1. ## Equation

Prove that the following integral equation has an integrable solution for all $\mu > 0$.

$\int_{0}^{x} \phi(t) dt = \mu x \phi(x), \qquad 0 \leq x \leq 1$

2. Hello,

I think that if you integrate by parts the first one, you could get something interesting :-)

3. Originally Posted by Moo
Hello,

I think that if you integrate by parts the first one, you could get something interesting :-)
But $\phi$ may not be differentiable.

4. Originally Posted by albi
But $\phi$ may not be differentiable.
Just putting my two cents in. The left hand side should be differentiable thanks to the Fundamental Theorem. So shouldn't that imply that the right hand side would have to be? (This seems to form a contradiction, but I don't know how to get out of it.)

-Dan

5. Originally Posted by topsquark
Just putting my two cents in. The left hand side should be differentiable thanks to the Fundamental Theorem. So shouldn't that imply that the right hand side would have to be? (This seems to form a contradiction, but I don't know how to get out of it.)

-Dan
So $\phi$ may be discontinuous so the integral may not be differentiable.

The fundamental theorem says, if $\phi$ is Lebesgue integrable then the integral is differentiable almost everywere.

Probably we can integrate by parts almost everywere, but what's the point?

6. Originally Posted by albi
So $\phi$ may be discontinuous so the integral may not be differentiable.

The fundamental theorem says, if $\phi$ is Lebesgue integrable then the integral is differentiable almost everywere.

Probably we can integrate by parts almost everywere, but what's the point?
Well, I guess I was just "thinking out loud" but it would seem to me that the equation would require $\phi$ to be differentiable in this case. However, as you say, the condition of discontinuity in $\phi$ 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 $\phi$ and see what comes up.

-Dan

7. Originally Posted by topsquark
Well, I guess I was just "thinking out loud" but it would seem to me that the equation would require $\phi$ to be differentiable in this case. However, as you say, the condition of discontinuity in $\phi$ 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 $\phi$ and see what comes up.

-Dan
So, that's what I tried.

Let $\phi: [0, 1] \rightarrow \mathbb{R}$ be an Lebesgue integrable solution of the equation. Then by the fundamental theorem the function:
$\Phi(x) = \int_{0}^{x} \phi(t)dt$

is differentiable almost everywere.

Let $\Omega = \{x \in [0, 1]: \Phi \mbox{ is differentiable in } x\} - \{0\}$

Let $\psi = \phi |_{\Omega}$

Then for all $x \in \Omega$
$\int_{0}^{x} \phi(t)dt = \mu x \psi(x)$

$x\psi$ is differentiable in its domain, hence $\psi$is differentiable ( $x \not\in \Omega$). By fundamental theorem and definition of $\psi$ we can obtain:

$\psi(x) = \mu \frac{d}{dx} (x \psi(x))$

Hence:

$x\psi'(x) =(1-\mu)\psi(x)$

This differential equation has general solution:
$\psi(x) = Cx^{1-\mu}$

SO MY CONCLUSION IS.... $\phi(x) = Cx^{1-\mu}$ almost everywhere!

THE PROBLEM occurs when $\mu \geq 2$. For instance when $\mu = 2$ we get $\phi(x) = \frac{C}{x}$ (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?

8. Originally Posted by topsquark
Just putting my two cents in. The left hand side should be differentiable thanks to the Fundamental Theorem. So shouldn't that imply that the right hand side would have to be? (This seems to form a contradiction, but I don't know how to get out of it.)
Not true. Here is the actual statement.

Theorem: If $f:[a,b]\mapsto \mathbb{R}$ is integrable then the function defined as $g(x) = \int_a^x f(\xi) d\xi$ is continous on $[a,b]$. Furthermore, if $f$ is continous at $x_0\in (a,b)$ then $g$ shall be differenciable at $x_0$.

In general, the integral smoothens the function out. Meaning if you have a $\mathcal{C}^k$ function its integral shall be a $\mathcal{C}^{k+1}$ function.

9. Originally Posted by ThePerfectHacker
Not true. Here is the actual statement.

Theorem: If $f:[a,b]\mapsto \mathbb{R}$ is integrable then the function defined as $g(x) = \int_a^x f(\xi) d\xi$ is continous on $[a,b]$. Furthermore, if $f$ is continous at $x_0\in (a,b)$ then $g$ shall be differenciable at $x_0$.

In general, the integral smoothens the function out. Meaning if you have a $\mathcal{C}^k$ function its integral shall be a $\mathcal{C}^{k+1}$ function.
Please read my previous posts. There you shall find another version of this theorem.

And maybe if you have some time you can think a little about the problem. Because it seems to me that it has no integrable solutions for $\mu \geq 2$