Functional Equation $f(x+y)-f(x-y)=2f'(x)f'(y)$

Question

I am trying to solve the equation $f(x+y)-f(x-y)=2f'(x)f'(y)$ for all $f:\mathbb{R}\to\mathbb{R}$ non-constant, differentiable functions.

Here is my progress:
Any solution must be an even function and therefore zero is a critical point. Further if $f(x)$ is a solution, then $f(x)+c$ is also a solution for any constant $c.$

In addition $\dfrac{x^2}2, -\cos x$ and $\cosh x$ are some solutions of our functional equation.

How can I solve this equation completely?

Answer 1

1. OP has already derived

   • $f(2x)-f(0)~=~2f^{\prime}(x)^2~\geq~0$.

   • $f^{\prime}(0)~=~0$.

   • $f(y)-f(-y)~=~0$ even function.

   • $x\mapsto f(x)$ solution $\quad\Rightarrow\quad$ $x\mapsto f(x)+c$ solution.

2. Assume that $f$ is not a constant function. Then there exists a point $y_0\in\mathbb{R}$ such that $f^{\prime}(y_0)\neq 0$. From the bootstrap/recurrence relation $$ f^{(n+1)}(x)~=~\frac{f^{(n)}(x+y_0)-f^{(n)}(x-y_0)}{2f^{\prime}(y_0)}, \qquad n~\in~\mathbb{N}_0,$$ we can deduce inductively that $f\in C^{\infty}(\mathbb{R})$ is infinitely many times differentiable.

3. Under the assumption that $f$ is 4 times differentiable, one can derive the ODE $$f^{(3)}(x)~=~f^{\prime}(x)f^{(4)}(0).$$

4. Combined with what OP already has derived, it follows that the full solution is

   $$ f(x)~=~\lambda^{-2}\cosh(\lambda x)+ c\qquad\vee\qquad f(x)~=~\frac{x^2}{2}+c, \qquad\vee\qquad f(x)~=~c, $$ $$ \lambda~\in~\mathbb{C}\backslash\{0\}, \qquad c ~\in~\mathbb{C}.$$

Answer 2

Assuming function is twice differentiable,

Differentiating w.r.t $y$ and evaluating at $y=0$, I got $ f''(0)=1 $.

Differentiating w.r.t. $y$ and then w.r.t. $x$ gives $$ f''(x+y) + f''(x-y) = 2f''(x)f''(y) $$ which at $y=x$ gives $ g(2x) + 1 = 2(g(x))^2 $ where $g(x) = f''(x)$.

Does this help?

Answer 3

This is not a complete answer and just a discussion about the differentiability of $f$.

We have that

$$ 0=f(x+0)-f(x-0)=2f'(x)f'(0)\implies f'(0)=0$$ because in other case $f$ is constant.

Since if $f$ is a solution then $f+c$ is also a solution we assume $f(0)=0$ WLOG.

We have that

$$f(2x)-0=f(x+x)-f(x-x)=2(f'(x))^2\implies f'(x)=\pm \sqrt{\dfrac{f(2x)}{2}}.$$ So, $$f(2x)\ne 0\implies \exists f''(x).$$

Let $x_0\ne 0$ be the smallest positive value such that $f(2x_0)=0.$ Then it is $$f(x_0+y)=f(x_0-y), \forall y.$$ Since $$f(y)=f(-y),\forall y$$ we have that $f$ must be periodic with period $2x_0$. Indeed

$$f(2x_0+y)=f(x_0+x_0+y)=f(x_0-x_0-y)=f(-y)=f(y).$$

So, $f''(x)$ exists on $\mathbb{R}\setminus\{kx_0|k\in\mathbb{Z}.\}$

Let $y_0$ be a point such that $f'(y_0)\ne 0.$ Then

\begin{align} f''(0) &= \lim_{h\to 0}\dfrac{f'(h)}{h}\\&= \lim_{h\to 0}\dfrac{f(h+y_0)-f(h-y_0)}{2hf'(y_0)} \\&= \dfrac{1}{2f'(y_0)}\left(\lim_{h\to 0}\dfrac{f(h+y_0)-f(y_0)}{h}+\lim_{h\to 0}\dfrac{f(y_0)-f(h-y_0)}{h}\right) \\&=1. \end{align} This shows that $f''(x)$ exists $\forall x\in \mathbb{R}.$

Now, we have

\begin{align} f'''(x)&=\lim_{h\to 0}\dfrac{f''(x+h)-f''(x)}{h}\\&= \lim_{h\to 0}\dfrac{f'(x+h+y)-f'(x+h-y)-f'(x+y)+f'(x-y)}{2hf'(y)}\\&=\dfrac{f''(x+y)}{f'(y)}. \end{align} This shows that $\exists f'''(x),\forall x.$

In a similar way we can show the existence of $f^{(4)}$ using that

$$f''(x+y)-f''(x-y)=2f'''(x)f'(y).$$

Thus the answer given by @Qmechanic includes all possible solutions.