# first-order linear differential equation with boundary conditions

let be the differential equation

$-ixDf(x)-if(x)/2= E_{n}f(x)$

with the boundary conditions $f(x)=f(p^{k}x)$ for 'p' prime and $k=...,-2,-1,0,1,2,...$

is this possible to solve this eigenvalue problem ?? thanks

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here D is the derivative operator with respect to 'x' and the boundary conditions apply to ALL the primes $p=2,3,5,7,....$ –  Mathman Apr 5 '12 at 16:15
Your "boundary condition" does not appear to be a boundary condition in the usual sense. Is this supposed to be for all $x$? If $f(x) = f(p^kx)$ for all $x$ and all primes $p$ and integers $k$, then $f(x) = f(rx)$ for all positive rationals $r$, and then if $f$ is continuous it is constant. –  Robert Israel Apr 5 '12 at 17:34
As for the differential equation, its general solution is $f(x) = c x^{iE_n - 1/2}$ where $c$ is an arbitrary constant. –  Robert Israel Apr 5 '12 at 17:37
wouldn't the function $F(x)= \sum_{q}f(qx)$ with a sum taken over all the positive rational would satisfy the boundary conditions ?? with $F(0)=0$ and $\int_{0}^{\infty} F(x)dx =0$ –  Mathman Apr 5 '12 at 21:00
If $f$ is continuous, such a sum won't converge unless $f=0$ everywhere. –  Robert Israel Apr 5 '12 at 21:09

As I understand the question, you first fix $p$, and then you search the solutions of $$−ixDf(x)−if(x)/2=E_nf(x)$$ which satisfy $f(px)=f(x)$ (by reccurence, your $p^k$ conditions is automatically satisfied). So we solve the differential equation and we find $cx^{iE_n-1/2}$ and your condition gives $p^{iE_n-1/2}=1$. This is satisfied if and only if it exists some integer $n\in\mathbb{Z}$ such that $(iE_n-1/2)\ln p=2in\pi$. So the possible values of $E_n$ and then eigenfunctions are $$E_n=\frac{2in\pi}{\ln p}-i/2\qquad f_n(x)=ce^{\frac{2in\pi\ln x}{\ln p}}$$ If you intend not to fix first the value of $p$, then we need to consider only the solutions valid for all prime $p$. Then only solution is then for $n=0\;$ which gives $$E=-i/2\qquad f(x)=c$$