Elasticity/Equilibrium example 2

Example 2
 Given: The displacement equation of equilibrium for an isotropic inhomogeneous linear elastic material can be written as

\boldsymbol{\nabla} \bullet (\mathbf{C} : \boldsymbol{\nabla} \mathbf{u}) + \mathbf{b} = 0 $$ where

\mathbf{C} = \lambda \mathbf{1}^{(2)}\otimes\mathbf{1}^{(2)} + 2\mu\mathbf{1}^{(4s)} $$ and $$\lambda(\mathbf{x})$$ and $$\mu(\mathbf{x})$$ are the Lamé moduli.

Show:

Show that the displacement equation of equilibrium can be expressed as

\mu \boldsymbol{\nabla}\bullet(\boldsymbol{\nabla}\mathbf{u}) + (\lambda+\mu) \boldsymbol{\nabla}(\boldsymbol{\nabla}\bullet\mathbf{u}) + (\boldsymbol{\nabla}\mathbf{u}+\boldsymbol{\nabla}\mathbf{u}^T) \boldsymbol{\nabla}{\mu} + (\boldsymbol{\nabla}\bullet\mathbf{u}) \boldsymbol{\nabla}{\lambda} + \mathbf{b} = 0 $$

Solution
The skew part of the tensor $$\boldsymbol{\nabla}\mathbf{u}$$ does not affect the stress because it leads to a rigid displacement field. Therefore, the displacement equation of equilibrium may be written as

\boldsymbol{\nabla} \bullet \left[\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u})\right] + \mathbf{b} = 0 $$ where

\text{symm}(\boldsymbol{\nabla}\mathbf{u}) = \frac{1}{2}(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T) $$ In index notataion,

\text{symm}(\boldsymbol{\nabla}\mathbf{u}) = \boldsymbol{\varepsilon} \equiv \varepsilon_{kl} = \frac{1}{2}(u_{k,l} + u_{l,k}) $$ and

\mathbf{C} \equiv C_{ijkl} = \lambda\delta_{ij}\delta_{kl} + \mu(\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}) $$ Therefore,
 * $$\begin{align}

\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u}) \equiv C_{ijkl}~\varepsilon_{kl} & = \lambda\delta_{ij}\delta_{kl}~\varepsilon_{kl} + \mu\delta_{ik}\delta_{jl}~\varepsilon_{kl} + \mu\delta_{il}\delta_{jk}~\varepsilon_{kl} \\ & = \lambda~\varepsilon_{mm}\delta_{ij} + \mu~\varepsilon_{ij} + \mu~\varepsilon_{ij} \\ & = \lambda~\varepsilon_{mm}\delta_{ij} + 2\mu~\varepsilon_{ij} \\ & \equiv \lambda~(\text{tr}~\boldsymbol{\varepsilon})\mathbf{1} + 2\mu~\boldsymbol{\varepsilon} \end{align}$$ Now,

\text{tr}~\boldsymbol{\varepsilon} \equiv \varepsilon_{mm} = \frac{1}{2}(u_{m,m} + u_{m,m}) = u_{m,m} \equiv \boldsymbol{\nabla}\bullet\mathbf{u} $$ Hence,

\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u}) = \lambda~(\boldsymbol{\nabla}\bullet\mathbf{u})\mathbf{1} + \mu~(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T) $$ Taking the divergence,
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left[\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u})\right]} & = \boldsymbol{\nabla}\bullet{\left[\lambda~(\boldsymbol{\nabla}\bullet\mathbf{u})\mathbf{1} + \mu~(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^T)\right]} \\ & = \boldsymbol{\nabla}\bullet{\left[\lambda~(\boldsymbol{\nabla} \bullet\mathbf{u})\mathbf{1}\right]} + \boldsymbol{\nabla}\bullet{\left(\mu~\boldsymbol{\nabla}\mathbf{u}\right)} + \boldsymbol{\nabla}\bullet{\left(\mu~\boldsymbol{\nabla}\mathbf{u}^T\right)} \end{align}$$ Recall that
 * $$\begin{align}

\boldsymbol{\nabla}{\phi} & = \phi_{,j} \\ \boldsymbol{\nabla}{\mathbf{v}} & = v_{i,j} \\ \boldsymbol{\nabla}\bullet{\mathbf{v}} & = v_{j,j} \\ \boldsymbol{\nabla}\bullet{\mathbf{T}} & = T_{ij,j} \end{align}$$ Therefore,
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left[\lambda~(\boldsymbol{\nabla} \bullet\mathbf{u})\mathbf{1}\right]} & \equiv \left(\lambda~u_{k,k}\delta_{ij}\right)_{,j} \\ & = \lambda_{,i}~u_{k,k} + \lambda~u_{k,ki} \\ & \equiv \boldsymbol{\nabla}{\lambda}(\boldsymbol{\nabla}\bullet\mathbf{u}) + \lambda\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} \end{align}$$
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left(\mu~\boldsymbol{\nabla}\mathbf{u}\right)} & \equiv \left(\mu~u_{i,j}\right)_{,j} \\ & = \mu_{,j}~u_{i,j} + \mu~u_{i,jj} \\ & \equiv \boldsymbol{\nabla}{\mu} \boldsymbol{\nabla}\mathbf{u} + \mu\boldsymbol{\nabla}\bullet{(\boldsymbol{\nabla}\mathbf{u})} \end{align}$$
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left(\mu~\boldsymbol{\nabla}\mathbf{u}^{T}\right)} & \equiv \left(\mu~u_{j,i}\right)_{,j} \\ & = \mu_{,j}~u_{j,i} + \mu~u_{j,ij} \\ & \equiv \boldsymbol{\nabla}{\mu} \boldsymbol{\nabla}\mathbf{u}^{T} + \mu\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} \end{align}$$ Hence,
 * $$\begin{align}

\boldsymbol{\nabla}\bullet{\left[\mathbf{C} : \text{symm}(\boldsymbol{\nabla}\mathbf{u})\right]} & = \boldsymbol{\nabla}{\lambda}(\boldsymbol{\nabla} \bullet\mathbf{u}) + \lambda\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} + \boldsymbol{\nabla}{\mu} \boldsymbol{\nabla}\mathbf{u} + \mu\boldsymbol{\nabla}\bullet{(\boldsymbol{\nabla}\mathbf{u})} + \boldsymbol{\nabla}{\mu} \boldsymbol{\nabla}\mathbf{u}^{T} + \mu\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} \\ & = \mu\boldsymbol{\nabla}\bullet{(\boldsymbol{\nabla}\mathbf{u})} + (\lambda+\mu)\boldsymbol{\nabla}{(\boldsymbol{\nabla}\bullet\mathbf{u})} + \boldsymbol{\nabla}{\mu}\left(\boldsymbol{\nabla}\mathbf{u} + \boldsymbol{\nabla}\mathbf{u}^{T}\right) + \boldsymbol{\nabla}{\lambda}(\boldsymbol{\nabla}\bullet \mathbf{u}) \end{align}$$ Therefore, the displacement equation of equilibrium can be expressed as required, i.e,

\mu \boldsymbol{\nabla}\bullet(\boldsymbol{\nabla}\mathbf{u}) + (\lambda+\mu) \boldsymbol{\nabla}(\boldsymbol{\nabla}\bullet\mathbf{u}) + (\boldsymbol{\nabla}\mathbf{u}+\boldsymbol{\nabla}\mathbf{u}^T) \boldsymbol{\nabla}{\mu} + (\boldsymbol{\nabla}\bullet\mathbf{u}) \boldsymbol{\nabla}{\lambda} + \mathbf{b} = 0 $$