LEC DQN 2

$X^{\prime }=AX$.
Uncoupled systems correspond to diagonal matrices. If the matrix is not diagonal but is diagonalizable (for example by having all distinct real eigenvalues), we can write $Y=P^{-1}X$. Then, $Y^{\prime }=P^{-1}X=P^{-1}AX=(P^{-1}AP)(P^{-1}X)$, so $Y^{\prime }=(P^{-1}AP)Y$. It follows that $X(t)=PE(t)P^{-1}X(0)$.

[!Lemma]
Let $X^{\prime }=AX$. Suppose $V_0$ is an eigenvector for $A$ with eigenvalue $\lambda \in \mathbb{R}$. Then $X(t)=e^{\lambda t}{V}_0$ is a solution to the system.

[!Proof]-

$$X^{\prime }(t)=\lambda e^{\lambda t}{V}_0=e^{\lambda t}(\lambda V_0)=e^{\lambda t}(A{V}_0)=AX(t).$$

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Equilibrium solutions: If $\det A\ne 0$, then $(0,0,\dots ,0)$ is the only equilibrium solution. If $\det A=0$, they will correspond to a kernel of $A$.

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[!Definition]
If $A$ has negative eigenvalues ${\lambda }_1,\dots ,{\lambda }_k$ and positive eigenvalues ${\lambda }_{k+1},\dots ,{\lambda }_n$ all distinct, let $\{v_1,\dots ,v_n\}$ be corresponding eigenvectors. Then, $E^S=\text{span}\{v_1,\dots ,v_k\}$ is called the stable subspace, and $E^{UN}=\text{span}\{v_{k+1},\dots ,v_n\}$ is called the unstable subspace.

Theorem 388.1(Theorem).

Given $T\in L({\mathbb{R}}^n)$ and $t_0⩾0$,

$$\sum _{k=0}^{\infty }\frac{T^k{t}^k}{k!}$$

converges absolutely and uniformly for all $|t|⩽t_0$.

the book shows absolute convergence. Convergence follows form the fact that all norms on ${\mathbb{R}}^n$ are equivalent and Prp 169.1.

Use $M$-test.

[!Definition]
For a matrix $A$, define

$$e^{At}:=\sum _{k=0}^{\infty }\frac{A^k{t}^k}{k!}$$

[!proposition]
If $S=PT{P}^{-1}$, then $e^S=P{e}^T{P}^{-1}$.

[!proposition]
If $S,T\in L({\mathbb{R}}^n)$ commute, then $e^{S+T}=e^S{e}^T$.

[!Proof]-

If $ST=TS$, then by the binomial theorem

$$(S+T{)}^n=n!\sum _{j+k=n}\frac{S^j{T}^k}{j!k!}.$$

Therefore,

$$e^{S+T}=\sum _{n=0}^{\infty }\sum _{j+k=n}\frac{S^j{T}^k}{j+k}=e^S{e}^T$$

[!Corollary]
$(e^T{)}^{-1}=e^{-T}$.

[!Corollary]
If

$$A=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right],$$

then

$$\begin{array}{r}{e}^A=e^a\left[\begin{array}{cc}\cos b& -\sin b\\ \sin b& \cos b\end{array}\right]\end{array}$$

Proof: write $\lambda =a+ib$. Write $A^k$ in terms of $\text{Re}({\lambda }^k)$ and $\text{Im}({\lambda }^k)$.

[!Corollary]
If $A=\left[\begin{array}{cc}a& b\\ 0& 0\end{array}\right]$, then $e^A=e^a\left[\begin{array}{cc}1& b\\ 0& 1\end{array}\right]$.

[!Proof]-
Use the fact that $A=aI+B$ and that $aI$ and $B$ commute.