Boundary integral of a fast decreasing test function in the evaluation of Helmholtz equation's Green function

Question

Evaluating the Green function for the Helmholtz equation in three dimensions I face the following identity which i cannot understand: $$\int_{R^3}d^3x(\nabla^2+k^2)\frac{exp(ik\|{x-x_0}\|)}{\||x-x_0\|}f(x)=\int_{R^3}d^3x\frac{exp(ik\|x-x_0\|)}{\|x-x_0\|}(\nabla^2+k^2)f(x)$$ where $f(x)$ is a fast decreasing test function and $\frac{exp(ik\|{x-x_0}\|)}{\||x-x_0\|}$ is my candidate green function, up to normalization constant, solving $(\nabla^2+k^2)G(x-x_0)=\delta(x-x_0)$. I know that for a domain $D$ and two function $u(x)$ and $f(x)$, i have $$\int_{D}d^3x[v(\nabla^2+k^2)u-u(\nabla^2+k^2)v]=\oint_{\partial D}[u\nabla v -v\nabla u ]\cdot \vec{n}ds$$ Now, is this relation valid also if $D=R^3$? If so, how can i show that the right-hand member of the last identity is equal to $0$ with $u=\frac{exp(ik\|{x-x_0}\|)}{\||x-x_0\|}$ and $v=f(x)$?

Answer 1

Using the vector identity $\nabla (\phi \vec A)=\phi \nabla \cdot \vec A+\nabla \phi \cdot \vec A $, we have

$$\begin{align} f(\vec x)\nabla^2G(\vec x|\vec x_0) &=\nabla\cdot\left(f(\vec x)\nabla G(\vec x|\vec x_0)\right)-\nabla f(\vec x)\cdot \nabla G(\vec x|\vec x_0)\\\\ &=\nabla\cdot\left(f(\vec x)\nabla G(\vec x|\vec x_0)\right)-\nabla\cdot\left(G(\vec x|\vec x_0)\nabla f(\vec x) \right)+G(\vec x|\vec x_0)\nabla^2 f(\vec x) \tag 1 \end{align}$$

Integrating $(1)$ over a sphere of radius $R$, centered at $\vec x_0$, and applying the divergence theorem yields

$$\begin{align}\int_{|\vec x-\vec x_0|\le R} f(\vec x)\nabla^2G(\vec x|\vec x_0)\,d^3\vec x&=\int_{|\vec x-\vec x_0|\le R} G(\vec x|\vec x_0)\nabla^2 f(\vec x)\,d^3\vec x\\\\ &+\oint_{|\vec x-\vec x_0|=R} \left(f(\vec x)\nabla G(\vec x|\vec x_0)-G(\vec x|\vec x_0)\nabla f(\vec x)\right)\cdot \hat n\,dS \tag 2 \end{align}$$

Now, for $|\vec x-\vec x_0|=R$, we have the following:

$$\begin{align} \hat n &=\frac{\vec x-\vec x_0}{R}=\hat R\\\\ dS&=R^2\,\sin(\theta)\,d\theta\,d\phi\\\\ G(\vec x|\vec x_0)&=\frac{e^{ikR}}{4\pi R}\\\\ \nabla G(\vec x|\vec x_0)\cdot \hat n&=-\frac{e^{ikR}}{4\pi R^2}\left(1-ikR\right)\\\\ f(\vec x)&= f(\vec x_0+\hat RR)\\\\ \nabla f(\vec x)\cdot \hat n&=\frac{\partial f(\vec x_0+\hat RR)}{\partial R} \end{align}$$

Therefore, we can write the integrand of the surface integral in $(2)$ as

$$\begin{align} R^2\left(f(\vec x)\nabla G(\vec x|\vec x_0)-G(\vec x|\vec x_0)\nabla f(\vec x)\right)\cdot \hat n&=-\frac{e^{ikR}}{4\pi }\left(f(\vec x_0+\hat RR)+R\left(\frac{\partial f(\vec x_0+\hat RR)}{\partial R}-ikf(\vec x_0+\hat RR)\right)\right) \tag 3 \end{align}$$

Provided that $f(\vec x)$ satisfies the Sommerfeld Radiation Condition, the right-hand side of $(3)$ approaches zero as $R\to \infty$. Thus, we find

$$\int_{\mathbb{R}^3} f(\vec x)\nabla^2G(\vec x|\vec x_0)\,d^3\vec x=\int_{\mathbb{R}^3} G(\vec x|\vec x_0)\nabla^2 f(\vec x)\,d^3\vec x$$

Answer 2

By the comments to the question: the fact that $f(x)$ is a fast decreasing test function (hence it and its derivatives $\to 0$ as $\|x\|\to0$) makes the integral $$\lim_{\epsilon \to \infty}\oint_{\|x\|=\epsilon}[u\nabla v -v\nabla u ]\cdot \vec{n}ds\to 0$$ as the boundary goes to infinity, where $u=\frac{exp(ik\|{x-x_0}\|)}{\||x-x_0\|}$ and $v=f(x)$.