ベクトル解析 divの問題
問題
① \(\mathrm{div} \boldsymbol{r}\) ② \(\mathrm{div} \displaystyle\frac{\boldsymbol{r}}{r}\) ③ \(\mathrm{div} \displaystyle\frac{\boldsymbol{r}}{r^3}\)
※\(r=\sqrt{x^2+y^2+z^2}\)
解答
\(\mathrm{div} \boldsymbol{A}=\displaystyle\frac{\partial A_{x}}{\partial x}+\displaystyle\frac{\partial A_{y}}{\partial y}+\displaystyle\frac{\partial A_{z}}{\partial z}\)
1番
\(\mathrm{div} \boldsymbol{r}=\displaystyle\frac{\partial x}{\partial x}+\displaystyle\frac{\partial y}{\partial y}+\displaystyle\frac{\partial z}{\partial z}=3\)
\(\displaystyle\frac{\partial x}{\partial x}\)というのは \(x\) を \(x\) で微分するということ。つまり、\(1\) となる。
2番
\(\mathrm{div} \displaystyle\frac{\boldsymbol{r}}{r}\)
\(=\displaystyle\frac{\partial}{\partial x}\biggl(\displaystyle\frac{x}{\sqrt{x^2+y^2+z^2}}\biggr)+\displaystyle\frac{\partial}{\partial y}\biggl(\displaystyle\frac{y}{\sqrt{x^2+y^2+z^2}}\biggr)+\displaystyle\frac{\partial}{\partial z}\biggl(\displaystyle\frac{z}{\sqrt{x^2+y^2+z^2}}\biggr)\)
ここで
\(\displaystyle\frac{\partial}{\partial x}\biggl(\displaystyle\frac{x}{\sqrt{x^2+y^2+z^2}}\biggr)\)
\(=\displaystyle\frac{1}{\sqrt{x^2+y^2+z^2}}+x\cdot \biggl(-\displaystyle\frac{1}{2}\biggr)(x^2+y^2+z^2)^{-\frac{3}{2}}\cdot 2x\)
\(=\displaystyle\frac{1}{\sqrt{x^2+y^2+z^2}}-\displaystyle\frac{x^2}{(x^2+y^2+z^2)^{\frac{3}{2}}}\)
同様に計算すると
\(\displaystyle\frac{\partial}{\partial y}\biggl(\displaystyle\frac{y}{\sqrt{x^2+y^2+z^2}}\biggr)=\displaystyle\frac{1}{\sqrt{x^2+y^2+z^2}}-\displaystyle\frac{y^2}{(x^2+y^2+z^2)^{\frac{3}{2}}}\)
\(\displaystyle\frac{\partial}{\partial z}\biggl(\displaystyle\frac{z}{\sqrt{x^2+y^2+z^2}}\biggr)=\displaystyle\frac{1}{\sqrt{x^2+y^2+z^2}}-\displaystyle\frac{z^2}{(x^2+y^2+z^2)^{\frac{3}{2}}}\)
なので、 \(\mathrm{div} \displaystyle\frac{\boldsymbol{r}}{r}\)
\(=\displaystyle\frac{\partial}{\partial x}\biggl(\displaystyle\frac{x}{\sqrt{x^2+y^2+z^2}}\biggr)+\displaystyle\frac{\partial}{\partial y}\biggl(\displaystyle\frac{x}{\sqrt{x^2+y^2+z^2}}\biggr)+\displaystyle\frac{\partial}{\partial z}\biggl(\displaystyle\frac{x}{\sqrt{x^2+y^2+z^2}}\biggr)\)
\(=\displaystyle\frac{3}{\sqrt{x^2+y^2+z^2}}-\displaystyle\frac{x^2+y^2+z^2}{(x^2+y^2+z^2)^{\frac{3}{2}}}=\displaystyle\frac{3}{r}-\displaystyle\frac{1}{r}\)\(=\displaystyle\frac{2}{r}\)
3番
\(\mathrm{div} \displaystyle\frac{\boldsymbol{r}}{r^3}\)
\(=\displaystyle\frac{\partial}{\partial x}\biggl(\displaystyle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)+\displaystyle\frac{\partial}{\partial y}\biggl(\displaystyle\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)+\displaystyle\frac{\partial}{\partial z}\biggl(\displaystyle\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)\)
ここで
\(\displaystyle\frac{\partial}{\partial x}\biggl(\displaystyle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)\)
\(=\displaystyle\frac{1}{(x^2+y^2+z^2)^{\frac{3}{2}}}+x\cdot \biggl(-\displaystyle\frac{3}{2}\biggr)(x^2+y^2+z^2)^{-\frac{5}{2}}\cdot 2x\)
\(=\displaystyle\frac{1}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\displaystyle\frac{3x^2}{(x^2+y^2+z^2)^{\frac{5}{2}}}\)
同様に計算すると
\(\displaystyle\frac{\partial}{\partial y}\biggl(\displaystyle\frac{y}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)=\displaystyle\frac{1}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\displaystyle\frac{3y^2}{(x^2+y^2+z^2)^{\frac{5}{2}}}\)
\(\displaystyle\frac{\partial}{\partial z}\biggl(\displaystyle\frac{z}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)=\displaystyle\frac{1}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\displaystyle\frac{3z^2}{(x^2+y^2+z^2)^{\frac{5}{2}}}\)
なので、 \(\mathrm{div} \displaystyle\frac{\boldsymbol{r}}{r^3}\)
\(=\displaystyle\frac{\partial}{\partial x}\biggl(\displaystyle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)+\displaystyle\frac{\partial}{\partial y}\biggl(\displaystyle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)+\displaystyle\frac{\partial}{\partial z}\biggl(\displaystyle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)\)
\(=\displaystyle\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\displaystyle\frac{3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}=\displaystyle\frac{3}{r^3}-\displaystyle\frac{3}{r^3}\)\(=0\)