Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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

Non-linear waves do not superimpose to each other, but why? What characteristics give this property?

share|cite|improve this question
Ummm... they don't by definition. If you have some other definition of "nonlinear" in mind, you should state it. Otherwise this is equivalent to asking "what characteristic of positive numbers makes them greater than 0?" – Chris White Jul 10 '13 at 0:40
@ChrisWhite I think this comment should be an answer. The question is "naive" (I don't mean that disparagingly) but good in the sense that I'm sure many of us have had the same thoughts when beginning to think characteristics of solutions. So I'm sure Raisa's qusetion will get hits, then your answer will beget the aha! moment - "so that's what linear really means!". Too many physics lectures bandy the word "linear" about without defining it properly, and too many maths lectures have all the definitions but leave a sense of bewilderdness - until a simple example shows "what it's all for". – WetSavannaAnimal aka Rod Vance Jul 10 '13 at 1:12
up vote 8 down vote accepted

Let's back up for a second. Before going into the complexities of nonlinear waves, let's ask what a linear wave is. Actually, let's go even further back and ask "what do we mean when we say linear?"

"Linear" comes from the study of things like vector spaces. We have objects (call them vectors, or arrows, or whatever) that can be both added together and scaled by a number, with the result being another object of the same type. Any collection of objects that satisfies certain conditions (which basically boil down to "addition and scalar multiplication behave as expected") can be considered a vector space.

Now let's talk about waves. But to keep things simple, let's just talk about the effect of some waves at a single point, where the effect can change in time. One wave might have a value $\psi_1(t) = \sin(\omega_1 t)$ at this point. Another might have a different frequency: $\psi_2(t) = \sin(\omega_2 t)$. Suppose we scale the waves by factors of $a$ and $b$, and suppose we have them both affect the point together. If the waves' effects just scale and add in the sensible way, then the value of the combined wave at the point will be $$ \psi_{(a\otimes1)\oplus(b\otimes2)}(t) \equiv a \sin(\omega_1 t) + b \sin(\omega_2 t) = a \psi_1(t) + b \psi_2(t). $$ Here I am using the symbol "$\otimes$" to mean "physically scaled by the preceding factor" and "$\oplus$" to mean "combined physically." In this particular case, $\otimes$ and $\oplus$ reduced to sensible scalar multiplication of the the wave value and regular addition of the values of two waves. We call these "linear" waves. One of their characteristics is that you can think of the waves as noninteracting $\psi_2$ will add its effect to the total in the same way, regardless of how much amplitude $\psi_1$ has already contributed.

But I didn't have to have that structure. In some cases, driving a physical displacement with twice the force does not result in twice the displacement, and having two different driving forces work together does not result in a force that gives a displacement that is is the sum of the independent displacements. For instance, perhaps the rule is $$ \psi_{(a\otimes1)\oplus(b\otimes2)}(t) \equiv \sqrt{a \sin(\omega_1 t) + b \sin(\omega_2 t)} \neq a \psi_1(t) + b \psi_2(t). $$ This then would be a nonlinear wave. They are defined by having the definition of how disturbances scale ($\otimes$) and combine ($\oplus$) be incompatible with scalar multiplication and regular addition of the waves' values. That is, our physical definitions of $\otimes$ and $\oplus$ did not yield the structure of a vector space - at least not in any obvious way.

The physics question remaining then is whether or not this situation is ever actually realized. The above discussion defines nonlinear waves, but it does not prove any such things exist. As it turns out, though, many waves important to physics show nonlinear behavior if you push them far enough. The classic example in optics is when the amplitude of an electromagnetic wave is so great that electrons in nearby atoms (thinking classically here) are pushed and pulled quite far from the "sweet spot" distance they want to have from their nuclei. Then the restoring force that pushes them back to that sweet spot is not simply directly proportional to their displacement, their motion is anharmonic, and the wave becomes nonlinear.

share|cite|improve this answer
Got you mostly, okay then how do we differentiate two nonlinear waves "called oscillons" and "solitions"? – Raisa Jul 10 '13 at 19:20
@Raisa That sounds like the beginning of a whole other question you could post - I'm not an expert in either of those phenomena. – Chris White Jul 11 '13 at 1:22
@Raisa - Oscillons and solitons are two specific types of nonlinear waves. Solitons are considered to be nonlinear partly because their phase speed depends upon amplitude and their spatial scale. I think the main differences between the two are that oscillons are solutions in granular media, they can change form when colliding with other oscillons, and oscillons can be stationary waves. Generally speaking, solitons are solutions to KdV-like equations. – honeste_vivere Oct 9 '14 at 18:15

Nonlinear waves do not superpose with a trivial rule like $f(a)+f(b)=f(a+b)$, but they do obviously superpose nonlinearly. For certain nonlinear wave equations, a Lie group can be defined for solutions, such that for solutions $\psi_a$ and $\psi_b$, a new solution $G(\psi_a, \psi_b)$ can be constructed. This new solution can be understood as a superposition. The Lie algebra is then used in the nonlinear variant of Fourier transform, which is called Inverse Scattering Transform. This is an active, but highly specialised area of research

share|cite|improve this answer

A simple example of why linear functions do not obey superposition is the associative property:


ax + bx = (a+b)x


f(a) + f(b) = f(a+b)


sin(ax) + sin(bx) != sin(ax+bx)


f(a) + f(b) != f(a+b)

share|cite|improve this answer

Let's start with linear waves. They stem from an differential equation of the form

$$\hat W\, \psi(\vec x,t)$$

where $\hat W$ is a linear operator, e.g. $(\partial_t^2-\Delta)$. This means that for the sum of two functions $\psi_1$ and $\psi_2$ you can simply use the distribution rule

$$\hat W(\psi_1+\psi_2) = \hat W\psi_1+\hat W\psi_2$$

So if $\psi_{1,2}$ each solve the equation, their sum also does and thus you obtain superposition.

Now a nonlinear system is any system that cannot be written in this simple form. This means that in general the sum of two solutions of a nonlinear equation is no longer a solution as well.

However, there are often special solutions called Solitons which not only solve the nonlinear equation by themselves, but also in their sum. This is basically due to friction effects compensating nonlinearity effects for those solutions. So it is not entirely correct to state that nonlinear system don't show superposition at all - the correct statement is

In nonlinear systems not all possible solutions can be superimposed to obtain new solutions, but there may be Solitons

As a side-note, this usually means that you cannot even just rescale a nonlinear system's solution to obtain a new solution.

share|cite|improve this answer
okay then how do we differentiate two nonlinear waves "called oscillons" and "solitions"? – Raisa Jul 10 '13 at 19:24
@Raisa That could almost be a separate new question... Well, as the Oscillon Wikipedia article states: "an oscillon is a soliton-like phenomenon" and "Whereas solitons occur as travelling waves in a fluid or as electromagnetic waves in a waveguide oscillons may be stationary", i.e. basically a soliton moves while an oscillon does not – Tobias Kienzler Jul 11 '13 at 6:29

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.