$f=x+\phi(y,z)=y+\psi(x,z)=z+\varphi(x,y)$ so $f(x,y,z)=x+y+z$?

Question

This seems to be a very easy problem. Given that: $$f=x+\phi(y,z)$$

$$f=y+\psi(x,z)$$

$$f=z+\varphi(x,y)$$

We want to solve for $f$, which is a real function with three variables.

First without loss of generality, we write $\varphi=h(x)+g(y)+k(x,y)$,

by the first equation, we have $h=x$,

by the second equation, we have $g=y$.

So repeating this method, I can conclude that $f=x+y+z$ but I think my proof is wrong...

Is it true?

What are some strategy to solve this kind of equations with a function as unknown?

Answer 1

Let $y=0$ in $y+\psi(x,z)=x+\phi(y,z)$.

Then $\psi(x,z)=x+\phi(0,z)$ and so $f=x+y+\phi(0,z)$, where $\phi(0,z)$ is a function of $z$ alone.

Similarly, $f=x+z+\phi(y,0)$.

Now let $y=0$ in $x+y+\phi(0,z)=x+z+\phi(y,0)$.

Then $\phi(0,z)=z+\phi(0,0)$, where $\phi(0,0)$ is a constant.

So the correct solution is $$f(x,y,z)=x+y+z+c,$$ where $c$ is a constant.