a) Consider the initial value problem $\dot{x} = f(x) = x^{3/2}$ and $x(0) = 2$. Find the solution and show that the maximum interval of definition is of the form $(-\infty, A)$, with $A \in \mathbb{R}$ a constant.
b) Let $g(x)$ be a $C^1$-function $\mathbb{R} \to \mathbb{R}$ such that $g(x) > f(x)$ for all $x>0$. Show that the solution of $\dot{x}=g(x)$ and $x(0)=2$ goes in finite time to infinity.
So I did solve part a) and found the solution $x(t) = \frac{4}{(\sqrt{2}-t)^2}$ and the maximum interval of definition $(\infty, \sqrt{2})$.
Now I get the intuitive idea why the solution of the second initial value problem has to go to infinity in finite time, because the solution has to be larger than $x(t)$ and $x(t)$ goes to infinity when $t$ approaches $\sqrt{2}$. But how can I prove that if $y(t)$ is a solution of the second initial value problem, that $y(t) \geq x(t)$, so $y(t)$ has to go to infinity in finite time?
Best Answer
Let $I\subset [0,\sqrt{2})$ be an interval on which $x, y$ are both defined.
Let $J = \{ t\in I |\ \ \ y(t)\ge x(t)\}$. We want to prove that $$\lim_{t\to (\sup J)^-} y = \infty$$
For $t\in J_1$ the maximal interval of $J$ containing $0$, $$ y'(t) - x'(t) = g(y(t)) - f(x(t)) > 0 $$because $g > f$, $f$ is increasing and so $$s\in J \implies y(s)\ge x(s) \implies g(y(t)) > f(y(t)) \ge f(x(t)) $$
This proves that the difference between both increases, and becomes $\neq 0$ as soon as $g>f$ (that is, $t>0$). As this difference is continuous, positive and keeps increasing, then $J_1 = J = I$.
And as $\lim_{t\to(\sqrt{2})^-} x(t) = \infty$, $y$ is not defined for $t>0$ outside $I$ and $$\lim_{t\to (\sup J)^-} y = \infty$$