On the extension of solutions to ODEs

I've been trying to grasp some theorems on the extension of solutions and I still have some questions.

The theorems say thing about the solution escaping compact sets, being unbounded, etc. but I'm having trouble applying them to concrete equations.

Is there any way to estimate the size of the interval of a solution?

E.g., given the equation:

$x' = 1 + x^2$

how can we know, without explicitly or numerically solving it, that the interval where its solutions are defined has finite longitude? This is rather easy to see if you solve the equation as you get

$x(t) = tan(c + t)$

which is only defined in intervals of finite longitude.

But how would one reach the same conclusion just studying $f(x) = 1 + x^2$ ? Is this even possible?

On a related note, how does one go about proving that a solution can be extended to infinity?

Thank you in advance.

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About the solution of $x'=1+x^{2}$, you probably meant $x(t)=tan(c+t)$? –  Thomas E. Jan 13 '12 at 13:13
I'm not sure how to do any of these things in general, although comparison theorems can be very helpful. Proving divergence on a finite interval can be done by showing absolute value being greater than that of a known ODE whose solution blows up. Similarly bounding absolute value by an ODE whose solution is infinite proves that a solution can be extended to infinity. –  davin Jan 13 '12 at 13:32
Yes, @ThomasE. , I meant c+t :) –  hpusall Jan 13 '12 at 13:56
You mean "length", not "longitude". The two words have completely different meanings. –  Robert Israel Jan 13 '12 at 23:54

There is a general theorem about the existence of global solutions that you can find in

• Wolfgang Walter: "Ordinary Differential Equations", Springer

Paragraph XIII Existence of Global Solutions

If for a differential equation $$x' = f(t, x)$$ on $[0, \infty]$ there are differential functions $v, w$ with $$v \le w$$ and $$P w \le 0 \le P v$$ where $P$ denotes the defect of a function: $$P w := w' - f(x, w)$$ then there is a global solution $x$ with $$v \le x \le w$$

This is a kind of comparison theorem. In our case of ordinary differential equations on $\mathbb{R}$ it is possible to be more concrete and prove that there is a global solution if the function $f(t, x)$ grows at most linearly in $x$ (but that this statement is false if one replaces linear growth with an upper bound $|x^{r}|$ with $r\gt 1$). I think this is an important theorem, but it is surprisingly not mentioned in most ODE textbooks that I know. There is an online source, however, here. Look at theorem 2.17 and proposition 2.18.

So, for $$f(t, x) = 1 + x^2$$ you can see that all solutions will blow up in finite time $t$, because it grows stronger than linearly in $x$.

I don't know anything about estimations of the maximal interval, however.

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Let's show for example that if $f(x)\ge x^2$, then any solution of $x'=f(x)$ with $x(\pi/4)=1$ cannot exist beyond $t=1+\pi/4$. This is certainly the case with your $x'=1+x^2$ and the solution $x=\tan t$.

You have $x'\ge x^2$. Thus $x^{-2}x'\ge1$. Integrate from $\pi/4$ to $t$ giving $-x^{-1}+1\ge t-\pi/4$. Rearrange this to get $$x\ge\frac{1}{1+\pi/4-t}$$ That does it unless I messed up somewhere.

For your question about extending solutions, start with the Cauchy Picard existence theorem in any good book, where a short time of existence is given (in the proof if not in the statement) depending on $f$ and the initial conditions. To go beyond that look at a case such as the logistic model $x'=x-x^2$, with $0<x(t_0)<1$. Cauchy-Picard guarantees an interval of existence of length 10 or so, significantly independent of the initial condition, and you argue, say from the slopefield, that the solution remains in $[0,1]$. Therefore you can apply Cauchy-Picard repeatedly to assert that the solution can be extended for all $t$.

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Another technique that's often useful is interchanging independent and dependent variables. Consider again $\frac{dy}{dx} = 1 + y^2$. If you make $x$ the dependent and $y$ the independent variable, you get $\frac{dx}{dy} = \frac{1}{1+y^2}$. Note that since $1 + y^2 > 0$, solutions of the first DE with $y(x_0) = y_0$ defined for $x \in [x_0, x_1]$ will correspond to solutions of the second DE with $x(y_0) = x_0$ defined for $y \in [y_0, y_1]$ where $x(y_0) = x_1$. Now for such a solution, $x(y) = x_0 + \int_{y_0}^{y_1} \frac{dy}{1+y^2}$. But since $\int_{y_0}^\infty \frac{dy}{1+y^2} < \infty$, this second equation will have solutions for all $y>0$, but with $x(y)$ staying bounded. Turning that around, it says the interval $[x_0, b)$ where the first equation is defined must be bounded, with length $\int_{y_0}^\infty \frac{dy}{1+y^2}$.

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