example of solving a functional equation

example of solving a functional equation

Let’s determine all twice

differentiable

real functions

f

which satisfy the functional equation

f

⁢

(

x

+

y

)

⋅

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

′

⁢

(

x

+

y

)

⁢

f

⁢

(

x

-

y

)

-

f

′

⁢

(

x

-

y

)

⁢

f

⁢

(

x

+

y

)

=

-

2

⁢

f

⁢

(

y

)

⁢

f

′

⁢

(

y

)

f

′′

⁢

(

x

+

y

)

⁢

f

⁢

(

x

-

y

)

+

f

′

⁢

(

x

-

y

)

⁢

f

′

⁢

(

x

+

y

)

-

f

′′

⁢

(

x

-

y

)

⁢

f

⁢

(

x

+

y

)

-

f

′

⁢

(

x

+

y

)

⁢

f

′

⁢

(

x

-

y

)

=

0

The result is simplified to

f

′′

⁢

(

x

+

y

)

⁢

f

⁢

(

x

-

y

)

=

f

′′

⁢

(

x

-

y

)

⁢

f

⁢

(

x

+

y

)

,  i.e.

f

′′

⁢

(

x

+

y

)

/

f

⁢

(

x

+

y

)

=

f

′′

⁢

(

x

-

y

)

/

f

⁢

(

x

-

y

)

.

Denoting

x

+

y

:=

u

,

x

-

y

:=

v

we obtain the equation

f

′′

⁢

(

u

)

f

⁢

(

u

)

=

f

′′

⁢

(

v

)

f

⁢

(

v

)

for all real values of

u

and

v

.  This is not possible unless the proportion

f

′′

⁢

(

u

)

f

⁢

(

u

)

has a on

u

.  Thus the

homogeneous linear differential equation

f

′′

⁢

(

t

)

/

f

⁢

(

t

)

=

±

k

2

or

f

′′

⁢

(

t

)

=

±

k

2

⁢

f

⁢

(

t

)

,

with

k

some , is valid.

There are three cases:

1.

k

=

0

.  Now

f

′′

⁢

(

t

)

≡

0

and consequently

f

⁢

(

t

)

≡

C

⁢

t

.  If one especially

C

equal to 1, the solution is the

identity function

(

http://planetmath.org/IdentityMap

)

f

:

t

↦

t

.  This yields from (1) the well-known “memory formula”

(

x

+

y

)

⁢

(

x

-

y

)

=

x

2

-

y

2

.

2.

f

′′

⁢

(

t

)

=

-

k

2

⁢

f

⁢

(

t

)

with

k

≠

0

.  According to the oddness one obtains for the

general solution

the

sine function

f

:

t

↦

C

⁢

sin

⁡

k

⁢

t

.  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

′′

⁢

(

t

)

=

k

2

⁢

f

⁢

(

t

)

with

k

≠

0

.  According to the oddness we obtain for the general solution the

hyperbolic sine

(

http://planetmath.org/HyperbolicFunctions

)

function

f

:

t

↦

C

⁢

sinh

⁡

k

⁢

t

.  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

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