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For a scalar $c$, the function: $$f(x) = cx$$ Is linear in that it satisfies the superposition property. If: $$f(x_1)=cx_1$$ $$f(x_2)=cx_2$$ Then: $$\alpha f(x_1) + \beta f(x_2) = \alpha(cx_1)+\beta(cx_2)=c(\alpha x_1 + \beta x_2) = f(\alpha x_1 + \beta x_2)$$ However, for a scalar $b$, the function: $$g(x)=cx+b$$ Is not linear but is affine, since: $$\alpha g(x_1) + \beta g(x_2) = \alpha(cx_1+b)+\beta(cx_2+b)=c(\alpha x_1 + \beta x_2)+b(\alpha + \beta) \neq g(\alpha x_1 + \beta x_2)$$ How, then, can I apply the same reasoning to show that a differential equation is linear?

For example, the linear (or affine?) differential equation for a driven harmonic oscillator is:

$$\frac{d^2x}{dt^2}+2 \zeta \omega_0 \frac{dx}{dt} + \omega_0^2 x - \frac{F(t)}{m} = 0$$

This equation can be re-written as: $$\mathbf{w}^T\mathbf{x} - \frac{F(t)}{m} = 0$$ Where: $$\mathbf{w}=\begin{bmatrix} 1\\ 2 \zeta \omega_0\\ \omega_0^2 \end{bmatrix}$$ $$\mathbf{x}=\begin{bmatrix} \frac{d^2x}{dt^2}\\ \frac{dx}{dt}\\ x \end{bmatrix}$$

Which suggests that this is indeed affine. I am not sure if this reasoning is correct so would appreciate some feedback. Also, is there a special name for $\mathbf{x}$?

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  • $\begingroup$ X is just a degree of freedom in this case time dependent coordinate. $\endgroup$ – morbo Oct 11 '20 at 12:06
  • $\begingroup$ Hi @morbo I am not exactly sure what you mean, could you please explain? $\endgroup$ – mhdadk Oct 11 '20 at 12:50
  • $\begingroup$ Ahh, which x do you mean? Your vector x or the x in your diff equation? $\endgroup$ – morbo Oct 11 '20 at 13:06
  • $\begingroup$ I meant the vector $\mathbf{x}$ at the very end. $\endgroup$ – mhdadk Oct 11 '20 at 13:17
  • $\begingroup$ That as far as i know has no specific name, it is simply the vector descriptor of the system, if it was transformed into n first order der. Then it‘d be the statespace vector $\endgroup$ – morbo Oct 11 '20 at 13:26
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A system is some kind of function that maps an input as a function of t to an output. $$y(t) = H(u(t))$$ This system is linear if the following holds: $$y_1 = H(u_1), \quad y_2 = H(u_2)$$ $$\alpha y_1 + \beta y_2 = H(\alpha u_1 + \beta u_2)$$ for any scalar value $\alpha$, $\beta$. Your driven harmonic oscillator is currently described in such a way the combination of the force input $F(t)$ and the position $x(t)$ equals 0. As such, you could describe the system as the following: $$ 0 = H(F(t), x(t))$$ Apply the theorem of linearity on this function: $$0 = H(\alpha F_1 + \beta F_2, \alpha x_1 + \beta x_2)$$ $$0 = \frac{d^2}{dt^2}(\alpha x_1 + \beta x_2) + 2\zeta\omega_0\frac{d}{dt}(\alpha x_1 + \beta x_2) + \omega_0^2(\alpha x_1 + \beta x_2) - \frac{\alpha F_1 + \beta F_2}{m}$$ $$0 = \alpha\left(\frac{d^2x_1}{dt^2} + 2\zeta\omega_0\frac{dx_1}{dt} + \omega_0^2x_1 - \frac{F_1}{m}\right) + \beta\left(\frac{d^2x_2}{dt^2} + 2\zeta\omega_0\frac{dx_2}{dt} + \omega_0^2x_2 - \frac{F_2}{m}\right)$$ $$ 0 = \alpha H(F_1, x_1) + \beta H(F_2, x_2)$$ Because it is not intentional to have both the position and the force as inputs, the system could be described as the following: $$m\frac{d^2x}{dt^2} + 2\zeta\omega_0m\frac{dx}{dt} + \omega_0^2mx = F(t)$$ In this case, the position is the input of the system and the force the output. If you apply the properties of linearity on this system you will notice it is linear as well.

If a constant offset in a system that is logically not really an input (for instance its not controllable), the system is indeed not linear. For example: $$y(t) = F(x(t)) = cx(t) + b$$ This is often solved by linearizing the system such that the input takes this offset into account (supposing its known or can be approximated): $$y(t) = F_l(u(t)) = cu(t), \quad u(t) = x(t) + \frac{b}{c}$$ With this, the function is still not linear in $x(t)$, but is linear in $u(t)$.

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  • $\begingroup$ "In this case, the position is the input of the system and the force the output." Surely it is the other way around? This is a driven harmonic oscillator, so the force is the input and the displacement is the output, correct? $\endgroup$ – mhdadk Oct 13 '20 at 14:10
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    $\begingroup$ yes, it is usually the other way around. But in the current Differential Equation notation, I wanted to keep logic in the same matter. If you transform it to a Laplace equation: $$H(s) = \frac{X(s)}{F(s)} = \frac{1}{m(s^2+2\zeta\omega_0s+\omega_0^2)}$$ You'll notice the definition is literally inverted. Sadly, this notation is a bit hard to perform without using the Laplace notation since you cannot simply divide out the $x$ from a differential equation. $\endgroup$ – Petrus1904 Oct 14 '20 at 10:23

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