# 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 $f(x)=x/2+C$. – k1next 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. – k1next 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
The answer of Henning Makholm is sufficiently clean and algebraic (I wonder what do Optional exactly means by algebraic?). The problem itself does not require some super complex notions. It is quite simple. And the answer that solutions are $f(x)=0$ and $f(x)=\frac{1}{2} x+c$ is correct. – Tomas Feb 5 '13 at 13:41

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) $S_f$ has some bounded open interval as a component, (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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