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Given the following text description, write Python code to implement the functionality described below step by step
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<USER_TASK:>
Description:
 Matrix generation
Init symbols for sympy
Step1:  Lame params
Step2:  Metric tensor
${\displaystyle \hat{G}=\sum_{i,j} g^{ij}\vec{R}_i\vec{R}_j}$
Step3:  ${\displaystyle \hat{G}=\sum_{i,j} g_{ij}\vec{R}^i\vec{R}^j}$
Step4:  Christoffel symbols
Step5:  Gradient of vector
$ 
\left( 
\begin{array}{c} 
\nabla_1 u_1 \ \nabla_2 u_1 \ \nabla_3 u_1 \
\nabla_1 u_2 \ \nabla_2 u_2 \ \nabla_3 u_2 \
\nabla_1 u_3 \ \nabla_2 u_3 \ \nabla_3 u_3 \
\end{array} 
\right)
=
B \cdot
\left( 
\begin{array}{c} 
u_1 \
\frac { \partial u_1 } { \partial \alpha_1} \
\frac { \partial u_1 } { \partial \alpha_2} \
\frac { \partial u_1 } { \partial \alpha_3} \
u_2 \
\frac { \partial u_2 } { \partial \alpha_1} \
\frac { \partial u_2 } { \partial \alpha_2} \
\frac { \partial u_2 } { \partial \alpha_3} \
u_3 \
\frac { \partial u_3 } { \partial \alpha_1} \
\frac { \partial u_3 } { \partial \alpha_2} \
\frac { \partial u_3 } { \partial \alpha_3} \
\end{array} 
\right) 
= B \cdot D \cdot
\left( 
\begin{array}{c} 
u^1 \
\frac { \partial u^1 } { \partial \alpha_1} \
\frac { \partial u^1 } { \partial \alpha_2} \
\frac { \partial u^1 } { \partial \alpha_3} \
u^2 \
\frac { \partial u^2 } { \partial \alpha_1} \
\frac { \partial u^2 } { \partial \alpha_2} \
\frac { \partial u^2 } { \partial \alpha_3} \
u^3 \
\frac { \partial u^3 } { \partial \alpha_1} \
\frac { \partial u^3 } { \partial \alpha_2} \
\frac { \partial u^3 } { \partial \alpha_3} \
\end{array} 
\right) 
$
Step6:  Physical coordinates
$u_i=u_{[i]} H_i$
Step7:  Strain tensor
$ 
\left( 
\begin{array}{c} 
\varepsilon_{11} \ 
\varepsilon_{22} \ 
\varepsilon_{33} \ 
2\varepsilon_{12} \ 
2\varepsilon_{13} \ 
2\varepsilon_{23} \ 
\end{array} 
\right)
=
\left(E + E_{NL} \left( \nabla \vec{u} \right) \right) \cdot 
\left( 
\begin{array}{c} 
\nabla_1 u_1 \ \nabla_2 u_1 \ \nabla_3 u_1 \
\nabla_1 u_2 \ \nabla_2 u_2 \ \nabla_3 u_2 \
\nabla_1 u_3 \ \nabla_2 u_3 \ \nabla_3 u_3 \
\end{array} 
\right)$
Step8:  Virtual work
Step9:  Tymoshenko theory
$u_1 \left( \alpha_1, \alpha_2, \alpha_3 \right)=u\left( \alpha_1 \right)+\alpha_3\gamma \left( \alpha_1 \right) $
$u_2 \left( \alpha_1, \alpha_2, \alpha_3 \right)=0 $
$u_3 \left( \alpha_1, \alpha_2, \alpha_3 \right)=w\left( \alpha_1 \right) $
$ \left( 
\begin{array}{c} 
u_1 \
\frac { \partial u_1 } { \partial \alpha_1} \
\frac { \partial u_1 } { \partial \alpha_2} \
\frac { \partial u_1 } { \partial \alpha_3} \
u_2 \
\frac { \partial u_2 } { \partial \alpha_1} \
\frac { \partial u_2 } { \partial \alpha_2} \
\frac { \partial u_2 } { \partial \alpha_3} \
u_3 \
\frac { \partial u_3 } { \partial \alpha_1} \
\frac { \partial u_3 } { \partial \alpha_2} \
\frac { \partial u_3 } { \partial \alpha_3} \
\end{array} 
\right) = T \cdot 
\left( 
\begin{array}{c} 
u \
\frac { \partial u } { \partial \alpha_1} \
\gamma \
\frac { \partial \gamma } { \partial \alpha_1} \
w \
\frac { \partial w } { \partial \alpha_1} \
\end{array} 
\right) $
Step10:  Square theory
$u^1 \left( \alpha_1, \alpha_2, \alpha_3 \right)=u_{10}\left( \alpha_1 \right)p_0\left( \alpha_3 \right)+u_{11}\left( \alpha_1 \right)p_1\left( \alpha_3 \right)+u_{12}\left( \alpha_1 \right)p_2\left( \alpha_3 \right) $
$u^2 \left( \alpha_1, \alpha_2, \alpha_3 \right)=0 $
$u^3 \left( \alpha_1, \alpha_2, \alpha_3 \right)=u_{30}\left( \alpha_1 \right)p_0\left( \alpha_3 \right)+u_{31}\left( \alpha_1 \right)p_1\left( \alpha_3 \right)+u_{32}\left( \alpha_1 \right)p_2\left( \alpha_3 \right) $
$ \left( 
\begin{array}{c} 
u^1 \
\frac { \partial u^1 } { \partial \alpha_1} \
\frac { \partial u^1 } { \partial \alpha_2} \
\frac { \partial u^1 } { \partial \alpha_3} \
u^2 \
\frac { \partial u^2 } { \partial \alpha_1} \
\frac { \partial u^2 } { \partial \alpha_2} \
\frac { \partial u^2 } { \partial \alpha_3} \
u^3 \
\frac { \partial u^3 } { \partial \alpha_1} \
\frac { \partial u^3 } { \partial \alpha_2} \
\frac { \partial u^3 } { \partial \alpha_3} \
\end{array} 
\right) = L \cdot 
\left( 
\begin{array}{c} 
u_{10} \
\frac { \partial u_{10} } { \partial \alpha_1} \
u_{11} \
\frac { \partial u_{11} } { \partial \alpha_1} \
u_{12} \
\frac { \partial u_{12} } { \partial \alpha_1} \
u_{30} \
\frac { \partial u_{30} } { \partial \alpha_1} \
u_{31} \
\frac { \partial u_{31} } { \partial \alpha_1} \
u_{32} \
\frac { \partial u_{32} } { \partial \alpha_1} \
\end{array} 
\right) $
Step11:  Mass matrix
<ASSISTANT_TASK:>
Python Code:
from sympy import *
from geom_util import *
from sympy.vector import CoordSys3D
N = CoordSys3D('N')
alpha1, alpha2, alpha3 = symbols("alpha_1 alpha_2 alpha_3", real = True, positive=True)
init_printing()
%matplotlib inline
%reload_ext autoreload
%autoreload 2
%aimport geom_util
Explanation: Matrix generation
Init symbols for sympy
End of explanation
H1=symbols('H1')
H2=S(1)
H3=S(1)
H=[H1, H2, H3]
DIM=3
dH = zeros(DIM,DIM)
for i in range(DIM):
    for j in range(DIM):
        if (i == 0 and j != 1):
            dH[i,j]=Symbol('H_{{{},{}}}'.format(i+1,j+1))
dH
    
Explanation: Lame params
End of explanation
G_up = getMetricTensorUpLame(H1, H2, H3)
Explanation: Metric tensor
${\displaystyle \hat{G}=\sum_{i,j} g^{ij}\vec{R}_i\vec{R}_j}$
End of explanation
G_down = getMetricTensorDownLame(H1, H2, H3)
Explanation: ${\displaystyle \hat{G}=\sum_{i,j} g_{ij}\vec{R}^i\vec{R}^j}$
End of explanation
DIM=3
G_down_diff = MutableDenseNDimArray.zeros(DIM, DIM, DIM)
for i in range(DIM):
    for j in range(DIM):
        for k in range(DIM):
                
            G_down_diff[i,i,k]=2*H[i]*dH[i,k]
                
GK = getChristoffelSymbols2(G_up, G_down_diff, (alpha1, alpha2, alpha3))
GK
Explanation: Christoffel symbols
End of explanation
def row_index_to_i_j_grad(i_row):
    return i_row // 3, i_row % 3
        
B = zeros(9, 12)
B[0,1] = S(1)
B[1,2] = S(1)
B[2,3] = S(1)
B[3,5] = S(1)
B[4,6] = S(1)
B[5,7] = S(1)
B[6,9] = S(1)
B[7,10] = S(1)
B[8,11] = S(1)
for row_index in range(9):
    i,j=row_index_to_i_j_grad(row_index)
    B[row_index, 0] = -GK[i,j,0]
    B[row_index, 4] = -GK[i,j,1]
    B[row_index, 8] = -GK[i,j,2]
B
Explanation: Gradient of vector
$ 
\left( 
\begin{array}{c} 
\nabla_1 u_1 \ \nabla_2 u_1 \ \nabla_3 u_1 \
\nabla_1 u_2 \ \nabla_2 u_2 \ \nabla_3 u_2 \
\nabla_1 u_3 \ \nabla_2 u_3 \ \nabla_3 u_3 \
\end{array} 
\right)
=
B \cdot
\left( 
\begin{array}{c} 
u_1 \
\frac { \partial u_1 } { \partial \alpha_1} \
\frac { \partial u_1 } { \partial \alpha_2} \
\frac { \partial u_1 } { \partial \alpha_3} \
u_2 \
\frac { \partial u_2 } { \partial \alpha_1} \
\frac { \partial u_2 } { \partial \alpha_2} \
\frac { \partial u_2 } { \partial \alpha_3} \
u_3 \
\frac { \partial u_3 } { \partial \alpha_1} \
\frac { \partial u_3 } { \partial \alpha_2} \
\frac { \partial u_3 } { \partial \alpha_3} \
\end{array} 
\right) 
= B \cdot D \cdot
\left( 
\begin{array}{c} 
u^1 \
\frac { \partial u^1 } { \partial \alpha_1} \
\frac { \partial u^1 } { \partial \alpha_2} \
\frac { \partial u^1 } { \partial \alpha_3} \
u^2 \
\frac { \partial u^2 } { \partial \alpha_1} \
\frac { \partial u^2 } { \partial \alpha_2} \
\frac { \partial u^2 } { \partial \alpha_3} \
u^3 \
\frac { \partial u^3 } { \partial \alpha_1} \
\frac { \partial u^3 } { \partial \alpha_2} \
\frac { \partial u^3 } { \partial \alpha_3} \
\end{array} 
\right) 
$
End of explanation
P=zeros(12,12)
P[0,0]=H[0]
P[1,0]=dH[0,0]
P[1,1]=H[0]
P[2,0]=dH[0,1]
P[2,2]=H[0]
P[3,0]=dH[0,2]
P[3,3]=H[0]
P[4,4]=H[1]
P[5,4]=dH[1,0]
P[5,5]=H[1]
P[6,4]=dH[1,1]
P[6,6]=H[1]
P[7,4]=dH[1,2]
P[7,7]=H[1]
P[8,8]=H[2]
P[9,8]=dH[2,0]
P[9,9]=H[2]
P[10,8]=dH[2,1]
P[10,10]=H[2]
P[11,8]=dH[2,2]
P[11,11]=H[2]
P=simplify(P)
P
B_P = zeros(9,9)
for i in range(3):
    for j in range(3):
        
        row_index = i*3+j
        
        B_P[row_index, row_index] = 1/(H[i]*H[j])
        
Grad_U_P = simplify(B_P*B*P)
Grad_U_P
Explanation: Physical coordinates
$u_i=u_{[i]} H_i$
End of explanation
E=zeros(6,9)
E[0,0]=1
E[1,4]=1
E[2,8]=1
E[3,1]=1
E[3,3]=1
E[4,2]=1
E[4,6]=1
E[5,5]=1
E[5,7]=1
E
StrainL=simplify(E*Grad_U_P)
StrainL
def E_NonLinear(grad_u):
    N = 3
    du = zeros(N, N)
    #    print("===Deformations===")
    for i in range(N):
        for j in range(N):
            index = i*N+j
            du[j,i] = grad_u[index]
    #    print("========")
    
    I = eye(3)
    a_values = S(1)/S(2) * du * G_up
    E_NL = zeros(6,9)
    E_NL[0,0] = a_values[0,0]
    E_NL[0,3] = a_values[0,1]
    E_NL[0,6] = a_values[0,2]
    E_NL[1,1] = a_values[1,0]
    E_NL[1,4] = a_values[1,1]
    E_NL[1,7] = a_values[1,2]
    E_NL[2,2] = a_values[2,0]
    E_NL[2,5] = a_values[2,1]
    E_NL[2,8] = a_values[2,2]
    E_NL[3,1] = 2*a_values[0,0]
    E_NL[3,4] = 2*a_values[0,1]
    E_NL[3,7] = 2*a_values[0,2]
    E_NL[4,0] = 2*a_values[2,0]
    E_NL[4,3] = 2*a_values[2,1]
    E_NL[4,6] = 2*a_values[2,2]
    E_NL[5,2] = 2*a_values[1,0]
    E_NL[5,5] = 2*a_values[1,1]
    E_NL[5,8] = 2*a_values[1,2]
    return E_NL
%aimport geom_util
u=getUHat3DPlane(alpha1, alpha2, alpha3)
# u=getUHatU3Main(alpha1, alpha2, alpha3)
gradu=B*u
E_NL = E_NonLinear(gradu)*B
E_NL
%aimport geom_util
u=getUHatU3MainPlane(alpha1, alpha2, alpha3)
gradup=Grad_U_P*u
# e=E*gradup
# e
E_NLp = E_NonLinear(gradup)*gradup
simplify(E_NLp)
w
Explanation: Strain tensor
$ 
\left( 
\begin{array}{c} 
\varepsilon_{11} \ 
\varepsilon_{22} \ 
\varepsilon_{33} \ 
2\varepsilon_{12} \ 
2\varepsilon_{13} \ 
2\varepsilon_{23} \ 
\end{array} 
\right)
=
\left(E + E_{NL} \left( \nabla \vec{u} \right) \right) \cdot 
\left( 
\begin{array}{c} 
\nabla_1 u_1 \ \nabla_2 u_1 \ \nabla_3 u_1 \
\nabla_1 u_2 \ \nabla_2 u_2 \ \nabla_3 u_2 \
\nabla_1 u_3 \ \nabla_2 u_3 \ \nabla_3 u_3 \
\end{array} 
\right)$
End of explanation
%aimport geom_util
C_tensor = getIsotropicStiffnessTensor()
C = convertStiffnessTensorToMatrix(C_tensor)
C
StrainL.T*C*StrainL*H1
Explanation: Virtual work
End of explanation
T=zeros(12,6)
T[0,0]=1
T[0,2]=alpha3
T[1,1]=1
T[1,3]=alpha3
T[3,2]=1
T[8,4]=1
T[9,5]=1
T
D_p_T = StrainL*T
simplify(D_p_T)
u = Function("u")
t = Function("theta")
w = Function("w")
u1=u(alpha1)+alpha3*t(alpha1)
u3=w(alpha1)
gu = zeros(12,1) 
gu[0] = u1
gu[1] = u1.diff(alpha1)
gu[3] = u1.diff(alpha3)
gu[8] = u3
gu[9] = u3.diff(alpha1)
gradup=Grad_U_P*gu
# E_NLp = E_NonLinear(gradup)*gradup
# simplify(E_NLp)
# gradup=Grad_U_P*gu
# o20=(K*u(alpha1)-w(alpha1).diff(alpha1)+t(alpha1))/2
# o21=K*t(alpha1)
# O=1/2*o20*o20+alpha3*o20*o21-alpha3*K/2*o20*o20
# O=expand(O)
# O=collect(O,alpha3)
# simplify(O)
StrainNL = E_NonLinear(gradup)*gradup
StrainL*gu+simplify(StrainNL)
Explanation: Tymoshenko theory
$u_1 \left( \alpha_1, \alpha_2, \alpha_3 \right)=u\left( \alpha_1 \right)+\alpha_3\gamma \left( \alpha_1 \right) $
$u_2 \left( \alpha_1, \alpha_2, \alpha_3 \right)=0 $
$u_3 \left( \alpha_1, \alpha_2, \alpha_3 \right)=w\left( \alpha_1 \right) $
$ \left( 
\begin{array}{c} 
u_1 \
\frac { \partial u_1 } { \partial \alpha_1} \
\frac { \partial u_1 } { \partial \alpha_2} \
\frac { \partial u_1 } { \partial \alpha_3} \
u_2 \
\frac { \partial u_2 } { \partial \alpha_1} \
\frac { \partial u_2 } { \partial \alpha_2} \
\frac { \partial u_2 } { \partial \alpha_3} \
u_3 \
\frac { \partial u_3 } { \partial \alpha_1} \
\frac { \partial u_3 } { \partial \alpha_2} \
\frac { \partial u_3 } { \partial \alpha_3} \
\end{array} 
\right) = T \cdot 
\left( 
\begin{array}{c} 
u \
\frac { \partial u } { \partial \alpha_1} \
\gamma \
\frac { \partial \gamma } { \partial \alpha_1} \
w \
\frac { \partial w } { \partial \alpha_1} \
\end{array} 
\right) $
End of explanation
L=zeros(12,12)
h=Symbol('h')
p0=1/2-alpha3/h
p1=1/2+alpha3/h
p2=1-(2*alpha3/h)**2
L[0,0]=p0
L[0,2]=p1
L[0,4]=p2
L[1,1]=p0
L[1,3]=p1
L[1,5]=p2
L[3,0]=p0.diff(alpha3)
L[3,2]=p1.diff(alpha3)
L[3,4]=p2.diff(alpha3)
L[8,6]=p0
L[8,8]=p1
L[8,10]=p2
L[9,7]=p0
L[9,9]=p1
L[9,11]=p2
L[11,6]=p0.diff(alpha3)
L[11,8]=p1.diff(alpha3)
L[11,10]=p2.diff(alpha3)
L
D_p_L = StrainL*L
simplify(D_p_L)
h = 0.5
exp=(0.5-alpha3/h)*(1-(2*alpha3/h)**2)#/(1+alpha3*0.8)
p02=integrate(exp, (alpha3, -h/2, h/2))
integral = expand(simplify(p02))
integral
Explanation: Square theory
$u^1 \left( \alpha_1, \alpha_2, \alpha_3 \right)=u_{10}\left( \alpha_1 \right)p_0\left( \alpha_3 \right)+u_{11}\left( \alpha_1 \right)p_1\left( \alpha_3 \right)+u_{12}\left( \alpha_1 \right)p_2\left( \alpha_3 \right) $
$u^2 \left( \alpha_1, \alpha_2, \alpha_3 \right)=0 $
$u^3 \left( \alpha_1, \alpha_2, \alpha_3 \right)=u_{30}\left( \alpha_1 \right)p_0\left( \alpha_3 \right)+u_{31}\left( \alpha_1 \right)p_1\left( \alpha_3 \right)+u_{32}\left( \alpha_1 \right)p_2\left( \alpha_3 \right) $
$ \left( 
\begin{array}{c} 
u^1 \
\frac { \partial u^1 } { \partial \alpha_1} \
\frac { \partial u^1 } { \partial \alpha_2} \
\frac { \partial u^1 } { \partial \alpha_3} \
u^2 \
\frac { \partial u^2 } { \partial \alpha_1} \
\frac { \partial u^2 } { \partial \alpha_2} \
\frac { \partial u^2 } { \partial \alpha_3} \
u^3 \
\frac { \partial u^3 } { \partial \alpha_1} \
\frac { \partial u^3 } { \partial \alpha_2} \
\frac { \partial u^3 } { \partial \alpha_3} \
\end{array} 
\right) = L \cdot 
\left( 
\begin{array}{c} 
u_{10} \
\frac { \partial u_{10} } { \partial \alpha_1} \
u_{11} \
\frac { \partial u_{11} } { \partial \alpha_1} \
u_{12} \
\frac { \partial u_{12} } { \partial \alpha_1} \
u_{30} \
\frac { \partial u_{30} } { \partial \alpha_1} \
u_{31} \
\frac { \partial u_{31} } { \partial \alpha_1} \
u_{32} \
\frac { \partial u_{32} } { \partial \alpha_1} \
\end{array} 
\right) $
End of explanation
rho=Symbol('rho')
B_h=zeros(3,12)
B_h[0,0]=1
B_h[1,4]=1
B_h[2,8]=1
M=simplify(rho*P.T*B_h.T*G_up*B_h*P)
M
M_p = L.T*M*L
integrate(M_p, (alpha3, -h/2, h/2))
Explanation: Mass matrix
End of explanation
<END_TASK>