# Thread: Existence and uniqueness of global solutions

1. ## [SOLVED] Existence and uniqueness of global solutions

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

I am actually wondering to find this result for delay differential equations
$\displaystyle \begin{cases} x^{\prime}(t)=A(t)x(\alpha(t))+B(t),& t\geq t_{0}\\ x(t)=\varphi(t)&, t_{-1}\leq t\leq t_{0}, \end{cases}$
where $\displaystyle t_{-1}:=\min\{\alpha(t):t\geq t_{0}\}$, $\displaystyle \varphi\in C([t_{-1},t_{0}],\mathbb{T})$, $\displaystyle \alpha(t)\leq t$ for all $\displaystyle t\geq t_{0}$ and $\displaystyle \lim\nolimits_{t\to\infty}\alpha(t)=\infty$.

2. Hi

I know nothing of delay differential equation, but there is a theorem that ensures the existence and uniqueness of a solution : Picard?Lindelöf theorem - Wikipedia, the free encyclopedia

But it requires some conditions.

3. Originally Posted by Moo
Hi

I know nothing of delay differential equation, but there is a theorem that ensures the existence and uniqueness of a solution : Picard?Lindelöf theorem - Wikipedia, the free encyclopedia

But it requires some conditions.
For delay equations if we have a minimal delay, i.e., $\displaystyle \alpha(t)$ is bounded away from $\displaystyle t$ by a positive constant, then the Global E&U follows very easily by the method of steps.
But the problem is when the delay is not minimal.

4. I think I have solved the problem.
I will soon sketch the proof for the ones who wonder to know about it.

Thanks.

5. ## Proved.

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 $\displaystyle x_{0},t_{0}\in\mathbb{R}$ and $\displaystyle A,B\in C([t_{0},\infty),\mathbb{R})$ and consider the following differential equation
$\displaystyle \begin{cases} x^{\prime}(t)=A(t)x(t)+B(t),& t\geq t_{0}\\ x(t_{0})=x_{0}.& \end{cases}$
Which theorem ensures existence of global solutions to this initial value problem?
...
Consider the following initial value problem
$\displaystyle \begin{cases} x^{\prime}(t)=f(t,x(\tau(t))),&t\in[b,c]\\ x(t)=\varphi(t),&t\in[a,b], \end{cases}\quad(1)$
where $\displaystyle f:[b,c]\times\mathbb{R}\to\mathbb{R}$, $\displaystyle \varphi\in C([a,b],\mathbb{R})$ and $\displaystyle \tau\in C([b,c],[a,c])$ satisfies $\displaystyle \tau(t)\leq t$ for all $\displaystyle t\in[b,c]$.
Set $\displaystyle I(\varphi,\varepsilon):=\{x\in\mathbb{R}:x\in B(\varphi(t),\varepsilon)\ \text{for}\ t\in[a,b]\}$, where $\displaystyle B(x_{0},\varepsilon):=\{x\in\mathbb{R}:|x-x_{0}|\leq\varepsilon\}$.

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

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

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

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

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

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

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

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

proof by bkarpuz