東京工業大学 理学院 数学系 2019年度 午前 [3]
Author
Description
Kai
(1)
\[
\begin{aligned}
\frac{\partial}{\partial x} \ln | \vec{r} |
&=
\frac{1}{2}
\frac{\partial}{\partial x} \ln \left( x^2+y^2 \right)
=
\frac{x}{x^2+y^2}
=
\frac{x}{r^2}
\\
\frac{\partial}{\partial y} \ln | \vec{r} |
&=
\frac{1}{2}
\frac{\partial}{\partial y} \ln \left( x^2+y^2 \right)
=
\frac{y}{x^2+y^2}
=
\frac{y}{r^2}
\end{aligned}
\]
なので、
\[
\begin{aligned}
\vec{\nabla} \ln | \vec{r} |
=
\left( \frac{x}{r^2}, \frac{y}{r^2} \right)
=
\frac{\vec{r}}{r^2}
\end{aligned}
\]
(2)
\(x = r \cos \theta, y = r \sin \theta, r = \sqrt{x^2+y^2}, \tan \theta = y/x\) より、
\[
\begin{aligned}
\frac{\partial r}{\partial x}
&= \frac{x}{r}
= \cos \theta
\\
\frac{\partial r}{\partial y}
&= \frac{y}{r}
= \sin \theta
\\
\frac{\partial \theta}{\partial x}
&= - \frac{y}{x^2} \cos^2 \theta
= - \frac{\sin \theta}{r}
\\
\frac{\partial \theta}{\partial y}
&= \frac{1}{x} \cos^2 \theta
= \frac{\cos \theta}{r}
\end{aligned}
\]
であるから、
\[
\begin{aligned}
\frac{\partial}{\partial x}
&=
\frac{\partial r}{\partial x} \frac{\partial}{\partial r}
+ \frac{\partial \theta}{\partial x} \frac{\partial}{\partial \theta}
=
\cos \theta \frac{\partial}{\partial r}
- \frac{\sin \theta}{r} \frac{\partial}{\partial \theta}
\\
\frac{\partial}{\partial y}
&=
\frac{\partial r}{\partial y} \frac{\partial}{\partial r}
+ \frac{\partial \theta}{\partial y} \frac{\partial}{\partial \theta}
=
\sin \theta \frac{\partial}{\partial r}
+ \frac{\cos \theta}{r} \frac{\partial}{\partial \theta}
\end{aligned}
\]
であり、
\[
\begin{aligned}
\frac{\partial^2}{\partial x^2}
&=
\left(
\cos \theta \frac{\partial}{\partial r}
- \frac{\sin \theta}{r} \frac{\partial}{\partial \theta}
\right)
\left(
\cos \theta \frac{\partial}{\partial r}
- \frac{\sin \theta}{r} \frac{\partial}{\partial \theta}
\right)
\\
&=
\cos^2 \theta \frac{\partial^2}{\partial r^2}
+ \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta}
- \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta}
+ \frac{\sin^2 \theta}{r} \frac{\partial}{\partial r}
- \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial \theta \partial r}
\\
&\ \ \ \
+ \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta}
+ \frac{\sin^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2}
\\
&=
\cos^2 \theta \frac{\partial^2}{\partial r^2}
+2 \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta}
-2 \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta}
+ \frac{\sin^2 \theta}{r} \frac{\partial}{\partial r}
+ \frac{\sin^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2}
\\
\frac{\partial^2}{\partial y^2}
&=
\left(
\sin \theta \frac{\partial}{\partial r}
+ \frac{\cos \theta}{r} \frac{\partial}{\partial \theta}
\right)
\left(
\sin \theta \frac{\partial}{\partial r}
+ \frac{\cos \theta}{r} \frac{\partial}{\partial \theta}
\right)
\\
&=
\sin^2 \theta \frac{\partial^2}{\partial r^2}
- \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta}
+ \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta}
+ \frac{\cos^2 \theta}{r} \frac{\partial}{\partial r}
+ \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial \theta \partial r}
\\
&\ \ \ \
- \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta}
+ \frac{\cos^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2}
\\
&=
\sin^2 \theta \frac{\partial^2}{\partial r^2}
-2 \frac{\sin \theta \cos \theta}{r^2} \frac{\partial}{\partial \theta}
+2 \frac{\sin \theta \cos \theta}{r} \frac{\partial^2}{\partial r \partial \theta}
+ \frac{\cos^2 \theta}{r} \frac{\partial}{\partial r}
+ \frac{\cos^2 \theta}{r^2} \frac{\partial^2}{\partial \theta^2}
\end{aligned}
\]
したがって、
\[
\begin{aligned}
\Delta
&=
\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}
\\
&=
\frac{\partial^2}{\partial r^2}
+ \frac{1}{r} \frac{\partial}{\partial r}
+ \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2}
\\
&=
\frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{\partial r}
+ \frac{1}{r^2} \frac{\partial^2}{\partial \theta^2}
\end{aligned}
\]
を得る。
(3)
\(| \vec{r} | \neq 0\) のとき、(2) で得た式をつかって、
\[
\begin{aligned}
\Delta \ln | \vec{r} |
=
\frac{1}{r} \frac{\partial}{\partial r} r \frac{\partial}{\partial r} \ln r
=
\frac{1}{r} \frac{\partial}{\partial r} r \frac{1}{r}
=
0
\end{aligned}
\]
を得る。
(4)
領域 \(| \vec{r} | \leq a\) を \(S\) , 閉曲線 \(| \vec{r} | = a\) を \(C\) で表して、 次のように計算できる：
\[
\begin{aligned}
I
&=
\iint_S \Delta \ln | \vec{r} | dx dy
\\
&=
\iint_S \vec{\nabla} \cdot \frac{\vec{r}}{r^2} \ dx dy
\\
&=
\iint_S \left(
\frac{\partial}{\partial x} \frac{x}{r^2}
+
\frac{\partial}{\partial y} \frac{y}{r^2}
\right)
dx dy
\\
&=
\oint_C \left( - \frac{y}{r^2} dx + \frac{x}{r^2} dy \right)
\\
&=
\int_0^{2 \pi} \left(
- \frac{a \sin \theta}{a^2} \cdot (-a \sin \theta)
+ \frac{a \cos \theta}{a^2} \cdot (a \cos \theta)
\right) d \theta
\\
&=
\int_0^{2 \pi} d \theta
\\
&=
2 \pi
\end{aligned}
\]
(5)
(3), (4) より、
\[
\begin{aligned}
\Delta \ln | \vec{r} |
=
2 \pi \delta^2 ( \vec{r} )
\end{aligned}
\]
がわかる。 ただし、 \(\delta^2 ( \cdot )\) は、 2次元のディラックのデルタ関数である。