Is it a linear mass-spring system?

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.

• So where exactly do you see a possible non-linearity? – Peaceful Feb 23 '17 at 18:15
• Your possible values of $F(t)$ are wrong because as the notation suggests, $F$ is an explicit function of the time only. – Peaceful Feb 23 '17 at 18:16
• 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 $x^2$. – zlin Feb 23 '17 at 18:21
• @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. – zlin Feb 23 '17 at 18:23
• 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. – Diracology Feb 23 '17 at 18:25

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.

• 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. – zlin Feb 23 '17 at 19:43
• that is very crude approximation, but easy to understand. In fact, linearity means two things: $f(x+y)=f(x)+f(y)$ and $f(\alpha x)=\alpha f(x)$ – Oct18 is day of silence on SE Feb 23 '17 at 20:42