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Variational Principle and Vectorial Identities

  1. Nov 11, 2013 #1
    Hello there,

    I am struggling in proving the following.
    The principle of Minimum energy for an elastic body (no body forces, no applied tractions) says that the equilibrium state minimizes
    $$\int_{\Omega} \nabla^{(s)} u D (\nabla^{(s)}u)$$
    among all vectorial functions u satisfying the boundary conditions, where $$\nabla^{(s)} = \frac{1}{2} (u_{i,j}+u_{j,i})$$ and D is a constant tensor $$D_{ijkl}$$
    The principle is expressed in terms of displacements, so one would expect that its Euler Lagrange Equation coinciides with the equlibrium equation of elasticity expressed in terms of displacements, the Navier equations, $$A \nabla (\nabla \cdot u) + B \nabla^{2} u = 0$$, A e B constants.
    How to prove that? I am quite shaky in dimensions higher than 1.
    I tried writing the first variation, after introducing $$u_{var} = U + \epsilon u$$ as
    $$\int_{\Omega} \nabla^{(s)} u D (\nabla^{(s)}U)$$
    and now by integration by parts I recover a Laplacian, as in Navier's equation (second term), but not the term $$\nabla (\nabla \cdot u)$$, any help would be so appreciated, thanks
  2. jcsd
  3. Nov 11, 2013 #2
    Let me rephrase the question, to make it clearer.
    How to compute the Euler Lagrange equation of the functional
    $$\int_{\Omega} \nabla^{(s)} u D (\nabla^{(s)}u)$$
    where u is a vectorial function, $$\nabla^{(s)}u = \frac{1}{2} (u_{i,j}+u_{j,i})$$ and D is a (symmetric) constant tensor $$D_{ijkl}$$?
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