# Estimate on the interval of definition as parameter varies

Given $\alpha\in\mathbb R$ can we give an estimate on the interval of definition of the solution of $$x'=x^2(\alpha+\sin(x)),\quad x(0)=1?$$

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## 3 Answers

In order to get a global hold on the solutions of this equation one has to distinguish the seven cases $$\alpha<-1,\quad \alpha=-1,\quad -1<\alpha<0,\quad \alpha=0,\quad 0<\alpha<1,\quad \alpha=1,\quad \alpha>1\ .$$ For each of these cases you have to draw a figure of the following kind: Draw a horizontal $t$-axis and a vertical $x$-axis. All solution curves will be graphs in this plane. Then choose a "typical" $\alpha$ for the case at hand, e.g., $\alpha:=-{1\over2}$. Determine the zeros of the function $$\phi(x):=x^2(\alpha+\sin x)\ .$$ When $\alpha=-{1\over2}$ these are the points $0$, ${\pi\over6}+2k\pi$ and ${5\pi\over6}+2k\pi$ $\,(k\in{\mathbb Z})$. For each of these values $x_\iota$ the constant function $x(t)\equiv x_\iota$ is a solution. Therefore draw a horizontal line at each of the levels $x_\iota$. No other solution can cross such a line. Between two successive $x_\iota$s the function $\phi$ does not change sign. It follows that between two successive horizontal lines we see congruent solution curves rising from $t=-\infty$ on the lower line to $t=\infty$ on the upper line, or vice versa, and having both lines as asymptotes.

Once one has understood this idea it becomes obvious that all solutions are defined on all of ${\mathbb R}$, as soon as there are an infinity of $x_\iota$s. This is the case iff $|\alpha|\leq 1$.

In the case $|\alpha|>1$ there is only one zero of $\phi$, namely $x_0=0$. Here the solutions only have a finite life-span (or pre-history, depending on the sign of $\alpha$), which in addition depends on the initial condition.

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A straightforward but probably not optimal bound can be found as follows. Use $-1\leq\sin x\leq 1$ to get $$\alpha-1\leq \frac{d}{dt}\left(-x^{-1}\right)\leq \alpha+1.$$ Integrating over $[0,t]$, using the initial value and rearranging then gives $$1-(\alpha+1)t\leq\frac{1}{x}\leq1-(\alpha-1)t.$$ How useful this is depends on the value of $\alpha$.

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I'm afraid your calculations are not correct.. – guido giuliani Aug 21 '12 at 17:00
Thanks - I've corrected the calculation. – user12477 Aug 22 '12 at 8:32

Since a closed form solution is not possible. You can use the Taylor series expansion $\sin(x(t)) =x(t)-O(x(t)^3)$. If you take $\sin(x(t)) \approx x(t)$, then you still can not get a nice closed form solution. In this case you can use the fact that $\sin(x(0)) \approx x(0 ) = 1$, since you are looking for a solution in the neighbourhood of zero. In this case you can solve the ODE by the method of separation of variables. Then you find the particular solution by exploiting the given initial condition. Finally, determine the domain where the solution $x(t)$ exists.

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