0

I have a problem of the form

$$\int_0^{2\pi}dx_1\int_0^{\pi}dx_2\delta(-a+\cos(x_1)+\cos(x_2))$$ where $a\in (-2,0)$. I have reduced it to the following $$\int_0^{2\pi}dx_1\sum_{j\in I}\frac{1}{\sqrt{1-(a-\cos(x_1))^2}},$$ where $I=\{j; x_1^{(j)}\in[0,2\pi], 0\leq \arccos(a-\cos(x_1^{(j)}))\leq\pi\}$. This seems to be satisfied when $x_1\in[\arccos(a+1),2\pi-\arccos(a+1)]$ which is a continuum of solutions for any given value of $a$. I've been looking for a formula for the delta function which has a non-isolated zeros but I can't find any.

I guessed maybe it has the form: $$\int_{\arccos(a+1)}^{2\pi-\arccos(a+1)}dx_2\int_0^{2\pi}dx_1\frac{\delta(x_1-x_2)}{\sqrt{1-\cos(x_2)^2}}.$$ I'm sure this is zero... But I have no proof of this and I don't know if I'm on the right tracks here, could anybody help?

Lewis Proctor
  • 335
  • Since the double integral can be interpreted as $\int_\gamma ds/\left| \nabla f(x_1, x_2) \right|$ (see here), you shouldn't expect it to be zero. Your derivation is essentially correct. It's valid to treat this as an iterated integral; evaluating the inner integral gives $$\int_0^{2 \pi} \frac {[a - \cos x_1 > -1]} {\left| -\sqrt {1 - (a - \cos x_1)^2} \right|} dx_1.$$ There is a closed form in terms of the elliptic integral $K(k)$: $$\frac 8 {2 - a} K {\left( \frac {2 + a} {2 - a} \right)}.$$ – Maxim Jun 15 '19 at 13:41
  • Hi Thank you for your answer! Could you possibly expand a bit more on how you arrived at the first integral? Since $a-\cos(x_1^{(j)})\in [-1,1]$ I'm not sure where the $a-\cos x_1>-1$ came from? I need a bit more information about how to deal with the first integral and how to restrict the values such that the integral is real since in my example, it is the area of some space. A little hint at how to extend this for $a\in (0,2)$ would be advantageous too, but If I knew more about how to deal with the first integral in the current example, I'm sure I can do it myself. – Lewis Proctor Jun 17 '19 at 11:30
  • We're interested in $x_2 \in (0, \pi)$ s.t. $\cos x_2 = a - \cos x_1$. This doesn't have a solution for all $x_1$. $a - \cos x_1 < 1$ always holds, but $a - \cos x_1 > -1$ holds only for some $x_1$. Then you need to solve $0 < x_1 < 2 \pi \land a - \cos x_1 > -1$, but you already worked it out in your question. – Maxim Jun 17 '19 at 12:40
  • But there is a continuum of solutions, for example, when $a=-0.2$ we have $x\in[0.643501,5.63968]$ but this is a continuum of values that $x_1$ could take and $a-\cos(x_1)<-1$ is still satisfied. But you haven't incorporated that in your solution. The above integral with the summation only works for isolated zeros, that's the bit I'm having trouble with. If I have misunderstood again I apologise. – Lewis Proctor Jun 17 '19 at 15:59
  • Let $a = -1$. Take $x_1 = \pi/3$ and evaluate $\int_0^\pi \delta(\cos x_2 + 3/2) , dx_2$. Then take $x_1 = 2 \pi/3$ and evaluate $\int_0^\pi \delta(\cos x_2 + 1/2) , dx_2$. For each $x_1$, you get a real value $f(x_1)$. You need to integrate this $f$ over $[0, 2 \pi]$. – Maxim Jun 17 '19 at 16:20
  • The first value $x_1=\pi/3$ gives a complex $x_2$. There must be a better way than to evaluate it for each $a$. I will keep thinking. – Lewis Proctor Jun 17 '19 at 18:09
  • I think the formula $\int_{\mathbb{R}^n}f(x)\delta(g(x))dx=\int_{g^{-1}(0)}\frac{f(x)}{|\nabla g|}d\sigma(x)$ is helpful. The surface term seems to be tricky... I know $g^{-1}(0)=[\arccos(a+1),2\pi-\arccos(a+1)]$ and since it's a surface term it means it's a one-dimensional integral, i.e., for $x_2$, what happens to $|\nabla g|$? since this is $|\nabla g|=|\sqrt{\sin^2(x_1)+\sin^2(x_2)}|$. – Lewis Proctor Jun 20 '19 at 16:25

0 Answers0