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My question: How do I use Mathematica to find the solution to Case 2? As a cross check for the code, does it verify the solution in Case I?

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 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: . $$\int_{-\alpha_c}^{\alpha_c}d\beta \, \rho(\beta) =1, \quad \rho(\beta) \geq 0$$

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  • 1
    $\begingroup$ "I am fine with either an exact answer or an approximate one." By "approximate one" do you mean a symbolic solution like series solution, or numeric solution? $\endgroup$
    – xzczd
    Apr 10, 2020 at 4:53
  • $\begingroup$ @xzczd Fine with either, however in both cases I would like to see a plot of the $\rho(\beta)$ with $\beta$ for some values of $\lambda$. $\endgroup$
    – user35588
    Apr 10, 2020 at 19:28
  • $\begingroup$ I suspect the question itself is wrong. Assuming $\rho(\beta)$ can be expanded as Fourier sine series on $[-\alpha_c, \alpha_c]$, take $\alpha=\pi/2$, $\alpha_c=\pi$, and compute the first term of the series expansion, Integrate and NIntegrate both complain the integral doesn't converge: Integrate[(Cot[(Pi/2 - b)/2] Sin[b/2 + Pi/2])/Sin[(Pi/2 - b)/2], {b, -Pi, Pi}, PrincipalValue -> True], NIntegrate[(Cot[(Pi/2 - b)/2] Sin[b/2 + Pi/2])/Sin[(Pi/2 - b)/2], {b, -Pi, Pi/2, Pi}, Method -> PrincipalValue]. $\endgroup$
    – xzczd
    Apr 11, 2020 at 3:08
  • $\begingroup$ @xzczd Why this particular integral? It just shows that Sin[b/2 + Pi/2] isn't the solution, right? I would suggest to cross check any method with Case 1, it might be helpful. I don't think that the question itself is wrong. $\endgroup$
    – user35588
    Apr 11, 2020 at 6:02
  • $\begingroup$ As mentioned above, this is the first term of a Fourier sine series expansion: $\rho(\beta)=c_1 \sin(\beta/2 +\pi/2)+c_2 \sin(2(\beta/2+\pi/2))+…$ $\endgroup$
    – xzczd
    Apr 11, 2020 at 6:17

2 Answers 2

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I have no chance to read the paper "Possible third-order phase transition in the large-N lattice gauge theory" by David J. Gross and Edward Witten. But we can easily check that both analytical solutions are wrong, not satisfices to normalisation condition $\int_{-\alpha_c}^{\alpha_c}\rho(\beta)d\beta =1 $. Really, let take $\lambda \ge 2$, and therefore $\rho =\frac {1}{2\pi}(1+2 \cos (\alpha)/\lambda)$. Now we use a line of code

Integrate[1/2/Pi (1 + 2 Cos[x]/lambda), {x, -a, a}]

Out[]= (a lambda + 2 Sin[a])/(lambda \[Pi]) 

If we assume that it should be equal to 1, then we have

Solve[(a lambda + 2 Sin[a])/(lambda \[Pi]) == 1, {lambda}]

Out[]= {{lambda -> (2 Sin[a])/(-a + \[Pi])}} 

But this function always less then 2 and equal 2 only for $a =\pi$. Therefore, this solution valid only for $\lambda =2, \alpha_c =\pi$. Let consider a numerical solution to the case 1 for $\lambda =2, \alpha_c=\pi-\pi/7$, the code based on Haar wavelets is

ac = Pi - Pi/7.; L = 2; A = -ac; B = ac; jm = 4; M = 
 2^jm; dx = (B - A)/(2 M); 
h1[x_] := Piecewise[{{1, A <= x < B}, {0, True}}];
h[x_, k_, m_] := 
  Piecewise[{{1, A + 2 k M dx/m <= x < A + (2 k + 1) M dx/m}, {-1, 
     A + (2 k + 1) M dx/m <= x < A + 2 ( k + 1) M dx/m}, {0, True}}];

xl = Table[A + l dx, {l, 0, 2 M}]; xcol = 
 Table[(xl[[l - 1]] + xl[[l]])/2, {l, 2, 2 M + 1}];

var = Flatten[Table[co[i, j], {j, 0, jm, 1}, {i, 0, 2^j - 1, 1}]];
varM = Join[{co[-1, -1]}, var];

eq[x_] := 
  Sum[NIntegrate[Cot[(x - t)/2]*h[t, i, 2^j], {t, -ac, x, ac}, 
      Method -> {"InterpolationPointsSubdivision", 
        Method -> {"PrincipalValue", "SymbolicProcessing" -> 0}}]*
     co[i, j], {j, 0, jm, 1}, {i, 0, 2^j - 1, 1}] + 
   NIntegrate[Cot[(x - t)/2]*h1[t], {t, -ac, x, ac}, 
     Method -> {"InterpolationPointsSubdivision", 
       Method -> {"PrincipalValue", "SymbolicProcessing" -> 0}}]*
    co[-1, -1] ;
eqM = Flatten[Table[-eq[x] + Sin[x] == 0, {x, xcol}]];



{b, m} = N[CoefficientArrays[eqM, varM]];


sol1 = LinearSolve[m, -b];

sol[x_] := 
 Sum[co[i, j] h[x, i, 2^j], {j, 0, jm, 1}, {i, 0, 2^j - 1, 1}] + 
  co[-1, -1] h1[x]; lst = 
 Table[{x, 
   Evaluate[
    sol[x] /. 
     Table[varM[[j]] -> sol1[[j]], {j, Length[varM]}]]}, {x, -ac, 
   ac, .01}];

Now we check how it consistent with analytical solution

Show[ListPlot[Re[lst], PlotStyle -> Orange, 
  PlotLabel -> Row[{"M = ", 2^jm}]], 
 Plot[1/2/Pi (1 + Cos[x]), {x, -ac, ac}, AxesLabel -> {"x", "u"}, 
  PlotStyle -> Blue]]

Figure 1 So the numerical solution get close to the analytical solution with M increasing. We put $\lambda =1$ and calculate that for this case $\alpha_c=\frac{\pi}{2}$, therefore analytical solution has a form 2/Pi Cos[x/2] Sqrt[(1/2 - Sin[x/2]^2)]. Numerical solutions for jm=4 and jm=5togehter with the analytical solution are shown below

Figure 2

We see a large discrepancies at $\alpha \rightarrow \pm\frac {\pi}{2}$. Analytical solution completely wrong for $\lambda =1$. We can use some combination of solutions as (2/(Pi lambda) Cos[x/2] Sqrt[(lambda/2 - Sin[x/2]^2)] + 1/2/Pi (1 + 2 Cos[x]))/2. Then agreement is better - see Figure 3.

Figure 3

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  • $\begingroup$ In Case I, $\alpha_c = \pi$ for $\lambda \geq 2$ as indicated in the question. $\alpha_c = |2 \sin^{-1}\left( \frac{\lambda}{2}\right)^{1/2}|$ for $\lambda <2$. This is what normalizes the solution $\rho(\alpha)$ to $\int_{-\alpha_c}^{\alpha_c} d\alpha \, \rho(\alpha) = 1$. I want to see whether in Case2 I can get something similar to the two kinds of analytic behaviour above and below $\lambda =2$ in Case I. The solution in Case 1 is correct, no doubt about that. $\endgroup$
    – user35588
    Apr 16, 2020 at 8:09
  • $\begingroup$ @BruceLee There is a big difference if we take one solution with $\lambda=2, \alpha =\pi$, and any solution with constrains $\alpha \le \pi, \lambda \ge 2$. It looks like nobody even try to make a routine analysis of this analytical solutions. We can't get solution in a Case 2 without regularisation rules. What do you mean when you put $P$ before integral? $\endgroup$ Apr 16, 2020 at 10:59
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Consider first the seemingly trivial cases.

𝛼=0:

Sin[0]=0

Cotangent can be developed into a Taylor series around 0 and that looks

SeriesData[b, 0, {1, 0, 
Rational[-1, 3], 0, 
Rational[-1, 45]}, -1, 4, 1]

So for small 𝛼𝑐 there is on change for this very integral for a constant part even with the method PrincipalValue. But the given solution has both constant contributions and it can be developed into a Taylor series around 0. Products of Taylor series are still a Taylor series.

This is in need of compensation for the singularity caused by the Sine in the denominator of the Cotangent function. Assume that rho in case 1 is indeed developable into a Taylor series than then first non zero contribution is indeed that for the linear beta.

The same is valid in case 2. The singularity rises for each natural k, so such a Taylor series will start with and k higher coefficient.

For the complete Cotangent Mathematica does not solve the integral with rho time Cotangent but it does for both a constant and the Cosine with some conditions due to integration effort that are much more complicated than the simple Findroot solution.

The alpha shifts and distorts the symmetry of the Cotangent around beta=0.

Lambda is arbitrary for small alphac in some contradiction to other solutions presented here. There is more work to do and more cases are to be considered.

There is in complex function theory a theorem that correlates the degree of the enclose singularity to the number of residue. That can be applied here. It is somewhat like a turning number. This theorem offers the k in front of the principal value. It is somewhat the main theorem for analytic complex functions.

It proves the relation if case 1 holds. Case 1 is a premise of the question and the work is done.

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