If I have the following integral equation $$\phi(\vec{x})=\frac{1}{\pi}\int \left[\phi\frac{\partial (\ln r)}{\partial n} -\ln(r) \frac{\partial \phi}{\partial n}\right] ds$$
An approximate solution of $\phi$ is obtained numerically by dividing the boundary into a finite number of segments ,N.
So we can write $$\phi(\vec x_j)=\sum_{i=1}^{N}\left[\phi(\vec x_i)\frac{\partial \ln(r_{ij})}{\partial n} -\ln(r_{ij})\frac{\partial \phi}{\partial n}(\vec x_i)\right]\Delta s_i $$ Where $\Delta s_j$ represents the boundary segment length and $r_{ij}$ is the distance between the $i^{th}$ and the $j^{th}$ segment
So it's easy to write $$\frac{\partial \ln(r_{ij})}{\partial n}=\frac{1}{r_{ij}}\left[-\frac{(x_i -x_j)}{r_{ij}}(\frac{\Delta y}{\Delta x})_j +\frac{(y_i -y_j)}{r_{ij}}(\frac{\Delta x}{\Delta y})_j\right]\Delta s_j$$ And $$Z_{ij}=\ln(r_{ij})\Delta s_j$$
Prove that $$\lim_{j \to i} \frac{\partial \ln(r_{ij})}{\partial n}=\left[\frac{(-x_{ss} y_s + x_s y_{ss})_i}{2}\right]\Delta s_i$$
My try $$\lim_{j \to i} \frac{\partial \ln(r_{ij})}{\partial n}=\lim_{h\to 0}\frac{-(x_i -x_{i+h})(y'_{i+h})+(y_i - y_{i+h})(x'_{i+h})}{(x_i -x_{i+h})^2 +(y_i -y_{i+h})^2}$$
Using $$x_{i+h}=x_i +h x'_i +(h^2/2) x''_i$$ and $$x'_{i+h}=x'_i +hx''_i$$ Hence we get the required result
My question How to prove that $$\lim_{j \to i} Z_{ij} = \left[\ln(\frac{\Delta s_i}{2})-1\right]\Delta s_i$$
Thanks in advance .
For $(G_n)_{ii}$ Define $$r=\sqrt{(x-x_i)^2 +(y-y_i)^2}$$ $$\frac{\partial \ln r}{\partial r}=\frac{\partial \ln r}{\partial x}\frac{\partial x}{\partial n} + \frac{\partial \ln r}{\partial y}\frac{\partial y}{\partial n}$$ Equivalently $$\frac{\partial \ln r}{\partial r}=\frac{\partial \ln r}{\partial x}\frac{\partial y}{\partial s} - \frac{\partial \ln r}{\partial y}\frac{\partial x}{\partial s}$$
$$\frac{\partial \ln r}{\partial r}\approx\frac{\partial \ln r}{\partial x}\frac{\Delta y}{\Delta s} - \frac{\partial \ln r}{\partial y}\frac{\Delta x}{\Delta s}$$ Using $$x-x_i \approx (x_s)_i \Delta s +(x_{ss})_i \frac{\Delta s^2}{2}+...$$ $$\frac{\partial x}{\partial s} \approx (x_s)_i +(x_{ss})_i \Delta s +...$$ Similarly for $y-y_i$ and $\frac{\partial y}{\partial s}$
Now $$\lim_{r\to 0} \frac{\partial \ln r}{\partial r}=\lim_{\Delta s\to 0} \frac{\partial \ln r}{\partial r}$$
Which implies that $$\lim_{\Delta s \to 0}\frac{(-x_{ss} y_s +x_sy_{ss})_i \frac{(\Delta s)^2}{2}+O(\Delta s^3)}{(x_s^2 +y_s^2)_i (\Delta s)^2 +O(\Delta s^3)}$$
Hence The result $$(G_n)_{ii}=\frac{(-x_{ss} y_s +x_s y_{ss})_i}{2} \Delta s_i$$
For $G_{ii}$ The average of the function over the length of the line segment $$(G)_{ii}=\frac{2}{\Delta s-i} \int_{0}^{\frac{\Delta s_i}{2}}\frac{\partial \ln r}{\partial n} dr \ \Delta s_i$$
Which implies that $$(G)_{ii}=[\ln(\Delta s_i /2)-1] \Delta s_i$$