example of solving a functional equation

Let’s determine all twice differentiable real functions $f$ which satisfy the functional equation

$f(x+y)\cdot f(x-y)={[f(x)]}^2-{[f(y)]}^2$(1)

for all real values of $x$ and $y$.

Substituting first  $y=0$ in (1) we see that  $f{(x)}^2=f{(x)}^2-f{(0)}^2$  or  $f(0)=0$.  The substitution  $x=0$  gives  $f(y)f(-y)=-f{(y)}^2$,  whence  $f(-y)=-f(y)$.  So $f$ is an odd function.

We differentiate both sides of (1) with respect to $y$ and the result with respect to $x$:

$$f^{\prime }(x+y)f(x-y)-f^{\prime }(x-y)f(x+y)=-2f(y)f^{\prime }(y)$$

$$f^{\prime \prime }(x+y)f(x-y)+f^{\prime }(x-y)f^{\prime }(x+y)-f^{\prime \prime }(x-y)f(x+y)-f^{\prime }(x+y)f^{\prime }(x-y)=0$$

The result is simplified to  $f^{\prime \prime }(x+y)f(x-y)=f^{\prime \prime }(x-y)f(x+y)$,  i.e.

$$f^{\prime \prime }(x+y)/f(x+y)=f^{\prime \prime }(x-y)/f(x-y).$$

Denoting  $x+y:=u$,  $x-y:=v$  we obtain the equation

$$\frac{f^{\prime \prime }(u)}{f(u)}=\frac{f^{\prime \prime }(v)}{f(v)}$$

for all real values of $u$ and $v$.  This is not possible unless the proportion $\frac{f^{\prime \prime }(u)}{f(u)}$ has a on $u$.  Thus the homogeneous linear differential equation  $f^{\prime \prime }(t)/f(t)=\pm k^2$ or

$$f^{\prime \prime }(t)=\pm k^{2}f(t),$$

with $k$ some , is valid.

There are three cases:

1. $k=0$.  Now  $f^{\prime \prime }(t)\equiv 0$  and consequently  $f(t)\equiv Ct$.  If one especially $C$ equal to 1, the solution is the identity function (http://planetmath.org/IdentityMap)  $f:t\mapsto t$.  This yields from (1) the well-known “memory formula”

   $$(x+y)(x-y)=x^2-y^2.$$

2. $f^{\prime \prime }(t)=-k^{2}f(t)$  with  $k\ne 0$.  According to the oddness one obtains for the general solution the sine function  $f:t\mapsto C\sin kt$.  The special case  $C=k=1$  means in (1) the

   $$\sin (x+y)\sin (x-y)={\sin }^{2}x-{\sin }^{2}y,$$

   which is easy to verify by using the addition and subtraction formulae (http://planetmath.org/AdditionFormula) of sine.

3. $f^{\prime \prime }(t)=k^{2}f(t)$  with  $k\ne 0$.  According to the oddness we obtain for the general solution the hyperbolic sine (http://planetmath.org/HyperbolicFunctions) function  $f:t\mapsto C\sinh kt$.  The special case  $C=k=1$  gives from (1) the

   $$\sinh (x+y)\sinh (x-y)={\sinh }^{2}x-{\sinh }^{2}y.$$

The solution method of (1) is due to andik and perucho.

Title	example of solving a functional equation
Canonical name	ExampleOfSolvingAFunctionalEquation
Date of creation	2013-03-22 15:30:00
Last modified on	2013-03-22 15:30:00
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	14
Author	pahio (2872)
Entry type	Example
Classification	msc 34A30
Classification	msc 39B05
Related topic	ChainRule
Related topic	AdditionFormula
Related topic	SubtractionFormula
Related topic	DefinitionsInTrigonometry
Related topic	GoniometricFormulae
Related topic	DifferenceOfSquares
Related topic	AdditionFormulas