1
$\begingroup$

The equation for a uniform oscillator is: $\dot\theta = \omega$ which has a solution of $\theta(t) = \omega t +\theta_0$.

For a non-uniform oscillator, the equation is: $\dot\theta= \omega - a$ where $a(\theta) \ne 0$.

These definitions are from Strogatz's book. Now I don't understand the difference between non-uniform and non-linear. Isn't a non-uniform oscillator is also a kind of non-linear oscillator? What is the difference between them?

$\endgroup$
1
  • 1
    $\begingroup$ $\dot \theta = \omega$ has the solution $\theta(t) = \omega t + \theta_0$. $\endgroup$
    – TimWescott
    Jan 31 at 5:55

3 Answers 3

2
$\begingroup$

On a more fundamental level the difference is between a linear differential equation, homogeneous differential equation, and an equation with constant coefficients.

Linear (ordinary) differential equation
If we take an equation $$\mathcal{L}x(t)=f(t),$$ where $\mathcal{L}$ is an arbitrary differential operator, then it is linear if the differential operator is linear, i.e., if for any functions $x(t),y(t)$ and any coefficients $a,b$, the action of the operator on the linear combination of these functions, $z(t)=ax(t)+by(t)$ is just a linear combination of its action on each of the functions $$\mathcal{L}z(t)=a\mathcal{L}x(t)+b\mathcal{L}y(t).$$

Homogeneous differential equation
Further, we can write a linear operator as $$ \mathcal{L}=\sum_n a_n(t)\frac{d^n}{dt^n},$$ so that the corresponding linear equation can be written as $$\sum_n a_n(t)\frac{d^n x(t)}{dt^n}=f(t).$$ If the right-hand-side of thsi equation is zero, i.e., if we have only the derivative terms, then the equation is called homogeneous, otherwise - inhomogeneous.

Differential equation with constant coefficients
Finally, if the coefficients of the equation, $\{a_n(t)\}$ are constant (i.e., independent on $t$), it is called an equation with constant coefficients.

Oscillator
Oscillator is generally described by differential equation $$ \ddot{x}(t)+\omega_0^2x(t)=0\Leftrightarrow \mathcal{L}=\frac{d^2}{dt^2}+\omega_0^2, \mathcal{L}x(t)=f(t)$$ This is a homogeneous linear differential equation with constant coefficients. Adding a driving force makes it inhomogeneous. (Perhaps, this is what is meant by non-uniform, although it may also refer to non-constant coefficients.)

$\endgroup$
1
$\begingroup$

Without referring to the source material, the differential equation $$\theta = \omega - a(\theta)$$ is linear if $a(\theta)$ is a linear relationship, i.e. $a(\theta) = k \theta$, with $k$ constant. It's only nonlinear if $a(\theta)$ is a nonlinear relation.

$\endgroup$
3
  • $\begingroup$ So an oscillator can be non-uniform without being non-linear and vice-versa? $\endgroup$
    – Earmen
    Jan 31 at 6:02
  • $\begingroup$ Btw, in the example of the book, they used $a(\theta)=a \sin \theta$. So, it seems that is non-uniform and also non-linear. $\endgroup$
    – Earmen
    Jan 31 at 6:04
  • $\begingroup$ By the definition you give, yes. I'm going by the definition of "nonlinear" for differential equations and dynamic systems, though, and sometimes terminology can vary from discipline to discipline. I think this'll adhere to the standard, though. $\endgroup$
    – TimWescott
    Jan 31 at 6:07
0
$\begingroup$

lets look at those two examples:

$$\dot\phi+\sin(\phi)=\omega$$

the solution

enter image description here

this is non uniform oscillator and also non linear oscillator

and $$\dot\phi=\omega$$

the solution $~\phi(t)=\omega\,t$

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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