# A condition for uniqueness of solution

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 $$\int_0^c \frac{dx}{f(x)}= \infty.$$

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

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 $$I$$.

• 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 $x_0\cdot \min_{(0,x_0]}\frac{1}{f}=1.$ Dec 14, 2021 at 12:27
• 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 Dec 14, 2021 at 16:34