User:Egm6321.f11.team2.Xia/RP3.9

= R*3.9 - System of Coupled L1-ODE-VC =

Given
The following equation (1),(2),(3)p.15-4 $$ \displaystyle \mathbf{\Phi}(t,t_{0})_{n\times n}=\exp(\mathbf{A}(t-t_{0}))$$ (3.9.1) The fundamental or state transition matrix $$ \displaystyle \mathbf{\Phi}$$ is related to the integrating factor.

Properties of the state transition matrix $$ \displaystyle \mathbf{\Phi}$$ $$ \displaystyle \frac{d}{dt}\mathbf{\Phi}(t,t_{0})_{n\times n}=\mathbf{A}\mathbf{\Phi}(t,t_{0})_{n\times n}$$ (3.9.2)

$$ \displaystyle \mathbf{\Phi}(t_0,t_0)=\mathbf{I}$$ (3.9.1)

Find
Verify that (3.9.1) satisfies (3.9.2)-(3.9.3)

Solution
$$ \displaystyle \frac{d}{dt}\mathbf{\Phi}(t,t_{0})_{n\times n}=\frac{d}{dt}\exp(\mathbf{A}(t-t_{0}))=\mathbf{A}\exp(\mathbf{A}(t-t_{0}))=\mathbf{A}\mathbf{\Phi}(t,t_{0})_{n\times n} $$

(3.9.4) According to the formula of exponentiation of a matrix

$$ \displaystyle \exp(\mathbf{A})=\sum_{k=0}^{\infty}\mathbf{A}^{k}$$

(3.9.5) and $$ \displaystyle \exp(\mathbf{A}(t-t_{0}))=\sum_{k=0}^{\infty}\mathbf{A}^{k}(t-t_{0})^{k}$$ (3.9.6) Rebember $$ \displaystyle (t-t_0)^0=1$$

Hence

$$ \displaystyle \mathbf{\Phi}(t_0,t_0)=\mathbf{I}$$

-- Remark Sophus Lie is the first one who found the the solutions of (3.9.2) form a sort of group, we then call it Lie Group. Lie group equips both geometric characteristic (differentiable manifold) and algebraic characteristic (group).