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

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 $v=f(x)$?

• What are your boundary conditions for your problem, are they Dirichlet, Neumann, Robin...? This will tell you the value of the function or its normal derivatives on the boundary. – Triatticus Aug 27 '16 at 13:35
• @Triatticus i really don't know, the passage i'm asking for is in the proof of Sommerfeld radial condition. The author just takes $u=\frac{exp(ik\|{x-x_0}\|)}{\||x-x_0\|}$ and a fast decreasing test function $f(x)$ and shows that $u$ satisfies the relation $$(\nabla^2+k^2)u(x)=-4\pi\delta(x)$$ The only thing i know is that my green function must satisfy $G(x)=0$ for $\|x\|\to 0$, and $f(x)$ is a fast decreasing test function. I don't see in the text any other type of assumption. – pier94 Aug 27 '16 at 13:46
• Well for example, a fast decreasing test function means that f will go to zero sufficiently fast so as to be zero along a boundary infinitely far away (more or less imagine since your space is euclidean space the boundary is going to infinity). This means the second term of your boundary integral is zero. – Triatticus Aug 27 '16 at 13:58
• @Triatticus so it is sufficient to say that, since $f(x)$ and it's derivatives go to $0$ at infinity, the term $$\oint_{\partial D}[u\nabla v -v\nabla u ]\cdot \vec{n}ds$$ is 0? – pier94 Aug 27 '16 at 14:25
• this is correct – tired Aug 27 '16 at 14:46

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

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

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 $v=f(x)$.