202303111503

Type :Note Tags : Differential Equations

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Linear Systems

Note

ODEs where dependence on phase is linear. For example, $\dot{x}=kx\leftrightarrow x(t)=x(0)e^{kt}$ $\dot{x}=p(t)x(t)+q(t)$

Existence and uniqueness for 1st order Linear systems

Note

Let $I\subseteq \mathbb{R}$, be an interval and $\mathcal{A}:I\to M_n(\mathbb{R})$ $b:I\to {\mathbb{R}}^n$ be continuous on $I$. Let $(t_0,x_0)\in I\times {\mathbb{R}}^n$. Then the IVP $\dot{\bar{x}}(t)=\mathcal{A}(t)+b(t);\bar{x}(t_0)=x_0$ has a unique solution on $I$.

Proof:

1. WLOG assume that $I$ is compact. Since if not, then we can take a sequence $I_n$ of compact subsets of $I$, which converge to $I$, such that on each of the $I_n$‘s the solution is unique, and then the solution on $I$ is given by patching together the solutions on these $I_n$‘s.
2. Take $I=[\alpha ,\beta ]$. Define $v:I\times {\mathbb{R}}^n\to {\mathbb{R}}^n$ $(t,\bar{x})\to A(t)\bar{x}+b(t)$
3. Check that $v$ satisfies the conditions of local existence and uniqueness thm (Maximal interval of existence theorem), i.e., check that $v$ is continuous and is lipschitz w.r.t. $x$.
4. To apply the theorem, we need open domain, so we extend $I$ to $I_{\eta }=(\alpha -\eta ,\beta +\eta )$ for some $\eta >0$,

\widetilde{A}(t) := \begin{cases} A(\alpha) & \mathrm{if} & \alpha-\eta < t \le \alpha; \ A(t) & \mathrm{if} & \alpha<t<\beta;\ A(\beta) & \mathrm{if} & \beta \le t < \beta + \eta \end{cases}

Similarly for $\widetilde{b}$. 5. We have $\widetilde{v} : I_{\eta} \times \mathbb{R}^{n} \to \mathbb{R}^{n}$. $$ \widetilde{v}(t,\bar{x}) = \widetilde{A}(t)\bar{x}+ \widetilde{b}(t)

being lipschitz with respect to $x$, now apply the local existence and uniqueness theorem which gives a unique solution on a maximal interval $({\omega }_{-},{\omega }_{+})$. 6. Show that $({\omega }_{-},{\omega }_{+})=I_{\eta }⟹{\varphi }_{max}{|}_I$ is unique solution of the given IVP. For $t\in [t_0,{\omega }_{+})$, we have (here $M=\underset{s\in I}{\sup }|b(s)|$, and $\mid \mid \tilde{A}(s)\mid \mid \le L$ for all $s\in I$)

$$\begin{array}{rl}{\varphi }_{max}(t)& =x_0+\underset{t_0}{\overset{t}{\int }}\tilde{v}(s,{\varphi }_{max}(s))ds\\ & \\ & =x_0+\underset{t_0}{\overset{t}{\int }}(\tilde{A}(s){\varphi }_{max}(s)+\tilde{b}(s))ds\\ ⟹|{\varphi }_{max}(t)|& \le |x_0|+\underset{t_0}{\overset{t}{\int }}\mid \mid \tilde{A}(s)\mid \mid \mid {\varphi }_{max}(s)\mid ds+M(t-t_0)\\ & \le |x_0|+L\underset{t_0}{\overset{t}{\int }}\mid {\varphi }_{max}(s)\mid ds+M(t-t_0)\end{array}$$

7. Take $K=|x_0|+M(t-t_0)$, then using Gronwall’s Inequality, we get $|{\varphi }_{max}(t)|\le K{e}^{L(t-t_0)}\forall t\in [t_0,{\omega }_{+})$, hence it is bounded. But we know $(t,{\varphi }_{max}(t))$ escapes every compact subset of $I_{\eta }\times {\mathbb{R}}^n$ as $t\to {\omega }_{+}$, hence ${\omega }_{+}=\beta +\eta$, similarly ${\omega }_{-}=\alpha -\eta$. Therefore, ${\varphi }_{max}$ exists on $I$ and is the unique solution to the required IVP.

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Related Problems

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References

Gronwall’s Inequality Maximal Interval of Existence