# acoustic dipole volume integral w/ dirac delta?

I have an acoustic research problem that leads to the following integral formulation: \begin{align} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}p(\mathbf{y},\tau)\frac{\partial}{\partial y_i}\left(n_i \delta(f) |\nabla f|\right) g(\mathbf{x} - \mathbf{y}, t-\tau) d^3\mathbf{y} d\tau, \end{align} where $p$ is a scalar function, $n_i=\nabla f / |\nabla f|$ is the normal vector of my control surface, and $\delta(f)$ is the dirac delta with my level set function, $f$, the zero set of which describes the control surface. $g$ is a numerical green's function. If legal, I would like to use the vector property, \begin{align} \nabla \cdot (\psi \mathbf{v}) = \nabla \psi \cdot \mathbf{v} + \psi \nabla \cdot \mathbf{v}, \end{align} to convert the integrand to a form such as \begin{align} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}p(\mathbf{y},\tau)\left[ \frac{\partial n_i}{\partial y_i} \left(\delta(f) |\nabla f|\right) + n_i\frac{\partial}{\partial y_i} \left(\delta(f) |\nabla f|\right) \right] g(\mathbf{x} - \mathbf{y}, t-\tau) d^3\mathbf{y} d\tau. \end{align} However, since $\delta(f)$ is a distribution, I am not quite sure if
(1) this is a correct manipulation;
(2) how to deal with the second term in the bracket?

I want to convert this to a surface integral to evaluate my boundary sources, and the first term gives me just that. Thanks in advance.

First do an integration by parts:

$$\int F(\mathbf{y})\frac{\partial}{\partial y_i}\left(n_i\delta(f)|\nabla f|\right)\,d^n\mathbf{y}= -\int \delta(f)|\nabla f|n_i\frac{\partial}{\partial y_i}F(\mathbf{y})\,d^n\mathbf{y}.$$

Then use this identity,

$$\int A(\mathbf{y})\delta(f(\mathbf{y}))d^n \mathbf{y}=\int_{S}\frac{A(\mathbf{y})}{|\nabla f|}\,d^{n-1}\mathbf{y},$$

with $S$ the surface defined by $f(\mathbf{y})=0$, to arrive at

$$\int F(\mathbf{y})\frac{\partial}{\partial y_i}\left(n_i\delta(f)|\nabla f|\right)\,d^n\mathbf{y}= -\int_S n_i\frac{\partial}{\partial y_i}F(\mathbf{y})\,d^{n-1}\mathbf{y}.$$

• I understand this. Is it possible to do it in the formulation I proposed? Or is handling the distribution like that is just not recommended? Apr 6 '15 at 22:09
• your formula is correct, but if you want to end up with a pure surface integral, it's not helpful. Apr 7 '15 at 6:16
• The first term can be transformed to surface integral using the same layer integral technique you used, but is there a way to deal with the gradient of the distribution? Is it even well-behaved? I am just trying to use this as an opportunity to learn more. Thanks! Apr 7 '15 at 14:19
• Ideally, I would not want to evaluate grad(F), since it has a Green's function and doing what you suggested would be pretty expensive computationally. It requires O(N^2) discrete gradient evaluation, whereas using the formulation I can compute curvature only once (but still with the problem of approximating second term). Apr 7 '15 at 14:43