Solve functional equation: $f(x) + f(y) = f\left(\frac{x+y}{1-xy}\right) $

Question

The equation being : $$ f(x) + f(y) = f\left(\frac{x+y}{1-xy}\right) $$ The most obvious answer is, $$f(x) = C\arctan x$$ but there is another possible answer e.i $$ f(x) = \log\left(\frac{1-x}{1+x}\right) $$ which i am unable to get by the process i have provided. First substituting: $$ x = y = 0 \Rightarrow f(0) = 0$$ then partially differentiating the original equation wrt to $x$ and then $y$ individually (provided $x \ne y$), we'll get : $$ f'(x) = f'\left(\frac{x+y}{1-xy}\right)\frac{1+y^2}{(1-xy)^2} - (i)$$ $$ f'(y) = f'\left(\frac{x+y}{1-xy}\right)\frac{1+x^2}{(1-xy)^2} - (ii)$$ diving equations $i$ by $ii$ we get $$ \frac{f'(x)}{f'(y)} = \frac{1+y^2}{1+x^2}$$ $$ \Rightarrow f'(x)\left(1+x^2\right) = f'(y)\left(1+y^2\right)$$ which is only possible if the above equation is equal to some constant $c$, since $x \neq y$ $$ f'(x)\left(1+x^2\right) = C $$ $$ \Rightarrow f'(x) = \frac{C}{1+x^2} $$ $$ \Rightarrow f(x) = \int \frac{C}{1+x^2} = C \arctan x + K $$ since $f(0) = 0, k = 0$ $$ \therefore f(x) = C \arctan x $$

Also, when do i know that there is no more functions that satisfy the original equation ?

Answer 1

You should be able to say that there is only one differentiable solution (which you found) based on the uniqueness theorems for ODEs.

You can also check that $f$ is odd to conclude that the logarithm cannot be a solution.

As far as non-smooth solutions, I'm not sure there is any general way to decide if there are any. But perhaps someone else knows more about this topic.

Answer 2

Choosing $x=0$ and $y=0$ gives $f(0)+f(0)=f(0) \implies f(0)=0$.

Let $z$ be any real number other than $0$.
$1/z$ is defined since $z \ne 0$.

Substituting $z$ and $1/z$ into the functional equation we get:

$\large f(z) + f(1/z) = f\left(\frac{z+\frac{1}{z}}{0}\right)$

$\large f(z) = f\left(\frac{z+\frac{1}{z}}{0}\right) - f(1/z)$

but the right hand side is undefined because of the division by $0$.

So $f(z)$ is undefined.

In summary, $f(x)=0$ when $x=0$ but otherwise it is undefined.