# The second variation of displacement interpolation function in Finite Element Method

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.

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

Thus,

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

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

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

And

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

• Can you define what you mean by the "variation" of $u$? Do you mean the derivative? – Wolfgang Bangerth Apr 21 '16 at 16:21

• @Joe: I think you have it backwards. $\delta N_a$ should be constant, not $\delta u_a$. – Paul Apr 19 '16 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.