POINT
次の書籍の計算メモです．
- 「物理とグリーン関数（今村勤）」の計算メモ．
記法
Fourier変換
関数$f$のFourier変換を\begin{aligned}
\hat{f}(\boldsymbol{p})
=\mathcal{F}[f](\boldsymbol{p})
=\frac{1}{(2\pi)^n}\int f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} \,\mathrm{d}\boldsymbol{r}
\end{aligned}
と表す．\hat{f}(\boldsymbol{p})
=\mathcal{F}[f](\boldsymbol{p})
=\frac{1}{(2\pi)^n}\int f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} \,\mathrm{d}\boldsymbol{r}
\end{aligned}
よく使う式
- $\displaystyle \mathcal{F}[\Delta f](\boldsymbol{p}) = -p^2\cdot \mathcal{F}[f](\boldsymbol{p})$
\begin{aligned}
\mathcal{F}[\Delta f](\boldsymbol{p})
=\frac{1}{(2\pi)^n}\int \Delta f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} \,\mathrm{d}\boldsymbol{r}
\end{aligned}
は，\mathcal{F}[\Delta f](\boldsymbol{p})
=\frac{1}{(2\pi)^n}\int \Delta f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} \,\mathrm{d}\boldsymbol{r}
\end{aligned}
\begin{aligned}
&\Delta f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} \\
&=\boldsymbol{\nabla}\cdot [\boldsymbol{\nabla} f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} ]
- \boldsymbol{\nabla} f(\boldsymbol{r}) \cdot (-i\boldsymbol{p} e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} )
\end{aligned}
の第1項の積分がGaussの定理から&\Delta f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} \\
&=\boldsymbol{\nabla}\cdot [\boldsymbol{\nabla} f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} ]
- \boldsymbol{\nabla} f(\boldsymbol{r}) \cdot (-i\boldsymbol{p} e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} )
\end{aligned}
\begin{aligned}
\frac{1}{(2\pi)^n} \int_S
[\boldsymbol{\nabla} f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} ] \cdot\boldsymbol{n} \,\mathrm{d}S=0
\end{aligned}
となるとすれば，\frac{1}{(2\pi)^n} \int_S
[\boldsymbol{\nabla} f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} ] \cdot\boldsymbol{n} \,\mathrm{d}S=0
\end{aligned}
\begin{aligned}
\mathcal{F}[\Delta f](\boldsymbol{p})
=-\frac{1}{(2\pi)^n}
\int \boldsymbol{\nabla} f(\boldsymbol{r}) \cdot (-i\boldsymbol{p} e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} )
\,\mathrm{d}\boldsymbol{r}.
\end{aligned}
同様に，\mathcal{F}[\Delta f](\boldsymbol{p})
=-\frac{1}{(2\pi)^n}
\int \boldsymbol{\nabla} f(\boldsymbol{r}) \cdot (-i\boldsymbol{p} e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} )
\,\mathrm{d}\boldsymbol{r}.
\end{aligned}
\begin{aligned}
&\boldsymbol{\nabla} f(\boldsymbol{r}) \cdot (-i\boldsymbol{p} e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} ) \\
&=\boldsymbol{\nabla} \cdot [f(\boldsymbol{r}) (-i\boldsymbol{p} e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} ) ]
- f(\boldsymbol{r}) (-p^2 e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} )
\end{aligned}
第1項の積分がゼロになるとすれば，&\boldsymbol{\nabla} f(\boldsymbol{r}) \cdot (-i\boldsymbol{p} e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} ) \\
&=\boldsymbol{\nabla} \cdot [f(\boldsymbol{r}) (-i\boldsymbol{p} e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} ) ]
- f(\boldsymbol{r}) (-p^2 e^{-i\boldsymbol{p}\cdot\boldsymbol{r}} )
\end{aligned}
\begin{aligned}
\mathcal{F}[\Delta f](\boldsymbol{p})
&=\frac{1}{(2\pi)^n}
\int (-p^2)f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}}
\,\mathrm{d}\boldsymbol{r} \\
&=-p^2 \cdot \mathcal{F}[f](\boldsymbol{p})
\end{aligned}
であることがわかる．//\mathcal{F}[\Delta f](\boldsymbol{p})
&=\frac{1}{(2\pi)^n}
\int (-p^2)f(\boldsymbol{r}) e^{-i\boldsymbol{p}\cdot\boldsymbol{r}}
\,\mathrm{d}\boldsymbol{r} \\
&=-p^2 \cdot \mathcal{F}[f](\boldsymbol{p})
\end{aligned}
Helmholtz型方程式
$\text{\sect} 3.2$ a)\begin{aligned}
(\Delta + k^2)G_n(\boldsymbol{r},\boldsymbol{r}^\prime)
=-\delta(\boldsymbol{r} - \boldsymbol{r}^\prime)
\end{aligned}
の両辺をFourier変換すると，(\Delta + k^2)G_n(\boldsymbol{r},\boldsymbol{r}^\prime)
=-\delta(\boldsymbol{r} - \boldsymbol{r}^\prime)
\end{aligned}
\begin{aligned}
(-p^2 + k^2)\hat{G}_n(\boldsymbol{r},\boldsymbol{r}^\prime)
=-\frac{1}{(2\pi)^n}
\end{aligned}
(-p^2 + k^2)\hat{G}_n(\boldsymbol{r},\boldsymbol{r}^\prime)
=-\frac{1}{(2\pi)^n}
\end{aligned}
波動方程式
$\text{\sect} 3.4$2次元
\begin{aligned}
&G_2(\rho,t)\\
&=\int_{-\infty}^{\infty} \frac{1}{4\pi}\frac{1}{\sqrt{\rho^2+z^2}} \delta\Biggl(t-\frac{\sqrt{\rho^2+z^2}}{c}\Biggr)\,\mathrm{d}z \\
&=\frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}}\theta(ct-\rho)
\end{aligned}
&G_2(\rho,t)\\
&=\int_{-\infty}^{\infty} \frac{1}{4\pi}\frac{1}{\sqrt{\rho^2+z^2}} \delta\Biggl(t-\frac{\sqrt{\rho^2+z^2}}{c}\Biggr)\,\mathrm{d}z \\
&=\frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}}\theta(ct-\rho)
\end{aligned}
【方法1】
\begin{aligned}
\frac{\mathrm{d}}{\mathrm{d}x}\theta\bigl(g(x)\bigr)
&=\theta^\prime\bigl(g(x)\bigr) g^\prime(x) \\
&=\delta\bigl(g(x)\bigr) g^\prime(x)
\end{aligned}
なので，\frac{\mathrm{d}}{\mathrm{d}x}\theta\bigl(g(x)\bigr)
&=\theta^\prime\bigl(g(x)\bigr) g^\prime(x) \\
&=\delta\bigl(g(x)\bigr) g^\prime(x)
\end{aligned}
\begin{aligned}
&\frac{\mathrm{d}}{\mathrm{d}x} \bigl[f(x)\theta\bigl(g(x)\bigr)\bigr] \\
&=f^\prime(x) \theta\bigl(g(x)\bigr) + f(x)\delta\bigl(g(x)\bigr) g^\prime(x).
\end{aligned}
ここで&\frac{\mathrm{d}}{\mathrm{d}x} \bigl[f(x)\theta\bigl(g(x)\bigr)\bigr] \\
&=f^\prime(x) \theta\bigl(g(x)\bigr) + f(x)\delta\bigl(g(x)\bigr) g^\prime(x).
\end{aligned}
\begin{aligned}
f(z)=-\frac{c}{4\pi}\frac{1}{z} ,\quad
g(z)=t-\frac{\sqrt{\rho^2+z^2}}{c}
\end{aligned}
とするとf(z)=-\frac{c}{4\pi}\frac{1}{z} ,\quad
g(z)=t-\frac{\sqrt{\rho^2+z^2}}{c}
\end{aligned}
\begin{aligned}
g^\prime(z)&=-\frac{1}{c}\frac{z}{\sqrt{\rho^2+z^2}}
\end{aligned}
なのでg^\prime(z)&=-\frac{1}{c}\frac{z}{\sqrt{\rho^2+z^2}}
\end{aligned}
\begin{aligned}
&G_2(\rho,t) \\
&=\int_{-\infty}^{\infty} f(z)g^\prime(z)\delta\bigl(g(z)\bigr) \,\mathrm{d}z \\
&=\cancel{\bigl[f(z)\theta\bigl(g(z)\bigr)\bigr]_{-\infty}^{\infty}}
-\int_{-\infty}^{\infty} f^\prime(z)\theta\bigl(g(z)\bigr) \,\mathrm{d}z \\
&=-\theta(ct-\rho) \int_{-\sqrt{c^2t^2-\rho^2}}^{\sqrt{c^2t^2-\rho^2}} f^\prime(z) \mathrm{d}z \\
&=-\theta(ct-\rho) \bigl[f(z)\bigr]_{-\sqrt{c^2t^2-\rho^2}}^{\sqrt{c^2t^2-\rho^2}} \\
&=\frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}} \theta(ct-\rho)
\end{aligned}
となる．ここで&G_2(\rho,t) \\
&=\int_{-\infty}^{\infty} f(z)g^\prime(z)\delta\bigl(g(z)\bigr) \,\mathrm{d}z \\
&=\cancel{\bigl[f(z)\theta\bigl(g(z)\bigr)\bigr]_{-\infty}^{\infty}}
-\int_{-\infty}^{\infty} f^\prime(z)\theta\bigl(g(z)\bigr) \,\mathrm{d}z \\
&=-\theta(ct-\rho) \int_{-\sqrt{c^2t^2-\rho^2}}^{\sqrt{c^2t^2-\rho^2}} f^\prime(z) \mathrm{d}z \\
&=-\theta(ct-\rho) \bigl[f(z)\bigr]_{-\sqrt{c^2t^2-\rho^2}}^{\sqrt{c^2t^2-\rho^2}} \\
&=\frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}} \theta(ct-\rho)
\end{aligned}
\begin{aligned}
&\bigl\{z\in(-\infty,\infty)\,|\, g(z) > 0\bigr\} \\
&=
\begin{cases}
\,\emptyset &(ct \leq \rho) \\
\, -\sqrt{c^2t^2-\rho^2} < z < \sqrt{c^2t^2-\rho^2}& (ct > \rho)
\end{cases}
\end{aligned}
を用いた．&\bigl\{z\in(-\infty,\infty)\,|\, g(z) > 0\bigr\} \\
&=
\begin{cases}
\,\emptyset &(ct \leq \rho) \\
\, -\sqrt{c^2t^2-\rho^2} < z < \sqrt{c^2t^2-\rho^2}& (ct > \rho)
\end{cases}
\end{aligned}
【方法2：$ct = \rho$の場合が不完全な方法（1解微分0の場合に公式を拡張する必要があるが，未検討）】
\begin{aligned}
g(z)=t-\frac{\sqrt{\rho^2+z^2}}{c}
\end{aligned}
とするとg(z)=t-\frac{\sqrt{\rho^2+z^2}}{c}
\end{aligned}
\begin{aligned}
g^{-1}(0)
&=\begin{cases}
\, \{z_\pm=\pm\sqrt{c^2t^2-\rho^2} \}& (ct > \rho)\\
\, \{0\} & (ct = \rho)\\
\, \{z^\prime_\pm=\pm i\sqrt{\rho^2 - c^2t^2}\} & (ct < \rho)
\end{cases} \\
g^\prime(z)&=-\frac{1}{c}\frac{z}{\sqrt{\rho^2+z^2}}
\end{aligned}
なので，デルタ関数の公式からg^{-1}(0)
&=\begin{cases}
\, \{z_\pm=\pm\sqrt{c^2t^2-\rho^2} \}& (ct > \rho)\\
\, \{0\} & (ct = \rho)\\
\, \{z^\prime_\pm=\pm i\sqrt{\rho^2 - c^2t^2}\} & (ct < \rho)
\end{cases} \\
g^\prime(z)&=-\frac{1}{c}\frac{z}{\sqrt{\rho^2+z^2}}
\end{aligned}
\begin{aligned}
&G_2(\rho,t) \\
&=\int_{-\infty}^{\infty}
\frac{1}{4\pi}\frac{1}{\sqrt{\rho^2+z^2}} \sum_{i=\pm}\frac{1}{|g^\prime(z_i)|} \delta(z-z_i)\,\mathrm{d}z \\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\times\theta(ct-\rho) \\
&=\frac{1}{4\pi} \sum_{i=\pm}
\int_{-\infty}^{\infty}\frac{1}{\sqrt{\rho^2+z^2}}
\frac{c \sqrt{\rho^2+z_i^2}}{|z_i|} \delta(z-z_i)
\,\mathrm{d}z \\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\times\theta(ct-\rho) \\
&=\frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}} \theta(ct-\rho)
\end{aligned}
（$ct = \rho$の場合に等号が成り立つことは未確認）&G_2(\rho,t) \\
&=\int_{-\infty}^{\infty}
\frac{1}{4\pi}\frac{1}{\sqrt{\rho^2+z^2}} \sum_{i=\pm}\frac{1}{|g^\prime(z_i)|} \delta(z-z_i)\,\mathrm{d}z \\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\times\theta(ct-\rho) \\
&=\frac{1}{4\pi} \sum_{i=\pm}
\int_{-\infty}^{\infty}\frac{1}{\sqrt{\rho^2+z^2}}
\frac{c \sqrt{\rho^2+z_i^2}}{|z_i|} \delta(z-z_i)
\,\mathrm{d}z \\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\times\theta(ct-\rho) \\
&=\frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}} \theta(ct-\rho)
\end{aligned}
1次元
\begin{aligned}
&G_1(|x|,t)\\
&=\int_{-\infty}^{\infty} \frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}}\theta(ct-\rho) \,\mathrm{d}y \\
&=\frac{c}{2}\theta(ct-|x|)
\end{aligned}
&G_1(|x|,t)\\
&=\int_{-\infty}^{\infty} \frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}}\theta(ct-\rho) \,\mathrm{d}y \\
&=\frac{c}{2}\theta(ct-|x|)
\end{aligned}
\begin{aligned}
&\bigl\{y\in(-\infty,\infty)\,|\, ct-\sqrt{x^2+y^2} > 0\bigr\} \\
&=
\begin{cases}
\,\emptyset &(ct \leq |x|) \\
\, -\sqrt{c^2t^2-x^2} < y < \sqrt{c^2t^2-x^2}& (ct > |x|)
\end{cases}
\end{aligned}
だから積分公式から&\bigl\{y\in(-\infty,\infty)\,|\, ct-\sqrt{x^2+y^2} > 0\bigr\} \\
&=
\begin{cases}
\,\emptyset &(ct \leq |x|) \\
\, -\sqrt{c^2t^2-x^2} < y < \sqrt{c^2t^2-x^2}& (ct > |x|)
\end{cases}
\end{aligned}
\begin{aligned}
&G_1(|x|,t)\\
&=\int_{-\infty}^{\infty} \frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}}\theta(ct-\rho) \,\mathrm{d}y \\
&=\theta(ct-|x|)\frac{c}{2\pi}
\int_{-\sqrt{c^2t^2-x^2}}^{\sqrt{c^2t^2-x^2}} \frac{\mathrm{d}y}{\sqrt{c^2t^2-(x^2+y^2)}}\\
&=\theta(ct-|x|)\frac{c}{2\pi}
\cdot \bigl[\arcsin(y/a) \bigr]_{-a}^{a=\sqrt{c^2t^2-x^2}} \\
&=\frac{c}{2}\theta(ct-|x|)
\end{aligned}
となる．//&G_1(|x|,t)\\
&=\int_{-\infty}^{\infty} \frac{c}{2\pi}\frac{1}{\sqrt{c^2t^2-\rho^2}}\theta(ct-\rho) \,\mathrm{d}y \\
&=\theta(ct-|x|)\frac{c}{2\pi}
\int_{-\sqrt{c^2t^2-x^2}}^{\sqrt{c^2t^2-x^2}} \frac{\mathrm{d}y}{\sqrt{c^2t^2-(x^2+y^2)}}\\
&=\theta(ct-|x|)\frac{c}{2\pi}
\cdot \bigl[\arcsin(y/a) \bigr]_{-a}^{a=\sqrt{c^2t^2-x^2}} \\
&=\frac{c}{2}\theta(ct-|x|)
\end{aligned}
境界面のあるGreen関数
$\text{\sect} 5.1$より．境界面のあるGreen関数を求める方法
境界がない場合のGreen関数を$G^{\infty}(\boldsymbol{r}-\boldsymbol{r}^\prime)$と表す．
- [I]適当な同次方程式の解$g(\boldsymbol{r},\boldsymbol{r}^\prime)$を用いて，$G^{\infty}(\boldsymbol{r}-\boldsymbol{r}^\prime)+g(\boldsymbol{r},\boldsymbol{r}^\prime)$が境界条件を満たすようにする．
- [Ia] 鏡像法：対称性から$g(\boldsymbol{r},\boldsymbol{r}^\prime)$を定める．
- [Ib] $G^{\infty}(\boldsymbol{r}-\boldsymbol{r}^\prime)$の適切な表式を用いる．
- [II]固有関数系による展開，または積分変換を用いる．
- [IIa] 直接求める：境界条件に適合した固有関数系で展開．
- [IIb] 低次方程式のGreen関数に帰着：固有関数系展開 or 積分変換．
- 境界条件を自動的に取り入れる．
- 低次Green関数を作るときに，境界条件を考慮．
