## 0

Before we start, some basic mathematic theorem.

### Divergence

$∇·v = v_{i,i}$

$∇·T = T_{ij,i} e_j$

### Gradient (dim+1)

$∇\phi = \phi_{,i} e_i$

$∇v = v_{i,j} e_i⊗e_j$

$∇T = T_{ij,k} e_i⊗e_j⊗e_k$

### Curl (dim=)

$∇ × v =ε_{ijk} v_{j,i} e_k$

$∇ × T =ε_{ijk} T_{mj,i} e_k⊗e_m$

### The product rule of differentiation

$\because \bm{σ}·\bm{\nu} = \bm{σ}_{ij}\bm{\nu}_j e_i$

$\therefore ∇·(\bm{σ}·\bm{\nu}) = (\bm{σ}_{ij}\bm{\nu}_j)_{,i} = \bm{σ}_{ij,i}\bm{\nu}_j + \bm{σ}_{ij}\bm{\nu}_{j,i}$

also

$(∇·\bm{σ})·\bm{\nu} = \bm{σ}_{ij,i} \bm{\nu}_j$

$∇\bm{\nu} : \bm{σ} = \bm{\nu}_{i,j}\bm{σ}_{ji}$

or as Prof. Zohdi suggested

$∇\bm{\nu} : \bm{σ} = \bm{\nu}_{i,j}\bm{σ}_{ij}$

but since $\bm{σ}$ is symmetric

$\bm{σ}_{ij} = \bm{σ}_{ji}$

therefore

$\tag{1} ∇·(\bm{σ}·\bm{\nu}) = (∇·\bm{σ})·\bm{\nu} + ∇\bm{\nu} : \bm{σ}$

### Divergence Theorem

$∫_{∂R} \phi n dA = ∫_{R} ∇·\phi dV$

$\tag{2} ∫_{∂R} \bm{v}·n dA = ∫_{R} ∇·\bm{v} dV$

$∫_{∂R} \bm{T} n dA = ∫_{R} ∇·\bm{T} dV$

or

$∫_{∂R} \phi n_i dA = ∫_{R} \phi_{,i} dV$

$∫_{∂R} v_i n_i dA = ∫_{R} v_{i,i} dV$

$∫_{∂R} T_{ij}n_j dA = ∫_{R} T_{ij,j} dV$

### Stokes theorem

$∫_{C} \phi d\bm{x} = ∫_{S} n × ∇\phi dA$

$∫_{C} v·d\bm{x} = ∫_{S} n·∇×v dA$

$∫_{C} T d\bm{x} = ∫_{S} (∇×T)^T n dA$

or

$∫_{C} \phi dx_i = ∫_{S} ε_{ijk} n_j \phi_{,k} dA$

$∫_{C} v_i dx_i = ∫_{S} n_i ε_{ijk} v_{k,j} dA$

$∫_{C} T_{ij} dx_j = ∫_{S} ε_{jpq} T_{iq,p} n_j dA$

## 1

The SMALL deformation of the body is governed by (strong form):

$∇⋅(\bm{IE}:∇\bm{u})+ρ\bm{g}=\bm{0}$

since

$\bm{σ} = \bm{IE}:∇\bm{u}$

let

$\bm{f} = ρ\bm{g}$

thus

$∇·\bm{σ} + \bm{f} = 0$

so we can write

$∫_Ω (∇·\bm{σ} + \bm{f} ) · \bm{\nu} dΩ = 0$

using $(1)$

$∫_Ω (∇·(\bm{σ}·\bm{\nu})-∇\bm{\nu}:\bm{σ}) dΩ + ∫_Ω \bm{f}·\bm{\nu} dΩ = 0$

using $(2)$

$∫_Ω ∇\bm{\nu}:\bm{σ} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{∂Ω} \bm{σ}·\bm{\nu}·n dA$

because

$\bm{t} = \bm{σ}·\bm{n}$

therefore

$∫_Ω ∇\bm{\nu}:\bm{σ} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{Γ_t} \bm{t}·\bm{\nu} dA$

so we have the weak form

$\begin{array}{lcr} \text{Find }\bm{u}, \bm{u}|_{Γ_u} = \bm{u}^* \text{, such that }∀\bm{\nu}, \bm{\nu}|_{Γ_u}= 0 \\ \displaystyle∫_Ω ∇\bm{\nu}:\bm{IE}:∇\bm{u} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{Γ_t} \bm{t}·\bm{\nu} dA \end{array}$

where $u^*$ is the applied boundary displacement on $Γ_u$, $t = t^*$ on $Γ_t$

since the integrals must be finite, we improve the weak form as below

$\begin{array}{lcr} \text{Find }\bm{u} ∈ \bm{H}^1(Ω), \bm{u}|_{Γ_u} = u^* \text{, such that }∀\bm{\nu} ∈ \bm{H}^1(Ω), \bm{\nu}|_{Γ_u}= 0 \\ \displaystyle∫_Ω ∇\bm{\nu}:\bm{IE}:∇\bm{u} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{Γ_t} \bm{t}·\bm{\nu} dA \end{array}$

where

$\bm{H}^1(Ω) \vcentcolon= [H^1(Ω)]^3$

explained as below

similar to the definition of $u ∈ H^1(Ω)$ which states

$u ∈ H^1(Ω) \text{ if } ||u||_{H^1(Ω)}^2 \vcentcolon= ∫_Ω u_{,j}u_{,j} dΩ + ∫_Ω uudΩ < \infty$

the definition of $\bm{u} ∈ \bm{H}^1(Ω)$ states as below

$\bm{u} ∈ \bm{H}^1(Ω) \text{ if } ||\bm{u}||_{\bm{H}^1(Ω)}^2 \vcentcolon= ∫_Ω u_{i,j}u_{i,j} dΩ + ∫_Ω u_iu_idΩ < \infty$

## 2

should $\bm{u}^*$ be different from the applied boundary displacement on $Γ_u$, weak form can be stated as below

$\tag{3} \begin{array}{lcr} \text{Find }\bm{u} ∈ \bm{H}^1(Ω) \text{, such that }∀\bm{\nu} ∈ \bm{H}^1(Ω) \\ \displaystyle∫_Ω ∇\bm{\nu}:\bm{IE}:∇\bm{u} dΩ = ∫_Ω \bm{f}·\bm{\nu} dΩ + ∫_{Γ_t} \bm{t}·\bm{\nu} dA + P^\star∫_{Γ_u} (\bm{u}^*-\bm{u})·\bm{\nu} dA \end{array}$

or

$\begin{array}{lcr} \text{Find }\bm{u} ∈ \bm{H}^1(Ω) \text{, such that }∀\bm{\nu} ∈ \bm{H}^1(Ω) \\ \displaystyle∫_Ω \nu_{i,j} IE_{ijkl} u_{k,l} dΩ = ∫_Ω f_i \nu_i dΩ + ∫_{Γ_t} t_i \nu_i dA + P^\star∫_{Γ_u} u^*_i \nu_i - u_i \nu_i dA \end{array}$

where the (penalty) parameter $P^\star$ is a large positive number.

## 3

To have

$\begin{cases} \hat{\phi}_i(\zeta_1,\zeta_2,\zeta_3) = 1 &\text{ , } \zeta_1=\zeta_{i1},\zeta_2=\zeta_{i2},\zeta_3=\zeta_{i3} \\ \hat{\phi}_i(\zeta_1,\zeta_2,\zeta_3) = 0 &\text{ , } \text{others} \end{cases}$

for each $\hat{\phi}_i$ 8 functions (for each of the 8 points) with 8 unknowns, aka. $\zeta_1^m \zeta_2^n \zeta_3^k$ where $m,n,k=0,1$ can be derived, leading to the only solution as following

$\begin{cases} \hat{\phi}_1 = \frac{1}{8} (1-\zeta_1) (1-\zeta_2) (1-\zeta_3) \\ \hat{\phi}_2 = \frac{1}{8} (1+\zeta_1) (1-\zeta_2) (1-\zeta_3) \\ \hat{\phi}_3 = \frac{1}{8} (1+\zeta_1) (1+\zeta_2) (1-\zeta_3) \\ \hat{\phi}_4 = \frac{1}{8} (1-\zeta_1) (1+\zeta_2) (1-\zeta_3) \\ \hat{\phi}_5 = \frac{1}{8} (1-\zeta_1) (1-\zeta_2) (1+\zeta_3) \\ \hat{\phi}_6 = \frac{1}{8} (1+\zeta_1) (1-\zeta_2) (1+\zeta_3) \\ \hat{\phi}_7 = \frac{1}{8} (1+\zeta_1) (1+\zeta_2) (1+\zeta_3) \\ \hat{\phi}_8 = \frac{1}{8} (1-\zeta_1) (1+\zeta_2) (1+\zeta_3) \end{cases}$

## 4

$\bm{IE}$ has the following symmetries

$IE_{ijkl} = IE_{klij} = IE_{jikl} = IE_{ijlk}$

allowing us to rewrite $(3)$ in a more compact matrix form

$\tag{4} \begin{aligned} ∫_Ω ([\bm{D}]\{\bm{\nu}\})^T [\bm{IE}] ([\bm{D}]\{\bm{u}\}) dΩ = &∫_Ω \{\bm{\nu}\}^T\{\bm{f}\} dΩ \\ &+ ∫_{Γ_t} \{\bm{\nu}\}^T\{\bm{t}^*\} dA \\ &+ P^\star∫_{Γ_u} \{\bm{\nu}\}^T\{\bm{u}^*-\bm{u}\} dA \end{aligned}$

where

$\tag{5} [\bm{D}] \vcentcolon = \begin{bmatrix} \cfrac{∂}{∂x_1} & 0 & 0 \\ 0 & \cfrac{∂}{∂x_2} & 0 \\ 0 & 0 & \cfrac{∂}{∂x_3} \\ \cfrac{∂}{∂x_2} & \cfrac{∂}{∂x_1} & 0 \\ 0 & \cfrac{∂}{∂x_3} & \cfrac{∂}{∂x_2} \\ \cfrac{∂}{∂x_3} & 0 & \cfrac{∂}{∂x_1} \\ \end{bmatrix}, \{\bm{u}\} \vcentcolon = \begin{Bmatrix} u_1 \\ u_2 \\ u_3 \\ \end{Bmatrix}, \{\bm{f}\} \vcentcolon = \begin{Bmatrix} f_1 \\ f_2 \\ f_3 \\ \end{Bmatrix}, \{\bm{t}^*\} \vcentcolon = \begin{Bmatrix} t^*_1 \\ t^*_2 \\ t^*_3 \\ \end{Bmatrix},$