Originally Posted by
topsquark Well, I guess I was just "thinking out loud" but it would seem to me that the equation would require $\displaystyle \phi$ to be differentiable in this case. However, as you say, the condition of discontinuity in $\displaystyle \phi$ may well lead to other kinds of solutions. I know I'm not on solid ground here. However, it might be fruitful to consider a differentiable $\displaystyle \phi$ and see what comes up.
-Dan
So, that's what I tried.
Let $\displaystyle \phi: [0, 1] \rightarrow \mathbb{R}$ be an Lebesgue integrable solution of the equation. Then by the fundamental theorem the function:
$\displaystyle \Phi(x) = \int_{0}^{x} \phi(t)dt $
is differentiable almost everywere.
Let $\displaystyle \Omega = \{x \in [0, 1]: \Phi \mbox{ is differentiable in } x\} - \{0\}$
Let $\displaystyle \psi = \phi |_{\Omega}$
Then for all $\displaystyle x \in \Omega$
$\displaystyle \int_{0}^{x} \phi(t)dt = \mu x \psi(x) $
$\displaystyle x\psi$ is differentiable in its domain, hence $\displaystyle \psi$is differentiable ($\displaystyle x \not\in \Omega$). By fundamental theorem and definition of $\displaystyle \psi$ we can obtain:
$\displaystyle \psi(x) = \mu \frac{d}{dx} (x \psi(x))$
Hence:
$\displaystyle x\psi'(x) =(1-\mu)\psi(x) $
This differential equation has general solution:
$\displaystyle \psi(x) = Cx^{1-\mu}$
SO MY CONCLUSION IS.... $\displaystyle \phi(x) = Cx^{1-\mu}$ almost everywhere!
THE PROBLEM occurs when $\displaystyle \mu \geq 2$. For instance when $\displaystyle \mu = 2$ we get $\displaystyle \phi(x) = \frac{C}{x}$ (a.e.). But I think that 1/x is not L-integrable (correct me if I'm wrong), so the equation should not have integrable solution!!!
Of course the point was to prove that the solution exists so where is the error?