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

$U[u(x)]=\underset{x_0}{\overset{x_1}{\int }}F(x,u,u^{\text{'}})dx$

is a minimum.

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

$\frac{\partial F}{\partial u}-\frac{d}{dx}\left(\frac{\partial F}{\partial u^{\text{'}}}\right)=0$

with the associated boundary conditions

$\eta (x_0)=0\text{and}\eta (x_1)=0$

or,

${\frac{\partial F}{\partial u^{\text{'}}}|}_{x_0}=0\text{and}{\frac{\partial F}{\partial u^{\text{'}}}|}_{x_1}=0$

Minimizing a Functional in 3-D

In 3-D, the equivalent minimization problem can be stated as

Find $𝐮(𝐱)$ such that

$U[𝐮(𝐱)]=\underset{ℛ}{\int }F(𝐱,𝐮,\mathrm{\nabla }𝐮)dV$

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 $({\widehat{𝐞}}_i)$.

Then,

$𝐱=x_i{\widehat{𝐞}}_i;𝐮=u_i{\widehat{𝐞}}_i;\mathrm{\nabla }𝐮=u_{i,j}{\widehat{𝐞}}_i\otimes {\widehat{𝐞}}_j$

and

$U[𝐮(𝐱)]=\underset{ℛ}{\int }\tilde{F}(x_i,u_i,u_{i,j})dV$

Taking the first variation of $U$, we get

$\delta U=\underset{ℛ}{\int }(\frac{\partial \tilde{F}}{\partial u_i}\delta u_i+\frac{\partial \tilde{F}}{\partial u_{i,j}}\delta u_{i,j})dV$

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,

$\begin{array}{cc}\underset{ℛ}{\int }\frac{\partial \tilde{F}}{\partial u_{i,j}}\delta u_{i,j}dV& =\underset{ℛ}{\int }\frac{\partial }{\partial x_j}\left(\frac{\partial \tilde{F}}{\partial u_{i,j}}\delta u_i\right)dV-\underset{ℛ}{\int }\frac{\partial }{\partial x_j}\left(\frac{\partial \tilde{F}}{\partial u_{i,j}}\right)\delta u_{i}dV\\ & =\underset{\partial ℛ}{\int }\frac{\partial \tilde{F}}{\partial u_{i,j}}\delta u_i{n}_{j}dA-\underset{ℛ}{\int }\frac{\partial }{\partial x_j}\left(\frac{\partial \tilde{F}}{\partial u_{i,j}}\right)\delta u_{i}dV\end{array}$

Substituting in the expression for $\delta U$, we have,

$\begin{array}{cc}\delta U& =\underset{ℛ}{\int }\frac{\partial \tilde{F}}{\partial u_i}\delta u_{i}dV+\underset{\partial ℛ}{\int }\frac{\partial \tilde{F}}{\partial u_{i,j}}\delta u_i{n}_{j}dA-\underset{ℛ}{\int }\frac{\partial }{\partial x_j}\left(\frac{\partial \tilde{F}}{\partial u_{i,j}}\right)\delta u_{i}dV\\ & =\underset{ℛ}{\int }[\frac{\partial \tilde{F}}{\partial u_i}-\frac{\partial }{\partial x_j}\left(\frac{\partial \tilde{F}}{\partial u_{i,j}}\right)]\delta u_{i}dV+\underset{\partial ℛ}{\int }\frac{\partial \tilde{F}}{\partial u_{i,j}}\delta u_i{n}_{j}dA\end{array}$

For $U$ to be minimum, a necessary condition is that $\delta U=0$ for all variations $\delta 𝐮$.

Therefore, the Euler equation for this problem is

$\frac{\partial \tilde{F}}{\partial u_i}-\frac{\partial }{\partial x_j}\left(\frac{\partial \tilde{F}}{\partial u_{i,j}}\right)=0\mathrm{\forall }𝐱\in ℛ$

and the associated boundary conditions are

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

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