In isogeometric analysis, state variables(e.g. displacement) are defined in the parametric domain, which can be mapped to the physical domain by $\boldsymbol{\xi}\mapsto \boldsymbol{x}$ as shown beneath. However the quantity related to displacment such as stress, strain are spacial derivatives of displacement. The following procedure is commonly used for solving those derivatives. $\blacksquare$

Let $u$ be one component of displacement vector $\boldsymbol{u}$

\begin{equation} u(\xi,\eta) = \sum_{i} c_iN^i(\xi,\eta) \end{equation} with geometric mapping from the parametric domain to the physical domain $$x(\xi,\eta) = \sum_{i} x_i N^i(\xi,\eta), \quad y(\xi,\eta) = \sum_{i} y_i N^i(\xi,\eta),$$ where $c_i,x_i,y_i$ are constants, with assumption that $(\xi,\eta)\mapsto(x,y)$ is bijective, i.e. inverse exists, $$J :=[\frac{\partial x_i}{\partial \xi_j}],\: |J| \neq 0\quad (\text{where }x_2 = y,\,\xi_2 = \eta).$$

By chain rule, $$\frac{\partial u}{\partial \xi_j} = \frac{\partial u}{\partial x_i}\frac{\partial x_i}{\partial \xi_j}$$


$$ \begin{bmatrix} \frac{\partial u}{\partial \xi}\\ \frac{\partial u}{\partial \eta} \end{bmatrix} = \begin{bmatrix} \frac{\partial x}{\partial \xi} & \frac{\partial y}{\partial \xi}\\ \frac{\partial x}{\partial \eta} & \frac{\partial y}{\partial \eta} \end{bmatrix} \begin{bmatrix} \frac{\partial u}{\partial x}\\ \frac{\partial u}{\partial y} \end{bmatrix} = J^T \begin{bmatrix} \frac{\partial u}{\partial x}\\ \frac{\partial u}{\partial y} \end{bmatrix}. $$

Hence, $$ \begin{bmatrix} \frac{\partial u}{\partial x}\\ \frac{\partial u}{\partial y} \end{bmatrix} = (J^T)^{-1} \begin{bmatrix} \frac{\partial u}{\partial \xi}\\ \frac{\partial u}{\partial \eta} \end{bmatrix} $$

This is the common procedure used in isogemetric analysis for computing stiffness matrix.

However, when working on sensitivity analysis, I need to compute the second derivatives $$\frac{\partial^2 u}{\partial x_i\partial x_j},\quad i \text{ and }j \in \{1,2\}.$$

One may think displacement $\boldsymbol{u} = [u, v]^T$.

Having searched on Scirus site, I failed to find any useful papers. But I believe it has to be done by some groups already. And reference would be appreciated. I have many accesses to scientific database, so only a link to the paper would be sufficient.

Thanks a lot.

A little bit explanation

I have encountered this problem when working on sensitivity analysis, especially adjoint method with the approach of material derivatives. Let assume we have an objective functional $$\phi = \int_{\Omega}f(\sigma,\epsilon,p)\,\mathrm{d}\Omega$$ where $p$ is a design parameter(e.g. coordinate of one control point). With linear elastostatics, $\sigma,\,\epsilon$ are functions of $\nabla \boldsymbol{u}$, so we can write $f = f(\nabla u)$.

And the material derivative of the domain functional is $$\dot{\phi} = \frac{\mathrm{d}}{\mathrm{d} p}\phi = \int_{\Omega}\frac{\partial}{\partial \nabla \boldsymbol{u}}f:\nabla\dot{\boldsymbol{u}}-\frac{\partial}{\partial \nabla \boldsymbol{u}}f:\nabla((\nabla\boldsymbol{u})\boldsymbol{v})\,\mathrm{d}\Omega + \int_{\partial\Omega}f(\boldsymbol{v}\cdot\boldsymbol{n})\,\mathrm{d}\Gamma. $$

The first term in the domain integral is taken care of by adjoint formulation. From the second term in the domain integral, the second derivative is required.


1 Answer 1


You're on the right track already. To calculate the first derivatives you wanted using the results you calculated, you used the chain rule.

To calculate the second derivatives you need using the information you have, you're essentially using the chain rule twice. Faà di Bruno's formula is a formula for the chain rule that you can use in this situation.

  • $\begingroup$ Thanks for the reply. Let me summarize the question: $u,x,y$ are given as functions of $(\xi,\eta)$ which leads to the difficult of computing $\partial^2 u/\partial x_i\partial x_j$, because $\xi(x,y),\:\eta(x,y)$ are not given. I doubt if the chain rule is directly applicable to this question. $\endgroup$
    – newbie
    Commented May 20, 2013 at 7:17
  • $\begingroup$ It is as Geoff says, a further use of the chain rule. We derive these rules in a paper we have submitted, and which I just uploaded to arxiv. I will post a link when it clears the submission process. For now: dl.dropboxusercontent.com/u/5711447/PetIGA.pdf $\endgroup$ Commented May 20, 2013 at 10:16
  • $\begingroup$ @NathanCollier I think it's because I'm too stupid to apply the rule then.XD Thanks for the paper. It seems to be really helpful. $\endgroup$
    – newbie
    Commented May 20, 2013 at 10:40
  • $\begingroup$ Well, albeit a further use of the chain rule, higher order spatial derivatives are not typically discussed in finite element books and absent in the isogeometric literature. We had to scratch our heads about it which is why we are publishing them. Glad the paper is helpful! $\endgroup$ Commented May 20, 2013 at 10:49
  • $\begingroup$ @NathanCollier I have added some background information. I was advised that it can be taken care of by converting into boundary integral, I doubt about it because it is multiplied with $\partial (f)/\partial \nabla\boldsymbol{u}$. Besides, I have also experienced that boundary integral sometimes delivers poor results. $\endgroup$
    – newbie
    Commented May 20, 2013 at 17:09

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