# Homework #2 Solutions

1. Identify the following tensor notation quantities. The first one is done to demonstrate what's requested.

Tensor NotationNameVector NotationExpanded
A.$$a_i b_i$$Vector Dot Product$${\bf a} \cdot {\bf b}$$$$a_x b_x + a_y b_y + a_z b_z$$
B.$$\epsilon_{rst} a_r b_t$$Vector Cross Product$${\bf b} \times {\bf a}$$No need to expand
C.$$A_{rs} B_{ts}$$Matrix Multiplication$${\bf A} \cdot {\bf B}^T$$No need to expand
D.$$A_{ii}$$Trace of Matrix$$\text{tr}({\bf A})$$$$A_{11} + A_{22} + A_{33}$$
E.$$\boldsymbol{\sigma}_{ij,j}$$Divergence of Stress Tensor$$\nabla \cdot \boldsymbol{\sigma}$$No need to expand
F.$$f_{,kk}$$Laplacian$$\nabla^2 f$$$${\partial^2 f \over \partial x^2} + {\partial^2 f \over \partial y^2} + {\partial^2 f \over \partial z^2}$$

2. Demonstrate that starting with: $$\qquad \epsilon_{ijk} \epsilon_{lmn} = \delta_{il} (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}) + \delta_{im} (\delta_{jn} \delta_{kl} - \delta_{jl} \delta_{kn}) + \delta_{in} (\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl})$$

and multiplying both sides through by $$\delta_{il}$$ produces: $$\qquad \epsilon_{ijk} \epsilon_{imn} = \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}$$

$\begin{eqnarray} \epsilon_{ijk} \epsilon_{lmn} \; & = & \; \delta_{il} (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}) + \delta_{im} (\delta_{jn} \delta_{kl} - \delta_{jl} \delta_{kn}) + \delta_{in} (\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}) \\ \\ \\ \\ \\ \delta_{il} \epsilon_{ijk} \epsilon_{lmn} \; & = & \; \delta_{il} \Big[ \delta_{il} (\delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km}) + \delta_{im} (\delta_{jn} \delta_{kl} - \delta_{jl} \delta_{kn}) + \delta_{in} (\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}) \Big] \\ \\ \\ \\ \\ \delta_{il} \epsilon_{ijk} \epsilon_{lmn} \; & = & \; \delta_{il} \Big[ \delta_{il} \delta_{jm} \delta_{kn} - \delta_{il} \delta_{jn} \delta_{km} + \delta_{im} \delta_{jn} \delta_{kl} - \delta_{im} \delta_{jl} \delta_{kn} + \delta_{in} \delta_{jl} \delta_{km} - \delta_{in} \delta_{jm} \delta_{kl} \Big] \\ \\ \\ \\ \\ \delta_{il} \epsilon_{ijk} \epsilon_{lmn} \; & = & \; \delta_{il} \delta_{il} \delta_{jm} \delta_{kn} - \delta_{il} \delta_{il} \delta_{jn} \delta_{km} + \delta_{il} \delta_{im} \delta_{jn} \delta_{kl} - \delta_{il} \delta_{im} \delta_{jl} \delta_{kn} + \delta_{il} \delta_{in} \delta_{jl} \delta_{km} - \delta_{il} \delta_{in} \delta_{jm} \delta_{kl} \\ \\ \\ \\ \\ \epsilon_{ijk} \epsilon_{imn} \; & = & \; \delta_{ii} \delta_{jm} \delta_{kn} - \delta_{ii} \delta_{jn} \delta_{km} + \delta_{lm} \delta_{jn} \delta_{kl} - \delta_{lm} \delta_{jl} \delta_{kn} + \delta_{ln} \delta_{jl} \delta_{km} - \delta_{ln} \delta_{jm} \delta_{kl} \\ \\ \\ \\ \\ \epsilon_{ijk} \epsilon_{imn} \; & = & \; 3 \, \delta_{jm} \delta_{kn} - 3 \, \delta_{jn} \delta_{km} + \delta_{jn} \delta_{km} - \delta_{jm} \delta_{kn} + \delta_{jn} \delta_{km} - \delta_{jm} \delta_{kn} \\ \\ \\ \\ \\ \epsilon_{ijk} \epsilon_{imn} \; & = & \; 3 \, \delta_{jm} \delta_{kn} - 3 \, \delta_{jn} \delta_{km} + 2 \, \delta_{jn} \delta_{km} - 2 \, \delta_{jm} \delta_{kn} \\ \\ \\ \\ \\ \epsilon_{ijk} \epsilon_{imn} \; & = & \; \delta_{jm} \delta_{kn} - \delta_{jn} \delta_{km} \\ \end{eqnarray}$

3. Show that $\nabla \times ({\bf u} \times {\bf v}) = ({\bf v} \cdot \nabla){\bf u} - {\bf v} (\nabla \cdot {\bf u}) + {\bf u} (\nabla \cdot {\bf v}) - ({\bf u} \cdot \nabla){\bf v}$

$\begin{eqnarray} {\bf u} \times {\bf v} & = & \epsilon_{ijk} u_j v_k \\ \\ \\ \\ \nabla \times ({\bf u} \times {\bf v}) & = & \epsilon_{mni} ( \epsilon_{ijk} u_j v_k )_{,n} \\ \\ \\ \\ & = & \epsilon_{mni} \epsilon_{ijk} ( u_j v_k )_{,n} \\ \\ \\ \\ & = & \epsilon_{mni} \epsilon_{ijk} ( u_{j,n} v_k + u_j v_{k,n} ) \\ \\ \\ \\ & = & \epsilon_{imn} \epsilon_{ijk} ( u_{j,n} v_k + u_j v_{k,n} ) \\ \\ \\ \\ & = & (\delta_{mj} \delta_{nk} - \delta_{mk} \delta_{nj}) ( u_{j,n} v_k + u_j v_{k,n} ) \\ \\ \\ \\ & = & \delta_{mj} \delta_{nk} u_{j,n} v_k - \delta_{mk} \delta_{nj} u_{j,n} v_k + \delta_{mj} \delta_{nk} u_j v_{k,n} - \delta_{mk} \delta_{nj} u_j v_{k,n} \\ \\ \\ \\ & = & u_{m,n} v_n - u_{n,n} v_m + u_m v_{n,n} - u_n v_{m,n} \\ \\ \\ \\ & = & ({\bf v} \cdot \nabla){\bf u} - {\bf v} (\nabla \cdot {\bf u}) + {\bf u} (\nabla \cdot {\bf v}) - ({\bf u} \cdot \nabla){\bf v} \\ \end{eqnarray}$

4. We've done the simpler version of this in class. This time, invert Hooke's Law for strain as a function of stress, to get stress as a function of strain, but with thermal expansion thrown into the mix.

The starting point is

$\epsilon_{ij} = {1 \over E} \left[ (1+\nu) \sigma_{ij} - \delta_{ij} \, \nu \, \sigma_{kk} \right] + \delta_{ij} \, \alpha (T - T_{ref})$
where $$\alpha$$ is the thermal expansion coefficient, and $$T$$ and $$T_{ref}$$ are temperatures.

Solve for $$\sigma_{ij}$$

$(1+\nu) \sigma_{ij} = E \, \epsilon_{ij} - \delta_{ij} \, E \, \alpha (T - T_{ref}) + \delta_{ij} \, \nu \, \sigma_{kk}$
Multiply thru by $$\delta_{ij}$$

$(1+\nu) \delta_{ij} \sigma_{ij} = E \, \delta_{ij} \epsilon_{ij} - \delta_{ij} \delta_{ij} \, E \, \alpha (T - T_{ref}) + \delta_{ij} \delta_{ij} \, \nu \, \sigma_{kk}$
Simplify

$(1+\nu) \sigma_{jj} = E \, \epsilon_{jj} - 3 \, E \, \alpha (T - T_{ref}) + 3 \, \nu \, \sigma_{kk}$
Regroup - note that $$\sigma_{jj} = \sigma_{kk}$$

$\sigma_{jj} = { E \, \epsilon_{jj} - 3 \, E \, \alpha (T - T_{ref}) \over 1-2\nu }$
Substitute back in earlier equation for $$\sigma_{kk}$$

$(1+\nu) \sigma_{ij} = E \, \epsilon_{ij} - \delta_{ij} \, E \, \alpha (T - T_{ref}) + {\delta_{ij} \, \nu \, E \, \epsilon_{kk} - 3 \, \delta_{ij} \, \nu \,E \, \alpha (T - T_{ref}) \over 1-2\nu }$
Regroup the thermal terms

$(1+\nu) \sigma_{ij} = E \, \epsilon_{ij} + {\delta_{ij} \, \nu \, E \, \epsilon_{kk} - \delta_{ij} \, (1 + \nu) \,E \, \alpha (T - T_{ref}) \over 1-2\nu }$
Simplify some more

$\sigma_{ij} = { E \over (1 + \nu) } \left[ \epsilon_{ij} + \delta_{ij} {\nu \over 1 - 2\nu } \epsilon_{kk} \right] - \delta_{ij} { E \, \alpha (T - T_{ref}) \over 1-2\nu }$