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I have a really simple differential equation: $\frac{dx}{dt} = x^2.t$ with initial value $x(0) = x_0$. Determine the maximal interval where it exists, depending on $x_0$

The maximal interval needs to be written in the form $(t^-,t^+)$, but I don't understand how you determine $t^+$ and $t^-$? please could someone explain if there is a method to do this?


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marked as duplicate by Christopher A. Wong, mrf, Ron Gordon, Chris Eagle, Henry T. Horton Feb 9 '13 at 23:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Have you tried solving the equation? – Christopher A. Wong Feb 9 '13 at 22:43
Did you not just ask this question:… – Ron Gordon Feb 9 '13 at 22:52
1/x0 − 1/x = 1/2.t^2 but how do i find the maximal interval? please can someone explain the method used to find this?! – camilla Feb 9 '13 at 22:52
I found this: – 1015 Feb 9 '13 at 23:06
@camilla: No "method" except looking at the solution and find those values of $t$ that avoid $1/0$ and square roots of negative numbers. – Ron Gordon Feb 9 '13 at 23:10

1 Answer 1

An initial value problem $x' = f(x)$ with $x(0) = x_0$, has a unique solution defined on some interval $(-a, a)$.

This IVP has a unique solution $x(t)$ defined on a maximal interval of existence $(\alpha,~\beta)$.

Furthermore, if $\beta \lt \infty$ and if the limit

$$x_1 = \lim_{t\rightarrow \beta^{-}} x(t)$$

exists then $x_1 \in \dot E$, the boundary of $E$. The boundary of the open set $E$, $\dot E = \overline E$ ~ $E$ where $\overline E$ denotes the closure of $E$.

On the other hand, if the above limit exists and $x_1 \in E$, then $\beta = \infty$, $f(x_1) = 0$ and $x_1$ is an equilibrium point of the IVP.

The domain of a particular solution to a differential equation is the largest open interval containing the initial value on which the solution satisfies the differential equation.

Theorem (Maximal Interval of Existence). An IVP has a maximal interval of existence, and it is of the form $(t^{-}, t^{+})$, with $t^{-} \in [-\infty, \infty)$ and $t^{+} \in (-\infty, \infty]$. There is a unique solution $x(t)$ on $(t^{-}, t^{+})$, and $(t, x(t))$ leaves every compact subset $\mathcal K$ of $\mathcal D$ as $t \downarrow t^{-}$ and as $t \uparrow t^{+}$.

Proof See ODE Notes.

More Examples of Domains See the very readable section More Examples of Domains.


$$x' = x^2, ~x(0) = 1$$

has the solution

$$x(t) = \frac{1}{1-t}$$

defined on its maximal interval of existence $(\alpha, ~ \beta) = (-\infty, 1)$.

Furthermore, $x_1 = \displaystyle \lim_{t\rightarrow 1^{-}} x(t) = \infty$. You can do the other side.

Original Problem

For your problem, we have:

$$\tag 1 x' = x^{2}t, ~ x(0) = x_0$$

Solving $(1)$, yields: $x(t) = \large -\frac{2}{c + t^{2}}$

Using the IC, $x(0) = x_0$, yields,

$$\tag 2 \large x(t) = -\frac{2}{t^{2} - \frac{2}{x_0}}$$

Depending on $c = \large \frac{2}{x_0}$, where $c \in \mathbb{R}$, there are several cases:

  • if $c \lt 0$, then $x(t) = \large x(t) = -\frac{2}{t^{2} - \frac{2}{x_0}}$ is a global solution on $\mathbb{R}$,

  • if $c \gt 0$, the solutions are defined on $(-\infty, -\sqrt{c})$, $(-\sqrt{c}, \sqrt{c})$, $(\sqrt{c}, \infty)$. The solutions are maximal solutions on $\mathbb{R}$, but are not global solutions.

  • if $c = 0$, then the maximal non global solutions on $\mathbb{R}$ are defined on $(-\infty, 0)$ and $(0, \infty)$.

Note for completeness, that there is another solution $x(t) = 0$, which is a global solution on $\mathbb{R}$.

I would strongly suggest:

$(1)$ That you review the solution and the different results for varying "c" and plot those to make sure you understand them.

$(2)$ That you use the initial definition I gave above, which works for $t^{-}$ and $t^{+}$ and make sure you can do it the way I showed and using that argument as per the theorem.


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+1000 why was this detailed answer left without any up-votes? – Babak S. Feb 16 '13 at 6:30
+1 Amzoti, which I could upvote $\times 1000$, like Babak! – amWhy May 3 '13 at 0:30

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