# Solving $f=(f^2)'$

Find all differentiable mappings $f:\mathbb{R}\to \mathbb{R}$ so that $f=(f^2)'=2ff'$. My problem is that $f$ may very well be $0$ at some points ($f=0$ is for example a solution and so is $\frac12x$) so I can't simplify. Any hints?

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Ok, then you can write up a zero product instead of simplifying: $f(2f'-1)=0$. –  Berci Feb 5 '13 at 12:35
@Berci Even if I do that this doesn't mean $f=0$ or $f'=\frac12$. –  Optional Feb 5 '13 at 12:36

Let us consider different cases for the problem.

1. $f\equiv 0$. This is the trivial solution.
2. $\exists! x^*$ such that $f(x^*)=0$. Here we can take the domain $\Omega = \{x |f(x)\not = 0\}= \mathbb R-\{x^*\}$. Then we can still solve $f=2ff'$ on $\Omega$ and get $f=2ff' \Rightarrow 1/2=f'\Rightarrow f(x) = x/2+c$ on $\Omega$. To get $f$ on $\mathbb R$ we have to define the value at $x^*$ to be $f(x^*)=0$ to get a continous function. Then we have $f(x) = x/2-x^*/2$ on $\mathbb R$ which is also a solution of the problem.
3. Assume there exist two different points (or more) $x_*$ and $x^*$, with both $f(x_*)=f(x^*)=0$. As before we have $f(x)=x/2+c$ on $\Omega = \{x |f(x)\not = 0\}$. Again we have to fulfill $f(x_*)=0$ and $f(x^*)=0$ to get $f$ continous on $\mathbb R$. But this is not possible if $x_* \not = x^*$ as we only have one parameter $c$. Therefore, a solution of the problem can only have one root or be the zero function.

By this we conclude, that there exist no other solution, different from $f\equiv 0$ or $f=x/2+c$ for the given problem.

Edit: If we assume $f(x)\not =0,\ \forall x$ we would come to a contradiction, as the solution would also be $f(x)=x/2+c$, which has one root.

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You can't simply have $f'=\frac12\implies f(x)=\frac x2+c$ on $\Omega$ as $\Omega$ is not an interval –  Optional Feb 5 '13 at 13:36
Let me argue this way: Assume $\exists x^*$ such that $f(x^*) \not = 0$. Let $f(x^*)>0$. Then there exists $\epsilon$ such that $f(x)>0,\ \forall x\in (x^*-\epsilon, x^*+\epsilon)$ as $f$ is continous. On this interval, the solution of the ODE is $f(x)=x/2+C$. –  sonystarmap Feb 5 '13 at 13:43
So if $f(x^*)\neq 0$ then locally $f(x)=\frac x2+c$ is the only solution to the ODE. I see –  Optional Feb 5 '13 at 13:50
Yes, otherwise $f=2ff'$ can't hold. –  sonystarmap Feb 5 '13 at 13:52

Just split it up in two cases: for every $x$ you need either $f(x)=0$ or $1=2f'(x)$.

Then figure out (this will be an ad hoc argument) how it is possible for a function to satisfy at least one of these at every point.

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And so? I have $f'(x)=\frac12$ for some $x$. This doesn't mean $f(x)=\frac12x+c$ –  Optional Feb 5 '13 at 12:40
@Optional: If you have $f'(x)=1/2$ for some $x$ and $f(x)$ is nonzero, then you can't get away from the $f'(x)=1/2$ condition until you reach a root where $f(x_0)=0$. But then $f$ is a linear function on an entire open interval that ends at $f(x_0)$, so the only way $f'(x_0)$ can exist is if it is $1/2$. And that means that just on the other side of $x_0$, $f$ will be nonzero again, and therefore continue to $\pm\infty$ with slope $1/2$. –  Henning Makholm Feb 5 '13 at 12:44
"then you can't get away from the f′(x)=1/2 condition until you reach a root" why is that? –  Optional Feb 5 '13 at 12:50
@Optional: Because the other condition is that $f(x)=0$. –  Henning Makholm Feb 5 '13 at 13:13
I don't like this argument. Is there a cleaner method to do this? –  Optional Feb 5 '13 at 13:17

Since $f$ is differentiable, then it is continuous, and so (as a preimage of an open set) we have that $$S_f=\{x\in\Bbb R:f(x)\neq 0\}$$ is an open set. Let's suppose that $f$ isn't identically zero (since we've found that solution already), so $S_f$ is a nonempty open set, and so $S_f$ is either a disjoint union of at most countably many open intervals/rays or else is all of $\Bbb R$. I claim that $S_f=(-\infty,x_0)\cup(x_0,\infty)$ for some $x_0\in\Bbb R$. If not, then either (i) some component of $S_f$ is an interval, (ii) S_f has no more than one ray as a component, (iii) $S_f$ has as its components exactly two rays which don't share an endpoint, or (iv) $S_f=\Bbb R$. (Why?)

Suppose $(a,b)$ is a component of $S_f$ for some $a<b$. Since $f'\equiv\frac12$ on $(a,b)$, then there is some $c$ such that $f(x)=\frac12x+c$ for all $x\in(a,b)$. By continuity, we have $f(a)=\frac12a+c$ and $f(b)=\frac12b+c$, so $f(a)<f(b)$. But $a,b\notin S_f$, so $f(a)=f(b)=0$. Contradiction.

Suppose $S_f$ has at most one ray as a component--so exactly one ray, since it is nonempty and has no interval components--meaning that (WLOG) $S_f=(x_0,\infty)$ for some $x_0$. As above, there is some $c$ such that for all $x\in(x_0,\infty)$, we have $f(x)=\frac12x+c$, and continuity necessitates that $f(x_0)=0$. But then $$f(x)=\begin{cases}0 & x\leq x_0\\ \frac12x-\frac12x_0 & x>x_0,\end{cases}$$ which fails to be differentiable at $x=x_0$. Contradiction. We run into a similar problem if we suppose that $S_f=(-\infty,x_0)\cup(x_1,\infty)$ for some $x_0<x_1$, giving us another piecewise linear function that fails to be differentiable at $x=x_0,x_1$.

Finally, it's clear that $S_f\neq\Bbb R$, for if not, then we'd have $f'\equiv\frac12$, yielding $f(x)=\frac12x+c$ for some $c,$ but then $f(-2c)=0,$ so $-2c\notin S_f,$ and so $S_f\neq\Bbb R.$ Contradiction.

Thus, we do, indeed, have $S_f=(-\infty,x_0)\cup(x_0,\infty)$. There then exist $c,d$ such that $f(x)=\frac12x+c$ for $x<x_0$ and $f(x)=\frac12x+d$ for $x>x_0$. To obtain continuity, we need $c=d=-\frac12x_0$, and so $f(x)=\frac12x+c$ on all of $\Bbb R$.

Hence, among differentiable functions, only the constant $0$ function and functions of the form $f(x)=\frac12x+c$ satisfy the given differential equation.

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