I need to calculate the second variation of displacement interpolation function $u = \sum N_a u_a$ in Finite Element Analysis, where $N_a$ are the shape functions and $u_a$ are the nodal values. Someone told me the second variation is 0. The reason is as shown below. But I don't understand why $\delta u_a$ is a constant.

Could you guys please tell me if the second variation of displacement interpolation function really equals to zero? If it is, could you please explain to me why $\delta u_a$ equals constant? Actually, I think $\delta u$ is arbitrary, so I don't think the second variation of displacement is equal zero.

$$u = \sum_{a=1}^{n} N_a u_a$$


$$\delta u = \delta\sum_{a=1}^{n}N_a u_a = \sum_{a=1}^{n}N_a \delta u_a$$

As $\delta u$ is constant (This is what I don't understand). Thus

$$\delta^2 u_a = 0 \enspace .$$


$$\delta^2 u = \delta^2\sum_{a=1}^{n}N_a u_a = \sum_{a=1}^{n}N_a \delta^2 u_a = 0 \enspace .$$

  • $\begingroup$ Can you define what you mean by the "variation" of $u$? Do you mean the derivative? $\endgroup$ Apr 21, 2016 at 16:21

2 Answers 2


The variation operator applies to the functions, not the constants.

Variations work similarly to derivatives. If your interpolation functions are piecewise linear, the first variation becomes piecewise constant and the second variation becomes zero.

  • $\begingroup$ Hi Paul, Thank you so much for your response. In this variation problem, the variation variable is the nodal value rather than the variable in the shape function. So, we don't see derivatives of the shape function. so I don't think it matters with the order of shape functions or the interpolation functions. $\endgroup$
    – Joe
    Apr 19, 2016 at 17:12
  • $\begingroup$ @Joe: I think you have it backwards. $\delta N_a$ should be constant, not $\delta u_a$. $\endgroup$
    – Paul
    Apr 19, 2016 at 18:11

I am not sure I understand your question. This is how I understand second variation (skipping much of the details). Imagine you have a simple linear spring whose stored energy is $W = {1\over2} k \thinspace u^2$ for spring stiffness $k$ and displacement $u$. Then the first variation gives $\delta W = \delta u \thinspace k \thinspace u$, and the second variation gives $\text{d}\delta W = \delta u \thinspace k \thinspace \text{d}u.$

In this case, the second variation can be expressed as $\text{d}u = N_i \text{d}u_i$ depending on the chosen discretization and shape functions. Thus the quadratic nature of energy expression lends itself to computing second variation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.