Functional Equation with Real Numbers: Solving for f(x) on R->R

[DorelXD]

Homework Statement

Let [itex] a.b,c,d [/itex] be real numbers such that [itex] a ≠ b [/itex] and [itex] c ≠ 0 [/itex], find f:R->R for which this statement holds:

[tex] af(x+y) + bf(x-y) = cf(x) + dy [/tex] , for all x,y real numbers.

Homework Equations

Well this is a functional equation, that I know. I have less experience with those type of euqations.

The Attempt at a Solution

I tried to plug in some values:

for x = y: [tex] af(2x)=cf(x)+dx-bf(0) [/tex]

for x = y = 0 :[tex] af(0) + bf(0) = cf(0) [/tex]

I don't know what to do next, and how could these facts can help me. Can anyone help me solve this problem, please :D ?

 

[pasmith]

Homework Statement

Let [itex] a.b,c,d [/itex] be real numbers such that [itex] a ≠ b [/itex] and [itex] c ≠ 0 [/itex], find f:R->R for which this statement holds:

[tex] af(x+y) + bf(x-y) = cf(x) + dy [/tex] , for all x,y real numbers.

Homework Equations

Well this is a functional equation, that I know. I have less experience with those type of euqations.

The Attempt at a Solution

I tried to plug in some values:

for x = y: [tex] af(2x)=cf(x)+dx-bf(0) [/tex]

If you set [itex]y = -x[/itex] in the functional equation, you will get a second linear equation which [itex]f(x)[/itex] and [itex]f(2x)[/itex] must satisfy.

Do these linear simultaneous equations have a unique solution? Are the resulting expressions for [itex]f(x)[/itex] and [itex]f(2x)[/itex] consistent, or do you have to impose further conditions on [itex]a[/itex], [itex]b[/itex], [itex]c[/itex] and [itex]d[/itex]?

for x = y = 0 :[tex] af(0) + bf(0) = cf(0) [/tex]

This tells you that if [itex]a + b = c[/itex] then [itex]f(0)[/itex] is arbitrary, and if [itex]a + b \neq c[/itex] then [itex]f(0) = 0[/itex].

 

[Ray Vickson]

If you set [itex]y = -x[/itex] in the functional equation, you will get a second linear equation which [itex]f(x)[/itex] and [itex]f(2x)[/itex] must satisfy.

Do these linear simultaneous equations have a unique solution? Are the resulting expressions for [itex]f(x)[/itex] and [itex]f(2x)[/itex] consistent, or do you have to impose further conditions on [itex]a[/itex], [itex]b[/itex], [itex]c[/itex] and [itex]d[/itex]?



This tells you that if [itex]a + b = c[/itex] then [itex]f(0)[/itex] is arbitrary, and if [itex]a + b \neq c[/itex] then [itex]f(0) = 0[/itex].

Unless f is identically 0 we must have a+b = c. Just put y = 0 in the functional equation to see why.

 

[pasmith]

Unless f is identically 0 we must have a+b = c. Just put y = 0 in the functional equation to see why.

The interesting case is [itex]a + b \neq c[/itex] and [itex]d \neq 0[/itex].

 

[Ray Vickson]

The interesting case is [itex]a + b \neq c[/itex] and [itex]d \neq 0[/itex].

If ##a+b \neq c## then ##f(x) \equiv 0##. To see this, put y = 0 to get
[tex] (a+b) f(x) = c f(x)[/tex]
which is supposed to hold for all values of x. If ##f(x_0) \neq 0## then ##a+b = c##.

 

[DorelXD]

Thank you all for your hints! I managed to solved. You guys are the best!

 

[Ray Vickson]

Thank you all for your hints! I managed to solved. You guys are the best!

Just for interest: what is your solution?

 

[DorelXD]

First, I noticed a subtle "symmetry" in the equation, I don't know how to call it otherwise. So I plugged -y instead of y and I got the system:

[tex] af(x+y) + bf(x-y) = cf(x) +dy [/tex]
[tex] af(x-y) + bf(x+y) = cf(x) - dy [/tex]

Then, I played a little with the system. I won't post the whole the whole steps because it's only a little algebra. But, I will tell you what I tought: I wanted to get rid of f(x+y). So I mutliplyed the first equation by b and the second by a, the I substracted the second equation from the first, and it led me to:

[tex] (b^2-a^2)f(x-y) = c(b-a)f(x) + d(b+a)y [/tex] .

Next I plugged x , instead of y , and I got that:

[tex] f(x) = \frac{a+b}{c}f(0)+\frac{d(a+b)}{c(a-b)}x [/tex]

That function f(x) verifies the initial conditions. And that's it.. :D