ベクトル解析3 divの問題

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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、 \(\mathrm{div} \displaystyle\frac{\boldsymbol{r}}{r^3}\)

 

\(=\displaystyle\frac{\partial}{\partial x}\biggl(\displaystyle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)+\displaystyle\frac{\partial}{\partial y}\biggl(\displaystyle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)+\displaystyle\frac{\partial}{\partial z}\biggl(\displaystyle\frac{x}{(x^2+y^2+z^2)^{\frac{3}{2}}}\biggr)\)

 

\(=\displaystyle\frac{3}{(x^2+y^2+z^2)^{\frac{3}{2}}}-\displaystyle\frac{3(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{5}{2}}}=\displaystyle\frac{3}{r^3}-\displaystyle\frac{3}{r^3}\)\(=0\)

 

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