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Thread: Existence and uniqueness of global solutions

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation [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 $\displaystyle \lim\nolimits_{t\to\infty}\alpha(t)=\infty$.


    Thanks for your help.
    Last edited by bkarpuz; Sep 19th 2009 at 03:27 AM. Reason: delay equation is added.
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  2. #2
    Moo
    Moo is offline
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    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.
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  3. #3
    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by Moo View Post
    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.
    TY Lady Moo but I already have them read.
    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.
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  4. #4
    Senior Member bkarpuz's Avatar
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    I think I have solved the problem.
    I will soon sketch the proof for the ones who wonder to know about it.

    Thanks.
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  5. #5
    Senior Member bkarpuz's Avatar
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    Thumbs up Proved.

    Quote Originally Posted by bkarpuz View Post
    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
    Last edited by bkarpuz; Sep 20th 2010 at 12:51 PM.
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