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I have this system of ODEs and I'm trying to get a linearized version of it around the "operating point" $\overline{x}_1 = 1$

$$ \left\{\begin{matrix} \ddot{x_1}(t)+2\dot{x_1}(t)+2x_1^2(t)-2\dot{x_2}(t)=0 \\ 2\ddot{x_2}(t)+2\dot{x_2}(t)-2\dot{x_1}(t)=f(t) \end{matrix}\right. $$

So I define a small perturbation $\delta x_1$, $\delta x_2$ and $\delta f$ around the operating point $\overline{x}_1$, $\overline{x}_2$ and $\overline{f}$

$$\delta x_1 = x_1 - \overline{x}_1 \Rightarrow \dot{x_1} = \dot{\delta x_1} \Rightarrow \ddot{x_1} = \ddot{\delta x_1}$$

$$\delta x_2 = x_2 - \overline{x}_2 \Rightarrow \dot{x_2} = \dot{\delta x_2} \Rightarrow \ddot{x_2} = \ddot{\delta x_2}$$

$$\delta f = f - \overline{f} $$

I use Taylor polynomial to linearize $x_1^2(t)$ around $\overline{x}_1=1$ as

$$ x_1^2 \approx \overline{x}_1^2 + 2\overline{x}_1 \delta x_1 = 1 + 2\delta x_1$$

I replace all in the original equations:

$$\left\{\begin{matrix}\delta\ddot{x_1}(t)+2\delta\dot{x_1}(t)+2\left [1+2\delta x_1(t) \right ] - 2 \delta \dot{x_2}(t)=0 \\ 2\delta \ddot{x_2}(t)+2\delta \dot{x_2}(t)-2\delta \dot{x_1}(t)=\overline{f}+\delta f(t) \end{matrix}\right.$$

This system is "linear", but not homogeneous, because it has constant terms $2$ and $\overline{f}$. In fact, through force balance we get that $\overline{f}=2$, so the constant terms should mathematically cancel out somehow.

How can I get rid of this constant terms? Is there another (better) way to linearize this system of ODEs around $\overline{x}_1=1$

By the way, I got this systems of ODEs from this physical system:

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  • $\begingroup$ How did you determine that $x_1=1$ is the operating point? You will need a non-constant $\bar f$, as with a constant one you get $x_1=0$ as equilibrium point, just from physical considerations. Note that $$\frac{d}{dt}\left[\frac12 \dot x_1(t)^2+\dot x_2(t)^2+\frac23x_1(t)^3\right]=f(t)\dot x_2(t)-2(\dot x_1(t)-\dot x_2(t))^2,$$ where the last term continuously loses energy, leading to $x_1 \to 0$. You will need a very specific $f$ to continuously replace that lost energy. $\endgroup$ – LutzL May 23 '18 at 7:24
  • $\begingroup$ This is a problem from a textbook. It specifically ask to linearize about $x_1=1$. Thank you $\endgroup$ – user246185 May 23 '18 at 7:35
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When linearising a non-linear system of the form $\dot{x} = g(x,f)$ at an operating point $\bar{x}$ and $\bar{f}$ requires that $g(\bar{x},\bar{f})=0$. Since $\bar{x}_1$ is given and $g(x,f)$ is not a function of $x_2$, then $g(\bar{x},\bar{f})=0$ only has a solution when $x_2$ is omitted from the state space vector, so $x$ only contains $x_1$, $\dot{x}_1$ and $\dot{x}_2$ and no $x_2$. So $\bar{\dot{x}}_2$ can then be a non-zero constant, which can be chosen such that $g(\bar{x},\bar{f})=0$ can be satisfied.

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