your view on this discussion? (when you got some spare time), thanks. $\endgroup$
    – user929304
    Commented Dec 12, 2014 at 17:28

As the first question has received sufficient exposition, I would like to make a point with regard to the second one. First thing to understand is that integrability and non-linearity of a system are two different concepts. It is true though that all linear systems in classical mechanics (i.e those that are described by systems of linear equations, be them algebraic or differential) are integrable, but integrability does not imply linearity. Typically looking at the case of a system of differential equations describing a system, integrability is then a condition requiring analytical solvability when all the required initial and boundary conditions have been specified (the problem is then well-posed). In classical Hamiltonian systems (or the more general case being a finite system of ODEs), integrability is then specified by the Liouville-Arnold theorem . In all autonomous systems, the hamiltonian automatically becomes a constant of motion and hence in 1 dimension, all hamiltonian systems are integrable. Enlarging the phase space of dynamics, one now requires many more integrals of motion that specify iso-surfaces upon which dynamics occurs and the intersection of all these surfaces will give the required trajectory of the system, once the initial conditions are fixed.

The idea of requiring constants or invariants of an equation is a very broad one and the argument mostly goes through for most other cases too. For infinite dimensional systems (like PDEs, say), we in principle need an infinite set of constants, though it is not sufficient. The existence of such independent constants in involution with each other many times points to a deeper symmetry or degeneracy in the problem. Take for example the Kepler problem, which has three constants of motion
$$ H=\dfrac{\vec{p}^2}{2m}-\dfrac{k}{r}\quad(k>0) $$ The three constants of motion in this problem are $H,\vec{L}^2$ and $L_z$ and they all poisson commute with each other. The dynamical symmetry group of this problem turns out to be $SO(4)$ (this group contains all the canonical transformations that preserve the symmetry of the hamiltonian). An additional constant of motion is the Laplace-Runge-Lenz vector $\vec{A}$, $$ \vec{A}=\vec{p}\times\vec{L}-\dfrac{mk\vec{r}}{r} $$ which also poisson commutes with the other constants $(\{\vec{A},H\}=0,\{\vec{A},\vec{L}\}=0)$. It is these two vectors $(\vec{L}\pm\vec{A})$ modulo constant factors, that generate the group of dynamical symmetries. The fact that $\vec{A}$ is a constant implies that orbits in the Kepler problem do not precess. In the context of the hydrogen atom, this same constancy gives rise to the degeneracy of the energy levels with $\ell$ (the azimuthal quantum number). Degeneracy in $m$ (the magnetic quantum number) is demanded by the spherical symmetry of the potential. From this, we see that the existence of constants of motions that allow a system to be integrable almost always point to a deeper understanding of the problem by exposing hitherto unknown symmetries or degeneracies in the system.

There are a number of non-linear systems like the KdV equation that exhibit soliton and soliton-like solutions and this is a very active area of research in understanding the existence of solutions and integrability of nonlinear systems. One can be pretty sure that if a non-linear equation is found to be integrable, then it is definitely special and has some underlying properties that allow one to solve the equation. There is a nice write up on some of these things given in this document, that I suppose will be helpful in elucidating further details with regard to integrable systems.

  • $\begingroup$ +1 Thanks a lot for this great answer, so well written! I have two small (and probably silly) questions if you don't mind: 1) You said at some point "integrability is then a condition requiring solvability when all the required initial and boundary conditions have been specified", then by implication, for linear system that are generally integrable, one can easily define the initial and boundary conditions, but why is it so difficult to define them for non-linear systems? 2) In the worst case where the system as you describe is non-integrable, what does one do to proceed? (numerically?) Thanks $\endgroup$
    – user929304
    Commented Oct 17, 2014 at 6:12
  • 1
    $\begingroup$ @user929304 1) The initial and boundary conditions have to be specified independently, that is they are part of the problem specification and not part of the solution. In most cases, these are specified by the physics of the problem and are an additional input provided to you. The linearity or non-linearity of the equations do not constrain the initial/boundary conditions (the order of the equation will dictate only the number of specific conditions required for well-posedness). So, it is not more or less easy to define these conditions in either case. $\endgroup$ Commented Oct 17, 2014 at 21:33
  • $\begingroup$ @user929304 2) There was a minor miswording in my answer, which I have corrected now. Integrability corresponds to analytical solvability, which means that in principle one can write down the solution on paper. Mind you, it does not say that the solution will be representable in an explicit form (one could always define a "special function" that "solves" the implicit relation in the variables to give an explicit "answer", but that in most cases is not very enlightening), but only that an analytical solution exists. $\endgroup$ Commented Oct 17, 2014 at 21:45
  • $\begingroup$ Numerically, there is never an issue (but for errors). One can always numerically solve well posed equations and evolve the system to whatever extent required. That is the precise power of numerics! The only thing to be careful is in differential systems with some conserved quantities, one would like the numerical discretization to also explicitly conserve these quantities at the discrete level. This is a major consideration in a variety of situations, eg: energy preserving verlet integrators in MD simulations, implementing exact incompressibility in fluid simulations etc. $\endgroup$ Commented Oct 17, 2014 at 21:52

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