Assuming that $f$ is a continuous real function and $f(0)=0$ , $f(x)>0 $ when $x\neq 0$, prove that the differential equation $x'= f(x)$ with the initial value $x(0)=0$ has a unique solution if and only if $$\int_0^c \frac{1}{f(x)}dx$$ is not defined for all $c\in \mathbb{R}$. That is $$\int_0^c \frac{dx}{f(x)}= \infty.$$

  • 3
    $\begingroup$ Nice problem. What's your attempt for to solve the problem? $\endgroup$
    – user798113
    Feb 17, 2021 at 21:05
  • $\begingroup$ This sounds to be a classical exercise. I suspect we should use epsilon approximations to the solution in some way. $\endgroup$
    – NotaChoice
    Mar 2, 2021 at 4:55

1 Answer 1


I am sharing a solution for one direction :

We note first that since $f$ is positive the integral $\int_0^c \frac{dx}{f(x)}$ can only be $+\infty$ when $c>0$ and $-\infty$ when $c<0$ as $\int_c^0 \frac{dx}{f(x)}= +\infty $ , moreover $\frac{1}{f}$ is continuous over $\mathbb{R}\setminus\{0\}$ and $\lim\limits_0 \frac{1}{f}= +\infty$

Let $y_1, y_2$ be two solutions of the differential equation, we seek to prove that $y_1= y_2$ under the hypothesis $\int_0^c \frac{dx}{f(x)}= \infty $ :

We have $y_1(0)= 0= y_2(0)$ and $$y_1(t)= \int_0^t f(y_1(s))ds \ \ ,\ \ y_2(t)= \int_0^t f(y_2(s))ds $$ for all $t$ in an interval $I= [-a,a]$ centered at $0$
The function $t\mapsto |y_1(t)- y_2(t)| $ is continuous and satisfies $$\max\limits_{t\in I} |y_1(t)- y_2(t)| = \max\limits_{t\in I} |\int_0^t (\ f(y_1(s))- f(y_2(s))\ )ds|\leq \int_0^{|t_0|} |f(y_1(s))- f(y_2(s))|ds $$ for some $t_0 \in I$ as the maximum is reached.

By the hypothesis we have $\int_0^a \frac{dx}{f(x)}= +\infty$ , the divergence of this integral implies that $\forall M>0\ \exists\ x_0\in (0,a] $ such that $(\min\limits_{(0,x_0]} \frac{1}{f} )\times x_0> M$ (we can show it by considering first the limit of $\frac{1}{f}$ then the fact that $\int_\delta^a \frac{dx}{f(x)}$ can be as large as desired and taking the mean value)
So we can transform this to $(\frac{1}{\max\limits_{(0,x_0]} f})\times x_0> M \quad$ or $\quad \max\limits_{(0,x_0]} f < x_0/M $
Using this in the previous inequality $$\int_0^{|t_0|} |f(y_1(s))- f(y_2(s))|ds \leq |t_0|\max\limits_{[0,t_0]} |f(y_1(s))- f(y_2(s))| \leq a\times 2\max\limits_J |f(s)|$$ where $J$ is the union $y_1([0,t_0])\cup y_2([0,t_0]) $ , $J$ is an interval (why?).

Now take $M$ to be $2a^2 N$ so that $\exists\ s_0\in J\setminus \{0\}$ for which $ \max\limits_{(0,s_0]} f(s) < a/M= \frac{1}{2aN}$ and the inequality becomes $$\max\limits_{I} |y_1(t)- y_2(t)|\leq 2a\max\limits_{(0,s_0]} f(s)< 1/N$$ As this hold for all $N>0$ then it can only be true when $\max\limits_{I} |y_1(t)- y_2(t)|= 0$ and we deduce that $y_1= y_2$ over the existence interval $I$.

  • $\begingroup$ There are some obvious gaps. At first $t_0=\pm a$ is easy to see. Second, $x_0\cdot\min_{(0,x_0]}\frac{1}{f}>M$ is also wrong. An example is $f(x)=x.$ Thus $x_0\cdot \min_{(0,x_0]}\frac{1}{f}=1.$ $\endgroup$
    – Vstal
    Dec 14, 2021 at 12:27
  • $\begingroup$ Thanks for pointing that out, I'm not sure if $\max\limits_{(0,a]} f<x_0/M$ will do the trick, anyway there is a way simpler answer to that problem which is here in stack $\endgroup$
    – NotaChoice
    Dec 14, 2021 at 16:34

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