A calculus problem with functions such that $f''(x) = g(x)$ and $g''(x) = f(x)$

Question

Let:

• $f(x)$ and $g(x)$ be twice differentiable, non-decreasing functions.
• $f''(x) = g(x)$ and $g''(x) = f(x)$.
• $f(x) \cdot g(x)$ is a linear function.

Then we have to show that $f(x) = g(x) = 0$.

I am really not able to do anything useful, any help would be appreciated :)

Answer 1

Let $f(x)g(x)=ax+b$ for some reals $a,b$. Differentiate w.r.t. x gives $f^\prime(x)g(x)+f(x)g^\prime(x)=a$, and differentiate again gives $$ f^{\prime\prime}(x)g(x)+2f^\prime(x)g^\prime(x)+f(x)g^{\prime\prime}(x)=0. $$ So, $$ g^2(x)+2f^\prime(x)g^\prime(x)+f^2(x)=0 $$ Since $f$ and $g$ are non-decreasing functions, then $f^\prime(x)g^\prime(x)\ge 0$. It follows that $$ 0=g^2(x)+2f^\prime(x)g^\prime(x)+f^2(x)\ge f^2(x)+g^2(x) $$ which of course implies $f(x)=g(x)=0$.

Answer 2

We have

$$\left\{\begin{array}{ll} (i)&f(x)g(x)=Ax+B\\ (ii)&f''(x)=g(x)\\ (iii)&f(x)=g''(x). \end{array}\right.$$

Let us omit the first condition (i) and first consider the system of ODE represented in conditions (ii) and (iii).

(ii) implies $$f(x)=\left(\int\int g\right)(x)+\alpha x+\beta$$ for constants $\alpha,\beta$, and this with (iii) implies $$g''(x)=\left(\int\int g\right)(x)+\alpha x+\beta.$$ Differentiating twice then yields $$g^{(4)}(x)=g(x)$$ This linear (homogeneous) ODE has four independent solutions: $e^{x},e^{-x},\cos x,\sin x$ (upto multiplicative constants) in addition to the trivial solution $0$. Applying (iii) repeatedly then gives the following non-trivial solution pairs $(f,g)$: $$\left\{\begin{array}{l}(e^{x},e^{x})\\ (e^{-x},e^{-x})\\ (\cos x, -\cos x)\\ (\sin x, -\sin x)\end{array}\right.$$

It is clear that none of these combinations satisfy (i), and therefore only the trivial solution pair $(0,0)$ holds, which proves the claim.