Minimizing a functional in 1-D
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In 1-D, the minimization problem can be stated as

Find $u(x)$ such that

$U[u(x)]=\int _{x_{0}}^{x_{1}}F(x,u,u^{'})dx$
is a minimum.

We have seen that the minimization problem can be reduced down to the solution of an Euler equation

${\frac {\partial F}{\partial u}}-{\frac {d}{dx}}\left({\frac {\partial F}{\partial u^{'}}}\right)=0$
with the associated boundary conditions

$\eta (x_{0})=0~{\text{and}}~\eta (x_{1})=0$
or,

$\left.{\frac {\partial F}{\partial u^{'}}}\right|_{x_{0}}=0~{\text{and}}\left.{\frac {\partial F}{\partial u^{'}}}\right|_{x_{1}}=0$
Minimizing a Functional in 3-D
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In 3-D, the equivalent minimization problem can be stated as

Find $\mathbf {u} (\mathbf {x} )$ such that

$U[\mathbf {u} (\mathbf {x} )]=\int _{\mathcal {R}}F(\mathbf {x} ,\mathbf {u} ,{\boldsymbol {\nabla }}\mathbf {u} )~dV$
is a minimum.

We would like to find the Euler equation for this problem and the associated boundary conditions required to minimize $U$ .

Let us define all our quantities with respect to an orthonormal basis $({\widehat {\mathbf {e} }}_{i})$ .

Then,

$\mathbf {x} =x_{i}{\widehat {\mathbf {e} }}_{i}~~;~~~\mathbf {u} =u_{i}{\widehat {\mathbf {e} }}_{i}~~;~~~{\boldsymbol {\nabla }}\mathbf {u} =u_{i,j}{\widehat {\mathbf {e} }}_{i}\otimes {\widehat {\mathbf {e} }}_{j}$
and

$U[\mathbf {u} (\mathbf {x} )]=\int _{\mathcal {R}}{\tilde {F}}(x_{i},u_{i},u_{i,j})~dV$
Taking the first variation of $U$ , we get

$\delta U=\int _{\mathcal {R}}\left({\frac {\partial {\tilde {F}}}{\partial u_{i}}}\delta u_{i}+{\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\delta u_{i,j}\right)dV$
All the nine components of $\delta u_{i,j}$ are not independent. Why ?

The variation of the functional $U$ needs to be expressed
entirely in terms of $\delta u_{i}$ . We do this using the 3-D equivalent of integration by parts - the divergence theorem .

Thus,

${\begin{aligned}\int _{\mathcal {R}}{\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\delta u_{i,j}~dV&=\int _{\mathcal {R}}{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\delta u_{i}\right)dV-\int _{\mathcal {R}}{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\right)\delta u_{i}~dV\\&=\int _{\partial {\mathcal {R}}}{\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\delta u_{i}~n_{j}~dA-\int _{\mathcal {R}}{\frac {\partial }{\partial x_{j}}}{}{}\left({\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\right)\delta u_{i}~dV\end{aligned}}$
Substituting in the expression for $\delta U$ , we have,

${\begin{aligned}\delta U&=\int _{\mathcal {R}}{\frac {\partial {\tilde {F}}}{\partial u_{i}}}\delta u_{i}~dV+\int _{\partial {\mathcal {R}}}{\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\delta u_{i}~n_{j}~dA-\int _{\mathcal {R}}{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\right)\delta u_{i}~dV\\&=\int _{\mathcal {R}}\left[{\frac {\partial {\tilde {F}}}{\partial u_{i}}}-{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\right)\right]\delta u_{i}~dV+\int _{\partial {\mathcal {R}}}{\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\delta u_{i}~n_{j}~dA\end{aligned}}$
For $U$ to be minimum, a necessary condition is that $\delta U=0$ for all variations $\delta \mathbf {u}$ .

Therefore, the Euler equation for this problem is

${\frac {\partial {\tilde {F}}}{\partial u_{i}}}-{\frac {\partial }{\partial x_{j}}}\left({\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}\right)=0~~~~\forall ~~\mathbf {x} \in {\mathcal {R}}$
and the associated boundary conditions are

${\frac {\partial {\tilde {F}}}{\partial u_{i,j}}}=0~~~{\text{or,}}~~~\delta u_{i}=0~~~~\forall ~~\mathbf {x} \in \partial {\mathcal {R}}$