Math Help Forum - View Single Post - Existence and uniqueness of global solutions

bkarpuz · #5 June 11th, 2009, 12:27 PM

Quote:

Originally Posted by bkarpuz

Dear friends, I need some help with the existence and/or uniqueness of global solutions to first-order linear differential equations.
For instance, let $x_{0},t_{0}\in\mathbb{R}$ and $A,B\in C([t_{0},\infty),\mathbb{R})$ and consider the following differential equation
$\begin{cases}x^{\prime}(t)=A(t)x(t)+B(t),& t\geq t_{0}\\x(t_{0})=x_{0}.&\end{cases}$
Which theorem ensures existence of global solutions to this initial value problem?
...

Consider the following initial value problem
$\begin{cases}x^{\prime}(t)=f(t,x(\tau(t))),&t\in[b,c]\\x(t)=\varphi(t),&t\in[a,b],\end{cases}\quad(1)$
where $f:[b,c]\times\mathbb{R}\to\mathbb{R}$ , $\varphi\in C([a,b],\mathbb{R})$ and $\tau\in C([b,c],[a,c])$ satisfies $\tau(t)\leq t$ for all $t\in[b,c]$ .
Set $I(\varphi,\varepsilon):=\{x\in\mathbb{R}:x\in B(\varphi(t),\varepsilon)\ \text{for}\ t\in[a,b]\}$ , where $B(x_{0},\varepsilon):=\{x\in\mathbb{R}:|x-x_{0}|\leq\varepsilon\}$ .

Theorem 1. Let $f:[b,c]\times I(\varphi,\varepsilon)\to\mathbb{R}$ for some $\varepsilon>0$ . Assume that there exist $M>0$ and $L>0$ such that $|f(t,x)|\leq M$ for all $(t,x)\in[b,c]\times I(\varphi,\varepsilon)$ and $|f(t,u)-f(t,v)|\leq L|u-v|$ for all $(t,u),(t,v)\in[b,c]\times I(\varphi,\varepsilon)$ . Then (1) has a unique solution on $[a,b+\delta]$ , where $\delta:=\min\{c-b,\varepsilon/M\}$ .

Proof. The proof can be given by following exactly the same arguments in the proof of Picard-Lidelof theorem, and thus omitted here. $\rule{0.2cm}{0.2cm}$

Now, consider the following initial value problem
$\begin{cases}x^{\prime}(t)=p(t)x(\tau(t))+q(t),&t\in[b,c]\\x(t)=\varphi(t),&t\in[a,b],\end{cases}\quad(2)$
where $p,q\in C([b,c],\mathbb{R})$ , and the other arguments are same to that of (1).

Corollary 1. (2) admits a unique solution on $[a,c]$ .

Proof. Let $f(t,u):=p(t)u+q(t)$ for $(t,u)\in[b,c]\times\mathbb{R}$ , and $M_{1},M_{2}>0$ satisfy $|p(t)|\leq M_{1}$ and [math]|q(t)|\leq M_{2}[math] for all $t\in[b,c]$ (since $p,q$ are continuous, we may always find such constants). The Lipschitz condition holds on $[\xi_{0},\xi_{1}]\times\mathbb{R}$ with the Lipschitz constant $M_{1}>0$ . Let $b=\xi_{0}<\xi_{1}<\cdots<\xi_{k_{0}}=c$ satisfy $\xi_{k}-\xi_{k-1}\leq1/(2M_{1})$ for $k=1,2,\ldots,k_{0}$ . For convenience in the notation define $x_{0}:=\varphi$ and $N_{0}:=\max\nolimits_{t\in[a,\xi_{0}]}\{|x_{0}(t)|\}$ . We may pick $\varepsilon_{0}>0$ such that $\varepsilon_{0}/(M_{1}(\varepsilon_{0}+N_{0})+M_{2})\geq1/(2M_{1})$ , we see that $|f(t,u)|\leq M_{1}(\varepsilon_{0}+N_{0})+M_{2}$ for all $(t,u)\in[\xi_{0},\xi_{1}]\times B(0,\varepsilon_{0}+N_{0})$ . Applying Theorem 1, we see that
$\begin{cases}x^{\prime}(t)=p(t)x(\tau(t))+q(t),&t\in[\xi_{0},\xi_{1}]\\x(t)=x_{0}(t),&t\in[a,\xi_{0}]\end{cases}$
admits a unique solution $x_{1}$ on $[a,\xi_{1}]$ since $\min\{\xi_{1}-\xi_{0},\varepsilon_{0}/(M_{1}(\varepsilon_{0}+N_{0})+M_{2})\}\geq\min\{\xi_{1}-\xi_{0},1/(2M_{1})\}$ . Next, let $N_{1}:=\max\nolimits_{t\in[a,\xi_{1}]}\{|x_{1}(t)|\}$ . We may find $\varepsilon_{1}>0$ such that $\varepsilon_{1}/(M_{1}(\varepsilon_{1}+N_{1})+M_{2})\geq1/(2M_{1})$ . And we have $|f(t,u)|\leq M_{1}(\varepsilon_{1}+N_{1})+M_{2}$ for all $(t,u)\in[\xi_{1},\xi_{2}]\times B(0,\varepsilon_{1}+N_{1})$ . Applying Theorem 1, we see that
$\begin{cases}x^{\prime}(t)=p(t)x(\tau(t))+q(t),&t\in[\xi_{1},\xi_{2}]\\x(t)=x_{1}(t),&t\in[a,\xi_{1}]\end{cases}$
admits a unique solution $x_{2}$ on $[a,\xi_{2}]$ . Repeating in this manner, we obtain the unique solution $x_{k_{0}}$ of (2) on $[a,c]$ . $\rule{0.2cm}{0.2cm}$

Remark 1. If we need to obtain the unique global solution to
$\begin{cases}x^{\prime}(t)=p(t)x(\tau(t))+q(t),&t\in[b,\infty)\\x(t)=\varphi(t),&t\in[a,b],\end{cases}\quad(3)$
where in addition $\lim\nolimits_{t\to\infty}\tau(t)=\infty$ is assumed to hold, we may pick an increasing divergent sequence $\{\xi_{k}\}_{k\in\mathbb{N}}\subset[b,\infty)$ with the convenience $\xi_{-1}:=a$ and $\xi_{0}:=b$ such that $\xi_{k-1}\leq\min\nolimits_{t\in[\xi_{k},\infty)}\{\tau(t)\}$ for all $k\in\mathbb{N}$ , and apply Corollary 1 successively to obtain the unique solution $x_{k}$ on each of the intervals $[\xi_{k-1},\xi_{k}]$ for $k\in\mathbb{N}$ by assuming the solution $x_{k-1}$ obtained in the previous step as the initial function on the current interval with the convenience $x_{0}:=\varphi$ . Then, letting $x(t)=x_{k}(t)$ for $t\in[\xi_{k-1},\xi_{k}]$ for $k\in\mathbb{N}$ , we obtain the unique global solution to (3).

Appendix. It is clear that the function $t/(at+b)\to1/a\geq1/(2a)$ as $t\to\infty$ for any fixed $a,b>0$ .

Remark. The proof is very simple in the case $\inf\nolimits_{t\in[b,\infty)}\{t-\tau(t)\}>0$ .

proof by bkarpuz