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

[DorelXD]

Homework Statement

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

$$af(x+y) + bf(x-y) = cf(x) + dy$$ , 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: $$af(2x)=cf(x)+dx-bf(0)$$

for x = y = 0 :$$af(0) + bf(0) = cf(0)$$

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

$$af(x+y) + bf(x-y) = cf(x) + dy$$ , 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: $$af(2x)=cf(x)+dx-bf(0)$$

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

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

for x = y = 0 :$$af(0) + bf(0) = cf(0)$$

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

 

[Ray Vickson]

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

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



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

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 $a + b \neq c$ and $d \neq 0$.

 

[Ray Vickson]

The interesting case is $a + b \neq c$ and $d \neq 0$.

If $a+b \neq c$ then $f(x) \equiv 0$. To see this, put y = 0 to get
$$(a+b) f(x) = c f(x)$$
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:

$$af(x+y) + bf(x-y) = cf(x) +dy$$
$$af(x-y) + bf(x+y) = cf(x) - dy$$

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:

$$(b^2-a^2)f(x-y) = c(b-a)f(x) + d(b+a)y$$ .

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

$$f(x) = \frac{a+b}{c}f(0)+\frac{d(a+b)}{c(a-b)}x$$

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