Please look at this equation representing a mass-spring system:

${\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}+2\zeta \omega _{0}{\frac {\mathrm {d} x}{\mathrm {d} t}}+\omega _{0}^{\,2}x=F$

where the function of $F$ is unknown (i.e. it can possibly be $-2\zeta \omega _{0}{\frac {\mathrm {d} x}{\mathrm {d} t}}$ or $(-2\zeta \omega _{0}{\frac {\mathrm {d} x}{\mathrm {d} t}})^2$).

So the question: MUST this mass-spring system be a linear system?

It may look simple, but I am really confused.

Appreciate any help.

  • $\begingroup$ So where exactly do you see a possible non-linearity? $\endgroup$
    – Peaceful
    Feb 23, 2017 at 18:15
  • $\begingroup$ Your possible values of $F(t)$ are wrong because as the notation suggests, $F$ is an explicit function of the time only. $\endgroup$
    – Peaceful
    Feb 23, 2017 at 18:16
  • $\begingroup$ The nonlinear terms brought from F(t), which can possibly include $x^2$ etc. So do u think F(t) is not possible to bring in non-linearity? i.e. F(t) can not possibly include terms like $x^2$. $\endgroup$
    – zlin
    Feb 23, 2017 at 18:21
  • $\begingroup$ @Peaceful thanks for flagging. I have changed the equation. It is unknown that F is an explicit function of only t or not, it may also contain other terms. $\endgroup$
    – zlin
    Feb 23, 2017 at 18:23
  • $\begingroup$ It is pretty standard (from differential equations theory) that $F$ in this case represents a non-homogeneous term that depends only on $t$. This is the equation of a driven damped oscillator and it is linear since it satisfies the superposition principle. $\endgroup$
    – Diracology
    Feb 23, 2017 at 18:25

1 Answer 1


You can go from definition of linearity.

If $x$ and $y$ are solutions of differential equation defined by linear operator $H$, then $x+y$ is also solution.

In other words, if $Hx=0$ and $Hy=0$, then it should follow that $H(x+y)=0$

Another property of linear operators is scaling: $H(\alpha x)=\alpha H(x)$

In your case it is easy to see that operator defining equation is non-linear. Let's say that $x$ and $y$ are solutions to the equation. Let's assume that $(x+y)$ is also solution:

${\frac {\mathrm {d} ^{2}(x+y)}{\mathrm {d} t^{2}}}+2\zeta \omega _{0}{\frac {\mathrm {d} (x+y)}{\mathrm {d} t}}+\omega _{0}^{\,2}(x+y)=(-2\zeta \omega _{0}{\frac {\mathrm {d} (x+y)}{\mathrm {d} t}})^2$

If you expand it:

${\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}+{\frac {\mathrm {d} ^{2}y}{\mathrm {d} t^{2}}} + 2\zeta \omega _{0}{\frac {\mathrm {d} x}{\mathrm {d} t}}+ 2\zeta \omega _{0}{\frac {\mathrm {d} y}{\mathrm {d} t}}+ \omega _{0}^{\,2}x+ \omega _{0}^{\,2}y =4(\zeta \omega _{0}{\frac {\mathrm {d} x}{\mathrm {d} t}})^2+ 4(\zeta \omega _{0}{\frac {\mathrm {d} y}{\mathrm {d} t}})^2+ 8(\zeta \omega_{0})^2 {\frac {\mathrm {d} x}{\mathrm {d} t}}{\frac {\mathrm {d} y}{\mathrm {d} t}}$

You will see requirement for $(x+y)$ to be solution: $8(\zeta \omega_{0})^2 {\frac {\mathrm {d} x}{\mathrm {d} t}}{\frac {\mathrm {d} y}{\mathrm {d} t}}=0$

Which is not automatically follows from $x$ and $y$ being solutions.

Hence, original operator (or original equation) is non-linear.

  • $\begingroup$ Thank u. So I guess that is the definition of linear system: f(x+y)=f(x)+f(y) is true for any function x, y and any value of t. $\endgroup$
    – zlin
    Feb 23, 2017 at 19:43
  • $\begingroup$ that is very crude approximation, but easy to understand. In fact, linearity means two things: $f(x+y)=f(x)+f(y)$ and $f(\alpha x)=\alpha f(x)$ $\endgroup$ Feb 23, 2017 at 20:42

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.