1. 最小势能原理
一个小变形线弹性体静力学问题，共有 \(\sigma_{ij}\)，\(\varepsilon_{ij}\) 和 \(u_{i}\) 等15个待定函数，它们在域\(V\)中必须满足弹性力学的15个基本方程和6个边界条件。现在，我们尝试将上述问题等效为泛函的驻值问题，建立最小势能原理。
最小势能原理：在满足几何方程\({\varepsilon _{ij}} = \frac{1}{2}\left( {{u_{i,j}} + {u_{j,i}}} \right)\)和位移边界条件\(u_{i}=\overline {u}_{i}\)的所有允许位移函数中，实际的位移\(u_i\)必定可以使弹性体的总势能
(1.1)\[\begin{equation}
\Pi = \iiint\limits_V {A\left( {{\varepsilon _{ij}}} \right){\text{d}}V} - \iiint\limits_V {{f_i}{u_i}{\text{d}}V - \iint\limits_{{S_p}} {{{\bar p}_i}{u_i}{\text{d}}S}}
\label{eq:Pi_1}
\end{equation}\]
为最小。
为了证明最小位能原理，先求\(\Pi\)的一阶变分：
(1.2)\[\begin{equation}
\delta \Pi = \iiint\limits_V {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {\varepsilon _{ij}}{\text{d}}V} - \iiint\limits_V {{f_i}\delta {u_i}{\text{d}}V - \iint\limits_{{S_p}} {{{\bar p}_i}\delta {u_i}{\text{d}}S}}
\label{eq:delta_Pi}
\end{equation}\]
由于几何方程事先要满足，同时基于应力的对称性\(\sigma_{ij}=\sigma_{ji}\)，有
(1.3)\[\begin{equation}
\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {\varepsilon _{ij}} = \frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\frac{1}{2}\delta \left( {{u_{i,j}} + {u_{j,i}}} \right) = \frac{1}{2}{\sigma _{ij}}\delta {u_{i,j}} + \frac{1}{2}{\sigma _{ij}}\delta {u_{j,i}} = {\sigma _{ij}}\delta {u_{i,j}} = \frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {u_{i,j}}
\end{equation}\]
根据求导的链式法则有
(1.4)\[\begin{equation}
{\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {u_i}} \right)_{,j}} = {\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}} \right)_{,j}}\delta {u_i} + \frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {u_{i,j}}
\end{equation}\]
同时根据高斯散度定理可知
(1.5)\[\begin{equation}
\iiint\limits_V {{{\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {u_i}} \right)}_{,j}}{\text{d}}V} = \iint\limits_S {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {u_i}{n_j}{\text{d}}S}
\end{equation}\]
因此式(1.2)的右边第一项
(1.6)\[\begin{split}\begin{equation}
\begin{array}{*{20}{l}}
{\iiint\limits_V {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {\varepsilon _{ij}}{\text{d}}V}}&{ = \iiint\limits_V {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {u_{i,j}}{\text{d}}V}} \\
{\text{ }}&{ = \iiint\limits_V {\left[ {{{\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {u_i}} \right)}_{,j}} - {{\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}} \right)}_{,j}}\delta {u_i}} \right]{\text{d}}V}} \\
{\text{ }}&{ = \iint\limits_S {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}\delta {u_i}{n_j}{\text{d}}S} - \iiint\limits_V {{{\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}} \right)}_{,j}}\delta {u_i}{\text{d}}V}} \\
{\text{ }}&{ = \iint\limits_{{S_p}} {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}{n_j}\delta {u_i}{\text{d}}S} - \iiint\limits_V {{{\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}} \right)}_{,j}}\delta {u_i}{\text{d}}V}}
\end{array}
\end{equation}\end{split}\]
上式中我们应用了条件，在\(S_u\)边界上\(\delta u_i=0\)。带入式(1.2)得
(1.7)\[\begin{equation}
\delta \Pi = \iint\limits_{{S_p}} {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}{n_j}\delta {u_i}{\text{d}}S} - \iiint\limits_V {{{\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}} \right)}_{,j}}\delta {u_i}{\text{d}}V} - \iiint\limits_V {{f_i}\delta {u_i}{\text{d}}V - \iint\limits_{{S_p}} {{{\bar p}_i}\delta {u_i}{\text{d}}S}}
\end{equation}\]
整理可得
(1.8)\[\begin{equation}
\delta \Pi = - \iiint\limits_V {\left[ {{{\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}} \right)}_{,j}} + {f_i}} \right]\delta {u_i}{\text{d}}V} + \iint\limits_{{S_p}} {\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}{n_j} - {{\bar p}_i}} \right)\delta {u_i}{\text{d}}S}
\end{equation}\]
弹性体的总势能\(\Pi\)取极值的条件是\(\delta \Pi=0\)，因此必须有
(1.9)\[\begin{equation}
{\left( {\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}} \right)_{,j}} + {f_i} = 0 \quad \left( \text { 在 } V \text { 内 } \right)
\end{equation}\]
(1.10)\[\begin{equation}
\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}}{n_j} - {{\bar p}_i} = 0 \quad \left( \text { 在 } S_p \text { 上 } \right)
\end{equation}\]
应用物理方程\(\frac{{\partial A}}{{\partial {\varepsilon _{ij}}}} = {\sigma _{ij}}\)，可得
(1.11)\[\begin{equation}
{\sigma _{ij,j}} + {f_i} = 0 \quad \left( \text { 在 } V \text { 内 } \right)
\end{equation}\]
(1.12)\[\begin{equation}
{\sigma _{ij}}{n_j} - {{\bar p}_i} = 0 \quad \left( \text { 在 } S_p \text { 上 } \right)
\end{equation}\]
以上两式就是平衡方程和力边界条件，即满足了\(\delta \Pi=0\)，就等于满足了以上两个条件。因此，使泛函\(\Pi\)取极值的的位移\(u_i\)就是真实的解。进一步我们可以证明该极值为最小值。