# TD 2 : Calcul tensoriel

## Exercice 1
Soit $\mathbf{u}$ un tenseur de rang un de composantes
$$
u_1 = x_1 + x_2 + x_3, \quad
u_2 = x_1x_2 + x_2x_3 + x_3\,x_1, \quad
u_3\ =\ x_1\,x_2\,x_3.
$$

Calculer $\nabla\otimes\mathbf{u}$, $\frac{\partial\mathbf{u}}{\partial \mathbf{x}}$, $\operatorname{grad_g}(\mathbf{u})$, $\operatorname{grad_d}(\mathbf{u})$, $\nabla\cdot\mathbf{u}$, $\operatorname{div}(\mathbf{u})$ et donner leurs analogues matriciels.

### Correction

$$\nabla\otimes \mathbf{u} = \operatorname{grad_g}(\mathbf{u}) = \frac{\partial u_j}{\partial x_{i}} \mathbf{e}_i\otimes \mathbf{e}_j.$$

$$[\nabla\otimes \mathbf{u}] = \left(
\begin{array}{ccc}
1 & x_2 + x_3 & x_2 x_3 \\
1 & x_1 + x_3 & x_1 x_3 \\
1 & x_1 +x_2 & x_1 x_2
\end{array}
\right)$$

$$\operatorname{grad_d}(\mathbf{u}) = \frac{\partial\mathbf{u}}{\partial\mathbf{x}} = \frac{\partial u_i}{\partial x_j} \mathbf{e}_i\otimes\mathbf{e}_j = (\nabla\otimes \mathbf{u})^T.$$

$$\nabla\cdot \mathbf{u} = \operatorname{div}(\mathbf{u}) = \frac{\partial u_n}{\partial x_n} = 1+x_1+x_3+x_1x_2.$$

## Exercice 2
Soit $\mathbb{T}$ un tenseur de rang deux s'identifiant à la matrice
$$
[\mathbb{T}]\ =\
\left[
\begin{array}{ccc}
x_1 & x_2\,x_3 & x_1\,x_2\,x_3 \\
x_2 & x_3\,x_1 & x_2\,x_3\,x_1 \\
x_3 & x_1\,x_2 & x_3\,x_1\,x_2
\end{array}
\right].
$$
Calculer $\nabla\cdot\mathbb{T}$, $\operatorname{div_g}(\mathbb{T})$, $\operatorname{div_d}(\mathbb{T})$ et donner leurs analogues matriciels.

### Correction

$$\nabla\cdot \mathbb{T} = \operatorname{div_g}(\mathbb{T}) = \frac{\partial T_{ij}}{\partial x_i}\mathbf{e}_j = T_{ij,i} \mathbf{e}_j$$

$$[\nabla\cdot \mathbb{T}] = \left(
\begin{array}{c}
3 \\
0 \\
x_2x_3 + x_1 x_3 + x_1x_2
\end{array}
\right)$$

$$\operatorname{div_d}(\mathbb{T}) = \frac{\partial T_{ij}}{\partial x_j} \mathbf{e}_i$$

$$[\operatorname{div_d}(\mathbb{T})] = \left(
\begin{array}{c}
1 + x_3 + x_1 x_2 \\
x_1 x_2 \\
x_1 + x_1 x_2
\end{array}
\right)$$

## Exercice 3
Soit $\mathbf{U}$ et $\mathbb{A}$ deux tenseurs de rangs respectifs un et deux.
Démontrer que
\begin{align}
\nabla\cdot(\mathbb{A}\cdot\mathbf{U})\ &=\ (\operatorname{div_g}\mathbb{A})\cdot\mathbf{U}\ +\ \mathbb{A}\boldsymbol{:}
\operatorname{grad_d}\mathbf{U}\ =\ (\operatorname{div_d}\mathbb{A}^T)\cdot\mathbf{U}\ +\ \mathbb{A}\boldsymbol{:}
(\operatorname{grad_g}\mathbf{U})^T.
\end{align}

### Correction

\begin{align}
\nabla\cdot(\mathbb{A}\cdot\mathbf{U}) &= \frac{\partial}{\partial x_i} (A_{ij}U_j) = A_{ij,i} U_j + A_{ij} U_{j,i} = (\operatorname{div_g}A)\cdot U + A_{ij} (\operatorname{grad_d}U)_{ji} \\
& = (\operatorname{div_g}A)\cdot U + \mathbf{A}\boldsymbol{:}\operatorname{grad_d}U \\
&= A^T_{ji,i} U_j + A_{ij} U_{j,i} = (\operatorname{div_d}(\mathbb{A}^T))\cdot \mathbf{U} + \mathbf{A}\boldsymbol{:}(\operatorname{grad_g}U)^T
\end{align}

## Exercice 4
Soit $\mathbf{U}$ un tenseur de rang un. Démontrer que
\begin{align*}
(\nabla\otimes\mathbf{U})\cdot\mathbf{U}\ &=\ \frac{1}{2}\,\nabla\otimes(\mathbf{U}\cdot\mathbf{U}), \\
\nabla\cdot(\mathbf{U}\otimes\mathbf{U})\ &=\ (\nabla\cdot\mathbf{U})\,\mathbf{U}\ +\ (\mathbf{U}\cdot\nabla)\,\mathbf{U}, \\
\mathbf{U}\cdot(\nabla\otimes\mathbf{U})\ &=\ (\mathbf{U}\cdot\nabla)\,\mathbf{U}\ =\ \frac{1}{2}\,\nabla(\mathbf{U}\cdot\mathbf{U})\
-\ \mathbf{U}\wedge(\nabla\wedge\mathbf{U}).
\end{align*}

### Correction
$$(\nabla\otimes\mathbf{U})\cdot\mathbf{U} = U_{j,i} U_j \mathbf{e}_i = \frac{\partial U_j}{\partial x_i}U_j \mathbf{e}_i = \frac{1}{2}\frac{\partial}{\partial x_i}(U_jU_j)\mathbf{e}_i = \frac{1}{2}\nabla\otimes(\mathbf{U}\cdot\mathbf{U})$$

$$\nabla\cdot(\mathbf{U}\otimes\mathbf{U}) = \frac{\partial}{\partial x_i}(U_i U_j)\mathbf{e}_j = U_{i,i} U_j\mathbf{e}_j + U_i U_{j,i}\mathbf{e}_j = (\nabla\cdot \mathbf{U})\mathbf{U} + (\mathbf{U}\cdot \nabla) \mathbf{U} $$

$$\mathbf{U}\cdot(\nabla \otimes \mathbf{U}) = U_iU_{j,i} \mathbf{e}_j = (\mathbf{U}\cdot\nabla)\mathbf{U}$$

$$\mathbf{a}\wedge(\mathbf{b}\wedge\mathbf{c}) = (\mathbf{a}\cdot\mathbf{c})\mathbf{b}-(\mathbf{a}\cdot\mathbf{b})\mathbf{c} = a_ib_jc_i\mathbf{e}_j - a_ib_ic_j\mathbf{e}_j $$

\begin{align}
\mathbf{U}\wedge(\nabla\wedge\mathbf{U}) &= U_iU_{i,j}\mathbf{e}_j - U_iU_{j,i}\mathbf{e}_j = \frac{1}{2}\nabla(\mathbf{U}\cdot\mathbf{U}) - (\mathbf{U}\cdot\nabla)\mathbf{U} \\
&= (\nabla\otimes\mathbf{U})\cdot \mathbf{U} - \mathbf{U}\cdot(\nabla\otimes\mathbf{U})
\end{align}