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Case 1: We will look at an easier problem first. Let $|\alpha|, |\beta| \leq \alpha_c, \alpha_c \leq \pi$. I want to solve for $\rho(\beta)$ in the following equation, where $P$ denotes the principal value of the integral:

$$\frac{2\sin{\alpha}}{\lambda} = P\int_{-\alpha_c}^{\alpha_c} d\beta \, \rho(\beta) \, \cot{\frac{\alpha -\beta}{2}}$$

Note here that $\rho(\beta)$ satisfies the following constraint: . $$\int_{-\alpha_c}^{\alpha_c}d\beta \, \rho(\beta) =1, \quad \rho(\beta) \geq 0$$

As given from eqns 23 - 30 in Gross and Witten's paper, there are two separate analytic functions which solve this integral equation for $\lambda \geq 2$, and $\lambda \leq 2$. The physical input which helps in solving this equation is that for very large $\lambda$, $\rho(\beta)$ is constant and is spread over the whole circle $(-\pi, \pi)$. They find a solution using this which is valid till $\lambda =2$, after which they construct a different solution for $\lambda \leq 2$. The solution $\rho(\alpha)$ is given by:

\begin{align} \rho(\alpha) & =\frac{2}{\pi \lambda} \cos{\frac{\alpha}{2}} \left( \frac{\lambda}{2} - \sin^2 \frac{\alpha}{2}\right)^{1/2}, \quad \lambda \leq 2 \quad \text{with} \quad |\alpha| < 2 \sin^{-1}\left( \frac{\lambda}{2}\right)^{1/2}\\ & = \frac{1}{2\pi} \left( 1 + \frac{2}{\lambda}\cos{\alpha}\right), \quad \lambda \geq 2\quad |\alpha| \leq \pi. \end{align}

Case 2: Here's my problem. Let $|\alpha|, |\beta| \leq \alpha_c, \alpha_c \leq \pi$. I want to solve for $\rho(\beta)$ in the following equation, where $P$ denotes the principal value of the integral:

$$\frac{2\sin{\alpha}}{\lambda} = P \,k\int_{-\alpha_c}^{\alpha_c} d\beta \, \rho(\beta) \, \frac{\cot{\frac{\alpha -\beta}{2}}}{\left(\sin{\frac{\alpha -\beta}{2}}\right)^k} $$

Here $k \in \mathbb{N}$. As before, again $\rho(\beta)$ satisfies the following constraint: . $$\int_{-\alpha_c}^{\alpha_c}d\beta \, \rho(\beta) =1, \quad \rho(\beta) \geq 0$$

Note that here also for very large $\lambda$, $\rho(\beta)$ should become constant and go to $1/{2\pi}$. How do I solve this?

  • What makes you think that there is an analytic solution for $k\neq0$? – Conifold Apr 07 '20 at 22:52
  • @Conifold There does exist a solution in the $\lambda \to \infty$ limit, which is $\rho(\alpha) = 1/(2\pi)$. My question is, what is the solution in finite $\lambda$ limit? Besides, this models a physical system, so there should exist a solution. –  Apr 07 '20 at 23:48
  • Yes, but why would it be expressible in simple functions like sines and cosines? The only way to approximate it might be by numerical methods. Unless this is a problem from a textbook or something. – Conifold Apr 07 '20 at 23:52
  • @Conifold It's not from a textbook, and I don't know whether it is expressible in terms of simple trigonometric functions. –  Apr 07 '20 at 23:57
  • @Saad True, $\alpha_c$ depends on $\rho(\alpha)$ being positive everywhere. It is nicely demonstrated in Case 1, where the solution for $\lambda < 2$ has $\alpha_c = 2 \sin^{-1} (\lambda/2)$. For $\lambda >2$, $\alpha_c = \pi$. –  Apr 11 '20 at 05:54
  • @BruceLee Some preliminary thoughts: If $ρ$ is assumed to continuous, then such $ρ$ doesn't exist for odd $k$ by considering the first integrand around $β=α$. And for even $k$, it's promising to try $ρ$ of the following form:$$ρ(β)=C_1\left(\sin\fracβ2\right)^{2k+2}+C_2,$$where $C_1$ and $C_2$ are constants. – Ѕᴀᴀᴅ Apr 13 '20 at 03:49
  • Equivalently $\int_{\sin\frac{\alpha+\alpha_c}2}^{\sin\frac{\alpha-\alpha_c}2}\frac{\rho(\alpha-2\arcsin\gamma)}{\gamma^{k+1}},d\gamma=\frac{\sin\alpha}{\lambda k}$. – TheSimpliFire May 27 '21 at 19:33
  • @TheSimpliFire Right, but is there a way to solve this? –  May 31 '21 at 10:56

1 Answers1

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Does the following work ?: let $$\langle f(x)|g(x)\rangle=P\int_{-\alpha_c}^{\alpha_c}f(x)g(x)dx$$ and $$f(\beta)=k\frac{\cot(\frac{\alpha-\beta}{2})}{(\sin(\frac{\alpha-\beta}{2}))^k}-\frac{\sin\alpha}{\alpha_c\lambda}$$.

Suppose a function g not multiple of f, then $$\rho_0(\beta)=g(\beta)-\frac{\langle g(s)|f(s)\rangle}{\langle f(s)|f(s)\rangle}f(\beta)$$

Then $$\langle\rho_0(\beta)|f(\beta)\rangle=0$$

If the condition $A\int_{-\alpha_c}^{\alpha_c}\rho_0(s)ds=1$ to find A is set then $A\rho_0(\beta)$ were a solution.

Remains to set the parameters $\alpha_c,\lambda$ to get positiveness. ?

QuantumPotatoïd
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  • +1 Seems promising. Thanks for the proposed solution, I will let you know in a while if it works. –  May 31 '21 at 10:59