Solution of functional equation $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$

Question

If $x,y\in \mathbb{R}$ and $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$ and $f'(0)=0\;,$ Then $f(x)$ is

$\bf{My\; Try::}$ Using $$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+h\sqrt{f(x)}-f(x)}{h}$$

Now Put $x=y=0$ in $$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}\;,$$ We get $f(0)=0$

So we get $f(0)=0$

So $$f'(x) = \sqrt{f(x)}+\lim_{h\rightarrow 0}\frac{f(h)}{h}=\sqrt{f(x)}$$

So $$\int\frac{f'(x)}{\sqrt{f(x)}}dx = 1\int dx\Rightarrow 2\sqrt{f(x)}=x+c$$

Now Put $x=\;,$ We get $c=0$

So we get $2\sqrt{f(x)}=x\Rightarrow 4f(x)=x^2\Rightarrow \displaystyle f(x)=\frac{x^2}{4}$

Can we solve it some short way, If yes then please explain here, Thanks

Answer 1

$$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}=f(y+x)=f(y)+f(x)+x \sqrt{f(y)}$$ Subtracting $f(x)+f(y)$ from each side and squaring , we have that $$y^2f(x)=x^2f(y) \Leftrightarrow \frac{f(x)}{x^2}=\frac{f(y)}{y^2}$$ So we have $\frac{f(x)}{x^2}$ is a constant function. Now put $f(x)=cx^2$ in the original equation, where $c$ is a constant. Simplifying gives us that $$2cx=\sqrt{c} |x|$$

Note that if $c \neq 0$, $c$ will take different values depending on the value of $x$, This is a contradiction, as $c$ is a constant. So we have $c=0$. Thus, $f(x)=0$ is the only solution. In order for $\frac{x^2}{4}=f(x)$, to be a solution we must have a constraint that $x \ge 0$, or the functional equation should be altered so: $$f(x+y)=f(x)+f(y)+y\operatorname{sgn}(x)\sqrt{f(x)}$$

Answer 2

I like writing $$\frac{f(x+y)-f(x)}y=\frac{f(y)}y+\sqrt{f(x)}$$ Then take the limit as $y\rightarrow0$ to get $$f^{\prime}(x)=f^{\prime}(0)+\sqrt{f(x)}$$ (assuming differentiability and noting that $f(0)=0$ so we can apply L'Hopital's rule) so that if $f^{\prime}(0)=0$, then $$\frac{df}{\sqrt f}=dx$$ $$2\sqrt{f(x)}=x+C$$ From the original equation it can be seen that $f(0)=0$ so $C=0$ and it follows that $$f(x)=\frac{x^2}4$$ Notice that this technique allows you to get implicit solutions even if $f^{\prime}(0)\ne0$.

EDIT: But such solutions are invalid because if $f^{\prime}(0)\ne0$, then since $f(0)=0$, there would be some points near $x=0$ where $f(x)<0$ which would be wrong because then $\sqrt{f(x)}\notin\mathbb{R}$.