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Find the solution of $x' = x^2t$ with initial value $x(0) = x_{0}$. Determine the maximal interval where it exists, depending on $x_{0}$

Please help me find the maximal interval!

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Hint: The equation is separable. (Assuming you mean $x'(t) = t\cdot x(t)^2$.) – mrf Feb 9 '13 at 21:55
sorry i wasnt very clear! I know how to separate and solve it, but i dont understand how to find the maximal interval! Please could you help with that. Thanks :) – camilla Feb 9 '13 at 21:59
up vote 1 down vote accepted

By separating and solving, the solution may be expressed as

$$\frac{1}{x_0} - \frac{1}{x} = \frac{1}{2} t^2$$

The LHS must be $>0$ as the RHS is. Therefore the maximal range of $x(t)$ is $\{x:x>x_0\}$. The maximal interval for $t$ that meets this condition on $x$ is then

$$ \left\{t:0<t<\sqrt{\frac{2}{x_0}} \right \}$$

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how did you work the maximal interval out? and we have to write in in the form (t^-,t^+), i take it means (0,strt(2/x0))? – camilla Feb 9 '13 at 22:22
@camilla: yes, that is an equivalent way of expressing it. I found it by using the fact that $1/x_0-1/x>0$, but also we want $x$ finite. It is clear that $1/x=0$ when $1/x_0 = (1/2) t^2$, so $x \ne 0$. Since we want a continuous interval, the result follow. – Ron Gordon Feb 9 '13 at 22:39

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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+1 again without any?! – Babak S. Feb 16 '13 at 6:31
Hmmm, where'd the OP go? +1 for a job well done! – amWhy May 3 '13 at 0:25
@Amzoti Are you sure that $c$ could be equal to zero? In this case $\frac{2}{x_0}=0$ yields $2=0$ – Zbigniew Feb 9 at 8:01

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