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--- 수학:범함수 [2020/01/10 11:25] – [첫 번째 항] admin
+++ 수학:범함수 [2020/10/13 20:05] – [함께 보기] admin
@@ Line 49: / Line 49: @@
 \]
 그 시간 변화율이 언제나
-\[ \dot{F} = - \int \xi \rho |\vec{u}|^2 d\vec{r} \le 0 \]
+\[ \frac{dF}{dt} = - \int \xi \rho |\vec{u}|^2 d\vec{r} \le 0 \]
 임을 보일 수 있다.
@@ Line 90: / Line 90: @@
 \end{eqnarray*}
+====세 번째 항====
+이 계산이 가장 간단하다.
+\begin{eqnarray*}
+\frac{\delta}{\delta \rho(\vec{r})} \int \rho(\vec{r}') \frac{|\vec{u}(\vec{r}')|^2}{2} d\vec{r}' &=&
+\lim_{\epsilon\to 0} \frac{1}{\epsilon} \left\{ \int \left[ \rho(\vec{r}') + \epsilon \delta (\vec{r}-\vec{r}') \right] \frac{|\vec{u}(\vec{r}')|^2}{2} d\vec{r}'
+- \int \rho(\vec{r}') \frac{|\vec{u}(\vec{r}')|^2}{2} d\vec{r}'\right\}\\
+&=&
+\int \delta (\vec{r}-\vec{r}') \frac{|\vec{u}(\vec{r}')|^2}{2} d\vec{r}'\\
+&=& \frac{|\vec{u}(\vec{r})|^2}{2}
+\end{eqnarray*}
+====$F$의 미분====
+연쇄법칙(chain rule)에 의해
+\[ \frac{dF}{dt} = \int d\vec{r} \left( \frac{\delta F}{\delta \rho(\vec{r})} \frac{\partial \rho(\vec{r},t)}{\partial t}
++ \frac{\delta F}{\delta u_x(\vec{r})} \frac{\partial u_x(\vec{r},t)}{\partial t}
++ \frac{\delta F}{\delta u_y(\vec{r})} \frac{\partial u_y(\vec{r},t)}{\partial t}
++ \frac{\delta F}{\delta u_z(\vec{r})} \frac{\partial u_z(\vec{r},t)}{\partial t}
+\right). \]
+$\vec{u}$의 성분별로도 변분하는 과정이 필요하다. 예를 들어
+\begin{eqnarray*}
+\frac{\delta}{\delta u_x(\vec{r})} \int \rho(\vec{r}') \frac{u_x^2(\vec{r}') + u_y^2(\vec{r}') + u_z^2(\vec{r}')}{2} d\vec{r}' &=&
+\lim_{\epsilon \to 0} \frac{1}{\epsilon} \int \rho(\vec{r}') \frac{[u_x(\vec{r}') + \epsilon \delta(\vec{r}'-\vec{r})]^2 - u_x^2(\vec{r}')}{2} d\vec{r}'\\
+&\approx& \int \rho(\vec{r}') \delta(\vec{r}'-\vec{r}) u_x(\vec{r}') d\vec{r}' = \rho(\vec{r}) u_x(\vec{r}).
+\end{eqnarray*}
+====종합====
+앞의 결과들을 모두 모으면 다음과 같다:
+\begin{eqnarray*}
+\frac{dF}{dt} &=& \int \left( \Phi + \Psi + \frac{P}{\rho} + \frac{|\vec{u}|^2}{2} \right) \frac{\partial \rho}{\partial t} d\vec{r} + \int \rho \vec{u} \cdot \frac{\partial \vec{u}}{\partial t} d\vec{r}\\
+&=&
+\int \left( \Phi + \Psi + \frac{P}{\rho} + \frac{|\vec{u}|^2}{2} \right) \left[ -\nabla \cdot (\rho \vec{u}) \right] d\vec{r}
++ \int (\rho \vec{u}) \cdot \left[ -(\vec{u} \cdot \nabla) \vec{u} - \frac{1}{\rho} \nabla P - \nabla \Phi - \xi \vec{u} \right] d\vec{r}\\
+&=&
+\int \nabla \left( \Phi + \Psi + \frac{P}{\rho} + \frac{|\vec{u}|^2}{2} \right) \cdot (\rho \vec{u}) d\vec{r}
++ \int (\rho \vec{u}) \cdot \left[ -(\vec{u} \cdot \nabla) \vec{u} - \frac{1}{\rho} \nabla P - \nabla \Phi - \xi \vec{u} \right] d\vec{r}.
+\end{eqnarray*}
+마지막 줄로 넘어올 때에는 부분적분을 시행했다. 이제
+\begin{eqnarray*}
+\nabla \Psi &=& \nabla \int^{\rho(\vec{r})} \frac{P(\rho')}{\rho'^2} d\rho' = \frac{P(\rho)}{\rho^2} \nabla \rho\\
+\nabla \frac{P[\rho(\vec{r})]}{\rho(\vec{r})} &=& \frac{\nabla P}{\rho} - \frac{\nabla \rho}{\rho^2} P
+\end{eqnarray*}
+임을 이용할 것이다. 따라서
+\begin{eqnarray*}
+\frac{dF}{dt} &=& \int \left[ \nabla \Phi + \frac{P}{\rho^2} \nabla \rho + \frac{\nabla P}{\rho} - \frac{\nabla \rho}{\rho^2} P + \nabla \left( \frac{|\vec{u}|^2}{2} \right) \right] \cdot (\rho \vec{u}) d\vec{r}
++\int \left[ -(\vec{u}\cdot \nabla) \vec{u} - \frac{\nabla P}{\rho} - \nabla \Phi - \xi \vec{u} \right] \cdot (\rho \vec{u}) d\vec{r}\\
+&=& \int (\rho \vec{u}) \cdot \left[ \nabla \left( \frac{|\vec{u}|^2}{2} \right) - \xi \vec{u} - (\vec{u} \cdot \nabla) \vec{u} \right] d\vec{r}.
+\end{eqnarray*}
+그런데 Green의 벡터 항등식
+\[ \nabla^2 (\vec{A} \cdot \vec{B}) = \vec{A} \cdot \nabla^2 \vec{B} - \vec{B} \cdot \nabla^2 \vec{A} + 2\nabla \cdot [(\vec{B}\cdot \nabla) \vec{A} + \vec{B} \times (\nabla \times \vec{A})]\]
+을 활용하면
+\[ (\vec{u} \cdot \nabla) \vec{u} = \nabla \left( \frac{|\vec{u}|^2}{2} \right) -\vec{u}\times (\nabla \times \vec{u}) \]
+임을 보일 수 있다. 그러므로
+\begin{eqnarray*}
+\frac{dF}{dt} &=&
+\int (\rho \vec{u}) \cdot \left[ \nabla \left( \frac{|\vec{u}|^2}{2} \right) - \xi \vec{u} - \nabla \left( \frac{|\vec{u}|^2}{2} \right) + \vec{u} \times (\nabla \times \vec{u}) \right] d\vec{r}
+\end{eqnarray*}
+인데 $\vec{u} \times (\nabla \times \vec{u})$는 $\vec{u}$와 수직하므로 $(\rho \vec{u})$와 내적하면 사라진다. 따라서 다음 결과를 얻는다:
+\[ \frac{dF}{dt} = \int (\rho \vec{u}) \cdot (-\xi \vec{u}) d\vec{r} = - \int \rho \xi |\vec{u}|^2 d\vec{r} \le 0. \]
+======함께 보기======
+  *[[전산물리학:변분법]]
+  *[[전자기학:정전기학의 톰슨 정리]]
+  *[[물리:극소 곡면]]
 ======참고 문헌======
   * T. Lancaster and S. J. Blundell, //Quantum Field Theory for the Gited Amateur// (Oxford Univerty Press, 2014).
-  * P. H. Chavanis, Eur. Phys. J. B 62, 179 (2008) [[doi:10.1140/epjb/e2008-00142-9|(link)]].
+  * P. H. Chavanis, Eur. Phys. J. B 62, 179 (2008) [[http://dx.doi.org/10.1140/epjb/e2008-00142-9|(link)]].