Hello,
I think that if you integrate by parts the first one, you could get something interesting :-)
Well, I guess I was just "thinking out loud" but it would seem to me that the equation would require to be differentiable in this case. However, as you say, the condition of discontinuity in 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 and see what comes up.
-Dan
So, that's what I tried.
Let be an Lebesgue integrable solution of the equation. Then by the fundamental theorem the function:
is differentiable almost everywere.
Let
Let
Then for all
is differentiable in its domain, hence is differentiable ( ). By fundamental theorem and definition of we can obtain:
Hence:
This differential equation has general solution:
SO MY CONCLUSION IS.... almost everywhere!
THE PROBLEM occurs when . For instance when we get (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?
Not true. Here is the actual statement.
Theorem: If is integrable then the function defined as is continous on . Furthermore, if is continous at then shall be differenciable at .
In general, the integral smoothens the function out. Meaning if you have a function its integral shall be a function.