Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose a smooth vector field is given on plane. I want to find a global solution.

Is the following procedure ok, or is there a problem?

Since the vector field is Lipschitz continuous, we can find a unique local solution $\gamma(t)$ for an initial point $P=\gamma(0)$. If the solution is defined for $-\epsilon \le t \le \epsilon$ we can find a unique local solution for the point $\gamma(\epsilon)$, so we can extend $\gamma(t)$ for $-\epsilon \le t \le \epsilon+\delta$. By reapetedly doing this, we get either solution for $t \ge 0$ or solution for $0\le t < k$. The first case will be okay, and the second case will be that the solution blows up at $t=k$.

share|improve this question

4 Answers 4

up vote 3 down vote accepted

That is basically how it goes. But there are different ways to organize the proof. I think the best is to prove uniqueness first, for any solution defined on an open interval around $t=0$. Then you take the union of all open intervals on which a solution exists, and use uniqueness to conclude that there is a solution on this union. This is, by definition, the solution on the maximal interval of existence.

Say one end point of the maximal interval of existence is $\beta$. If $\gamma(t)$ remains bounded as $t\to\beta$, there is a sequence $t_n\to\beta$ so that $\gamma(t_n)\to b$, say. Now straighten out the vector field around $b$ in order to obtain a contradiction. Conclude that $\gamma(t)$ goes to infinity at $\beta$.

share|improve this answer

The procedure is fine.

I am assuming the system you are trying to solve is $\dot{x} = f(x)$, where $x\in \mathbb{R}^n$, and $f$ is smooth.

Let $C_K = \overline{B(0,K)}$. This is compact for all $k$, so choose a Lipschitz constant $L_K$ that would be valid on $C_{K+1}$. Choose $x_0 \in C_K$ which is the initial condition at some time $t_0 \in \mathbb{R}$. Then by the existence & uniqueness theorem for ODEs (see, eg, Kantorovich & Akilov, "Functional analysis", XVI, Section 4, Theorem 1) there exists $\epsilon>0$ such that a unique solution starting from $x_0$ exists on $[t_0,t_0+\epsilon]$. Furthermore, this $\epsilon$ is uniformly valid for all $x_0 \in C_K$, hence if $x(t_0+\epsilon) \in C_K$, the procedure can be repeated to extend the solution to $[t_0,t_0+2\epsilon],...,[t_0,t_0+(k+1)\epsilon]$, etc, as long as $x(t_0+k\epsilon) \in C_K$.

Let $T_K = \sup \{ t_1 | x(\xi) \in C_K, \ \forall \xi \in [t_0,t_1] \}$. $T_K$ is a non-decreasing sequence, so let $T_\infty=\lim_{K\to \infty} T_K$. If $T_\infty = \infty$, then a solution exists for all $t\geq t_0$, if not, then the solution becomes unbounded in finite time, ie, $\lim_{t\to T_\infty} \|x(t)\| = \infty$ (since $\|x(T_K)\|=K$).

A simple example of this unbounded behaviour in $\mathbb{R}$ is $\dot{x} = x^2$ with a positive initial condition.

Global uniqueness follows from local uniqueness. If $x,y$ are two solutions starting from $x_0$, then let $t^* = \sup \{ t\geq t_0 | x(t) = y(t) \}$. If $t^* < \infty$, then by considering the system starting from $x(t^*)$ (which is the same as $y(t^*)$), we see that there is a unique solution on some interval $[t^*, t^*+\delta]$, which contradicts the definition of $t^*$.

So we see that with a smooth field, you can always extend the solution in space, but not necessarily in time.

share|improve this answer

Much more details about the problem are needed. Smooth does not imply Lipschitz, only locally Lipschitz, which if I remember correctly only gives rise to a unique maximal solution, which is either defined for all time or blows up at some point. Your patchwork procedure would work to find this, compare with the Euler method for the numerical computation of ODE.

share|improve this answer

There are two issues to think about.

First there is the difference between a smooth manifold and smooth parametrisable manifold. Take the sphere for example, you can find an explicit parametrisation close to any point, but there is no global parametrisation.

Second, what do you mean by a global solution? Does this have to be in terms of elementary functions? Consider the vector field given by $dy - e^{x^2}dx = 0$. There is no elementary function that parametrises the integral curves because $e^{x^2}$ doesn't have an elementary anti-derivative. (Even though the Picard–Lindelöf theorem tells us there exists a unique solution given sufficient initial conditions.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.