긴 방정식 시스템의 형식을 지정하는 방법

Question

방정식을 깨야 하는 긴 방정식 시스템의 형식을 어떻게 지정할 수 있는지 궁금합니다. 나는 이것을 복제하고 싶다

다음은 입력된 3개의 방정식입니다.

    \begin{equation}
    \frac{\partial u_r}{\partial t}+u_r\frac{\partial u_r}{\partial r}+\frac{u_{\theta }}{r}\frac{\partial u_r}{\partial \theta }-\frac{u_{\theta }^2}{r}+u_z\frac{\partial u_r}{\partial z}=\frac{\mu }{\rho \:}\left[\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial }{\partial r}\left(ru_r\right)\right)+\frac{1}{r^2}\frac{\partial ^2u_r}{\partial \theta ^2}+\frac{\partial ^2u_r}{\partial z^2}-\frac{2}{r^2}\frac{\partial u_{\theta }}{\partial \theta }\right]-\frac{1}{\rho }\frac{\partial P}{\partial r}
\end{equation}

\begin{equation}
    \frac{\partial u_{\theta }}{\partial t}+u_r\frac{\partial u_{\theta }}{\partial r}+\frac{u_{\theta }}{r}\frac{\partial u_{\theta }}{\partial \theta }-\frac{u_ru_{\theta }}{r}+u_z\frac{\partial u_{\theta }}{\partial z}=\frac{\mu }{\rho \:}\left[\frac{\partial }{\partial r}\left(\frac{1}{r}\frac{\partial }{\partial r}\left(ru_{\theta }\right)\right)+\frac{1}{r^2}\frac{\partial ^2u_{\theta }}{\partial \theta ^2}+\frac{\partial ^2u_{\theta }}{\partial z^2}+\frac{2}{r^2}\frac{\partial u_{\theta }}{\partial \theta }\right]-\frac{1}{r\rho }\frac{\partial P}{\partial \theta }
\end{equation}

\begin{equation}
    \frac{\partial u_z}{\partial t}+u_r\frac{\partial u_z}{\partial r}+\frac{u_{\theta }}{r}\frac{\partial u_z}{\partial \theta }+u_z\frac{\partial u_z}{\partial z}=\frac{\mu }{\rho \:}\left[\frac{1}{r}\frac{\partial }{\partial r}\left(r\frac{\partial u_z}{\partial \:r}\right)+\frac{1}{r^2}\frac{\partial ^2u_z}{\partial \theta ^2}+\frac{\partial ^2u_z}{\partial z^2}\right]-\frac{1}{\rho }\frac{\partial P}{\partial \theta }+g_z
\end{equation}

방정식을 개별적으로 분해할 수 있지만 위에 표시된 형식으로 나열하는 방법을 모르겠습니다. 무엇이든 도움이 됩니다. 감사합니다.

Answer 1

이 같은:

나는align*​엄청난\notag첫 번째 + 기호 다음에 정렬된 패키지와 중간 방정식 번호를 억제하는 몇 가지 명령이 있습니다. OP와 달리 등호는 첫 번째 줄의 끝이 아니라 방정식의 두 번째 줄에서 시작해야 한다고 강력히 제안합니다.

코드는 다음과 같습니다.

\documentclass{article}
\usepackage{amsmath}
\begin{document}

   \begin{align}
      \frac{\partial u_r}{\partial t}+&u_r\frac{\partial u_r}{\partial r}
        +\frac{u_{\theta }}{r}\frac{\partial u_r}{\partial \theta }
        -\frac{u_{\theta }^2}{r}+u_z\frac{\partial u_r}{\partial z}
      \\ &=\frac{\mu }{\rho \:}\left[\frac{\partial }{\partial r}
          \left(\frac{1}{r}\frac{\partial }{\partial r}\left(ru_r\right)\right)
            +\frac{1}{r^2}\frac{\partial ^2u_r}{\partial \theta ^2}
            +\frac{\partial ^2u_r}{\partial z^2}
            -\frac{2}{r^2}\frac{\partial u_{\theta }}{\partial \theta }\right]
            -\frac{1}{\rho }\frac{\partial P}{\partial r}\notag\\
      \frac{\partial u_{\theta }}{\partial t}+&u_r\frac{\partial u_{\theta }}{\partial r}
           +\frac{u_{\theta }}{r}\frac{\partial u_{\theta }}{\partial \theta }
                -\frac{u_ru_{\theta }}{r}+u_z\frac{\partial u_{\theta }}{\partial z}\\
        &=\frac{\mu }{\rho \:}\left[\frac{\partial }{\partial r}
              \left(\frac{1}{r}\frac{\partial }{\partial r}\left(ru_{\theta }\right)\right)
              +\frac{1}{r^2}\frac{\partial ^2u_{\theta }}{\partial \theta ^2}
              +\frac{\partial ^2u_{\theta }}{\partial z^2}
              +\frac{2}{r^2}\frac{\partial u_{\theta }}{\partial \theta }\right]
              -\frac{1}{r\rho }\frac{\partial P}{\partial \theta }
      \notag\\
      \frac{\partial u_z}{\partial t}+&u_r\frac{\partial u_z}{\partial r}
          +\frac{u_{\theta }}{r}\frac{\partial u_z}{\partial \theta }
              +u_z\frac{\partial u_z}{\partial z}\\
          &=\frac{\mu }{\rho \:}\left[\frac{1}{r}\frac{\partial }{\partial r}
              \left(r\frac{\partial u_z}{\partial \:r}\right)
                  +\frac{1}{r^2}\frac{\partial ^2u_z}{\partial \theta ^2}
                      +\frac{\partial ^2u_z}{\partial z^2}\right]
                          -\frac{1}{\rho }\frac{\partial P}{\partial \theta }+g_z\notag
  \end{align}

\end{document}

Answer 2

align이 포함된 또 다른 코드(왼쪽 설명 포함)와 패키지의 더 간단한 구문 diffcoeff:

\documentclass{article}
\usepackage{geometry} 
\usepackage{amsmath}
\usepackage{diffcoeff} 

\begin{document}

\begin{align}
r\text{-momentum:} & & & \diffp{u_r}{t} + u_r\diffp{u_r}{r} + \frac{u_{\theta }}{r}\diffp{u_r}{\theta }-\frac{u_{\theta }^2}{r} + u_z\diffp{u_r}{z} = \\
\notag & & & \frac{\mu }{\rho \:}\left[\diffp{}{r}\left(\frac{1}{r}\diffp{}{r}\left(ru_r\right)\right) + \frac{1}{r^2}\diffp[2]{u_r}{\theta} + \diffp[2]{u_r}{z}-\frac{2}{r^2}\diffp{u_{\theta }}{\theta }\right]-\frac{1}{\rho }\diffp{P}{r}\\[2ex]
%
\theta\text{-momentum:} & & & \diffp{u_{\theta }}{t} + u_r\diffp{u_{\theta }}{r} + \frac{u_{\theta }}{r}\diffp{u_{\theta }}{\theta }-\frac{u_ru_{\theta }}{r} + u_z\diffp{u_{\theta }}{z} = \\
\notag & & & \frac{\mu }{\rho \:}\left[\diffp{}{r}\left(\frac{1}{r}\diffp{}{r}\left(ru_{\theta }\right)\right) + \frac{1}{r^2}\diffp[2]{u_{\theta }}{\theta} + \diffp[2]{u_{\theta }}{z} + \frac{2}{r^2}\diffp{u_{\theta }}{\theta }\right]-\frac{1}{r\rho }\diffp{P}{\theta } \\[2ex]
%
z\text{-momentum:} & & & \diffp{u_z}{t} + u_r\diffp{u_z}{r} + \frac{u_{\theta }}{r}\diffp{u_z}{\theta } + u_z\diffp{u_z}{z} = \\
 \notag & & & \frac{\mu }{\rho \:}\left[\frac{1}{r}\diffp{}{r}\left(r\diffp{u_z}{\:r}\right) + \frac{1}{r^2}\diffp[2]{u_z}{\theta} + \diffp[2]{u_z}{z}\right]-\frac{1}{\rho }\diffp{P}{\theta } + g_z
\end{align}

\end{document}