グリーンの定理(Green's theorem)
閉曲線\(C\)で囲まれた領域\(D\)を考える場合，\(C^1\)級凾数\(P(x, y), Q(x, y)\)について以下が成り立つ。
$$
\begin{eqnarray}
\iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y
&=&
\oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y
\end{eqnarray}
$$
面積分を周回積分へ変形し，区間毎の積分に展開する
点\((x,y)\)から点\((p,q)\)までの距離は
$$
\begin{eqnarray}
r(x, y; p, q)&=&\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}
\end{eqnarray}
$$
であり，多角形D上の点\(x,y\)から任意の点\(p,q\)への距離の平均値\(\bar{r}\)は，
$$
\begin{eqnarray}
\bar{r}&=&\iint_D \rho(x,y)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y
\\&=&\iint_D \frac{1}{A}\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y
\;\cdots\;密度は均一と仮定: \rho(x,y)=\frac{1}{A}\;(定数, A:Dの面積)
\\&=&\frac{1}{A}\iint_D \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y
\;\cdots\;\int_XcA(x)\mathrm{d}x=c\int_XA(x)\mathrm{d}x
\end{eqnarray}
$$
である．
これをグリーンの定理の式で満たすために例えば
$$
\begin{eqnarray}
Q(x, y)&=&\int \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x
\\P(x, y)&=&0
\end{eqnarray}
$$
とおく．これは
$$
\begin{eqnarray}
\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}&=&\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}-0
\\&=&\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}
\end{eqnarray}
$$
となり，グリーンの定理の面積分側の被積分凾数を表現できている．
この\(P,Q\)を用いて周回積分側の 被積分凾数を求めると以下のようになる．
$$
\begin{eqnarray}
\bar{r}&=&\frac{1}{A}\iint_D \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\mathrm{d}y
\\&=&\frac{1}{A}\iint_D \left(\frac{\partial Q(x, y)}{\partial x}-\frac{\partial P(x, y)}{\partial y}\right)\mathrm{d}x\mathrm{d}y
\\&=&\frac{1}{A}\oint_C P(x, y)\mathrm{d}x+Q(x,y)\mathrm{d}y
\\&=&\frac{1}{A}\oint_C 0\mathrm{d}x+\left(\int_D \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\right)\mathrm{d}y
\\&=&\frac{1}{A}\oint_C \left(\int \sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\;\mathrm{d}x\right)\mathrm{d}y
\\&=&\frac{1}{A}\oint_C \left(\int \sqrt{u^2+v^2}\;\mathrm{d}u\right)\mathrm{d}y
\;\cdots\;u=x-p, v=y-q
\\&=&\frac{1}{A}\oint_C \frac{1}{2}\left\{u\sqrt{u^2+v^2}+v^2\ln{\left|u+\sqrt{u^2+v^2}\right|}\right\}\mathrm{d}y
\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/x2a2.html}{\int \sqrt{x^2+a^2}\;\mathrm{d}x=\frac{1}{2}\left\{x\sqrt{x^2+a^2}+a^2\ln{\left|x+\sqrt{x^2+a^2}\right|}\right\} + 積分定数}
\\&=&\frac{1}{2A}\oint_C \left\{\left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}
+\left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|}\right\}\mathrm{d}y
\end{eqnarray}
$$
ここで周回積分を多角形として考える．多角形の各頂点の列を以下のように与えるとする．
$$
\begin{eqnarray}
(x_0, y_0), (x_1, y_1),(x_2, y_2), \cdots, (x_n, y_n)=(x_0, y_0)
\end{eqnarray}
$$
周回積分を多角形の区間ごとに分割し，区間毎の積分の和として以下のようになる．
$$
\begin{eqnarray}
\bar{r}&=&\frac{1}{2A}\oint_C \left\{\left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}
+\left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|}\right\}\mathrm{d}y
\\&=&\frac{1}{2A} \sum_{k=1}^n \int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left\{\left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}
+\left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|}\right\}\mathrm{d}y
\\&=&\frac{1}{2A} \sum_{k=1}^n\left\{
\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y
+\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y
\right\}
\\&=&\frac{1}{2A} \sum_{k=1}^n \left\{I_{1k}+ I_{2k}\right\}
\\&&\;\cdots\;I_{1k}=\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y
\\&&\;\cdots\;I_{2k}=\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y
\end{eqnarray}
$$
多角形の頂点間線分(辺)に重なる直線の式を求める
多角形の各頂点において今の頂点\((x_k, y_k)\)とその前の頂点\((x_{k-1}, y_{k-1})\)とを通る直線の式は以下のように求められる．
$$
\begin{eqnarray}
y-y_{k-1}&=&\frac{y_k-y_{k-1}}{x_k-x_{k-1}}\left(x-x_{k-1}\right)
\;\cdots\;傾き\thetaの直線における原点を，点(p,q)へ移動した式と同じ．(y-q)=\theta(x-p)
\\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)
\end{eqnarray}
$$
$$
\begin{eqnarray}
\\x-x_{k-1}&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)
\\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}\left(y-y_{k-1}\right)+x_{k-1}
\\x&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{\left(x_k-x_{k-1}\right)(-y_{k-1})+x_{k-1}\left(y_k-y_{k-1}\right)}{y_k-y_{k-1}}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y
+\frac{
-x_ky_{k-1}
+x_{k-1}y_{k-1}
+x_{k-1}y_k
-x_{k-1}y_{k-1}
}{y_k-y_{k-1}}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_ky_{k-1}+x_{k-1}y_k}{y_k-y_{k-1}}
\\&=&\frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}y_k-x_ky_{k-1}}{y_k-y_{k-1}}
\\&=&\epsilon_k y+\zeta_k
\end{eqnarray}
$$
区間毎の積分より距離の平均値\(\bar{r}\)を求める 第一項\(I_{1k}\)
$$
\begin{eqnarray}
I_{1k}&=&\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y
\\&=&I_{1k}(y_k)-I_{1k}(y_{k-1})
\\I_{1k}(y)&=&\int \left(\left(\epsilon_k y+\zeta_k\right)-p\right)
\sqrt{\left(\left(\epsilon_k y+\zeta_k\right)-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y
\;\cdots\;x=\epsilon_k y+\zeta_k
\\&=&\int \left(\left(\epsilon_k y+\zeta_k\right)-p\right)
\sqrt{\left(\left(\epsilon_k y+\zeta_k\right)^2-2p\left(\epsilon_k y+\zeta_k\right)+p^2\right)+\left(y^2-2qy+q^2\right)} \;\mathrm{d}y
\\&=&\int \left(\epsilon_k y+\zeta_k-p\right)
\sqrt{\left(\epsilon_k^2 y^2+2\epsilon_k\zeta_k y+\zeta_k^2\right)-2p\epsilon_k y-2p\zeta_k+p^2+y^2-2qy+q^2} \;\mathrm{d}y
\\&=&\int \left(\epsilon_k y+\zeta_k-p\right)
\sqrt{\epsilon_k^2 y^2+2\epsilon_k\zeta_k y+\zeta_k^2-2p\epsilon_k y-2p\zeta_k+p^2+y^2-2qy+q^2} \;\mathrm{d}y
\\&=&\int \left(\epsilon_k y+\zeta_k-p\right)
\sqrt{\left(\epsilon_k^2 +1 \right)y^2+2\left\{\epsilon_k\left(\zeta_k-p\right)-q\right\}y +\left(\zeta_k-p\right)^2+q^2} \;\mathrm{d}y
\\&=&\int \left(\epsilon_k y+\zeta_k-p\right)\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y
\;\cdots\;\alpha_k=\epsilon_k^2 +1,\;\beta_k=2\left\{\epsilon_k\left(\zeta_k-p\right)-q\right\},\;\gamma_k=\left(\zeta_k-p\right)^2+q^2
\\&=& \int \epsilon_k y\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y
+ \int \left(\zeta_k-p\right) \sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y
\\&=& \epsilon_k \int y\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y
+ \left(\zeta_k-p\right) \int \sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y
\\&=& \epsilon_k \int y\sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y
+ \left(\zeta_k-p\right) \int \sqrt{\alpha_k y^2+\beta_k y +\gamma_k} \;\mathrm{d}y
\\&=& \epsilon_k \left\{
\frac{1}{3\alpha_k}\left(\alpha_k y^2+\beta_k y+\gamma_k \right)^{\frac{3}{2}}
-\frac{\beta_k \left(2\alpha_k y+\beta_k \right)}{8\alpha_k^{2}}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k}
+\frac{\beta_k \left(\beta_k^2-4\alpha_k\gamma_k\right)}{16\alpha_k^{\frac{5}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k \right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| }
+C_0
\right\}
\\&&+ \left(\zeta_k-p\right) \left\{
\frac{2\alpha_k y+\beta_k}{4\alpha_k}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k}
+\frac{4\alpha_k \gamma_k-\beta_k^2}{8\alpha_k^{\frac{3}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k\right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| }
+C_1
\right\}
\\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/xa-x2b-x-c.html}{\int x\sqrt{a x^2+b x +c} \;\mathrm{d}x=
\frac{1}{3a}\left(ax^2+bx+c\right)^{\frac{3}{2}}
-\frac{b\left(2ax+b\right)}{8a^{2}}\sqrt{\left(ax^2+bx+c\right)}
+\frac{b\left(b^2-4ac\right)}{16a^{\frac{5}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
+C_0\;(C_0:積分定数)}
\\&&\;\cdots\;\href{https://shikitenkai.blogspot.com/2020/07/a-x2b-x-c.html}{\int \sqrt{a x^2+b x +c} \;\mathrm{d}x=
\frac{2ax+b}{4a}\sqrt{\left(ax^2+bx+c\right)}
+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln{ \left|\left(2ax+b\right)+2\sqrt{a\left(ax^2+bx+c\right)}\right| }
+C_1\;(C_1:積分定数)}
\\&=& \epsilon_k \left\{
\frac{1}{3\alpha_k}\left(\alpha_k y^2+\beta_k y+\gamma_k \right)^{\frac{3}{2}}
-\frac{\beta_k \left(2\alpha_k y+\beta_k \right)}{8\alpha_k^{2}}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k}
+\frac{\beta_k \left(\beta_k^2-4\alpha_k\gamma_k\right)}{16\alpha_k^{\frac{5}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k \right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| }
\right\}
\\&&+ \left(\zeta_k-p\right) \left\{
\frac{2\alpha_k y+\beta_k}{4\alpha_k}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k}
+\frac{4\alpha_k \gamma_k-\beta_k^2}{8\alpha_k^{\frac{3}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k\right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| }
\right\}
+\epsilon_k C_0
+\left(\zeta_k-p\right) C_1
\\&=& \epsilon_k \left\{
\frac{1}{3\alpha_k}\left(\alpha_k y^2+\beta_k y+\gamma_k \right)^{\frac{3}{2}}
-\frac{\beta_k \left(2\alpha_k y+\beta_k \right)}{8\alpha_k^{2}}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k}
+\frac{\beta_k \left(\beta_k^2-4\alpha_k\gamma_k\right)}{16\alpha_k^{\frac{5}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k \right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| }
\right\}
\\&&+ \left(\zeta_k-p\right) \left\{
\frac{2\alpha_k y+\beta_k}{4\alpha_k}\sqrt{\alpha_k y^2+\beta_k y+\gamma_k}
+\frac{4\alpha_k \gamma_k-\beta_k^2}{8\alpha_k^{\frac{3}{2}}}\ln{ \left|\left(2\alpha_k y+\beta_k\right)+2\sqrt{\alpha_k\left(\alpha_k y^2+\beta_k y+\gamma_k\right)}\right| }
\right\}
+ C
\;\cdots\;C=\epsilon_k C_0+\left(\zeta_k-p\right) C_1\;(C:積分定数)
\end{eqnarray}
$$
区間毎の積分より距離の平均値\(\bar{r}\)を求める 第二項\(I_{2k}\)
$$
\begin{eqnarray}
I_{2k}&=&\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y
\\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2\ln{\left|\left(\left(\epsilon_k y+\zeta_k\right)-p\right)+\sqrt{\left(\left(\epsilon_k y+\zeta_k\right)-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y
\\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2
\ln{\left|
\epsilon_k y+\zeta_k-p
+\sqrt{
\left(
\left(
\epsilon_k y+\zeta_k
\right)^2
-2p\left(
\epsilon_k y+\zeta_k
\right)+p^2
\right)
+\left(
y^2-2qy-q^2
\right)
}
\right|}
\;\mathrm{d}y
\\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2
\ln{\left|
\epsilon_k y+\zeta_k-p
+\sqrt{
\left(
\epsilon_k^2 y^2
+2\epsilon_k\zeta_k y
+\zeta_k^2
\right)
-2p\epsilon_k y
-2p\zeta_k
+p^2
+y^2-2qy-q^2
}
\right|}
\;\mathrm{d}y
\\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2
\ln{\left|
\epsilon_k y+\zeta_k-p
+\sqrt{
\epsilon_k^2 y^2
+y^2
+2\epsilon_k\zeta_k y
-2p\epsilon_k y
-2qy
+\zeta_k^2
-2p\zeta_k
+p^2
-q^2
}
\right|}
\;\mathrm{d}y
\\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2
\ln{\left|
\epsilon_k y+\zeta_k-p
+\sqrt{
\left(\epsilon_k^2+1\right) y^2
+2\left\{\epsilon_k\left(\zeta_k-p\right)-q\right\} y
+\left(\zeta_k-p\right)^2
-q^2
}
\right|}
\;\mathrm{d}y
\\&=&\int^{y_k}_{y_{k-1}} \left(y-q\right)^2
\ln{\left|
\epsilon_k y+\zeta_k-p
+\sqrt{\alpha_k y^2+\beta_k y+\gamma_k}
\right|}\;\mathrm{d}y
\;\cdots\;\alpha_k=\epsilon_k^2+1
,\;\beta_k=2\left\{\epsilon_k \left(\zeta_k-p\right) -q\right\}
,\;\gamma_k=\left(\zeta_k-p\right)^2-q^2
\end{eqnarray}
$$
→要:数値計算
区間毎の積分より距離の平均値\(\bar{r}\)を求める 第一項\(I_{1k}\)+第二項\(I_{2k}\)
$$
\begin{eqnarray}
\bar{r}&=&\frac{1}{2A} \sum_{k=1}^n\left\{
\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(x-p\right)\sqrt{\left(x-p\right)^2+\left(y-q\right)^2} \;\mathrm{d}y
+\int^{(x_k, y_k)}_{(x_{k-1},y_{k-1})} \left(y-q\right)^2\ln{\left|\left(x-p\right)+\sqrt{\left(x-p\right)^2+\left(y-q\right)^2}\right|} \;\mathrm{d}y
\right\}
\\&=&\frac{1}{2A} \sum_{k=1}^n \left\{I_{1k} + I_{2k}\right\}
\\&=&\frac{1}{2A} \sum_{k=1}^n \left\{I_{1k}(y_{k}) - I_{1k}(y_{k-1}) + I_{2k}\right\}
\end{eqnarray}
$$
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