I would like to solve that problem: $$ \int d^2 \mathbf{k} \, \delta(f(k,\phi)) = \int_{k_1}^{k_2} dk \, k \int_0^{2\pi} d\phi \, \delta(f(k,\phi)) \,, $$ where $f(k,\phi) = a - bk^2 - ck^3 |\sin(2\phi)|$ ($a$, $b$ and $c$ are positive constants, $k=\sqrt{k_x^2+k_y^2}$, $k_x=k\cos(\phi)$, $k_y=k\sin(\phi)$).

I know that $$\delta(g(x))=\sum_i \frac{\delta(x-x_i)}{|g'(x)|_{x=x_i}} \,$$ and that for 2D Dirac delta function in polar coordinates we have $$\delta(\mathbf{k}-\mathbf{k}_0) = \frac{1}{k_0}\delta(k-k_0)\delta(\phi-\phi_0) \,.$$ But I have doubt that I can write (if so what about $\frac{1}{k_0}$ from above formula): $$\delta(f(k,\phi))=\sum_i \frac{\delta(\phi-\phi_i)}{|g'(\phi)|_{\phi=\phi_i}} \sum_j \frac{\delta(k-k_j)}{|g'(k)|_{k=k_j}}\,.$$

1. First try

I begin with the first part of above formula (with $\delta(\phi-\phi_i)$). I can calculate that

$$|\frac{\partial f(k,\phi)}{\partial \phi}|_{\phi=\phi_i}= ck^3|\cos(2\phi_i)|\,,$$

in the region $\phi \in [0,2\pi]$ function $f(k,\phi)$ has 8 roots ($|\sin(2\phi_0)|=\frac{a-bk^2}{ck^3}$), so: $$\delta(f(\phi))=\frac{8\delta(\phi-\phi_0)}{ck^3|\cos(2\phi_0)|}= \frac{8\delta(\phi-\phi_0)}{\sqrt{c^2 k^6 - (a-bk^2)^2}} \,.$$ And now should I do the same for the second part (with $\delta(k-k_j)$), multiply and integrate it?

3. Second try

If I consider, for beginning, only integral over $\phi$ of function $f(k,\phi)$ and substitute it to the primarly equation: $$ \int_{k_1}^{k_2} dk \, k \int_0^{2\pi} d\phi \, \delta(f(k,\phi))= \int_{k_1}^{k_2} dk \, k \int_0^{2\pi} d\phi \, \frac{8\delta(\phi-\phi_0)}{\sqrt{c^2 k^6 - (a-bk^2)^2}}= \int_{k_1}^{k_2} dk \, \frac{8k}{\sqrt{c^2 k^6 - (a-bk^2)^2}} \,, $$

Based on above formula, I can try to define the limits of integration over $k$ (from requirement $c^2 k^6 - (a-bk^2)^2>0$) and just integrate it over $k$. But I think that is also not appropriate approach to solve this problem.

I do not understand it well and perhaps I do not see something obvious.

  • $\begingroup$ You need a concrete definition of integrals with $\delta$. For example $ \lim_{n \to \infty} \int_{\mathbb{R}^2} n e^{-\pi n^2 |f(k)|^2}d^2 k$. Here $f(k)$ is non-zero for $\|k\| \not \in [k_1,k_2]$ and $ \to \infty$ fast enough as $\|k\| \to \infty$ to state it is $=\lim_{n \to \infty} \int_{\mathbb{R}^2, \|k \| \in [k_1,k_2]} n e^{-\pi n^2 |f(k)|^2}d^2 k$. Then apply the radial change of variable $k = (r \cos \phi, r \sin \phi), d^2 k = d\phi r dr$. Conclude by looking for each $r$ at the finitely many zeros of $\phi \mapsto f(r \cos \phi, r \sin \phi)$ $\endgroup$ – reuns Mar 8 '19 at 16:42

You need to use the relationship: $$\int_{\Omega}{g(\omega)\delta(f(\omega))\,d\omega} = \int_{f^{-1}(0)}{g(x)/|\nabla f| \,d\sigma(x)}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.