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degree elevation operation, and k-refinement combining by knot insertion and degree elevation. The k-refinement operation is performed by elevating the degree of basis function to a desired order firstly followed by knot insertion thus obtaining the maximum available continuity. Compared with finite element analysis, the main advantage of refinement operations in isogeometric analysis is that the geometry of the computational domain can be kept while the degree of freedom increases. Fig. 2 presents an example to compare h-refinement, p-refinement and k-refinement in isogeometric analysis. Note that the number of control points increases during the refinement operation, and more degree of freedom can be achieved by k-refinement.

III. ERROR ASSESSMENT METHOD BASED ON k-REFINEMENT

Suppose that $U(\mathbf{x})$ is the exact solution, and $U_h(\mathbf{x})$ is the approximation solution obtained by isogeometric method in Section II, then the discrete error $e$ can be written as

$e = U - U_h. \quad (3)$

After performing Laplacian operator $\Delta$ on each side of (3), a posteriori error assessment can be obtained by resolving the following problem,

$\begin{align} \Delta e &= f - \Delta U_h && \text{in } \Omega \\ e &= 0 && \text{on } \partial\Omega \tag{4} \end{align}$

From (4), the crucial point of a posteriori error estimation is the computation of $\Delta U_h(\mathbf{x})$. The following proposition is presented to show the computation of $\Delta U_h(\mathbf{x})$ directly on the parametric domain:

Proposition 1. Given B-spline parameterization $\sigma(\xi, \eta) = (x(\xi, \eta), y(\xi, \eta))$ of the computational domain and the solution field $U_h(\mathbf{x}) = U_h(x(\xi, \eta), y(\xi, \eta)) = T(\xi, \eta)$ over $\sigma(\xi, \eta)$, then $\Delta U_h(\mathbf{x})$ has the following form,

$\begin{align*} \Delta U_h &= \frac{\partial^2 U_h}{\partial^2 x} + \frac{\partial^2 U_h}{\partial^2 y} \\ &= \frac{J L_{\xi} T_{\eta\eta} - J L_{\eta} T_{\xi\xi} + L_{\xi\xi} T_{\eta} - L_{\eta\eta} T_{\xi}}{J K}, \end{align*}$

where

$\begin{align*} J &= x_\xi y_\eta - x_\eta y_\xi, & K &= (x_\xi y_\eta)^2 - (x_\eta y_\xi)^2, \\ L_\xi &= x_\xi^2 - y_\xi^2, & L_\eta &= x_\eta^2 - y_\eta^2, \\ L_{\xi\xi} &= (L_\eta y_{\xi\xi} - L_\xi y_{\eta\eta}) x_\xi - (L_\eta x_{\xi\xi} - L_\xi x_{\eta\eta}) y_\xi, \\ L_{\eta\eta} &= (L_\eta y_{\xi\xi} - L_\xi y_{\eta\eta}) x_\eta - (L_\eta x_{\xi\xi} - L_\xi x_{\eta\eta}) y_\eta. \end{align*}$

Proof. The idea of isogeometric analysis is to use the same mathematical representation for the computational domain and

solution field. Suppose that the computational domain $\Omega$ is parameterized by the following planar B-spline surface:

$\sigma(\xi, \eta) = (x(\xi, \eta), y(\xi, \eta)) = \sum_{i=1}^{n_1} \sum_{j=1}^{n_j} N_i^{d_i}(\xi) N_j^{d_j}(\eta) c_{ij},$

In isogeometric analysis, the solution field of heat conduction problem (1) over the computational domain $\Omega$ has the following form,

$U_h(x(\xi, \eta), y(\xi, \eta)) = T(\xi, \eta) = \sum_{i=1}^{n_1} \sum_{j=1}^{n_j} N_i^{d_i}(\xi) N_j^{d_j}(\eta) T_{ij},$

Here $T_{ij}$ are the unknown variables in isogeometric analysis to be solved from the boundary condition and Eq. (1).

From $U_h(x(\xi, \eta), y(\xi, \eta)) = T(\xi, \eta)$, we have

$\begin{align*} \frac{\partial T}{\partial \xi} &= \frac{\partial U_h}{\partial x} \frac{\partial x}{\partial \xi} + \frac{\partial U_h}{\partial y} \frac{\partial y}{\partial \xi}, \\ \frac{\partial T}{\partial \eta} &= \frac{\partial U_h}{\partial x} \frac{\partial x}{\partial \eta} + \frac{\partial U_h}{\partial y} \frac{\partial y}{\partial \eta}, \end{align*}$

Then we can obtain

$\begin{align*} \frac{\partial U_h}{\partial x} &= (\frac{\partial T}{\partial \xi} y_{\eta} - \frac{\partial T}{\partial \eta} y_{\xi}) / J, \\ \frac{\partial U_h}{\partial y} &= (\frac{\partial T}{\partial \eta} x_{\xi} - \frac{\partial T}{\partial \xi} x_{\eta}) / J, \end{align*}$

where $J = x_\xi y_\eta - x_\eta y_\xi$.

Similarly,

$\begin{align*} \frac{\partial^2 T}{\partial^2 \xi} &= \frac{\partial^2 U_h}{\partial^2 x} \left( \frac{\partial x}{\partial \xi} \right)^2 + \frac{\partial U_h}{\partial x} \frac{\partial^2 x}{\partial^2 \xi} + \frac{\partial^2 U_h}{\partial^2 y} \left( \frac{\partial y}{\partial \xi} \right)^2 + \frac{\partial U_h}{\partial y} \frac{\partial^2 y}{\partial^2 \xi}, \\ \frac{\partial^2 T}{\partial^2 \eta} &= \frac{\partial^2 U_h}{\partial^2 x} \left( \frac{\partial x}{\partial \eta} \right)^2 + \frac{\partial U_h}{\partial x} \frac{\partial^2 x}{\partial^2 \eta} + \frac{\partial^2 U_h}{\partial^2 y} \left( \frac{\partial y}{\partial \eta} \right)^2 + \frac{\partial U_h}{\partial y} \frac{\partial^2 y}{\partial^2 \eta}, \end{align*}$

From above two equations, we have

$\begin{align*} \frac{\partial^2 U_h}{\partial^2 x} &= [y_{\eta}^2 (T_{\xi\xi} - \frac{\partial U_h}{\partial x} x_{\xi\xi} - \frac{\partial U_h}{\partial y} y_{\xi\xi})] - \\ &y_{\xi}^2 (T_{\eta\eta} - \frac{\partial U_h}{\partial x} x_{\eta\eta} - \frac{\partial U_h}{\partial y} y_{\eta\eta})] / K, \\ \frac{\partial^2 U_h}{\partial^2 y} &= [x_{\xi}^2 (T_{\eta\xi} - \frac{\partial U_h}{\partial x} x_{\eta\xi} - \frac{\partial U_h}{\partial y} y_{\eta\xi})] - \\ &x_{\eta}^2 (T_{\xi\xi} - \frac{\partial U_h}{\partial x} x_{\xi\xi} - \frac{\partial U_h}{\partial y} y_{\xi\xi})] / K, \end{align*}$

where $K = (x_\xi y_\eta)^2 - (x_\eta y_\xi)^2$.

Hence, we can obtain