Nonlinear finite elements/Rate form of hyperelastic laws

Rate equations

Let us now derive rate equations for a hyperelastic material.

First elasticity tensor

We start off with the relation

$𝑷={\rho }_0\frac{\partial W}{\partial 𝑭}$

Then the material time derivative of $𝑷$ is given by

First elasticity tensor $\dot{𝑷}=𝖠:\dot{𝑭};𝖠:={\rho }_0\frac{{\partial }^{2}W}{\partial 𝑭\partial 𝑭}$

where the fourth order tensor $𝖠$ is call the first elasticity tensor. This tensor has major symmetries but not minor symmetries. In index notation with respect to an orthonormal basis

${\dot{P}}_{iJ}={𝖠}_{iJkL}{\dot{F}}_{kL}={\rho }_0\frac{{\partial }^{2}W}{\partial F_{iJ}\partial F_{kL}}{\dot{F}}_{kL}$

Proof:

We have

$\dot{𝑷}={\rho }_0\frac{\partial }{\partial t}\left(\frac{\partial W}{\partial 𝑭}\right)={\rho }_0\frac{\partial }{\partial 𝑭}\left(\frac{\partial W}{\partial t}\right)={\rho }_0\frac{\partial }{\partial 𝑭}(\frac{\partial W}{\partial 𝑭}:\dot{𝑭})$

Using the product rule, we have

$\dot{𝑷}={\rho }_0\frac{{\partial }^{2}W}{\partial 𝑭\partial 𝑭}:\dot{𝑭}+{\rho }_0\frac{\partial W}{\partial 𝑭}:\frac{{\partial }^2𝑭}{\partial 𝑭\partial t}={\rho }_0\frac{{\partial }^{2}W}{\partial 𝑭\partial 𝑭}:\dot{𝑭}+{\rho }_0\frac{\partial W}{\partial 𝑭}:\frac{\partial }{\partial t}\left(\frac{\partial 𝑭}{\partial 𝑭}\right)$

Therefore,

$\dot{𝑷}={\rho }_0\frac{{\partial }^{2}W}{\partial 𝑭\partial 𝑭}:\dot{𝑭}=𝖠:\dot{𝑭}◻$

Second elasticity tensor

Similarly, if we start off with the relation

$𝑺=2{\rho }_0\frac{\partial W}{\partial 𝑪}$

the material time derivative of $𝑺$ can be expressed as

Second elasticity tensor $\dot{𝑺}=𝖢:\frac{1}{2}\dot{𝑪};𝖢=4{\rho }_0\frac{{\partial }^{2}W}{\partial 𝑪\partial 𝑪}$

where the fourth order tensor $𝖢$ is called the material elasticity tensor or the second elasticity tensor. Since this tensor relates symmetric second order tensors it has minor symmetries. It also has major symmetries because the two partial derivatives are with the same quantity and an interchange does not change things. In index notation with respect to an orthonormal basis

${\dot{S}}_{IJ}=\frac{1}{2}{𝖢}_{IJKL}{\dot{C}}_{KL}=\frac{1}{2}\left(4{\rho }_0\frac{{\partial }^{2}W}{\partial C_{IJ}\partial C_{KL}}\right){\dot{C}}_{KL}$

Proof:

We have

$\dot{𝑺}=2{\rho }_0\frac{\partial }{\partial t}\left(\frac{\partial W}{\partial 𝑪}\right)=2{\rho }_0\frac{\partial }{\partial 𝑪}\left(\frac{\partial W}{\partial t}\right)=2{\rho }_0\frac{\partial }{\partial 𝑪}(\frac{\partial W}{\partial 𝑪}:\dot{𝑪})$

Again using the product rule, we have

$\dot{𝑺}=2{\rho }_0\frac{{\partial }^{2}W}{\partial 𝑪\partial 𝑪}:\dot{𝑪}+2{\rho }_0\frac{\partial W}{\partial 𝑪}:\frac{{\partial }^2𝑪}{\partial 𝑪\partial t}=2{\rho }_0\frac{{\partial }^{2}W}{\partial 𝑪\partial 𝑪}:\dot{𝑪}$

Therefore,

$\dot{𝑺}=4{\rho }_0\frac{{\partial }^{2}W}{\partial 𝑪\partial 𝑪}:\frac{1}{2}\dot{𝑪}=𝖢:\frac{1}{2}\dot{𝑪}◻$

Relation between first and second elasticity tensors The first and second elasticity tensors are related by ${𝖠}_{iJkL}={\delta }_{ik}{S}_{JL}+F_{iN}{𝖢}_{NJML}{F}_{kM}$

Proof:

Recall that the first and second Piola-Kirchhoff stresses are related by

$𝑷=𝑭\cdot 𝑺$

Taking the material time derivative of both sides gives

$\dot{𝑷}=\dot{𝑭}\cdot 𝑺+𝑭\cdot \dot{𝑺}$

Using the expression for $\dot{𝑺}$ above, we get

$\dot{𝑷}=\dot{𝑭}\cdot 𝑺+\frac{1}{2}𝑭\cdot (𝖢:\dot{𝑪})$

Now

$𝑪={𝑭}^T\cdot 𝑭⟹\dot{𝑪}=\dot{{𝑭}^T}\cdot 𝑭+{𝑭}^T\cdot \dot{𝑭}$

Therefore,

$\dot{𝑷}=\dot{𝑭}\cdot 𝑺+\frac{1}{2}𝑭\cdot [𝖢:(\dot{{𝑭}^T}\cdot 𝑭+{𝑭}^T\cdot \dot{𝑭})]$

Now

$𝖢:(\dot{{𝑭}^T}\cdot 𝑭)\equiv {𝖢}_{IJKL}{\dot{F}}_{mK}{F}_{mL}\text{and}𝖢:({𝑭}^T\cdot \dot{𝑭})\equiv ={𝖢}_{IJLK}{F}_{mL}{\dot{F}}_{mK}={𝖢}_{IJKL}{F}_{mL}{\dot{F}}_{mK}$

That means

$𝖢:(\dot{{𝑭}^T}\cdot 𝑭)=𝖢:({𝑭}^T\cdot \dot{𝑭})$

which gives us

$\dot{𝑷}=\dot{𝑭}\cdot 𝑺+𝑭\cdot [𝖢:({𝑭}^T\cdot \dot{𝑭})]$

In index notation,

$\begin{array}{cc}{\dot{P}}_{iJ}& ={\dot{F}}_{iL}{S}_{LJ}+F_{iN}[{𝖢}_{NJML}{F}_{kM}{\dot{F}}_{kL}]\\ & =S_{LJ}{\delta }_{ik}{\dot{F}}_{kL}+{𝖢}_{NJML}{F}_{iN}{F}_{kM}{\dot{F}}_{kL}\\ & =({\delta }_{ik}{S}_{JL}+F_{iN}{𝖢}_{NJML}{F}_{kM}){\dot{F}}_{kL}& ={𝖠}_{iJkL}{\dot{F}}_{kL}\end{array}$

Therefore,

${𝖠}_{iJkL}={\delta }_{ik}{S}_{JL}+F_{iN}{𝖢}_{NJML}{F}_{kM}◻$

Fourth elasticity tensor

Now we will compute the spatial elasticity tensor for the rate constitutive equation for a hyperelastic material. This tensor relates an objective rate of stress (Cauchy or Kirchhoff) to the rate of deformation tensor. We can show that

Fourth elasticity tensor for the Kirchhoff stress ${ℒ}_{\phi }[\mathit{\tau }]=𝖼:𝒅\equiv {𝖼}_{\tau }:𝒅$ where ${𝖼}_{ijkl}=F_{iK}{F}_{jL}{F}_{kM}{F}_{lN}{𝖢}_{KLMN}=4{\rho }_0{F}_{iK}{F}_{jL}{F}_{kM}{F}_{lN}\frac{{\partial }^{2}W}{\partial C_{KL}\partial C_{MN}}.$

The fourth order tensor $𝖼$ is called the spatial elasticity tensor or the fourth elasticity tensor. Clearly, ${𝖼}_{ijkl}$ cannot be derived from the store energy function $W$ because of the dependence on the deformation gradient.

Proof:

Recall that the Lie derivative of the Kirchhoff stress is defined as

${ℒ}_{\phi }[\mathit{\tau }]=\stackrel{\circ }{\mathit{\tau }}=𝑭\cdot \dot{𝑺}\cdot {𝑭}^T$

We have found that

$\dot{𝑺}=𝖢:\frac{1}{2}\dot{𝑪}$

We also know from Continuum_mechanics/Time_derivatives_and_rates#Time_derivative_of_strain that

$\dot{𝑪}=2{𝑭}^T\cdot 𝒅\cdot 𝑭$

where $𝒅$ is the spatial rate of deformation tensor. Therefore,

${ℒ}_{\phi }[\mathit{\tau }]=𝑭\cdot [𝖢:({𝑭}^T\cdot 𝒅\cdot 𝑭)]\cdot {𝑭}^T$

In index notation,

$({ℒ}_{\phi }[\mathit{\tau }]{)}_{ij}=F_{iK}{𝖢}_{KLMN}{F}_{kM}{d}_{kl}{F}_{lN}{F}_{jL}=F_{iK}{F}_{jL}{F}_{kM}{F}_{lN}{𝖢}_{KLMN}{d}_{kl}={𝖼}_{ijkl}{d}_{kl}$

or,

${ℒ}_{\phi }[\mathit{\tau }]=𝖼:𝒅$

where

${𝖼}_{ijkl}=F_{iK}{F}_{jL}{F}_{kM}{F}_{lN}{𝖢}_{KLMN}◻$

Alternatively, we may define $𝖼$ in terms of the Cauchy stress $\mathit{\sigma }$, in which case the constitutive relation is written as

Fourth elasticity tensor for the Cauchy stress ${ℒ}_{\phi }[\mathit{\sigma }]=𝖼:𝒅=J^{-1}{𝖼}_{\tau }:𝒅$ where ${𝖼}_{ijkl}=J^{-1}{F}_{iK}{F}_{jL}{F}_{kM}{F}_{lN}{𝖢}_{KLMN}$

The proof of this relation between the spatial and material elasticity tensors is very similar to that for the rate of Kirchhoff stress. Many authors define this quantity ${𝖼}_{ijkl}$ as the spatial elasticity tensor. Note the factor of $J^{-1}$. This form of the spatial elasticity tensor is crucial for some of the calculations that follow.

Relation between first and fourth elasticity tensors The first and fourth elasticity tensors are related by ${𝖠}_{iJkL}=F_{Jj}^{-1}[{𝖼}_{ijkl}+{\tau }_{jl}{\delta }_{ik}]F_{Ll}^{-1}$ In the above equation ${𝖼}_{ijkl}$ is the elasticity tensor that relates the rate of Kirchhoff stress to the rate of deformation. Instead, if we use the Cauchy stress and the spatial elasticity tensor ${𝖼}_{ijkl}$ that relates the Cauchy stress to the rate of deformation), the above relation becomes ${𝖠}_{iJkL}=J{F}_{Jj}^{-1}[{𝖼}_{ijkl}+{\sigma }_{jl}{\delta }_{ik}]F_{Ll}^{-1}$

Proof:

Recall that

$\dot{𝑷}=𝖠:\dot{𝑭}$

Therefore,

$\dot{𝑷}\cdot {𝑭}^T=(𝖠:\dot{𝑭})\cdot {𝑭}^T$

Also recall that

${𝖠}_{iJkL}={\delta }_{ik}{S}_{JL}+F_{iN}{𝖢}_{NJML}{F}_{kM}$

Therefore, using index notation,

${\dot{P}}_{iJ}{F}_{lJ}={\delta }_{ik}{S}_{JL}{F}_{lJ}{\dot{F}}_{kL}+F_{iN}{𝖢}_{NJML}{F}_{kM}{F}_{lJ}){\dot{F}}_{kL}$

Now,

$𝒍=\dot{𝑭}\cdot {𝑭}^{-1}\text{or}\dot{𝑭}=𝒍\cdot 𝑭$

In index notation

${\dot{F}}_{kL}=l_{km}{F}_{mL}$

Using this we get

${\dot{P}}_{iJ}{F}_{lJ}={\delta }_{ik}{S}_{JL}{F}_{lJ}{l}_{km}{F}_{mL}+F_{iN}{𝖢}_{NJML}{F}_{kM}{F}_{lJ})l_{km}{F}_{mL}$

or,

${\dot{P}}_{iJ}{F}_{lJ}=({\delta }_{ik}{F}_{lJ}{S}_{JL}{F}_{mL}+F_{iN}{F}_{lJ}{F}_{kM}{F}_{mL}{𝖢}_{NJML})l_{km}$

Now,

$\mathit{\tau }=𝑭\cdot 𝑺\cdot {𝑭}^T\text{or}{\tau }_{lm}=F_{lJ}{S}_{JL}{F}_{mL}$

Therefore,

${\dot{P}}_{iJ}{F}_{lJ}=({\delta }_{ik}{\tau }_{lm}+F_{iN}{F}_{lJ}{F}_{kM}{F}_{mL}{𝖢}_{NJML})l_{km}$

Also,

${𝖼}_{ilkm}=F_{iN}{F}_{lJ}{F}_{kM}{F}_{mL}{𝖢}_{NJML}$

So we have

${\dot{P}}_{iJ}{F}_{lJ}=({\delta }_{ik}{\tau }_{lm}+{𝖼}_{ilkm})l_{km}={𝖺}_{ilkm}{l}_{km}$

Note:

The fourth order tensor

${𝖺}_{ijkl}={\delta }_{ik}{\tau }_{jl}+{𝖼}_{ijkl}$

which depends on the symmetry of $\mathit{\tau }$ is called the third elasticity tensor, i.e.,

$\dot{𝑷}\cdot {𝑭}^T=𝖺:𝒍$

Therefore, the relation between the first and third elasticity tensors is

$\dot{𝑷}\cdot {𝑭}^T=(𝖠:\dot{𝑭})\cdot {𝑭}^T=𝖺:𝒍=𝖺:(\dot{𝑭}\cdot {𝑭}^{-1})$

or,

$𝖠:\dot{𝑭}=[𝖺:(\dot{𝑭}\cdot {𝑭}^{-1})]\cdot {𝑭}^{-T}$

In index notation

${𝖠}_{iJkL}{\dot{F}}_{kL}={𝖺}_{ijkl}{\dot{F}}_{kL}{F}_{Ll}^{-1}{F}_{jJ}^{-1}$

Therefore,

${𝖠}_{iJkL}={𝖺}_{ijkl}{F}_{Ll}^{-1}{F}_{jJ}^{-1}=({\delta }_{ik}{\tau }_{jl}+{𝖼}_{ijkl})F_{Ll}^{-1}{F}_{jJ}^{-1}◻$

An isotropic spatial elasticity tensor?

An isotropic spatial elasticity tensor cannot be derived from a stored energy function if the constitutive relation is of the form

${ℒ}_{\phi }[\mathit{\tau }]=𝖼:𝒅$

where

$𝖼=\lambda 1\otimes 1+2\mu {𝖨}^s.$

Since a significant number of finite element codes use such a constitutive equation, (also called the equation of a hypoelastic material of grade 0) it is worth examining why such a model is incompatible with elasticity.

Start with a constant and isotropic material elasticity tensor

Let us start of with an isotropic elastic material model in the reference configuration. The simplest such model is the St. Venant-Kirchhoff hyperelastic model

$𝑺=\lambda \text{tr}(𝑬)1+2\mu 𝑬$

where $𝑺$ is the second Piola-Kirchhoff stress, $𝑬$ is the Lagrangian Green strain, and $\lambda ,\mu$ are material constants. We can show that this equation can be derived from a stored energy function.

Taking the material time derivative of this equation, we get

$\dot{𝑺}=\lambda \text{tr}(\dot{𝑬})1+2\mu \dot{𝑬}$

Now,

$\dot{𝑺}=𝖢:\frac{1}{2}\dot{𝑪}=𝖢:\dot{𝑬}$

where $𝖢$ is the second (material) elasticity tensor.

Therefore,

$𝖢:\dot{𝑬}=\lambda \text{tr}(\dot{𝑬})1+2\mu \dot{𝑬}$

which implies that

$𝖢=\lambda 1\otimes 1+2\mu {𝖨}^s.$

In index notation,

${𝖢}_{IJKL}=\lambda {\delta }_{IJ}{\delta }_{KL}+\mu ({\delta }_{IK}{\delta }_{JL}+{\delta }_{JK}{\delta }_{IL})$

Now, from the relations between the second elasticity tensor and the fourth (spatial) elasticity tensor, we have

${𝖼}_{ijkl}=F_{iI}{F}_{jJ}{F}_{kK}{F}_{lL}{𝖢}_{IJKL}$

Therefore, in this case,

${𝖼}_{ijkl}=F_{iI}{F}_{jJ}{F}_{kK}{F}_{lL}[\lambda {\delta }_{IJ}{\delta }_{KL}+\mu ({\delta }_{IK}{\delta }_{JL}+{\delta }_{JK}{\delta }_{IL})]$

or,

${𝖼}_{ijkl}=\lambda b_{ij}{b}_{kl}+\mu (b_{ik}{b}_{jl}+b_{il}{b}_{jk})$

where $𝒃=𝑭\cdot {𝑭}^T$. So we see that the spatial elasticity tensor $𝖼$ cannot be a constant tensor unless $𝒃=1$.

Alternatively, if we define

${𝖼}_{ijkl}=J^{-1}{F}_{iI}{F}_{jJ}{F}_{kK}{F}_{lL}{𝖢}_{IJKL}$

we get

${𝖼}_{ijkl}=J^{-1}\lambda b_{ij}{b}_{kl}+J^{-1}\mu (b_{ik}{b}_{jl}+b_{il}{b}_{jk})$

Start with a constant and isotropic spatial elasticity tensor

Let us now look at the situation where we start off with a constant and isotropic spatial elasticity tensor, i.e.,

$𝖼=\lambda 1\otimes 1+2\mu {𝖨}^s$

In index notation,

${𝖼}_{ijkl}=\lambda {\delta }_{ij}{\delta }_{kl}+\mu ({\delta }_{ik}{\delta }_{jl}+{\delta }_{jk}{\delta }_{il})$

Since

${𝖼}_{ijkl}=F_{iI}{F}_{jJ}{F}_{kK}{F}_{lL}{𝖢}_{IJKL}$

multiplying both sides by $F_{Ai}^{-1}{F}_{Bj}^{-1}{F}_{Ck}^{-1}{F}_{Dl}^{-1}$ we have,

$F_{Ai}^{-1}{F}_{Bj}^{-1}{F}_{Ck}^{-1}{F}_{Dl}^{-1}{𝖼}_{ijkl}={𝖢}_{ABCD}$

Therefore, substituting in the expression for a constant and isotropic $𝖼$, we have

${𝖢}_{ABCD}=F_{Ai}^{-1}{F}_{Bj}^{-1}{F}_{Ck}^{-1}{F}_{Dl}^{-1}[\lambda {\delta }_{ij}{\delta }_{kl}+\mu ({\delta }_{ik}{\delta }_{jl}+{\delta }_{jk}{\delta }_{il})]$

or,

${𝖢}_{ABCD}=\lambda F_{Aj}^{-1}{F}_{Bj}^{-1}{F}_{Cl}^{-1}{F}_{Dl}^{-1}+\mu (F_{Ak}^{-1}{F}_{Bl}^{-1}{F}_{Ck}^{-1}{F}_{Dl}^{-1}+F_{Al}^{-1}{F}_{Bk}^{-1}{F}_{Ck}^{-1}{F}_{Dl}^{-1})$

or,

${𝖢}_{ABCD}=\lambda ({𝑭}^{-1}\cdot {𝑭}^{-T}{)}_{AB}({𝑭}^{-1}\cdot {𝑭}^{-T}{)}_{CD}+\mu ({𝑭}^{-1}\cdot {𝑭}^{-T}{)}_{AC}({𝑭}^{-1}\cdot {𝑭}^{-T}{)}_{BD}+{𝑭}^{-1}\cdot {𝑭}^{-T}{)}_{AD}({𝑭}^{-1}\cdot {𝑭}^{-T}{)}_{BC})$

Since $𝑪={𝑭}^T\cdot 𝑭$ which gives us ${𝑪}^{-1}={𝑭}^{-1}\cdot {𝑭}^{-T}$, we can write

${𝖢}_{ABCD}=\lambda {𝑪}_{AB}^{-1}{𝑪}_{CD}^{-1}+\mu ({𝑪}_{AC}^{-1}{𝑪}_{BD}^{-1}+{𝑪}_{AD}^{-1}{𝑪}_{BC}^{-1})$

Alternatively, if we define

${𝖼}_{ijkl}=J^{-1}{F}_{iI}{F}_{jJ}{F}_{kK}{F}_{lL}{𝖢}_{IJKL}$

we get

${𝖢}_{ABCD}=J{F}_{Ai}^{-1}{F}_{Bj}^{-1}{F}_{Ck}^{-1}{F}_{Dl}^{-1}{𝖼}_{ijkl}$

and therefore,

${𝖢}_{ABCD}=J\lambda {𝑪}_{AB}^{-1}{𝑪}_{CD}^{-1}+J\mu ({𝑪}_{AC}^{-1}{𝑪}_{BD}^{-1}+{𝑪}_{AD}^{-1}{𝑪}_{BC}^{-1})$

Hypoelastic material of grade 0

A hypoelastic material of grade zero is one for which the stress-strain relation in rate form can be expressed as

${ℒ}_{\phi }[\mathit{\sigma }]=𝖼:𝒅$

where $𝖼$ is constant. When the material is isotropic we have

$𝖼=\lambda 1\otimes 1+2\mu {𝖨}^s$

We want to show that hypoelastic material models of grade 0 cannot be derived from a stored energy function. To do that, recall that

${𝖢}_{IJKL}=4{\rho }_0\frac{{\partial }^{2}W}{\partial C_{IJ}\partial C_{KL}}$

and

$S_{IJ}=2{\rho }_0\frac{\partial W}{\partial C_{IJ}}$

For a material elasticity tensor $𝖢$ to be derivable from a stored energy function it has to satisfy the Bernstein integrability conditions. We have

$\frac{\partial S_{IJ}}{\partial C_{KL}}=2{\rho }_0\frac{{\partial }^{2}W}{\partial C_{IJ}\partial C_{KL}}=\frac{1}{2}{𝖢}_{IJKL}$

Also, due to the interchangeability of derivatives,

$\frac{\partial }{\partial C_{MN}}\left(\frac{\partial S_{IJ}}{\partial C_{KL}}\right)=\frac{\partial }{\partial C_{KL}}\left(\frac{\partial S_{IJ}}{\partial C_{MN}}\right)$

Therefore,

$\frac{\partial {𝖢}_{IJKL}}{\partial C_{MN}}=\frac{\partial {𝖢}_{IJMN}}{\partial C_{KL}}$

These integrability conditions have to be satisfied by any material elasticity tensor.

At this stage we will use the relation

${𝖢}_{ABCD}=J\lambda {𝑪}_{AB}^{-1}{𝑪}_{CD}^{-1}+J\mu ({𝑪}_{AC}^{-1}{𝑪}_{BD}^{-1}+{𝑪}_{AD}^{-1}{𝑪}_{BC}^{-1})$

If we plug this into the integrability condition we will see that

$(\lambda +\mu )C_{KL}^{-1}(C_{IM}^{-1}{C}_{JN}^{-1}+C_{IN}^{-1}{C}_{JM}^{-1})=(\lambda +\mu )C_{MN}^{-1}(C_{IK}^{-1}{C}_{JL}^{-1}+C_{IL}^{-1}{C}_{JK}^{-1})$

If we multiply both sides by $C_{MN}{C}_{IK}{C}_{JL}$ we are left with

$\lambda +\mu =0$

This is an unphysical situation and hence shows that a hypoelastic material of grade zero requires that $\lambda =-\mu$ for it to be derivable from a stored energy function.

Proof:

Let us simplify the notation by writing $𝑨:={𝑪}^{-1}$. Then,

${𝖢}_{IJKL}=J\lambda A_{IJ}{A}_{KL}+J\mu (A_{IK}{A}_{JL}+A_{JK}{A}_{IL})$

Then,

$\begin{array}{cc}\frac{\partial {𝖢}_{IJKL}}{\partial C_{MN}}=& \lambda \frac{\partial J}{\partial C_{MN}}{A}_{IJ}{A}_{KL}+\\ & J\lambda [\frac{\partial A_{IJ}}{\partial C_{MN}}{A}_{KL}+A_{IJ}\frac{\partial A_{KL}}{\partial C_{MN}}]+\\ & \mu \frac{\partial J}{\partial C_{MN}}(A_{IK}{A}_{JL}+A_{JK}{A}_{IL})+\\ & J\mu [\frac{\partial A_{IK}}{\partial C_{MN}}{A}_{JL}+A_{IK}\frac{\partial A_{JL}}{\partial C_{MN}}+\frac{\partial A_{JK}}{\partial C_{MN}}{A}_{IL}+A_{JK}\frac{\partial A_{IL}}{\partial C_{MN}}]\end{array}$

and

$\begin{array}{cc}\frac{\partial {𝖢}_{IJMN}}{\partial C_{KL}}=& \lambda \frac{\partial J}{\partial C_{KL}}{A}_{IJ}{A}_{MN}+\\ & J\lambda [\frac{\partial A_{IJ}}{\partial C_{KL}}{A}_{MN}+A_{IJ}\frac{\partial A_{MN}}{\partial C_{KL}}]+\\ & \mu \frac{\partial J}{\partial C_{KL}}(A_{IM}{A}_{JN}+A_{JM}{A}_{IN})+\\ & J\mu [\frac{\partial A_{IM}}{\partial C_{KL}}{A}_{JN}+A_{IM}\frac{\partial A_{JN}}{\partial C_{KL}}+\frac{\partial A_{JM}}{\partial C_{KL}}{A}_{IN}+A_{JM}\frac{\partial A_{IN}}{\partial C_{KL}}]\end{array}$

As this stage we use the identities (see Nonlinear finite elements/Kinematics#Some_useful_results for proofs)

$\frac{\partial J}{\partial C_{IJ}}=\frac{J}{2}{C}_{IJ}^{-1}=\frac{J}{2}{A}_{IJ}$

and

$\frac{\partial C_{IJ}^{-1}}{\partial C_{KL}}=-\frac{1}{2}(C_{IK}^{-1}{C}_{JL}^{-1}+C_{JK}^{-1}{C}_{IL}^{-1})\text{or}\frac{\partial A_{IJ}}{\partial C_{KL}}=-\frac{1}{2}(A_{IK}{A}_{JL}+A_{JK}{A}_{IL})$

Therefore we have

$\begin{array}{cc}\frac{\partial {𝖢}_{IJKL}}{\partial C_{MN}}=& \frac{J\lambda }{2}{A}_{MN}{A}_{IJ}{A}_{KL}-\\ & \frac{J\lambda }{2}[(A_{IM}{A}_{JN}+A_{JM}{A}_{IN})A_{KL}+(A_{KM}{A}_{LN}+A_{LM}{A}_{KN})A_{IJ}]+\\ & \frac{J\mu }{2}{A}_{MN}(A_{IK}{A}_{JL}+A_{JK}{A}_{IL})-\\ & \frac{J\mu }{2}[(A_{IM}{A}_{KN}+A_{KM}{A}_{IN})A_{JL}+(A_{JM}{A}_{LN}+A_{LM}{A}_{JN})A_{IK}+\\ & (A_{JM}{A}_{KN}+A_{KM}{A}_{JN})A_{IL}+(A_{IM}{A}_{LN}+A_{LM}{A}_{IN})A_{JK}]\end{array}$

and

$\begin{array}{cc}\frac{\partial {𝖢}_{IJMN}}{\partial C_{KL}}=& \frac{J\lambda }{2}{A}_{KL}{A}_{IJ}{A}_{MN}-\\ & \frac{J\lambda }{2}[(A_{IK}{A}_{JL}+A_{JK}{A}_{IL})A_{MN}+(A_{KM}{A}_{LN}+A_{LM}{A}_{KN})A_{IJ}]+\\ & \frac{J\mu }{2}{A}_{KL}(A_{IM}{A}_{JN}+A_{JM}{A}_{IN})-\\ & \frac{J\mu }{2}[(A_{IK}{A}_{LM}+A_{KM}{A}_{IL})A_{JN}+(A_{JK}{A}_{LN}+A_{JL}{A}_{KN})A_{IM}+\\ & (A_{JK}{A}_{LM}+A_{JL}{A}_{KM})A_{IN}+(A_{IK}{A}_{LN}+A_{IL}{A}_{KN})A_{JM}]\end{array}$

Equating the two, we see that the terms that cancel out are

$\frac{J\lambda }{2}{A}_{KL}{A}_{IJ}{A}_{MN};\frac{J\lambda }{2}(A_{KM}{A}_{LN}+A_{LM}{A}_{KN})A_{IJ};$

and

$\begin{array}{cc}\frac{J\mu }{2}& [(A_{IM}{A}_{KN}+A_{KM}{A}_{IN})A_{JL}+(A_{JM}{A}_{LN}+A_{LM}{A}_{JN})A_{IK}+\\ & (A_{JM}{A}_{KN}+A_{KM}{A}_{JN})A_{IL}+(A_{IM}{A}_{LN}+A_{LM}{A}_{IN})A_{JK}]\end{array}$

Therefore,

$\frac{\partial {𝖢}_{IJKL}}{\partial C_{MN}}=\frac{\partial {𝖢}_{IJMN}}{\partial C_{KL}}$

implies that

$\begin{array}{cc}-\frac{J\lambda }{2}& (A_{IM}{A}_{JN}+A_{JM}{A}_{IN})A_{KL}+\frac{J\mu }{2}{A}_{MN}(A_{IK}{A}_{JL}+A_{JK}{A}_{IL})=\\ & -\frac{J\lambda }{2}(A_{IK}{A}_{JL}+A_{JK}{A}_{IL})A_{MN}+\frac{J\mu }{2}{A}_{KL}(A_{IM}{A}_{JN}+A_{JM}{A}_{IN})\end{array}$

or,

$(\lambda +\mu )A_{MN}(A_{IK}{A}_{JL}+A_{JK}{A}_{IL})=(\lambda +\mu )A_{KL}(A_{IM}{A}_{JN}+A_{JM}{A}_{IN})$

In other words,

$(\lambda +\mu )C_{KL}^{-1}(C_{IM}^{-1}{C}_{JN}^{-1}+C_{IN}^{-1}{C}_{JM}^{-1})=(\lambda +\mu )C_{MN}^{-1}(C_{IK}^{-1}{C}_{JL}^{-1}+C_{IL}^{-1}{C}_{JK}^{-1})$

Now, if we multiply both sides by $C_{MN}$ we get

$(\lambda +\mu )C_{KL}^{-1}({\delta }_{IN}{C}_{JN}^{-1}+{\delta }_{IM}{C}_{JM}^{-1})=(\lambda +\mu ){\delta }_{MM}(C_{IK}^{-1}{C}_{JL}^{-1}+C_{IL}^{-1}{C}_{JK}^{-1})$

or,

$2(\lambda +\mu )C_{KL}^{-1}{C}_{JI}^{-1}=3(\lambda +\mu )(C_{IK}^{-1}{C}_{JL}^{-1}+C_{IL}^{-1}{C}_{JK}^{-1})$

Next, multiplying both sides by $C_{IK}$ gives

$2(\lambda +\mu ){\delta }_{IL}{C}_{JI}^{-1}=3(\lambda +\mu )({\delta }_{KK}{C}_{JL}^{-1}+{\delta }_{KL}{C}_{JK}^{-1})$

or,

$2(\lambda +\mu )C_{JL}^{-1}=12(\lambda +\mu )C_{JL}^{-1}$

Finally, multiplying both sides by $C_{JL}$ gives

$(\lambda +\mu ){\delta }_{JJ}=6(\lambda +\mu ){\delta }_{JJ}$

Therefore,

$\lambda +\mu =0$