多角形D上の点から任意の点への距離の平均値 (グリーンの定理を用いて面積分を周回積分にして求める)

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。 $$\begin{array}{rcl}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y& =& {\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\end{array}$$

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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 $$\begin{array}{rcl}r(x,y;p,q)& =& \sqrt{{(x-p)}^2+{(y-q)}^2}\end{array}$$ 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$\overline{r}$は， $$\begin{array}{rcl}\overline{r}& =& {\iint }_D\rho (x,y)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\mathrm{d}y\\ & =& {\iint }_D\frac{1}{A}\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\mathrm{d}y\cdots \mathrm{密}\mathrm{度}\mathrm{は}\mathrm{均}\mathrm{一}\mathrm{と}\mathrm{仮}\mathrm{定}:\rho (x,y)=\frac{1}{A}(\mathrm{定}\mathrm{数},A:D\mathrm{の}\mathrm{面}\mathrm{積})\\ & =& \frac{1}{A}{\iint }_D\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\mathrm{d}y\cdots {\int }_{X}cA(x)\mathrm{d}x=c{\int }_{X}A(x)\mathrm{d}x\end{array}$$ 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 $$\begin{array}{rcl}Q(x,y)& =& \int \sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\\ P(x,y)& =& 0\end{array}$$ とおく．これは $$\begin{array}{rcl}\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y}& =& \sqrt{{(x-p)}^2+{(y-q)}^2}-0\\ & =& \sqrt{{(x-p)}^2+{(y-q)}^2}\end{array}$$ 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． $$\begin{array}{rcl}\overline{r}& =& \frac{1}{A}{\iint }_D\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}{\iint }_D(\frac{\partial Q(x,y)}{\partial x}-\frac{\partial P(x,y)}{\partial y})\mathrm{d}x\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_{C}P(x,y)\mathrm{d}x+Q(x,y)\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_{C}0\mathrm{d}x+\left({\int }_D\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x\right)\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_C(\int \sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}x)\mathrm{d}y\\ & =& \frac{1}{A}{\oint }_C(\int \sqrt{u^2+v^2}\mathrm{d}u)\mathrm{d}y\cdots u=x-p,v=y-q\\ & =& \frac{1}{A}{\oint }_C\frac{1}{2}\{u\sqrt{u^2+v^2}+v^2\ln |u+\sqrt{u^2+v^2}|\}\mathrm{d}y\cdots \int \sqrt{x^2+a^2}\mathrm{d}x=\frac{1}{2}\{x\sqrt{x^2+a^2}+a^2\ln |x+\sqrt{x^2+a^2}|\}+\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数}\\ & =& \frac{1}{2A}{\oint }_C\{(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}+{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\}\mathrm{d}y\end{array}$$ 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． $$\begin{array}{r}(x_0,y_0),(x_1,y_1),(x_2,y_2),\cdots ,(x_n,y_n)=(x_0,y_0)\end{array}$$ 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． $$\begin{array}{rcl}\overline{r}& =& \frac{1}{2A}{\oint }_C\{(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}+{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\}\mathrm{d}y\\ & =& \frac{1}{2A}\sum _{k=1}^n{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}\{(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}+{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\}\mathrm{d}y\\ & =& \frac{1}{2A}\sum _{k=1}^n\{{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}y+{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\mathrm{d}y\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\{I_{1k}+I_{2k}\}\\ & & \cdots I_{1k}={\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}y\\ & & \cdots I_{2k}={\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\mathrm{d}y\end{array}$$

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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． $$\begin{array}{rcl}y-y_{k-1}& =& \frac{y_k-y_{k-1}}{x_k-x_{k-1}}(x-x_{k-1})\cdots \mathrm{傾}\mathrm{き}\theta \mathrm{の}\mathrm{直}\mathrm{線}\mathrm{に}\mathrm{お}\mathrm{け}\mathrm{る}\mathrm{原}\mathrm{点}\mathrm{を}，\mathrm{点}(p,q)\mathrm{へ}\mathrm{移}\mathrm{動}\mathrm{し}\mathrm{た}\mathrm{式}\mathrm{と}\mathrm{同}\mathrm{じ}．(y-q)=\theta (x-p)\\ x-x_{k-1}& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})\end{array}$$ $$\begin{array}{r}\\ x-x_{k-1}& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}(y-y_{k-1})+x_{k-1}\\ x& =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_k-x_{k-1}}{y_k-y_{k-1}}(-y_{k-1})+x_{k-1}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{(x_k-x_{k-1})(-y_{k-1})+x_{k-1}(y_k-y_{k-1})}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_{k-1}+x_{k-1}{y}_k-x_{k-1}{y}_{k-1}}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{-x_k{y}_{k-1}+x_{k-1}{y}_k}{y_k-y_{k-1}}\\ & =& \frac{x_k-x_{k-1}}{y_k-y_{k-1}}y+\frac{x_{k-1}{y}_k-x_k{y}_{k-1}}{y_k-y_{k-1}}\\ & =& {ϵ}_{k}y+{\zeta }_k\end{array}$$

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区間毎の積分より距離の平均値$\overline{r}$を求める　第一項$I_{1k}$

$$\begin{array}{rcl}{I}_{1k}& =& {\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}y\\ & =& I_{1k}(y_k)-I_{1k}(y_{k-1})\\ I_{1k}(y)& =& \int (({ϵ}_{k}y+{\zeta }_k)-p)\sqrt{{(({ϵ}_{k}y+{\zeta }_k)-p)}^2+{(y-q)}^2}\mathrm{d}y\cdots x={ϵ}_{k}y+{\zeta }_k\\ & =& \int (({ϵ}_{k}y+{\zeta }_k)-p)\sqrt{({({ϵ}_{k}y+{\zeta }_k)}^2-2p({ϵ}_{k}y+{\zeta }_k)+p^2)+(y^2-2qy+q^2)}\mathrm{d}y\\ & =& \int ({ϵ}_{k}y+{\zeta }_k-p)\sqrt{({ϵ}_k^2{y}^2+2{ϵ}_k{\zeta }_{k}y+{\zeta }_k^2)-2p{ϵ}_{k}y-2p{\zeta }_k+p^2+y^2-2qy+q^2}\mathrm{d}y\\ & =& \int ({ϵ}_{k}y+{\zeta }_k-p)\sqrt{{ϵ}_k^2{y}^2+2{ϵ}_k{\zeta }_{k}y+{\zeta }_k^2-2p{ϵ}_{k}y-2p{\zeta }_k+p^2+y^2-2qy+q^2}\mathrm{d}y\\ & =& \int ({ϵ}_{k}y+{\zeta }_k-p)\sqrt{({ϵ}_k^2+1)y^2+2\{{ϵ}_k({\zeta }_k-p)-q\}y+{({\zeta }_k-p)}^2+q^2}\mathrm{d}y\\ & =& \int ({ϵ}_{k}y+{\zeta }_k-p)\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y\cdots {\alpha }_k={ϵ}_k^2+1,{\beta }_k=2\{{ϵ}_k({\zeta }_k-p)-q\},{\gamma }_k={({\zeta }_k-p)}^2+q^2\\ & =& \int {ϵ}_{k}y\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y+\int ({\zeta }_k-p)\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y\\ & =& {ϵ}_k\int y\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y+({\zeta }_k-p)\int \sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y\\ & =& {ϵ}_k\int y\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y+({\zeta }_k-p)\int \sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}\mathrm{d}y\\ & =& {ϵ}_k\{\frac{1}{3{\alpha }_k}{({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}^{\frac{3}{2}}-\frac{{\beta }_k(2{\alpha }_{k}y+{\beta }_k)}{8{\alpha }_k^2}\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}+\frac{{\beta }_k({\beta }_k^2-4{\alpha }_k{\gamma }_k)}{16{\alpha }_k^{\frac{5}{2}}}\ln |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|+C_0\}\\ & & +({\zeta }_k-p)\{\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 |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|+C_1\}\\ & & \cdots \int x\sqrt{a{x}^2+bx+c}\mathrm{d}x=\frac{1}{3a}{(a{x}^2+bx+c)}^{\frac{3}{2}}-\frac{b(2ax+b)}{8a^2}\sqrt{(a{x}^2+bx+c)}+\frac{b(b^2-4ac)}{16a^{\frac{5}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|+C_0(C_0:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & & \cdots \int \sqrt{a{x}^2+bx+c}\mathrm{d}x=\frac{2ax+b}{4a}\sqrt{(a{x}^2+bx+c)}+\frac{4ac-b^2}{8a^{\frac{3}{2}}}\ln |(2ax+b)+2\sqrt{a(a{x}^2+bx+c)}|+C_1(C_1:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\\ & =& {ϵ}_k\{\frac{1}{3{\alpha }_k}{({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}^{\frac{3}{2}}-\frac{{\beta }_k(2{\alpha }_{k}y+{\beta }_k)}{8{\alpha }_k^2}\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}+\frac{{\beta }_k({\beta }_k^2-4{\alpha }_k{\gamma }_k)}{16{\alpha }_k^{\frac{5}{2}}}\ln |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|\}\\ & & +({\zeta }_k-p)\{\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 |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|\}+{ϵ}_k{C}_0+({\zeta }_k-p)C_1\\ & =& {ϵ}_k\{\frac{1}{3{\alpha }_k}{({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}^{\frac{3}{2}}-\frac{{\beta }_k(2{\alpha }_{k}y+{\beta }_k)}{8{\alpha }_k^2}\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}+\frac{{\beta }_k({\beta }_k^2-4{\alpha }_k{\gamma }_k)}{16{\alpha }_k^{\frac{5}{2}}}\ln |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|\}\\ & & +({\zeta }_k-p)\{\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 |(2{\alpha }_{k}y+{\beta }_k)+2\sqrt{{\alpha }_k({\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k)}|\}+C\cdots C={ϵ}_k{C}_0+({\zeta }_k-p)C_1(C:\mathrm{積}\mathrm{分}\mathrm{定}\mathrm{数})\end{array}$$

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区間毎の積分より距離の平均値$\overline{r}$を求める　第二項$I_{2k}$

$$\begin{array}{rcl}{I}_{2k}& =& {\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |(({ϵ}_{k}y+{\zeta }_k)-p)+\sqrt{{(({ϵ}_{k}y+{\zeta }_k)-p)}^2+{(y-q)}^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{({({ϵ}_{k}y+{\zeta }_k)}^2-2p({ϵ}_{k}y+{\zeta }_k)+p^2)+(y^2-2qy-q^2)}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{({ϵ}_k^2{y}^2+2{ϵ}_k{\zeta }_{k}y+{\zeta }_k^2)-2p{ϵ}_{k}y-2p{\zeta }_k+p^2+y^2-2qy-q^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{{ϵ}_k^2{y}^2+y^2+2{ϵ}_k{\zeta }_{k}y-2p{ϵ}_{k}y-2qy+{\zeta }_k^2-2p{\zeta }_k+p^2-q^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{({ϵ}_k^2+1)y^2+2\{{ϵ}_k({\zeta }_k-p)-q\}y+{({\zeta }_k-p)}^2-q^2}|\mathrm{d}y\\ & =& {\int }_{y_{k-1}}^{y_k}{(y-q)}^2\ln |{ϵ}_{k}y+{\zeta }_k-p+\sqrt{{\alpha }_k{y}^2+{\beta }_{k}y+{\gamma }_k}|\mathrm{d}y\cdots {\alpha }_k={ϵ}_k^2+1,{\beta }_k=2\{{ϵ}_k({\zeta }_k-p)-q\},{\gamma }_k={({\zeta }_k-p)}^2-q^2\end{array}$$ →要:数値計算

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区間毎の積分より距離の平均値$\overline{r}$を求める　第一項$I_{1k}$+第二項$I_{2k}$

$$\begin{array}{rcl}\overline{r}& =& \frac{1}{2A}\sum _{k=1}^n\{{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}(x-p)\sqrt{{(x-p)}^2+{(y-q)}^2}\mathrm{d}y+{\int }_{(x_{k-1},y_{k-1})}^{(x_k,y_k)}{(y-q)}^2\ln |(x-p)+\sqrt{{(x-p)}^2+{(y-q)}^2}|\mathrm{d}y\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\{I_{1k}+I_{2k}\}\\ & =& \frac{1}{2A}\sum _{k=1}^n\{I_{1k}(y_k)-I_{1k}(y_{k-1})+I_{2k}\}\end{array}$$