D'Alembert's Functional Equation

This wiki is dedicated to d'Alembert's functional equation, which is a functional equation of the form $$f(x+y)+f(x-y)= 2f(x)f(y)$$ for any real values $x$ and $y.$ Obviously, an example of a solution of this equation is any function of the form $f(x)=\cos (kx),$ where $k$ is a constant real value. After exploring a little bit more, you can see that any function of the form $f(x)=\cosh (kx)$ is also a solution, when $k$ is a constant real value. The last function is called hyperbolic cosine and is defined by the formula $\cosh x =\frac{e^x+e^{-x}}{2}.$ Making $k=0,$ into $\cos {kx}$ or into $\cosh{kx},$ we get a constant function whose value is 1. There is another obvious constant solution that is $f(x)=0$ for all value of $x.$ Then the logical question to ask ourselves is whether the mentioned functions are the only solutions of the d'Alembert's functional equation if it is required 1) the differentiability up to order 2 of the solutions, or just 2) the continuity of the solutions.

When the differentiability of the solutions up to order 2 is required over the set of all real values, it easy to solve the functional equation completely, and the solutions are going to be all the functions mentioned above. If the solutions are continuous, then it can be proved that they must be infinitely differentiable at any real number, and then the solutions are going to be the same as in the previous case.

Make $x=y=0$ into d'Alembert functional equation. Then $2f(0)=2\big(f(0)\big)^2.$ Hence $f(0)=0$ or $f(0)=1.$ If $f(0)=0,$ the only possible solution for the given functional equation will be $f(x)\equiv 0,$ because by making $y=0$ into the d'Alembert's functional equation, it follows that $2f(x)=2f(x)f(0)=2f(x)\times 0=0.$ So for the rest of this wiki, we will assume that $$f(0)=1.\qquad (1)$$

Let us assume the a function $f(x)$ is twice-differentiable on the whole set of real numbers and satisfies the functional equation $$f(x+y)+f(x-y)= 2f(x)f(y).$$ We are going to prove that the only solutions of this equation are functions of the form $f(x)=\cos (kx),$ $f(x)=\cos (kx),$ and $f(x)=0$ for all real values of $x.$ The proof is as follows:

First, differentiating d'Alembert functional equation with respect to $y$ and making $x=y=0,$ and using (1), it can be obtained that $0=f'(0)-f'(0)= 2f(0)f'(0)=2f'(0).$ Therefore, $$f'(0)=0.\qquad (2)$$ We differentiate d'Alembert equation twice with respect to $y$ and make $y=0.$ So, we obtain the equation $f''(x)+f''(x)=2f''(0)f(x).$ This equation can be rewritten as $$f''(x)=f''(0)f(x).\qquad (3)$$ The number $f''(0)$ can be positive, negative, or zero.

Case 1: When $f''(0)$ is zero
The equation $(3)$ becomes $f''(x)=0,$ whose solutions are linear functions $f(x)= c_1x+c_2.$ By using the conditions (1) and (2), it can be found that $c_1 =0$ and $c_2=1.$ Then the only constant solution of d'Alembert's equations is $f(x)\equiv 0,$ obtained above, and $f(x)\equiv 1$ for all real values of $x.$

Case 2: When $f''(0)> 0$
The differential equation $(3)$ has the general solution $f(x)=c_1 \cosh{k x} +c_2 \sinh {k x},$ which together with the conditions (1) and (2) implies that $f(x)=\cosh{kx}.$

Case 3: When $f''(0)< 0$
The differential equation $(3)$ has the general solution $f(x)=c_1 \cos{k x} +c_2 \sin {k x},$ which together with the conditions (1) and (2) implies that $f(x)=\cos{kx}.$

Therefore, the proof is complete. $_\square$