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I've been doing some mathematical modelling in geophysics and the following integration came up. Are there any identities for the Dirac Delta that can be used to simplify this expression:

$$\iint \delta \left( t-\frac1c\left(\sqrt{(x_1-z_1)^2+z_2^2}-\sqrt{z_1^2+z_2^2}\right) \right)dz_1dz_2$$

Edit: To make matters more interesting/challenging (or worse) there are other factors in the integrend, i.e., the full expression is:

$$\iint \frac{\rho(z_1)}{\sqrt{(x_1-z_1)^2+z_2^2}\sqrt{z_1^2+z_2^2}}\delta \left( t-\frac1c\left(\sqrt{(x_1-z_1)^2+z_2^2}-\sqrt{z_1^2+z_2^2}\right) \right)dz_1dz_2$$

where $\rho$ is a some random function. So the question now boils down to: Will Fubini together with

$$\delta(f(x)) = \sum_{i}\frac{\delta(x - x_i)}{|f'(x_i)|}$$

(as pointed out in the answers below) suffice for a simplification still?

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  • $\begingroup$ the range of integration for $z_1$ is $(0, \infty)$ while for $z_2$ it is the entire real line $\endgroup$
    – Iconoclast
    Sep 22, 2016 at 17:18

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Judging from the setup of the problem, I'll assume $x_1, t > 0$ and only study that case.
Let $\lambda = \frac12x_1$ and $\mu = \frac12 ct$. Introduce a bunch of variables:

$$ \begin{cases} (x,y) = (z_1 - \lambda, z_2)\\ r_1 = \sqrt{(x+\lambda)^2 + y^2} = \sqrt{z_1^2+z_2^2}\\ r_2 = \sqrt{(x-\lambda)^2 + y^2} = \sqrt{(z_1-x_1)^2 + z_2^2}\\ (u,v) = \left(\frac{r_1 - r_2}{2},\frac{r_1 + r_2}{2}\right) \end{cases}$$ The actual integral you want to integral can be rewritten as

$$\mathcal{I} = \int_{[-\lambda,\infty)\times(-\infty,\infty)}\rho(x+\lambda)\delta\left(\frac2c(u+\mu)\right)\frac{dxdy}{r_1r_2}$$

Since the integrand is symmetric with respect to the $y$-axis, we can replace the integral by one over the upper $(x,y)$ plane: $$\int_{[-\lambda,\infty)\times(-\infty,\infty)}\cdots\quad\leftrightarrow\quad 2\int_{[-\lambda,\infty)\times [0,\infty)}\cdots$$ Change variables to $(u,v)$, the upper $(x,y)$ plane corresponds to the strip $[-\lambda,\lambda] \times [\lambda,\infty)$ in the $(u,v)$ plane. Notice $$\begin{align} 4uv &= (r_1-r_2)(r_1+r_2) = r_1^2 - r_2^2 = 4\lambda x\\ u^2 + v^2 &= \frac14\left((r_1-r_2)^2+(r_1+r_2)^2\right) = \frac12\left(r_1^2+r_2^2\right) = x^2+y^2+\lambda^2 \end{align} $$ A point $(x,y) \in [-\lambda,\infty) \times [0,\infty)$ is equivalent to corresponding $(u,v)$ belongs to following region: $$\Omega \stackrel{def}{=} \{ (u,v) \in [ -\lambda,\lambda ] \times [ \lambda, \infty ) : uv \ge -\lambda^2 \}$$

In terms of $(u,v)$, the area element has the form $$dx \wedge dy = dx \wedge \frac{dy^2}{2y} = \frac{1}{2\lambda y} d(\lambda x)\wedge d(x^2+y^2+\lambda^2) = \frac{1}{2\lambda y} d(uv)\wedge d(u^2+v^2)\\ = \frac{1}{\lambda y} (udv + vdu)\wedge(udu + vdv) = \frac{v^2-u^2}{\lambda y} du\wedge dv = \frac{r_1r_2}{\lambda y} du\wedge dv $$ Since $$y^2 = u^2 + v^2 - x^2 - \lambda^2 = u^2 + v^2 - \frac{u^2v^2}{\lambda^2} - \lambda^2 = \frac{1}{\lambda^2}(v^2 - \lambda^2)(\lambda^2-u^2)$$ We find $$\frac{dxdy}{r_1r_2} = \frac{dudv}{\lambda y} = \frac{dudv}{\sqrt{(v^2-\lambda^2)(\lambda^2-u^2)}} $$ and hence $$\mathcal{I} = 2\int_\Omega \rho(x+\lambda)\frac{\delta(\frac{2}{c}(u+\mu))dudv}{\sqrt{(v^2-\lambda)^2(\lambda^2-u^2)}}$$

Apply the formula mentioned in other answers, we have $$\delta\left(\frac2c(u+\mu)\right) = \frac{c}{2}\delta(u + \mu)$$

Integrate the delta function over $u$, the $u$ get fixed to $-\mu$.

When $\mu > \lambda$, the line $u = -\mu$ lies outside the strip $[-\lambda,\lambda] \times [\lambda,\infty)$ and $\mathcal{I}$ vanishes.

For $0 < \mu < \lambda$, the integral reduces to

$$\mathcal{I} = \frac{c}{\sqrt{\lambda^2-\mu^2}} \int_\lambda^{\lambda^2/\mu}\frac{\rho\left(\lambda - \frac{\mu v}{\lambda}\right) d v}{\sqrt{v^2-\lambda^2}} = \frac{c}{\sqrt{\lambda^2-\mu^2}}\int_0^{\lambda-\mu} \frac{\rho(z_1)dz_1}{\sqrt{(\lambda-z_1)^2-\mu^2}} $$ The rest of the integral depends on the actual form of $\rho(\cdot)$ and I'll leave that for you.

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  • $\begingroup$ Thank you. I'd like to ask: how can I modify the answer if the original integral looks like this $$\iiint \frac{\rho(z_1)}{\sqrt{(x_1-z_1)^2+z_2^2 +z_3^2}\sqrt{z_1^2+z_2^2+z_3^2}}\delta \left( t-\frac1c\left(\sqrt{(x_1-z_1)^2+z_2^2+z_3^2}-\sqrt{z_1^2+z_2^2+z_3^2}\right) \right)dz_1dz_2dz_3$$ $\endgroup$
    – Iconoclast
    Oct 7, 2016 at 12:54
  • $\begingroup$ @Iconoclast That case will be even simpler. If one parametrize $(z_1,z_2,z_3)$ as $(x+\lambda,y\cos\theta,y\sin\theta)$, then $$\frac{dz_1 dz_2 dz_3}{r_1r_2} = \frac{dx dy}{r_1 r_2}yd\theta = \frac{du dv}{\lambda y}yd\theta = \frac{1}{\lambda}du dv d\theta$$ Integrate over $\theta$ and then over $u$, you will get back essentially the same integral with the ugly square root at the denominator replaced by some constant. $\endgroup$ Oct 7, 2016 at 14:04
  • $\begingroup$ One last question, you are assuming that $z_1>0$ right? What happens if that assumption is invalid, i.e., $z_1 \in (-\infty, \infty)$? $\endgroup$
    – Iconoclast
    Oct 7, 2016 at 14:16
  • $\begingroup$ @Iconoclast If $z_1$ can be any value, the integral of delta function will be over the whole strip $[-\lambda,\lambda]\times[\lambda,\infty)$ in $uv$-plane and the final integral over $z_1$ will be over $(-\infty,\lambda - \mu)$ when $\mu < \lambda$ and identically zero when $\mu > \lambda$. $\endgroup$ Oct 7, 2016 at 14:24
  • $\begingroup$ how do you determine that the region $[-\lambda, \infty) \times (-\infty, \infty)$ in the $xy$-plane corresponds to the region $[-\lambda, \lambda] \times [\lambda, \infty)$ in the $uv$-plane? By plotting? $\endgroup$
    – Iconoclast
    Oct 13, 2016 at 16:19
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Hint:

$$ \delta \left [g(x) \right ] = \sum_{k=1}^N \frac{\delta(x-x_k)}{|g'(x_k)|}$$

where $g(x_k)=0$ for $k=1,2,\ldots,N$.

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  • $\begingroup$ I guess what i have is $\delta[g(x_1, x_2)]$. How would the above expression change in case of two variables. $\endgroup$
    – Iconoclast
    Sep 22, 2016 at 16:59
  • $\begingroup$ Well, you are still integrating over one of the variables first. Let that variable be $x$ in my notation. $\endgroup$
    – Ron Gordon
    Sep 22, 2016 at 17:01
  • $\begingroup$ Just a question: Shouldn't it be $|g'(x_k)|$? $\endgroup$
    – Laplacian
    Sep 22, 2016 at 17:03
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    $\begingroup$ @FourierTransform: yes, thanks. $\endgroup$
    – Ron Gordon
    Sep 22, 2016 at 17:03
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First of all use Fubinit's theorem to "change" the integration:

$$\int \text{d}z_1 \int \delta \left( t-\frac{\sqrt{(x_1-z_1)^2+z_2^2}-\sqrt{(z_1^2+z_2^2)}}{c} \right)\text{d}z_2$$

Now the integration of a Delta "function" of the type

$$\delta(f(x))$$

can be done via the definition of the delta "function" of a function:

$$\delta(f(x)) = \sum_{i}\frac{\delta(x - x_i)}{|f'(x_i)|}$$

Where $i$ runs over all the possible number of roots of the equation

$$f(x_i) = 0$$

Can you proceed?

Hint

Despite you have two variables of integration, thanks to Fubini you can first integrate in $z_2$ (or $z_1$) and then in the other one, so first of all, treat one of the two as a constant and find the roots in terms of $z_2$.

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