# example of solving a functional equation

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

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

$f^{\prime\prime}(t)=-k^{2}f(t)$  with  $k\neq 0$.  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. 3.

$f^{\prime\prime}(t)=k^{2}f(t)$  with  $k\neq{0}$.  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