# Nonlinear finite elements/Kinematics - spectral decomposition

## Spectral decompositions

Many numerical algorithms use spectral decompositions to compute material behavior.

### Spectral decompositions of stretch tensors

Infinitesimal line segments in the material and spatial configurations are related by

$d\mathbf {x} ={\boldsymbol {F}}\cdot d{\boldsymbol {X}}={\boldsymbol {R}}\cdot ({\boldsymbol {U}}\cdot d{\boldsymbol {X}})={\boldsymbol {V}}\cdot ({\boldsymbol {R}}\cdot d{\boldsymbol {X}})~.$

So the sequence of operations may be either considered as a stretch of in the material configuration followed by a rotation or a rotation followed by a stretch.

Also note that

${\boldsymbol {V}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\cdot {\boldsymbol {R}}^{T}~.$

Let the spectral decomposition of ${\boldsymbol {U}}$  be

${\boldsymbol {U}}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}$

and the spectral decomposition of ${\boldsymbol {V}}$  be

${\boldsymbol {V}}=\sum _{i=1}^{3}{\hat {\lambda }}_{i}~\mathbf {n} _{i}\otimes \mathbf {n} _{i}~.$

Then

${\boldsymbol {V}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\cdot {\boldsymbol {R}}^{T}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {R}}\cdot ({\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i})\cdot {\boldsymbol {R}}^{T}=\sum _{i=1}^{3}\lambda _{i}~({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})\otimes ({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})$

Therefore the uniqueness of the spectral decomposition implies that

$\lambda _{i}={\hat {\lambda }}_{i}\quad {\text{and}}\quad \mathbf {n} _{i}={\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i}$

The left stretch (${\boldsymbol {V}}$ ) is also called the spatial stretch tensor while the right stretch (${\boldsymbol {U}}$ ) is called the material stretch tensor.

### Spectral decompositions of deformation gradient

The deformation gradient is given by

${\boldsymbol {F}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}$

In terms of the spectral decomposition of ${\boldsymbol {U}}$  we have

${\boldsymbol {F}}=\sum _{i=1}^{3}\lambda _{i}~{\boldsymbol {R}}\cdot ({\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i})=\sum _{i=1}^{3}\lambda _{i}~({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})\otimes {\boldsymbol {N}}_{i}=\sum _{i=1}^{3}\lambda _{i}~\mathbf {n} _{i}\otimes {\boldsymbol {N}}_{i}$

Therefore the spectral decomposition of ${\boldsymbol {F}}$  can be written as

${\boldsymbol {F}}=\sum _{i=1}^{3}\lambda _{i}~\mathbf {n} _{i}\otimes {\boldsymbol {N}}_{i}$

Let us now see what effect the deformation gradient has when it is applied to the eigenvector ${\boldsymbol {N}}_{i}$ .

We have

${\boldsymbol {F}}\cdot {\boldsymbol {N}}_{i}={\boldsymbol {R}}\cdot {\boldsymbol {U}}\cdot {\boldsymbol {N}}_{i}={\boldsymbol {R}}\cdot \left(\sum _{j=1}^{3}\lambda _{j}~{\boldsymbol {N}}_{j}\otimes {\boldsymbol {N}}_{j}\right)\cdot {\boldsymbol {N}}_{i}$

From the definition of the dyadic product

$(\mathbf {u} \otimes \mathbf {v} )\cdot \mathbf {w} =(\mathbf {w} \cdot \mathbf {v} )~\mathbf {u}$

Since the eigenvectors are orthonormal, we have

$({\boldsymbol {N}}_{j}\otimes {\boldsymbol {N}}_{j})\cdot {\boldsymbol {N}}_{i}={\begin{cases}0&{\mbox{if}}~i\neq j\\{\boldsymbol {N}}_{i}&{\mbox{if}}~i=j\end{cases}}$

Therefore,

$\left(\sum _{j=1}^{3}\lambda _{j}~{\boldsymbol {N}}_{j}\otimes {\boldsymbol {N}}_{j}\right)\cdot {\boldsymbol {N}}_{i}=\lambda _{i}~{\boldsymbol {N}}_{i}{\text{no sum on}}~i$

${\boldsymbol {F}}\cdot {\boldsymbol {N}}_{i}=\lambda _{i}~({\boldsymbol {R}}\cdot {\boldsymbol {N}}_{i})=\lambda _{i}~\mathbf {n} _{i}$

So the effect of ${\boldsymbol {F}}$  on ${\boldsymbol {N}}_{i}$  is to stretch the vector by $\lambda _{i}$  and to rotate it to the new orientation $\mathbf {n} _{i}$ .

We can also show that

${\boldsymbol {F}}^{-T}\cdot {\boldsymbol {N}}_{i}={\cfrac {1}{\lambda _{i}}}~\mathbf {n} _{i}~;~~{\boldsymbol {F}}^{T}\cdot \mathbf {n} _{i}=\lambda _{i}~{\boldsymbol {N}}_{i}~;~~{\boldsymbol {F}}^{-1}\cdot \mathbf {n} _{i}={\cfrac {1}{\lambda _{i}}}~{\boldsymbol {N}}_{i}$

### Spectral decompositions of strains

Recall that the Lagrangian Green strain and its Eulerian counterpart are defined as

${\boldsymbol {E}}={\frac {1}{2}}~({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}-{\boldsymbol {\mathit {1}}})~;~~{\boldsymbol {e}}={\frac {1}{2}}~({\boldsymbol {\mathit {1}}}-\left({\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}\right)^{-1})$

Now,

${\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}={\boldsymbol {U}}\cdot {\boldsymbol {R}}^{T}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {U}}={\boldsymbol {U}}^{2}~;~~{\boldsymbol {F}}\cdot {\boldsymbol {F}}^{T}={\boldsymbol {V}}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {R}}^{T}\cdot {\boldsymbol {V}}={\boldsymbol {V}}^{2}$

Therefore we can write

${\boldsymbol {E}}={\frac {1}{2}}~({\boldsymbol {U}}^{2}-{\boldsymbol {\mathit {1}}})~;~~{\boldsymbol {e}}={\frac {1}{2}}~({\boldsymbol {\mathit {1}}}-{\boldsymbol {V}}^{-2})$

Hence the spectral decompositions of these strain tensors are

${\boldsymbol {E}}=\sum _{i=1}^{3}{\frac {1}{2}}(\lambda _{i}^{2}-1)~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}~;~~\mathbf {e} =\sum _{i=1}^{3}{\frac {1}{2}}\left(1-{\cfrac {1}{\lambda _{i}^{2}}}\right)~\mathbf {n} _{i}\otimes \mathbf {n} _{i}$

#### Generalized strain measures

We can generalize these strain measures by defining strains as

${\boldsymbol {E}}^{(n)}={\cfrac {1}{n}}~({\boldsymbol {U}}^{n}-{\boldsymbol {\mathit {1}}})~;~~{\boldsymbol {e}}^{(n)}={\cfrac {1}{n}}~({\boldsymbol {\mathit {1}}}-{\boldsymbol {V}}^{-n})$

The spectral decomposition is

${\boldsymbol {E}}^{(n)}=\sum _{i=1}^{3}{\cfrac {1}{n}}(\lambda _{i}^{n}-1)~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}~;~~\mathbf {e} ^{(n)}=\sum _{i=1}^{3}{\cfrac {1}{n}}\left(1-{\cfrac {1}{\lambda _{i}^{n}}}\right)~\mathbf {n} _{i}\otimes \mathbf {n} _{i}$

Clearly, the usual Green strains are obtained when $n=2$ .

#### Logarithmic strain measure

A strain measure that is commonly used is the logarithmic strain measure. This strain measure is obtained when we have $n\rightarrow 0$ . Thus

${\boldsymbol {E}}^{(0)}=\ln({\boldsymbol {U}})~;~~{\boldsymbol {e}}^{(0)}=\ln({\boldsymbol {V}})$

The spectral decomposition is

${\boldsymbol {E}}^{(0)}=\sum _{i=1}^{3}\ln \lambda _{i}~{\boldsymbol {N}}_{i}\otimes {\boldsymbol {N}}_{i}~;~~\mathbf {e} ^{(0)}=\sum _{i=1}^{3}\ln \lambda _{i}~\mathbf {n} _{i}\otimes \mathbf {n} _{i}$