Differential equation with periodic conditions

Question

Consider the differential operator eigenvalue problem $-D^{2}y(x)=E_{n}y(x)$, with boundary periodic conditions $y(x)=y(x+1)$. My question is, if there is a similar problem for the dilation operator $T:= xD$ where $D$ denotes the derivative respect to $x$ which appears in $T^{2}y(x)=E_{n}y(x)$ with the dilation periodic condition, then how can I impose periodic conditions not for the translation group $y(x+1)$ but for the dilation group?

I believe that some periodic conditions involving dilations would be $ y(x)=y(2x)$, but I am not sure.

How can I impose periodic conditions if the lie group is the dilation in 1D?

thanks.

Answer 1

You want to reduce symmetry from a continuous group to a discrete subgroup, and impose the condition that the solution is symmetric under the discrete subgroup only. You can do this for a group containing dilatations. If you consider the dilatation group around a center, a logarithm coverts scale factors into additive factors, so that the subgroup is equally spaced logarithms, or dilatations by powers of a fundamental constant $\alpha$.

The boundary condition you want then is the annulus condition $\phi(\alpha r) = \alpha^C \phi(r)$ where the fundamental region is the annulus between 1 and $\alpha$. The constant C is the scale dimension of $\phi$, it's representation under the dilatation group.

You can also have rotations broken, or unbroken. If they are unbroken, you impose the condition that $\phi$ forms a representation of the rotation group in addition to satisfying the above constraint under dilatations. In 2d this is known as conformally mapping the cylinder to the punctured plane, and it is realized by mapping functions of $e^z$, which are periodic under adding $2\pi i$, to the same functions of z instead.

Answer 2

You are right that $y(x)=y(2x)$ can be used as condition for your differential equation with the dilation operator, and that this is a form of boundary condition.

The important point is that if $y(x)$ satisfies $T^{2}y(x)=E_{n}y(x)$ then $\bar{y}(x):=y(2x)$ satisfies $T^{2}\bar{y}(x)=E_{n}\bar{y}(x)$. I convinced myself that this is the case by noting that $T:=x\frac{d}{dx}=2x\frac{d}{d(2x)}$, in the same sense that $\frac{d}{dx}=\frac{d}{d(x+1)}$.

Converting such a "periodic condition" into a "real" boundary conditions requires some care. Let's first look at the condition $y(x)=y(x+1)$. You could solve the eigenvalue problem on the interval $[0,1]$ with the boundary conditions $y(0)=y(1)$ and $y'(0)=y'(1)$. For the condition $y(x)=y(2x)$, you could solve the eigenvalue problem on the interval $[1,2]$ with the boundary conditions $y(1)=y(2)$ and $y'(1)=2y'(2)$.