Linearization of nonlinear systems, linear systems and their solution forms (some small thoughts on differential equations, Laplace transforms, and ordinary differential equations of linear systems)

1. General method of linearization

Generally, the linearization of nonlinear systems uses the mathematical tool of Taylor expansion . By expanding around a certain point, a function can be written in polynomial form. A function $f\left(x\right)$ at point$x_0$The general formula for the nearby Taylor expansion is:
$$f(x)=\sum _{k=0}^{\infty }\frac{1}{k!}{\left(x-x_0\right)}^k{f}^{(k)}\left(x_0\right)\text{\hspace{1em}( 1 )}$$ When Taylor expansion is applied to function linearization, only the first item of linearization is taken, and all subsequent high-order items are represented by high-order small quantities. At this time, formula (1) can be written as:
$$f(x)=f\left(x_0\right)+\left(x-x_0\right)f^{\prime }\left(x_0\right)+o\left(x-x_0\right)\text{\hspace{1em}( 2 )}$$ where$o\left(x-x_0\right)$ for higher-order small quantities.

The so-called linearization means to express the original function in a linearized way, so the above formula can only be expressed as:
$$f\left(x\right)\approx f\left(x_0\right)+\left(x-x_0\right)f^{\prime }\left(x_0\right)$$ Obviously, this representation ignores the high-order small quantities, causing errors. Therefore, it can be said thatin the process of linearization by Taylor expansion, high-order small quantities can be regarded as errors in the process.

2. Linearized systems and their solutions

After a system is linearized, it can be expressed by the following linear ordinary differential equation:
$$a_n{x}^{(n)}+a_{n-1}{x}^{(n-1)}+\cdots +a_1\dot{x}+a_{0}x=b_m{u}^{\left(n\right)}+b_{m-1}{u}^{(m-1)}+\cdots +b_1\dot{u}+b_{0}u\text{\hspace{1em}( 3 )}$$ uu$u$ is the control quantity of the system,$x$ is the output of the system. The "linear" here means that each differential term in the system becomes an independent term, that is, there is no$\dot{x}\stackrel{¨This}{x}$ type of item; "constant" refers to the constant coefficient, that is, the coefficient$a_i,b_j$is a constant that does not change with time.

Equation (3) is a typical linear ordinary differential equation, and its solution can be divided into two parts : general solution and special solution . According to the knowledge of advanced mathematics, the general solution is obtained by setting the right side of formula (3) to 0:
$$a_n{x}^{(n)}+a_{n-1}{x}^{(n-1)}+\cdots +a_1\dot{x}+a_{0}x=0\text{\hspace{1em}( 3–1 )}$$ As for the special solution, there is a set of fixed solutions, which are mentioned in advanced mathematics without going into details here.

For formula (3–1), generally, when solving, two methods are given:
(1) The solution method of advanced mathematics is to write out its characteristic equation according to the order of derivation in formula (3–1) :
$$a_n{p}^n+a_{n-1}{p}^{n-1}+\cdots +a_{1}p+a_0=0\text{\hspace{1em}( 3–2 )}$$ The equation is$n$ order equation, so$n$个解$p_i(i=1,2,\cdots ,n)$ . According to the obtained$n$ solutions, directly write the general solution:
$$x(t)=\sum _{i=1}^n{C}_i{e}^{p_{i}t}\text{\hspace{1em}( 3–2–1 )}$$ Among them,$C_i$is a constant, determined by the initial conditions.
(2) Laplace transform solution . After learning the basic theory of automatic control, you can perform Laplace transform on formula (3–1) :
$$\begin{gathered} a_n{s}^{n}X(\mathrm{s})+a_{n-1}{s}^{n-1}X(\mathrm{s})+\cdots +a_{1}sX(\mathrm{s})+a_{0}X(\mathrm{s})=0 \\ ⟹\left(a_n{s}^n+a_{n-1}{s}^{n-1}+\cdots +a_{1}s+a_0\right)X\left(s\right)=0\text{\hspace{1em}( 3–3 )} \end{gathered}$$ To make formula (3–3) valid, obviously
$$a_n{s}^n+a_{n-1}{s}^{n-1}+\cdots +a_{1}s+a_0\equiv 0\text{\hspace{1em}( 3–4 )}$$ It is easy to see that formula (3–4) and formula (3–2) have the same form, so the general solution (3–2–1) can also be obtained.

The above two methods solve the general solution , that is, the right side of the equation (3) is 0, and the situation of the equation (3–1). Comparing formula (3) and formula (3–1), it is not difficult to find that (3–1) is obtained by setting the right side of the equal sign to zero in (3), that is, there is no control variable uu in the systemObtained when$u$ . Formula (3–1) does not contain$Item\; u$ , so there is no external effect in the system, and the whole system moves completely spontaneously and autonomously. Therefore, the dynamic process represented by the general solution (3–2–1) obtained from (3–1) is also called “free motion”.

It should be noted that "free movement" does not mean that there is no external action at all, but that at $t=At\; time\; 0$ , after applying an external action to the system, remove it immediately, and under this action, the system changes from the original static state to a dynamic process. while at$t>0$ process,$u$ no longer acts on the system in any form, so the whole system except at$t=0$ moment is$Except\; for\; the\; u$ excitation, the rest of the time is spontaneously moving. Obviously,$t=$uuapplied at time$0$$u$ is a pulse signal.

Correspondingly, each component $C_i{e}^{p_{i}t}$ , are the parts that make up the free movement, also known asthe mode. For the concept of mode, please refer toModal Feedback Control.

In addition to the general solution, formula (3) also has a special solution, that is, uu on the right side of the equal signThe solution obtained when the polynomial of$u\; is\; not\; negligible.$Obviously, these solutions are the existence of external action$The\; dynamic\; performance\; when\; u$ is sustaineduuThe movement caused by$u$ is called " forced movement".

Then, the solution of formula (3) can be expressed in the following way:
$$x(t)=x_{}(t)+x_{}\left(t\right)=x_{}\left(t\right)+x_{}\left(t\right)$$ That is:the motion of the system consists of free motion and forced motion.

3. The relationship between the Taylor expansion of the nonlinear system and the solution of the Laplace transform

We know that a nonlinear system $f\left(x\right)$ can be Taylor expanded into the form of formula (1):
$$f(x)=\sum _{k=0}^{\infty }\frac{1}{k!}{\left(x-x_0\right)}^k{f}^{(k)}\left(x_0\right)\text{\hspace{1em}( 1 )}$$ Emphatically point out: in the above formula,$x$ is an independent variable, and$x_0,f^{\left(k\right)}\left(x_0\right)$ is a constant.

May wish to carry out the Laplace transform of the above formula . Before the transformation, the following Laplace transform characteristics need to be used:
If L { f ( t ) } = F ( s ) , then L { f ( t − τ ) } = e − s τ L { f ( t ) } = e − s τ F ( s ) If \quad \mathscr{L} \left\{ f(t) \right\} = F({\rm s}), \qquad then \quad \mathscr{L } \left\{ f(t - \tau) \right\} = e^{-{\rm s} \tau} \mathscr{L} \left\{ f(t) \right\} = e^{ -{\rm s} \tau} F ({\rm s})likeL{ f(t)}=F(s)，butL{ f(t−) } _=e−sτL{ f(t)}=e− s τ F(s)The above properties are called time-shift properties.

Then introduce the Laplace transform of the power function:
L { tn } = n ! sn + 1 \mathscr{L} \left\{ t^n \right\} = \frac{n!}{ s^{n+1 } }L{ tn}=sn+1n!​

Then, the Laplace transform of formula (1) is:
L { f ( x ) } = L { ∑ k = 0 ∞ 1 k ! ( x − x 0 ) k f ( k ) ( x 0 ) } = ∑ k = 0 ∞ L { 1 k ! ( x − x 0 ) k f ( k ) ( x 0 ) } = ∑ k = 0 ∞ 1 k ! f ( k ) ( x 0 ) L { ( x − x 0 ) k } = ∑ k = 0 ∞ 1 k ! f ( k ) ( x 0 ) L { x k } e − x 0 s = ∑ k = 0 ∞ 1 k ! f ( k ) ( x 0 ) k ! s k + 1 e − x 0 s = e − x 0 s ∑ k = 0 ∞ 1 k ! f ( k ) ( x 0 ) k ! s k + 1 = e − x 0 s [ f ( x 0 ) 1 s + f ′ ( x 0 ) 1 s 2 + 1 2 f ′ ′ ( x 0 ) 2 s 3 + ⋯   ] (4) \begin{aligned} \mathscr{L} \left\{ f(x) \right\} &= \mathscr{L} \left\{ \sum _{k=0}^\infty \frac{1}{k!} \left( x - x_0 \right) ^k f^{(k)} \left( x_0 \right) \right\} \\ &= \sum _{k=0}^\infty \mathscr{L} \left\{ \frac{1}{k!} \left( x - x_0 \right) ^k f^{(k)} \left( x_0 \right) \right\} \\ &= \sum _{k=0}^\infty \frac{1}{k!} f^{(k)} \left( x_0 \right) \mathscr{L} \left\{ \left( x - x_0 \right) ^k \right\} \\ &= \sum _{k=0}^\infty \frac{1}{k!} f^{(k)} \left( x_0 \right) \mathscr{L} \left\{ x^k \right\} e^{-x_0 {\rm s}} \\ &= \sum _{k=0}^\infty \frac{1}{k!} f^{(k)} \left( x_0 \right) \frac{k!}{ s^{k+1} } e^{-x_0 {\rm s}} \\ &= e^{-x_0 {\rm s}} \sum _{k=0}^\infty \frac{1}{k!} f^{(k)} \left( x_0 \right) \frac{k!}{ s^{k+1} } \\ &= e^{-x_0 {\rm s}} \left[ f \left( x_0 \right) \frac{1}{s} + f' \left( x_0 \right) \frac{1}{s^2} + \frac{1}{2} f'' \left( x_0 \right) \frac{2}{s^3} + \cdots \right] \end{aligned} \tag{4} L{ f(x)}​=L{ k=0∑∞​k!1​(x−x0​)kf(k)(x0​)}=k=0∑∞​L{ k!1​(x−x0​)kf(k)(x0​)}=k=0∑∞​k!1​f(k)(x0​)L{ (x−x0​)k}=k=0∑∞​k!1​f(k)(x0​)L{ xk}e−x0​s=k=0∑∞​k!1​f(k)(x0​)sk+1k!​e−x0​s=e−x0​sk=0∑∞​k!1​f(k)(x0​)sk+1k!​=e−x0​s[f(x0​)s1​+f′(x0​)s21​+21​f′′(x0​)s32​+⋯]​( 4 ) Note that there is a term$e^{-x_0\mathrm{s}}$，将其泰勒展开有：
$$\begin{gathered} e^x=1+x+\frac{1}{2}{x}^2+\cdots =\sum _{i=0}^{\infty }\frac{1}{i!}{x}^i⟹ \\ {e}^{-x_0\mathrm{s}}=1-x_{0}s+\frac{1}{2}{x}_0^2{s}^2+\cdots =\sum _{i=0}^{\infty }\frac{1}{i!}{\left(-x_{0}s\right)}^i \end{gathered}$$由此得到
L { f ( x ) } = e − x 0 s ∑ k = 0 ∞ 1 k ! f ( k ) ( x 0 ) k ! s k + 1 = ∑ i = 0 ∞ 1 i ! ( − x 0 s ) i ⋅ ∑ k = 0 ∞ 1 k ! f ( k ) ( x 0 ) k ! s k + 1 (5) \mathscr{L} \left\{ f(x) \right\} = e^{-x_0 {\rm s}} \sum _{k=0}^\infty \frac{1}{k!} f^{(k)} \left( x_0 \right) \frac{k!}{ s^{k+1} } = \sum_{i=0}^\infty \frac{1}{i!} \left( -x_0 s \right)^i \cdot \sum _{k=0}^\infty \frac{1}{k!} f^{(k)} \left( x_0 \right) \frac{k!}{ s^{k+1} } \tag{5} L{ f(x)}=e−x0​sk=0∑∞​k!1​f(k)(x0​)sk+1k!​=i=0∑∞​i!1​(−x0​s)i⋅k=0∑∞​k!1​f(k)(x0​)sk+1k!​(5)

Example: The original function is:
$$f(t)=(t-3{)}^2=t^2-6t+9$$ (1) If it is solved by ordinary Laplace transform, there is:
L { f ( t ) } = L { t 2 − 6 t + 9 } = 2 s 3 − 6 s 2 + 9 s = 9 s 2 − 6 s + 2 s 3 (6–1) \mathscr{L} \left\{ f(t) \right\} = \mathscr{L} \left\{ t^2-6t+9 \right\} = \ frac{2}{s^3} - \frac{6}{s^2} + \frac{9}{s} = \frac{9s^2 - 6s + 2}{s^3} \tag{6 --1}L{ f(t)}=L{ t2−6 t+9}=s32​−s26​+s9​=s39s _2−6 s+2​( 6–1 ) (2) According to the property of time shift,$f(t)=(t-3{)}^2$ as$g(t)=t^2$ , according to the time-shift property:
L { f ( t ) } = L { ( t − 3 ) 2 } = e − 3 s L { t 2 } = e − 3 s 2 s 3 (6 –2) \mathscr{L} \left\{ f(t) \right\} = \mathscr{L} \left\{ (t-3)^2 \right\} = e^{-3 {\rm s}} \mathscr{L} \left\{ t^2 \right\} = e^{-3 {\rm s}} \frac{2}{s^3} \tag{6--2}L{ f(t)}=L{ (t−3)2}=e− 3 s L{ t2}=e− 3 ss32​( 6–2 ) Then formula (6–1) and formula (6–2) should be equal:
$$\frac{9s{\_}^2-6s+2}{s^3}=\frac{2e{\_}^{-3}}{s^3}\text{\hspace{1em}( 6–3 )}$$ the right side$2e{\_}^{-3\mathrm{s}}$泰勒展开：
$$2e{\_}^{-3s}=2\sum _{i=0}^{\infty }\frac{1}{i!}{\left(-3s\right)}^i=2\left(1-3s+\frac{9s{\_}^2}{2}+o\left(s^3\right)\right)$$where$o\left(s^3\right)$is a high-order small quantity. It can be seen by expanding the brackets that it is the molecule on the left side of the equal sign in formula (6–3). So (6–3) holds.

The above analysis shows that the complex plane expression obtained by directly performing Laplace transform on a function is the same as the complex plane expression obtained by first performing Taylor expansion and then Laplace transform.