The question is basically about when to apply the variational operator...

Given the general functional representing the strain energy of a solid under a given stress state $\sigma$ and strain state $\varepsilon$, both representing vectors.

$$U=\frac12\int_{V}\sigma^T\varepsilon dV$$

The stress is related to the strain with:

$$\sigma=C \varepsilon$$

where C is a symmetric matrix with constant terms (for linear elastic materials).

Suppose I want the variation of $U$ and the functions that I am varying are the strain field functions. Which one of the following two possibilities should I do:

1-)$$\delta U=\frac12\int_V \sigma^T\delta\varepsilon dV$$

2-)$$\delta U=\frac12 \left( \int_V \delta \varepsilon^T C \varepsilon dV + \int_V \varepsilon^T C \delta\varepsilon dV\right)=\int_V \varepsilon^T C \delta \varepsilon dV=\int_V \sigma^T \delta \varepsilon dV$$

Another case is when we approximate the displacement field by a set of approximation functions (finite element, for example). Then it comes that:


Where $B$ is a matrix containing all the strain-displacement equations and the approximation functions (may be a function of $c$ for non-linear analysis). And $c$ is a vector containing all the amplitudes of each term in the approximation function (Ritz constants). The new energy functional becomes:

$$U=\frac12 \left( \int_V\sigma^T BdV \right) c$$

$c$ is the unkkown of the problem and the variational now should be applied to $c$, such that the following two options may be used:

1-)$$\delta U=\frac12 \left( \int_V\sigma^T BdV \right) \delta c +\frac12 \left( \int_V\sigma^T \delta BdV \right)c$$

2-)$$\delta U=\frac12 c^T \left( \int_V B^T C BdV \right) \delta c + \frac12 \delta c^T \left( \int_V B^T C BdV \right) c + \frac12 c^T \left( \int_V B^T C \delta BdV \right)c + \frac12 c^T \left( \int_V \delta B^T C BdV \right)c$$ $$=c^T \left( \int_V B^T C BdV \right) \delta c + c^T \left( \int_V B^T C \delta BdV \right)c$$ $$=\left( \int_V \sigma^T BdV \right) \delta c + \left( \int_V \sigma^T \delta BdV \right)c$$

In both cases the difference is the $\cfrac12$...

Which option (if any!) is the right one? This may be a basic question, but it remains not totally clear to me...

  • $\begingroup$ if $\sigma=C\epsilon$, then $U$ is a functional of the vector $\epsilon$, am I right? If yes, then you have to consider the variation of $U$ under small perturbations of epsilon, i.e. $U[\epsilon+\eta\rho]$, with $\eta$ small parameter etc...does it make sense? $\endgroup$
    – Avitus
    Commented Nov 15, 2013 at 10:33

1 Answer 1


I got the point here... based on the fact that the variation operator works similarly as the total differential operator with the independent variables constants, it comes that the chain rule should be applied to $\delta \sigma$.

Option 2-) was correct and the following shows how option 1-) should be corrected to render the same result:

  • directly applied to the strain vector: $$\delta U=\frac12\int_V \sigma^T\delta\varepsilon dV + \frac12\int_V \delta \sigma^T\varepsilon dV$$

  • applied to the approximated strain field (corrected option 1-)): $$\delta U=\frac12 \left( \int_V\sigma^T BdV \right) \delta c +\frac12 \left( \int_V\sigma^T \delta BdV \right)c +\frac12 \left( \int_V \delta \sigma^T BdV \right)c$$


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