corresponding paper

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Question1: how the equatoin 3 is linearized in the neighborhood of the equilibrium point $x = x_{sp}, \dot x =0 $

4 Question2: how the equatoin 12 is linearized in the neighborhood of the equilibrium point $y = y_r$ ? what operations on $y \ \dot y \ {\dot y}^2 \ \ddot y$ and the constant ?

let $f_1 = \dot y,$ and $f_2 = \ddot y= -(\lambda/2) (\dot y + 2 r_0) \dot y - U \lambda y - 2 U - (\lambda r_0^2) /2$

let $ d[y,\dot y]/dt=[f_1,f_2]=0 $ then get the equilibrium point $$ \dot y_r =0,y_r=-2/\lambda-r_0^2/(2U) $$ then I linearized $f_2$ around $[y_r,\dot y_r]$

get the two partial derivatives

$\frac{\partial f_2}{\partial \dot{y}} = -\lambda \dot y-\lambda r_0$

$\frac{\partial f_2}{\partial y} = -U \lambda$

get the value of the two partial derivatives at $[y_r,\dot y_r]$

$\frac{\partial f_2}{\partial \dot{y}}|_{y_r,\dot y_r} = -\lambda r_0$

$\frac{\partial f_2}{\partial y}|_{y_r,\dot y_r} =-U \lambda $

obviously $f_2(y_r,\dot y_r)=0$

linearized differential equation in the neighborhood becomes:

$\ddot y = f_2 \approx 0 +(- \lambda r_0)(\dot y- \dot y_r) +(-U \lambda) (y-y_r) = - \lambda r_0 \dot y- U \lambda y + U \lambda y_r$

final result is

$\ddot y + \lambda r_0 y + U \lambda y = U \lambda y_r \ne 0$

not the same as in the paper.


1 Answer 1


Consider the formula for the linearization of a function $f(\mathbf{x})$ about a point $\mathbf{x}_{e}$, where $\mathbf{x}, \mathbf{x}_{e} \in \mathbb{R}^{n}$:

$$f(\mathbf{x}) \approx f(\mathbf{x}_{e}) + \nabla f|_{\mathbf{x}_{e}} \cdot (\mathbf{x}-\mathbf{x}_{e})$$

I will only write out the math for problem 1, but problem 2 is hardly any different. So, we have:

$$\ddot{x} = f(x, \dot{x})$$ $$\ddot{x} = -\lambda \frac{|\dot{x}|\dot{x}}{2} - U \lambda (x-x_{sp})$$ Thus, when considering the linearization about the point $\mathbf{x}_{e} = [x_e \ \dot{x_e}]^T = [x_{sp}\ 0]^T$ , we have: $$f(\mathbf{x}_e) = 0$$ $$\nabla f|_{\mathbf{x}_e} \cdot (\mathbf{x} - \mathbf{x}_e) = \frac{\partial f}{\partial x}|_{x_{sp}} \cdot (x-x_{sp}) + \frac{\partial f}{\partial \dot{x}}|_{0} \cdot (\dot{x} - 0) $$ We can see the first term $\frac{\partial f}{\partial x}|_{x_{sp}} \cdot (x-x_{sp})$ is simply $-U\lambda (x-x_{sp})$. As the second term is the point of interest in this question, we will work it out in a more verbose manner: $$\frac{\partial f}{\partial \dot{x}} = -\lambda \frac{2sgn(\dot{x})\dot{x}}{2} = - \lambda|\dot{x}|$$ $$\frac{\partial f}{\partial \dot{x}}|_{0} = -\lambda |0| = 0$$ $$\frac{\partial f}{\partial \dot{x}}|_{0} \cdot (\dot{x} - 0) = 0$$ Thus, our linearized differential equation in the neighborhood of $\mathbf{x}_{e}$ becomes: $$\ddot{x} + U\lambda(x-x_{sp}) \approx 0 $$ The intuition as to why we see certain terms disappear is that as we approach neighborhoods of the origin, terms with degree $\gt 1$ (i.e. $|\dot{x}|\dot{x}$), or more generally first order Taylor expansions that are still functions of the original variable (for the most part), have derivatives which are negligible in a neighborhood of the origin - unlike a linear term. Remember that $f(\mathbf{x}_{e}) = 0$, so we should not see any $i^{th}$ order term where $\lim\limits_{\mathbf{x} \to 0} f^{(i)}(\mathbf{x}) = \infty$, so this removes the possibility of a first-order term like $1/\mathbf{x}$.

Finally, consider a point $\epsilon \approx 0$. Given functions $g(\epsilon) = \epsilon$ and $h(\epsilon) = \epsilon^{2}$, we note $g'(\epsilon) = 1$ and $h'(\epsilon) = 2\epsilon$. Since we know $\epsilon \approx 0$, $h'(\epsilon) \approx 0$ as well. Thus if we have the equation $f(x) = g(x) + h(x)$ modeling our dynamics, we would see that the linear term "dominates" close to the origin.

Edit: Let $y_{r} = -\frac{1}{U\lambda}(2U + \frac{\lambda r_{0}^{2}}{2})$ and $\dot{y}_{r} = 0$

Consider (12) rewritten as:

$$\ddot{y} = f(y, \dot{y})$$ $$\ddot{y} = -\frac{\lambda}{2}(\dot{y}+2r_{0})\dot{y} - U\lambda y - 2U - \frac{\lambda r_{0}^{2}}{2}$$ $$\ddot{y} = -\frac{\lambda}{2}\dot{y}^{2} - \lambda r_{0}\dot{y} - U\lambda y - 2U - \frac{\lambda r_{0}^{2}}{2}$$

Let us now consider $\frac{\partial f}{\partial y}$ and $\frac{\partial f}{\partial \dot{y}}$:

$$\frac{\partial f}{\partial y} = -U\lambda$$ $$\frac{\partial f}{\partial \dot{y}} = -\lambda\dot{y} - \lambda r_{0}$$

Now, let us consider the equilibrium point $\mathbf{y}_e = [y_{r} \ \dot{y}_{r}]^{T}$. So, using the linearization formula from part one, we get:

$$f(y_{r}, \dot{y}_{r}) = 0$$ $$\frac{\partial f}{\partial y}|_{y_{r}} \cdot (y-y_{r}) = -U\lambda (y-y_{r}) = -U\lambda y - 2U - \frac{\lambda r_{0}^{2}}{2}$$ $$\frac{\partial f}{\partial \dot{y}}|_{\dot{y}_{r}} \cdot (\dot{y}-\dot{y}_{r}) = -(\lambda \dot{y}_{r} - \lambda r_{0})\cdot (\dot{y}-\dot{y}_{r}) = -\lambda r_{0} (\dot{y}-0) = -\lambda r_{0}\dot{y}$$

Now, I would assume the author drops the constant terms as they have no impact on the stability characteristics at the point $\mathbf{y}_e$ (they only determine where the equilibrium point will be, not what the convergence actually looks like). After dropping these terms, we are left with:

$$\ddot{y} + \lambda r_{0}\dot{y} + U \lambda y \approx 0$$

in a neighborhood of the point $\mathbf{y}_e$.


  • $\begingroup$ thanks for your answer. I did the second question as your process, but I got a different result, the constant should be omitted? $\endgroup$
    – eason
    May 24, 2022 at 3:21
  • $\begingroup$ After looking it over problem two does have some notable differences with the original problem - mainly the change of variables. I've edited my answer to include work for part two, hopefully it helps! $\endgroup$ May 25, 2022 at 12:04
  • $\begingroup$ And yes, the constant yr can be assumed to be 0, but it's always nice to be told explicitly that this is the case. $\endgroup$ May 25, 2022 at 12:07
  • $\begingroup$ The paper says that "that is, the position error is constant and the velocity error is zero", so the $y_r$ is not to be zero but $ \dot y_r$ is . Maybe the constant does not matter in differential equation . $\endgroup$
    – eason
    May 25, 2022 at 13:53
  • $\begingroup$ I believe you are correct there. I have added in the definition they gave for yr, and it seems like the only way they could come to the conclusion reached was if the constant terms were dropped. $\endgroup$ May 25, 2022 at 19:49

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