Elasticity/Minimizing a functional

Minimizing a functional in 1-D
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
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{align}

\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{align}$$ Substituting in the expression for $$\delta U$$, we have,
 * $$\begin{align}

\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{align}$$ For $$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} $$