Notes: Linearization

I. Literature

II. Notes

1. Note of [LIT1]最適制御入門.pdf

• Solution of simultaneous homogeneous ordinary differential equations (pp.2)

$\left\{\begin{array}{l}{\dot{x}}_1=x_2 (1.A)\\ {\dot{x}}_2=-2x_1-3x_2+1 (1.B)\end{array}\right.$

Initial state is $x_1\left(0\right), x_2\left(0\right)$ when $t=0$ .

Substituting equation (1.A) into equation (1.B) , following equation may be obtained as:

${\ddot{x}}_1+3{\dot{x}}_1+2x_1=1 \left(2\right)$

The characteristic polynomial of equation (2) via Laplace transform is:

$s^2+3s+2=0 \left(3\right)$

eigenvalues of equation (3) are -1, -2, general solution of homogeneous part of equation (2) is $c_1{e}^{-t}+c_2{e}^{-2t}+c_3$ , where $c_1$ , $c_2$ and $c_3$ are arbitrary constant.
One of the particular solution may be obtained by substituting general solution into equation (2) as:

$c_1{e}^{-t}+4c_2{e}^{-2t}+3(-c_1{e}^{-t}-2c_2{e}^{-2t})+2(c_1{e}^{-t}+c_2{e}^{-2t}+c_3)=1 \left(4\right)$

Particular solution is $x_1=c_3=\frac{1}{2}$ , general solution is updated as

$x_1=c_1{e}^{-t}+c_2{e}^{-2t}+\frac{1}{2} \left(5\right)$

Which may be substituted into equation (1.A) :

$x_2=-c_1{e}^{-t}-2c_2{e}^{-2t} \left(6\right)$

The initial state is derived by setting $t=0$ into equation (5) (6) as:

$\left\{\begin{array}{l}{c}_1+c_2+\frac{1}{2}=x_1\left(0\right)\\ -c_1-2c_2=x_2\left(0\right)\end{array}\right. \left(7\right)$

The roots of simultaneous functions equation (7) are

$\left\{\begin{array}{l}{c}_1=2x_1\left(0\right)+x_2\left(0\right)-1\\ c_2=-x_1\left(0\right)-x_2\left(0\right)+\frac{1}{2}\end{array}\right. \left(8\right)$

, which may be substituted into equation (5) (6) and rearranged as:

$\left[\begin{array}{c}{x}_1\left(t\right)\\ x_2\left(t\right)\end{array}\right]=\left[\begin{array}{cc}2e^{-t}-e^{-2t}& e^{-t}-e^{-2t}\\ -2e^{-t}+2e^{-2t}& -e^{-t}+2e^{-2t}\end{array}\right]\left[\begin{array}{c}{x}_1\left(0\right)\\ x_2\left(0\right)\end{array}\right]+\left[\begin{array}{c}-e^{-t}+\frac{1}{2}{e}^{-2t}+\frac{1}{2}\\ e^{-t}-e^{-2t}\end{array}\right] \left(9\right)$

Some other method see [ref1] . Note equation (1) may be rewitten into state-space format as:

$\dot{\mathit{x}}\left(t\right)=\left[\begin{array}{cc}0& 1\\ -2& -3\end{array}\right]\mathit{x}\left(t\right)+\underset{regard as input}{\underbrace{\left[\begin{array}{c}0\\ 1\end{array}\right]}} \left(10\right)$

-> Solve by Matlab using dsolve ^[ref2] :

    syms x1(t) x2(t) x10 x20
    ode1 = diff(x1) == x2;
    ode2 = diff(x2) == -2*x1 - 3*x2 +1;
    odes = [ode1; ode2]
    cond1 = x1(0) == x10;
    cond2 = x2(0) == x20;
    conds = [cond1; cond2];
    [x1gsol(t),x2gsol(t)] = dsolve(odes)
    [x1sol(t),x2sol(t)] = dsolve(odes,conds)

We may find that xgsol is not coincide totally to equation (5) (6) , because general solution is a set of simultaneous polynomials. Meanwhile xsol is coincide to equation (9) .

Using transition matrix for examination

    syms t
    A = [0,1;-2,-3];
    expm(A*t)*[x10;x20]

We may find the result is totally the same with the first term of equation (9) .