# Estimate blow-up time.

How am I supposed to find the blow-up time of this ODE solution? $$y'=e^x + y^2 \qquad y(0)=0$$

The fact that it blows up it's granted by the fact that $$y' \geq y^2$$ which solution explodes. But how to estimate the time of explosion? I suppose I should use Gronwall or things like this, but don't actually know.

If we set $$y(x)=e^{x/2}f(e^{x/2})$$ and $$e^{x/2}=t$$ we are left with
$$e^{-x/2}f(e^{x/2})+f'(e^{x/2}) = 2 + 2\,f(e^{x/2})^2$$ $$\frac{f(t)}{t}+f'(t) = 2 + 2 f(t)^2,\qquad f(1)=0 \tag{A}$$ and $$f(t)$$ is greater than the solution of $$g(t)+g'(t) = 2+2\,g(t)^2,\qquad g(1)=0 \tag{B}$$ which is a separable DE. The blow-up time of $$g(t)$$ is given by $$1+\frac{\pi}{\sqrt{15}}+\frac{2}{\sqrt{15}}\arctan\frac{1}{\sqrt{15}}$$ hence the lifetime of $$y(x)$$ is bounded by $$2\log\left(1+\frac{\pi}{\sqrt{15}}+\frac{2}{\sqrt{15}}\arctan\frac{1}{\sqrt{15}}\right)<\color{red}{\frac{4}{3}}.$$ It is interesting to point out that $$(A)$$ can be solved in terms of a ratio of linear combinations of Bessel-J and -Y functions. The actual lifetime is twice the logarithm of the solution of $$\frac{Y_0(2t)}{Y_1(2)}=\frac{J_0(2t)}{J_1(2)}$$ which is closest to $$2$$, i.e. approximately $$\color{red}{1.27081}$$.
Since $$\forall x>0 \;\;y'(x)>0$$, there exists an inverse function $$x(y)$$ defined on $$[0,+\infty)$$. Its derivative is $$\tag{1} x'(y)=\frac1{e^{x(y)}+y^2}.$$ Integrating both sides of the equation (1), we obtain $$x(y)=\int_0^{y}\frac{dy}{e^{x(y)}+y^2}.$$ Since $$\forall x>0\;\; e^x>1$$, $$\int_0^{y}\frac{dy}{e^{x(y)}+y^2}<\int_0^{y}\frac{dy}{1+y^2} =\arctan y\Big|_0^{y}= \arctan y.$$ Finally, $$x(y)<\arctan y$$ implies $$\forall y>0\;\;x(y)<\frac{\pi}2$$. Hence, the blow-up time is less than $$\frac{\pi}2$$.
• Simple and nice and $\to +1$ – Claude Leibovici Jun 2 at 13:27