When I read about the superposition principle, it says that it works only on linear systems, my problem is that I cannot really understand the difference between a linear and a non-linear system. I cannot understand whether it means the order of the corresponding differential expression for that certain fields, or something else. So if anybody can help me get the essence of linear systems and differentiate between a non-linear one, it would be really helpful.


2 Answers 2


So if anybody can help me get the essence of linear systems and differentiate between a non linear one

Consider a simple system to be a black box with an input (stimulus) $x(t)$ and an output (response) $y(t)$.

Let $y_1(t)$ be the output given an input $x_1(t)$ and $y_2(t)$ be the output given an input $x_2(t)$.

The question is now what is the output $y_3(t)$ given the input

$$x_3(t) = a_1x_1(t) + a_2x_2(t)$$

One possibility is that the output is

$$y_3(t) = a_1y_1(t) + a_2y_2(t)$$

and then this system is a linear system (this more or less defines a linear system). If this doesn't hold, then the system is not a linear system.

Let's work some examples to help you get the essence of this. First, let $y(t) = Ax(t)$ and then

$$y_1(t) = Ax_1(t)$$

$$y_2(t) = Ax_2(t)$$

$$a_1y_1(t) + a_2y_2(t) = A\left(a_1x_1(t) + a_2x_2(t)\right) $$

$$y_3(t) = Ax_3(t) = A\left(a_1x_1(t) + a_2x_2(t)\right)$$

and so this is a linear system. But, somewhat surprisingly, $y(t) = Ax(t) + B$ is not a linear system:

$$y_1(t) = Ax_1(t) + B$$

$$y_2(t) = Ax_2(t) + B$$

$$a_1y_1(t) + a_2y_2(t) = A\left(a_1x_1(t) + a_2x_2(t)\right) + (a_1 + a_2)B $$

$$y_3(t) = Ax_3(t) + B = A\left(a_1x_1(t) + a_2x_2(t)\right) + B$$

Not surprisingly, the system $y(t) = x^2(t)$ is not a linear system:

$$y_1(t) = x^2_1(t)$$

$$y_2(t) = x^2_2(t)$$

$$a_1y_1(t) + a_2y_2(t) = a_1x^2_1(t) + a_2x^2_2(t)$$

$$y_3(t) = x^2_3(t) = \left(a_1x_1(t) + a_2x_2(t)\right)^2 = a^2_1x^2_1(t) + a^2_2x^2_2(t) + 2a_1a_2x_1(t)x_2(t)$$

Finally, let's look at $y(t) = \frac{dx}{dt}$:

$$y_1(t) = \frac{dx_1}{dt}$$

$$y_2(t) = \frac{dx_2}{dt}$$

$$a_1y_1(t) + a_2y_2(t) = a_1\frac{dx_1}{dt} + a_2\frac{dx_2}{dt}$$

$$y_3(t) = \frac{dx_3}{dt} = a_1\frac{dx_1}{dt} + a_2\frac{dx_2}{dt}$$

and so, a differentiator is a linear system.

These examples should be enough help for you to get a clear 'picture' of what the label linear system entails.

  • $\begingroup$ Thanks a Lot Sir, i had a question That something like the inverse square function, doesnt satisfy the condition to be linear, but for example electric fields can still be superposed..am i missing something here? $\endgroup$ Jun 19, 2018 at 15:54
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    $\begingroup$ @SidharthGiri, I think I see what is confusing to you. Note that the $x(t)$ and $y(t)$ above do not need to be linear functions of time $t$. The inverse square law electric field isn't the system, it is the 'output' of the linear system (that is Maxwell's equations) to the 'input' of a point charge at the origin. $\endgroup$
    – Hal Hollis
    Jun 19, 2018 at 18:36

This can sound somewhat circular, but the essence of it is this:

a linear system is one that obeys the superposition system,

by definition of the former. This means that the superposition principle holds in all linear systems, but it also means that this is a relatively trivial property, and it shifts the bulk of the work into determining whether a given system is linear or not.

In more specific terms, this has nothing to do with the order of the differential operator $\mathcal L$ that enacts the system's equations of motion, but only with its linearity: i.e. we demand that $$ \mathcal L (x+y) = \mathcal L(x) + \mathcal L(y) $$ for all pairs of potential solutions $x,y$. This does rule out certain systems (such as e.g. $\mathcal L(x) = \ddot x(t) +x(t)^3$, for which the relation above doesn't hold), and it is in checking this linearity that the bulk of the work occurs.

  • $\begingroup$ Also need to have as a requirement L(ax) = aL(x). Or maybe that follows from your single requirement. $\endgroup$
    – user45664
    Jun 19, 2018 at 16:37
  • $\begingroup$ Thanks a Lot, although i dont have much experience with operators, Can you please help Maam , like demonstrate how this applies to a harmonic function..? $\endgroup$ Jun 19, 2018 at 17:06
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    $\begingroup$ @user45664 No, it's an independent postulate, but I figured it would complicate things instead of simplifying them. $\endgroup$ Jun 19, 2018 at 17:41

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