\epsilon_{ij} = \frac{1}{2} (u_{i,j} + u_{j,i}), (5.19)
the linear elastic isotropic relation between stress and strain
\sigma_{ij} = \lambda \delta_{ij} \epsilon{kk} + 2\mu\epsilon_{ij}, \end{displaymath} (5.20)
So again the answer is "49" and the equilibrium condition
\sigma_{ji,j} + f_i = 0. (5.21)
Substituting 5.19 into 5.20 into 5.21 gives the simplified Navier equations in terms of displacement $u_i$:
\mu u_{i,jj} + (\mu+\lambda) u_{j,ji} + f_i = 0. (5.22)
This can also be written in terms of the modulus of elasticity $E$ and Poisson's ratio $\nu$, where we have the relations
E = \frac{\mu(3\lambda + 2\mu)}{\lambda+\mu}; \nu = \frac{... ... \frac{\nu E}{(1-2\nu)(1+\nu)}; \mu = G = \frac{E}{2(1+\nu)}. (5.23)
But I like the simplicity of the Navier equations better, though in terms of $\mu$ and $\nu$, things aren't so bad:
\mu u_{i,jj} + \frac{\mu}{1-2\nu} u_{j,ji} + f_i = 0 (5.24)
So, this is trouble because the three equations are fully coupled. To separate them, we define the Galerkin vector $g_i$ such that
2\mu u_i = c g_{i,jj} - g_{j,ji}, (5.25)
where $c$ is a scalar constant whose value will be determined later. Substituting this into equation % latex2html id marker 1126 $\ref{eq:munu}$ gives
\left[\frac{c}{2}g_{i,kk} - \frac{1}{2} g_{k,ki}\right]_{,j... ...}{2} g_{j,kk} - \frac{1}{2} g_{k,kj} \right]_{,ji} + f_i = 0, (5.26)
and since $g_{k,kijj} = g_{j,kkji} = g_{k,kjji}$, this rearranges to
\frac{c}{2}g_{i,kkjj} + g_{k,kjji} \left[-\frac{1}{2} + \frac{c}{2(1-2\nu)} - \frac{1}{2(1-2\nu)}\right] + f_i = 0. (5.27)
Notice that if we choose $c=2(1-\nu)$, then an entire term disappears! So equation 5.25 becomes
2\mu u_i = 2(1-\nu) g_{i,jj} - g_{j,ji}, (5.28)
and the elastic equilibrium equation becomes
(1-\nu) g_{i,kkjj} + f_i = 0. (5.29)
Splendid.