limit of a region of integration in $\mathbb{R}^2$ approaches a line

I am trying to follow the derivation of derivatives in a paper published in some japanese journal but there seems to be a mistake in the proof. I will present the problem in 2D and in 2 variables so that it's easier to visualize.

Let $c_1=(x_1,y_1)$ and $c_2 = (x_2,y_2)$ be two points in $\mathbb{R^2}$ which I will denote as centers. The domain will be a unit square with vertices at $(0,0),(1,1),$ etc. Define $V_1 = \{z=(x,y):\|z-c_1\| \le ||z-c_2||\}$ and $V_2 = \{z=(x,y): ||z-c_2|| \le ||z-c_1||\}$.

Define the function

$$G(c_1,c_2) = \displaystyle \int_{V_1}2f'(||z-c_1||^2)(x_1-z_1)\phi(z)\,dz$$ where $z = (z_1,z_2)$. For ease of notation, I will denote $h(z,c_1)=2f'(\|z-c_1\|^2)(x_1-z_1)\phi(z).$

My goal is to compute $$\frac{\partial G}{\partial{x_2}}.$$ Notice that the integrand does not depend on $x_2$ but rather $V_1$ does.

So a first step in this derivation is to consider $$\Delta G=G(c_1,c_2+he_1)-G(c_1,c_2)$$ where $e_1 = (1,0)$ and $h \in \mathbb{R}$ and subsequently compute the divided difference as $h \rightarrow 0$.

If $V_1' = \{z=(x,y): \|z-c_1\| \le \|z - (c_2+he_1)\|\}$ and $V_2' = \{z - (x,y): \|z-(c_2+he_1)\| \le \|z-c_1\|\}$ denote the modified regions about the shifted centers, we have that

$$\Delta G = \displaystyle \int_{V_1'\cap V_2}h(z,c_1)\, dz-\int_{V_1 \cap V_2'} h(z,c_1) \,dz$$

as can be seen from, for example: . In the picture, the red region corresponds to $V_1' \cap V_2$ while the yellow region corresponds to $V_2' \cap V_1$.

The derivative, as is presented in other sources (legitimate) but without proof is:

$$\frac{\partial G}{\partial x_2} = \displaystyle \int_W \frac{x_2-z_2}{\|c_1-c_2\|}h(z,c_1) \,dz$$ where $W$ is the hyperplane $V_1 \cap V_2 = \partial V_1 \cap \partial V_2$, i.e. the common line segment of the regions $V_1,V_2.$

I am trying to do reverse engineering, i.e. trying to make sense of the result to see how to proceed with the derivation. But what throws me off is the presence of a "unit-like normal vector" in the form of $\frac{x_2 - z_2}{\|c_1-c_2\|}$. If it helps, it's a standard property that the vector $c_1-c_2$ is orthogonal to $\partial V_1 \cap \partial V_2$.

Because of this, I've been thinking that maybe the divergence theorem could be invoked in some way but I just don't see it. I am also aware that as $h \rightarrow 0, (V_1'\cap V_2) \cup (V_2'\cap V_2) = \partial V_1 \cap \partial V_2$.

Any ideas how to proceed?

• This shows work, and research effort. Why isn't anyone upvoting this? – JohnWO Jul 26 '15 at 1:15
• Just a note, but it might be interesting if you included a link to the paper, for those curious about the original statement of the problem. – izœc Jul 26 '15 at 3:09