# how the two equatoin are linearized

corresponding paper

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

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.

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 $$\mathbf{y}_e$$.

Resources:

• thanks for your answer. I did the second question as your process, but I got a different result, the constant should be omitted? May 24, 2022 at 3:21
• 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! May 25, 2022 at 12:04
• 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. May 25, 2022 at 12:07
• 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 . May 25, 2022 at 13:53
• 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. May 25, 2022 at 19:49