Solving functional equations $2f(x) g(y) \pm f(x)^2 \pm g(y)^2 = 2 \sqrt{a} h(x \pm y) \pm (a+b)$

Question

Problem: Fix $a, b \in \mathbb{R}$ with $a > b > 0$. Find $f, g, h: \mathbb{R} \to \mathbb{R}$ such that $$ 2f(x) g(y) + f(x)^2 + g(y)^2 = 2 \sqrt{a} \ h(x+y) + (a+b) \\ 2 f(x) g(y) - f(x)^2 - g(y)^2 = 2 \sqrt{a} \ h(x-y) - (a + b) $$ It's also possible that $f = g = h$.

Attempt. I tried to consider $f(x) = \cos(x)$ and $g(y) = \sqrt{a}\cos(y)$ which would satisfy the addition of both these equations given by $$ 2 f(x) g(y) = \sqrt{a} \left( h(x+y) + h(x-y) \right) $$ with $h(z) = \cos(z)$ as well due to product-to-sum identities (i.e. $\cos$ and $\cosh$ satisfy d'Alembert's equation) but they won't necessarily satisfy each equation before addition..

Any ideas?

Answer 1

We have $$ (f(x) +g(y))^2 = 2 \sqrt{a} \ h(x+y) + (a+b) \\ -( f(x) -g(y))^2 = 2 \sqrt{a} \ h(x-y) - (a + b)$$

$$ f(x) +g(y) = \sqrt{2 \sqrt{a} \ h(x+y) + (a+b)} \\ f(x) -g(y) = \sqrt{-2 \sqrt{a} \ h(x-y) + (a + b)} $$ Add and subtract: $$ 2f(x) = \sqrt{2 \sqrt{a} \ h(x+y) + (a+b)} + \sqrt{-2 \sqrt{a} \ h(x-y) + (a + b)} \tag5$$ $$2g(y) = \sqrt{2 \sqrt{a} \ h(x+y) + (a+b)} - \sqrt{-2 \sqrt{a} \ h(x-y) + (a + b)}\tag 6 $$ Put y=0 in (6): $$2g(0) = \sqrt{2 \sqrt{a} \ h(x) + (a+b)} - \sqrt{-2 \sqrt{a} \ h(x) + (a + b)} \tag7$$ Assuming $a\ne0$, this means that h(x) must be constant for all x. So we have $$ 2f(x) = \sqrt{2 \sqrt{a} \ h + (a+b)} + \sqrt{-2 \sqrt{a} \ h + (a + b)} \tag8$$ $$2g(y) = \sqrt{2 \sqrt{a} \ h + (a+b)} - \sqrt{-2 \sqrt{a} \ h + (a + b)}\tag 9 $$ and so f,g,h are all constant functions.

Answer 2

I'll denote by $(S)$ the following system of equations, which is just the system from the statement but with the second equation multiplied by $-1$ for my convenience: $$(S) :\begin{cases}(f(x) + g(y))^2 = 2 \sqrt{a} \ h(x+y) + (a+b) \\ (f(x) - g(y))^2 = - 2 \sqrt{a} \ h(x-y) + (a + b)\end{cases}$$ I also won't use qualificators like "$\forall (x,y) \in \mathbb{R}^2, \dots$" at all times but they're implied whenever I use $x$ and $y$ instead of using another name like $y_0$.

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We have, by summing the equations of $(S)$: $$\begin{split}f(x)^2 + g(y)^2 &= \frac{1}{2}\left((f(x) + g(y))^2 + (f(x) - g(y))^2\right)\\ &= \sqrt{a}(h(x+y) - h(x-y)) + (a + b) \end{split}\tag{1}$$ Similarly, by substracting the second line from the first line, it follows that, as observed by OP: $$2 f(x) g(y) = \sqrt{a} \left( h(x+y) + h(x-y) \right) \tag{2}$$ By substituting $y = 0$ in $(1)$, we gain: $$f(x)^2 + g(0)^2 = a + b \tag{3}$$ Thus, there are no solutions if $g(0)^2 > a + b$ since $f(x)^2 \geq 0$, hence we can assume that $g(0)^2 \leq a + b$ from now on.
Let $C := a + b - g(0)^2 \geq 0$ such that $f(x)^2 \equiv C$.
Then, we get, after squaring $(2)$ and replacing $f(x)^2$ by $C$: $$C g(y)^2 = \frac{a}{4}(h(x+y) + h(x-y))^2 \tag{4}$$ By respectively setting $x = 0$ or $y = 0$ in $(3)$, it ensues that: $$\begin{align*} C g(y)^2 &= \frac{a}{4}(h(2y) + h(0))^2 \tag{5} \\ h(x)^2 &= \frac{Cg(0)^2}{a} =: K\tag{6}\end{align*}$$

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Therefore, as a result of $(3)$, $(6)$, and then $(5)$ in light of $(6)$, we know that all three functions $f$, $g$ and $h$ take a finite amount of values, with: $$\begin{cases}f(x)^2 \equiv C \\ h(x)^2 \equiv K \\ g(y)^2 = \begin{cases} g(0)^2 \quad\text{if } h(2y) = h(0)\\ 0 \quad\quad\,\,\,\,\text{if } h(2y) = - h(0) \end{cases}\end{cases}$$ This is just to sum things up mid-reasoning though. This is not sufficient yet as they're only necessary conditions.

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Let's start by looking at the critical case where $K = 0$.
First, assume that $C = 0$. This is equivalent to $g(0)^2 = a+b$, and to $f \equiv 0$.
From the expression of $K$ in $(6)$, $K = 0$, and thus $h \equiv 0$.
Inputting that in $(1)$ gives $g(y)^2 \equiv a + b$. Conversely, one can check that $f \equiv 0$, $g^2 \equiv a + b$, $h \equiv 0$ is solution of $(S)$ (in particular, $g$ doesn't have to be constant).
On the other hand, if $K = 0$ but $C \neq 0$, then, since $aK = C g(0)^2$, we have $g(0)^2 = 0$, thus $g \equiv 0$ due to our past observations. Conversely, $f^2 \equiv a + b$, $g \equiv 0$ and $h \equiv 0$ is solution of $(S)$ (in particular, $f$ didn't have to be constant either), and we are done with the critical case.

Let's assume moving forward that $K \neq 0$, which is equivalent to having $C \neq 0$ and $g(0) \neq 0$.
Take a fixed $y_0$. Going back to $(1)$ with the context of $(3)$ and $(4)$, and using that $C \neq 0$ by assumption, we can see that, if $h(x + y_0) = - h(x-y_0) \in \left\{\pm\sqrt{K}\right\}$, which is equivalent to $h(2y_0) = - h(0) \in \left\{ \pm\sqrt{K}\right\}$ thanks to $(4)$, then: $$C + g(y_0)^2 = a + b \pm 2\sqrt{aK}$$ hence, since $a + b - C = g(0)^2$: $$g(y_0)^2 = g(0)^2 \pm 2\sqrt{aK}$$ But we already have $g(y_0)^2 \in \left\{g(0)^2, 0\right\}$, thus this cannot happen unless $g(y_0)^2 = g(0)^2 - 2\sqrt{aK} = 0$, which would impose specifically that $$h(x + y_0) = - h(x - y_0) = h(x' + y_0) = - h(x' - y_0) = -\sqrt{K}$$ for all $(x, x') \in \mathbb{R}^2$.
Yet, if we consider $z \in \mathbb{R}$ and look at $(x, x') = (z - y_0, z + y_0)$, this would imply that: $$h(z) = h(x + y_0) = - h(x' - y_0) = - h(z)$$ absurd since $K \neq 0$ by assumption, hence this subcase cannot happen at all, and we always have: $$h(x + y) = h(x - y)$$ which consequently provides, because of $(4)$: $$g(y)^2 = \frac{a}{4C} \cdot \left(2\sqrt{K}\right)^2 = \frac{aK}{C} = g(0)^2$$ and also lets us deduce that: $$h(x) = h\left(\frac{x}{2} + \frac{x}{2}\right) = h\left(\frac{x}{2} - \frac{x}{2}\right) = h(0)$$ We can now end this by returning to equation $(2)$ which has no squares and nothing is equal to $0$ to see that, by setting $y = 0$ or $x = 0$ respectively we obtain: $$f(x) = \frac{\sqrt{a}h(0)}{g(0)} \\ g(y) = g(0)$$ where the second line comes from the fact that $f(0) = \frac{\sqrt{a}h(0)}{g(0)}$, hence $g(y) = \frac{\sqrt{a}h(0)}{\frac{\sqrt{a}h(0)}{g(0)}} = g(0)$.
Conversely, $g \equiv \alpha \in (-\sqrt{a+b}, \sqrt{a+b}) \setminus \{0\}, h \equiv \beta := \pm\sqrt{\frac{(a + b - \alpha^2)\alpha^2}{a}}, f \equiv \frac{\sqrt{a}\beta}{\alpha}$ is solution of $(S)$, and we are finished.

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Considering this was quite long to write, I would not be surprised if there were mistakes in there, but hopefully this answers your question correctly.