Solve functional equation $f(x^4+y)=x^3f(x)+f(y)$

Question

I need help solving this equation, please:

$$f(x^4+y)=x^3f(x)+f(y),$$ such that $ f:\Bbb{R}\rightarrow \Bbb{R},$ and differentiable

I've found that $f(0)=0$ and $f(y+1)=f(1)+f(y)$, but I couldn't continue, I think the solution is $f(x)=ax$.

Thanks for your help

Answer 1

Note that $$f'(y)=\lim_{x\to 0} \frac{f(y+x^4)-f(y)}{x^4}.$$ By the given relationship this is independent of $y$.

So $f'(y)$ is a constant $c$, and therefore $f(y)=cy+d$ for all $y$ and some constant $d$. You showed that $f(0)=0$, so $d=0$.

Conversely, it is easy to verify that $f(y)=cy$ works.

Remark: Instead of going back to the definition of derivative, we can differentiate both sides with respect to $y$, treating $x$ as constant. We get $f'(y+x^4)=f'(y)$ for all $y$. From this we can show that $f't)=f'(0)$ for all $t$, and now we argue that $f(t)=f(0)t$ as above.

Answer 2

We only need that $f$ is continuos.

With $x=1$, we find $f(y+1)=f(y)+f(1)$, in particular for $y=0$, $f(1)=f(1)+f(0)$, i.e., $f(0)=0$. Let $$ S=\{\,t\in\Bbb R\mid f(t)=tf(1)\,\}.$$ We have found that $0\in S$, and $$\tag1y\in S\iff y+1\in S,$$ hence $\Bbb Z\subseteq S$. Also, (letting $y=0$) we have $$\tag2 x\in S\iff x^4\in S.$$ Let $x\in (1,2)$ and $\epsilon>0$. We want to show that $(x-\epsilon,x+\epsilon)$ intersects $S$. If $x<1+\epsilon$, we are done. For $n$ big enough, $(x+\epsilon)^{4^n}$ exceeds $(x-\epsilon)^{4^k}$ by more than $1$, hence $\bigl((x-\epsilon)^{4^n},(x+\epsilon)^{4^n}\bigr)$ intersects $S$ and so by $(2)$, $(x-\epsilon,x+\epsilon)$ intersects $S$ as well. As $S$ is a closed set (per continuity of $f$), we conclude $[1,2]\subseteq S$, and then from $(1)$ by induction $S=\Bbb R$.

Answer 3

Differentiate the equation with respect to $x$. Then with respect to $y$. Each of these results has the term $f'(x^4 + y)$, which you can eliminate. Then solve for $f(x)$.

Answer 4

To help clarify or expand the hint by @xidgel and avoid any possible confusion.

First introduce $t(x,y) = x^4+y$, then we can write: $$f(t) = x^3f(x)+f(y)$$

Now we can express differentiation with respect to x and y with the chain rule:

$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial t} \cdot \frac{\partial t}{\partial x} \text{ and } \frac{\partial f}{\partial y} = \frac{\partial f}{\partial t} \cdot \frac{\partial t}{\partial y}$$

Now proceed as in the hint by calculating these differentials on both sides of the equation above.

Answer 5

This is an alternative proof when $f$ is required to only be continuous. Taking $x:=1$ and $y:=0$, we get $$f(0)=0\,.$$ Setting $y=0$, we have $$f\left(x^4\right)=x^3\,f(x)\,.$$ Therefore, $$f(x+y)=x^{3/4}\,f\left(x^{1/4}\right)+f(y)=f(x)+f(y)\,,$$ if $x\geq 0$. Setting $y:=-x$ in the equation above shows that $f(-x)=-f(x)$ for all $x\geq 0$. Thus, if $x<0$, we have $$f(y)=f\left(-x+(x+y)\right)=f(-x)+f(x+y)=-f(x)+f(x+y)\,,$$ or $$f(x+y)=f(x)+f(y)$$ in this case too. Thence, $f$ satisfies Cauchy's functional equation. Given that $f$ is continuous, there exists $a\in\mathbb{R}$ such that $f(x)=ax$ for all $x\in\mathbb{R}$.