Show that $f'(x) = f^2(x)$ and $f(0) = 0$ implies $f$ is the zero function

Question

Let $f : \mathbb{R} \to \mathbb{R}$ differentiable be such that $$f'(x)=f^2(x)$$ for all $x$ and $f(0)=0$. Show that $f(x)=0$ for all $x$.

My attempt:

Notice that $f'(x)=f^2(x)\geq 0$, thus $f$ is increasing. Since $f(0)=0, f(x)\geq0\; \forall x \geq 0$.
Since $f$ is differentiable, $f$ is continous, therefore given $1> \epsilon >0$, there exists $\delta<1$ such that for $x \in (-\delta,\delta)\implies f(x)<\epsilon<1.$ Take $x=\delta/2$. Then $f(x) = f'(c)\delta/2=f^2(c)\delta/2$. Then $f^2(c) \geq f(x)$. But since $c\in (-\delta,\delta)$, $0\leq f (c)<1 \implies f^2(c) \leq f(c)\implies f(c)\geq f(x)$.
But $c<x$, therefore $f(c)\geq f(x)$ which means $f(c)=f(x)$. Therefore for all $y \in (c,x)$, $f(y)=f(x) \implies f'(y)=f^2(y)=0 \implies f(y)=0$.
Where do I go from here?

Answer 1

Different solution: $$f'(x)=f^2(x)$$ $f(x)=0$ is a trivial solution. If $f(x)\neq 0$, then we can divide by $f^2(x)$: $$\frac{f'(x)}{f^2(x)}=1$$ Now integrate both sides with respect to $x$: $$\int \frac{f'(x)}{f^2(x)} \mathrm{d}x=\int 1 \mathrm{d}x$$ $$-\frac{1}{f(x)}=x+C$$ $$\frac{1}{f(x)}=C-x$$ $$f(x)=\frac{1}{C-x}$$ From this, we have: $f(0)=\frac{1}{C}=0$. But $\nexists C \in \mathbb{R}$ so that $\frac{1}{C}=0$, so there can't be a solution in the form of $f(x)=\frac{1}{C-x}$.
Now let's check the trivial solution: $f(x)=0$ is $0$ at $x=0$. It's the only solution that satisfies the initial conditions, so $f(x)=0$ is the only solution.

Answer 2

Hint for a brief solution:

$y=0$ is an obvious solution of $y'=y^2$ that satisfies the initial condition $y(0)=0$.

Ist is easy to prove that the other solutions are $y=\frac{1}{c-x}$ that cannot be $0$.

Answer 3

A solution that proceeds by showing that $(1/f)'=-1$, when we're trying to show that $f=0$, seems unfortunate. It may well be fixable. Here's a somewhat similar solution where division by zero doesn't come anywhere nearby:

Let $A=\{x:f(x)=0\}$. Then $A$ is closed and nonempty, so if we can show that $A$ is open then $A=\Bbb R$, qed.

Suppose that $x_0\in A$. Then we certainly have $1+xf(x)\ne0$ near $x_0$, so near $x_0$ we have $$\left(\frac {f(x)}{1+xf(x)}\right)'=0.$$So $$f(x)=C(1+xf(x))$$near $x_0$. Since $f(x_0)=0$ we must have $C=0$; hence $f=0$ near $x_0$, which is to say that $A$ is open.

Answer 4

The function $y(x):\equiv0$ obviously solves the given IVP. By the standard existence and uniqueness theorem, which is applicable in this situation, this is the only solution in a neighborhood of $x=0$.

If you do not want to make reference to this theorem you have to prove the uniqueness at hoc.

If $x\mapsto y(x)$ $\>(x\geq0)$ is a solution of the given IVP which is not $\equiv0$ define $$\xi:=\inf\,\{x>0\>|\> y(x)\ne0\}\ .$$ We may asssume $\xi=0$. As $y(\cdot)$ is continuous there is a $\rho\in \>\bigl]0,{1\over2}\bigr]\>$ such that $|y(t)|\leq{1\over2}$ for $0\leq t\leq\rho$. We now have $$|y(x)|\leq\int_0^\rho|y'(t)|\>dt=\int_0^\rho y^2(t)\>dt\leq \rho M^2\quad(0\leq x\leq\rho)\ ,\tag{1}$$ whereby $M:=\sup_{0\leq t\leq\rho}\bigl|y(t)\bigr|\leq{1\over2}$. Since $(1)$ is true for all $x\in[0,\rho]$ we have $M\leq {1\over2}M^2$, and this enforces $M=0$, hence in turn the nonexistence of such a $\xi$.