# f(x^2+y^2)=g(x)g(y) ...

• May 26th 2011, 04:18 AM
Also sprach Zarathustra
f(x^2+y^2)=g(x)g(y) ...
Hello everybody!

A problem:

Find functions $\varphi , \psi$ which are fulfilling:

$\varphi(x^2+y^2)=\psi(x) \psi(y)$ for all $x,y$.

Prove that if $\varphi , \psi$ are fulfilling the above equation then $\psi$ determined by $\varphi$. How?

Thank you.
• May 26th 2011, 04:26 AM
girdav
For example $\phi=\psi =0$. A more interesting example is given by $\phi :x\mapsto e^x$ and $\psi:x\mapsto e^{x^2}$.
• May 26th 2011, 04:59 AM
Also sprach Zarathustra
Quote:

Originally Posted by girdav
For example $\phi=\psi =0$. A more interesting example is given by $\phi :x\mapsto e^x$ and $\psi:x\mapsto e^{x^2}$.

What you wrote and the family of functions: a^x where a>1 ?

What about the second part of the question?

Thanks!
• May 27th 2011, 07:45 AM
Jester
This is how I might attempt to do this problem. I will assume that both $\psi$ and $\phi$ are smooth. Taking the natural log of both sides gives

$\ln \phi(x^2+y^2) = \ln \psi(x) + \ln \psi(y)$.

Call the term on the RHS $F(x^2+y^2)$. Differentiating both side wrt $x$ and $y$ gives

$F''(x^2+y^2) = 0$.

Thus, $F(x^2+y^2) = a(x^2+y^2) + \ln(b)$ where $a$ and $b$ are constant.

So $\ln \phi(x^2+y^2) = a(x^2+y^2) + \ln(b)$

so $\phi(x^2+y^2) = b e^{a(x^2+y^2)} = \psi(x)\psi(y)$.

Now set $y = 0$ and this gets you the form for $\psi(x) = k e^{ax^2}.$ Then substitute into the original functional equation to determine $k$.
• May 29th 2011, 02:05 PM
awkward
Quote:

Originally Posted by Also sprach Zarathustra
Hello everybody!

A problem:

Find functions $\varphi , \psi$ which are fulfilling:

$\varphi(x^2+y^2)=\psi(x) \psi(y)$ for all $x,y$.

Prove that if $\varphi , \psi$ are fulfilling the above equation then $\psi$ determined by $\varphi$. How?

Thank you.

Maybe there is something missing from the problem statement. Otherwise
$\phi(x) = e^x$ and
$\psi(x) = - e^{x^2}$,
combined with girdav's solution, provides a counterexample to the second part (uniqueness).