0

Before we start, some basic mathematic theorem^[1][2].

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 ^[2:1][3]

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

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

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

and $[\bm{IE}]$ is the elastic stiffness matrix of the material.

$$[\bm{IE}] \vcentcolon = \begin{bmatrix} IE_{1111} & IE_{1122} & IE_{1133} & IE_{1112} & IE_{1123} & IE_{1113} \\ IE_{2211} & IE_{2222} & IE_{2233} & IE_{2212} & IE_{2223} & IE_{2213} \\ IE_{3311} & IE_{3322} & IE_{3333} & IE_{3312} & IE_{3323} & IE_{3313} \\ IE_{1211} & IE_{1222} & IE_{1233} & IE_{1212} & IE_{1223} & IE_{1213} \\ IE_{2311} & IE_{2322} & IE_{2333} & IE_{2312} & IE_{2323} & IE_{2313} \\ IE_{1311} & IE_{1322} & IE_{1333} & IE_{1312} & IE_{1323} & IE_{1313} \\ \end{bmatrix}$$

Further rewrite $\{\bm{u}^h\}$

$$\tag{6} \{\bm{u}^h\} = [\phi]\{\bm{a}\}$$

where

$$\{\bm{a}\} \vcentcolon = \begin{Bmatrix} a_1 \\ a_2 \\ a_3 \\ .\\ .\\ .\\ a_{3N}\\ \end{Bmatrix}, [\phi] \vcentcolon = \begin{bmatrix} \phi_1 & \phi_2 & ... & \phi_N & & & & & & & & \\ & & & & \phi_1 & \phi_2 & ... & \phi_N & & & & \\ & & & & & & & & \phi_1 & \phi_2 & ... & \phi_N \\ \end{bmatrix},$$

choose $\bm{\nu}$ with the same basis, but a different linear combination

$$\tag{7} \{\bm{\nu}^h\} = [\phi]\{\bm{b}\}$$

with $(6),(7)$ we can rewrite $(4)$

$$\tag{8} \begin{aligned} ∫_Ω ([\bm{D}][\phi]\{\bm{b}\})^T [\bm{IE}] ([\bm{D}][\phi]\{\bm{a}\}) dΩ = &∫_Ω ([\phi]\{\bm{b}\})^T\{\bm{f}\} dΩ \\ & + ∫_{Γ_t} ([\phi]\{\bm{b}\})^T\{\bm{t}^*\} dA \\ & + P^\star∫_{Γ_u} ([\phi]\{\bm{b}\})^T\{\bm{u}^*-[\phi]\{\bm{a}\}\} dA \end{aligned}$$

We can rewrite $(8)$

$$\{\bm{b}\}^T \{ [\bm{K}] \{\bm{a}\} - \{\bm{R}\} \} = 0$$

where

$$\tag{9} [\bm{K}] \vcentcolon = ∫_Ω ([\bm{D}][\phi])^T [\bm{IE}] ([\bm{D}][\phi]) dΩ + P^\star∫_{Γ_u} [\phi]^T[\phi] dA$$

$$\tag{10} \{\bm{R}\} \vcentcolon = ∫_Ω [\phi]^T\{\bm{f}\} dΩ + ∫_{Γ_t} [\phi]^T\{\bm{t}^*\} dA + P^\star∫_{Γ_u} [\phi]^T\{\bm{u}^*\} dA$$

Consider

$$[\bm{K}] = \sum_{e=1}^{N_e}{[\bm{K}]^e}$$

we can rewrite $(9),(10)$ for $e=1,2..,N_e$

$$\tag{11} [\bm{K}]^e \vcentcolon = ∫_{Ω_e} ([\bm{D}][\phi])^T [\bm{IE}] ([\bm{D}][\phi]) dΩ_e + P^\star∫_{Γ_{u,e}} [\phi]^T[\phi] dA_e$$

$$\tag{12} \{\bm{R}\}^e \vcentcolon = ∫_{Ω_e} [\phi]^T\{\bm{f}\} dΩ_e + ∫_{Γ_{t,e}} [\phi]^T\{\bm{t}^*\} dA_e + P^\star∫_{Γ_{u,e}} [\phi]^T\{\bm{u}^*\} dA_e$$

where

$$Γ_{u,e} = Γ_u ∩ ∂Ω_e$$

$$Γ_{t,e} = Γ_t ∩ ∂Ω_e$$

5

In 3D,

$$[\hat\phi] \vcentcolon = \begin{bmatrix} \hat\phi_1 & \hat\phi_2 & ... & \hat\phi_8 & & & & & & & & \\ & & & & \hat\phi_1 & \hat\phi_2 & ... & \hat\phi_8 & & & & \\ & & & & & & & & \hat\phi_1 & \hat\phi_2 & ... & \hat\phi_8 \\ \end{bmatrix}$$

express $[\bm{D}]$ in terms $ζ_1, ζ_2, ζ_3$

$$[D(\phi(x1, x2, x3))] = [\hat D \phi(M_{x_1}(ζ1, ζ2, ζ3),M_{x_2}(ζ1, ζ2, ζ3),M_{x_3}(ζ1, ζ2, ζ3))]$$

so that

$$[\hat \bm{D}] = \begin{bmatrix} \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_1} & 0 & 0 \\ 0 & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_2} & 0 \\ 0 & 0 & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_3} \\ \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_1} & 0 \\ 0 & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_2} \\ \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_3} & 0 & \cfrac{∂}{ζ_i}\cfrac{ζ_i}{∂x_1} \\ \end{bmatrix}$$

therefore

$$[\hat \bm{D}][\hat\phi] = \begin{bmatrix} \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_1} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_1} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_1} & & & & & & & \\ & & & & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_2} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_2} & & & \\ & & & & & & & & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_3} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_3} \\ \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_2} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_1} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_1} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_1} & & & \\ & & & & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_3} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_2} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_2} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_2} \\ \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_3} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_3} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_3} & & & & & \cfrac{∂\hat\phi_1}{ζ_i}\cfrac{ζ_i}{∂x_1} & \cfrac{∂\hat\phi_2}{ζ_i}\cfrac{ζ_i}{∂x_1} & ... & \cfrac{∂\hat\phi_8}{ζ_i}\cfrac{ζ_i}{∂x_1} \\ \end{bmatrix}$$

with Gaussian Quadrature

$$\int_0^L{F(x)dx} = \int_{-1}^1{F(x(\zeta))J(\zeta)d\zeta} = \sum_1^G{w_i}F(\zeta_i)J(\zeta_i)$$

we can rewrite $(11),(12)$

$$[\bm{K}]^e = \sum_1^G{\sum_1^G{\sum_1^G{w_q w_r w_s ([\hat \bm{D}][\hat\phi])^T[\hat{\bm{IE}}][\hat \bm{D}][\hat\phi]J}}}+ P^\star\sum_1^G{\sum_1^G{w_q w_r [\phi]^T[\phi] J_s}}$$

$$\{\bm{R}\}^e = \sum_1^G{\sum_1^G{\sum_1^G{w_q w_r w_s [\phi]^T\{\bm{f}\} J}}} + \sum_1^G{\sum_1^G{w_q w_r[\phi]^T\{\bm{t}^*\} J_s}} + P^\star\sum_1^G{\sum_1^G{w_q w_r [\phi]^T\{\bm{u}^*\} J_s}}$$

where

$$\bm{F} \vcentcolon= ∇x(ζ_1,ζ_2,ζ_3)$$

$$J \vcentcolon= |\bm{F}|$$

and

$$J_s = J\bm{F}^{-T}·\bm{N}·\bm{n}$$

derived as below

Nanson’s formula

$$\bm{n} dA_e = J\bm{F}^{-T}·\bm{N} d\hat{Ae}$$

where $A_e$ is an area of a region in the current configuration, $\hat{Ae}$ is the same area in the reference configuration, and $\bm{n}$ is the outward normal to the area element in the current configuration while $\bm{N}$ is the outward normal in the reference configuration.

multiply $\bm{n}$ on both sides

$$dA_e = J\bm{F}^{-T}·\bm{N}·\bm{n} d\hat{Ae}$$

therefore

$$J_s = J\bm{F}^{-T}·\bm{N}·\bm{n}$$

6

6.1

$$∇⋅(\bm{IK}⋅∇θ)+ρs=0$$

let the heat flux density^[4] $q$

$$\bm{q} = \bm{IK}⋅∇θ$$

and let

$$f = ρs$$

therefore

$$∇⋅\bm{q} + f = 0$$

so we can write

$$\tag{13} ∫_Ω (∇·\bm{q} + f ) · \nu dΩ = 0$$

now consider

$$(∇⋅\bm{q})·\nu = q_{i,i}\nu$$

$$∇⋅(\bm{q}\nu) = (q_i \nu)_{,i} = q_{i,i}\nu + q_i \nu_{,i}$$

$$(∇\nu)·\bm{q} = q_i \nu_{,i}$$

therefore we can write $(13)$ as

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

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

using $(2)$

$$∫_Ω (∇\nu)·\bm{q} dΩ = ∫_Ω f\nu dΩ + ∫_{∂Ω} \bm{q}⋅\bm{n} \nu dA$$

so we have the weak form

$$\begin{array}{lcr} \text{Find }θ ∈ H^1(Ω), θ|_{Γ_θ} = θ^* \text{, such that }∀\nu ∈ H^1(Ω), \nu|_{Γ_θ}= 0 \\ \displaystyle∫_Ω (∇\nu)·\bm{IK}⋅∇θ dΩ = ∫_Ω f \nu dΩ + ∫_{∂Ω} \bm{q}⋅\bm{n} \nu dA \end{array}$$

6.2

should $θ^*$ be different from the applied boundary displacement on $Γ_θ$, weak form can be stated as below

$$\tag{14} \begin{array}{lcr} \text{Find }θ ∈ H^1(Ω), θ|_{Γ_θ} = θ^* \text{, such that }∀\nu ∈ H^1(Ω), \nu|_{Γ_θ}= 0 \\ \displaystyle∫_Ω (∇\nu)·\bm{IK}⋅∇θ dΩ = ∫_Ω f \nu dΩ + ∫_{∂Ω} \bm{q}⋅\bm{n} \nu dA + P^\star∫_{Γ_θ} (θ^*-θ) \nu dA \end{array}$$

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

6.3

same as 3

6.4

rewrite $θ^h$

$$\tag{15} θ^h = [\phi]\{\bm{a}\} = \phi_i a_i$$

where

$$\{\bm{a}\} \vcentcolon = \begin{Bmatrix} a_1 \\ a_2 \\ a_3 \\ .\\ .\\ .\\ a_{N}\\ \end{Bmatrix}, [\phi] \vcentcolon = \begin{bmatrix} \phi_1 & \phi_2 & ... & \phi_N \end{bmatrix},$$

choose $\bm{\nu}$ with the same basis, but a different linear combination

$$\tag{16} \nu^h = [\phi]\{\bm{b}\} = \phi_j b_j$$

with $(15),(16)$ we can rewrite $(14)$

$$\tag{17} \begin{aligned} ∫_Ω (∇[\phi]\{\bm{b}\})·\bm{IK}⋅∇[\phi]\{\bm{a}\} dΩ = & ∫_Ω f [\phi]\{\bm{b}\} dΩ \\ & + ∫_{∂Ω} \bm{q}⋅\bm{n} [\phi]\{\bm{b}\} dA \\ & + P^\star∫_{Γ_θ} (θ^*-[\phi]\{\bm{a}\})·[\phi]\{\bm{b}\} dA \end{aligned}$$

We can rewrite $(17)$

$$\{\bm{b}\}^T \{ K \{\bm{a}\} - \{\bm{R}\} \} = 0$$

where

$$\tag{18} K \vcentcolon = ∫_Ω (∇[\phi])·\bm{IK}⋅∇[\phi] dΩ + P^\star∫_{Γ_θ} [\phi]^T[\phi] dA$$

$$\tag{19} \{\bm{R}\} \vcentcolon = ∫_Ω [\phi]^Tf dΩ + ∫_{∂Ω} [\phi]^T\bm{q}⋅\bm{n} dA + P^\star∫_{Γ_θ} [\phi]^Tθ^* dA$$

Consider

$$K= \sum_{e=1}^{N_e}K^e$$

we can rewrite $(18),(19)$ for $e=1,2..,N_e$

$$K^e \vcentcolon = ∫_{Ω_e} (∇[\phi])·\bm{IK}⋅∇[\phi] dΩ_e + P^\star∫_{Γ_{u,e}} [\phi]^T[\phi] dA_e$$

$$\{\bm{R}\}^e \vcentcolon = ∫_{Ω_e} [\phi]^Tf dΩ_e + ∫_{∂Ω,e} [\phi]^T\bm{q}⋅\bm{n} dA_e + P^\star∫_{Γ_{u,e}} [\phi]^Tθ^* dA_e$$

where

$$Γ_{θ,e} = Γ_θ ∩ ∂Ω_e$$

7

Key difference is that the diemsion of the equations derived for thermodynamics are generally lower than those for linear elasticity.

This is because $θ$ and $s$ are scalers rather than vectors as $\bm{u}$ and $\bm{g}$, and $\bm{IK}$ is a second order tensor rather than a forth order tensor as $\bm{IE}$.

Reference