# Compute spatial second derivatives in Isogeometric analysis

Motivation:

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, $$\frac{\partial u}{\partial \xi_j} = \frac{\partial u}{\partial x_i}\frac{\partial x_i}{\partial \xi_j}$$

or

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

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.

• 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. May 20, 2013 at 7:17
• @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. May 20, 2013 at 17:09