## 3 Adaptive Subinterval Finite Element Method Based on Dynamic Sensitivity Analysis
### Dynamic-Sensitivity-Based Subinterval Finite Element Analysis Method  
This section proposes an enhanced subinterval method based on dynamic sensitivity analysis in order to improve computational efficiency. A two-dimensional problem is used as an example in order to illustrate the method. The intervals \(a_{1}^{I}\) and \(a_{2}^{I}\) are divided into subintervals according to the same method. For ease of expression, the number of subintervals for both parameters is taken to be \(m\left(m=2^{k},k=0,1,2,\cdots\right)\), which can be determined using the adaptive method proposed later. As shown in Figure 2, the traditional subinterval method requires solving the equilibrium equations for all subintervals. However, in the enhanced subinterval method, only a portion of the subintervals needs to be solved. For a \(d\)-dimensional displacement vector, where each element represents a degree of freedom of the structure, these elements may take on their maximum or minimum values in different combinations of subintervals. Therefore, the interval \(u_{r}^{I}\) of the \(d\)-th (\(r=1,2,\cdots,d\)) component of the displacement vector \(\mathbf{u}^{I}\) is calculated separately in the following. Ultimately, they form the displacement response interval for the entire structure. First, the finite element equilibrium equation is written for the initial subinterval \(\mathbf{a}_{0}^{I}\) as follows:  
\[\mathbf{K}\left(\mathbf{a}_{0}^{I}\right)\mathbf{u}_{0}^{I}=\mathbf{f}\left( \mathbf{a}_{0}^{I}\right) \tag{40}\]  
where \(\mathbf{a}_{0}^{C}=\mathbf{a}^{C}\), \(\mathbf{a}_{0}^{I}=\left[\mathbf{a}_{0}^{C}-\mathbf{a}^{R}/m\,\mathbf{a}_{0}^{C}+\mathbf{a}^{R}/m\ \right]\) represents a subinterval domain and \(\mathbf{u}_{0}^{I}\) represents an interval vector of the displacement response in the initial subinterval domain \(\mathbf{a}_{0}^{I}\). Then, using Equations (28)-(30) to solve the Equation (40), we obtain the vectors \(\mathbf{\Xi}_{0,i}\) and \(\mathbf{\overline{\varepsilon}}_{0,i}\) as follows:  
\[\mathbf{\Xi}_{0,i}=-sign\Big{(}\Delta\mathbf{u}_{i}\Big{(}\mathbf{a}_{0}^{I}\Big{)} \Big{)},\mathbf{\overline{\varepsilon}}_{0,i}=sign\Big{(}\Delta\mathbf{u}_{i} \Big{(}\mathbf{a}_{0}^{I}\Big{)}\Big{)},(i=1,2,\cdots,n) \tag{41}\]  
According to Equation (31), the upper and lower bounds of the displacement interval vectors \(\mathbf{u}_{0}^{I}\) in the subinterval domain \(\mathbf{a}_{0}^{I}\) can be obtained. \[ \begin{split}\underline{\mathbf{u}}_{0}&=\mathbf{u} ^{C}\Big{(}\mathbf{a}_{0}^{I}\Big{)}-\sum_{i=1}^{n}\Big{|}\Delta\mathbf{u}_{i} \Big{(}\mathbf{a}_{0}^{I}\Big{)}\Big{|}+\ \sum_{i=1}^{n}\sum_{j=1}^{n}\Delta\mathbf{u}_{i,j} \Big{(}\mathbf{a}_{0}^{I}\Big{)}\odot\mathbf{\Xi}_{0,i}\odot\mathbf{\Xi}_{0,j},\\ \overline{\mathbf{u}}_{0}&=\mathbf{u}^{C}\Big{(} \mathbf{a}_{0}^{I}\Big{)}+\sum_{i=1}^{n}\Big{|}\Delta\mathbf{u}_{i}\Big{(} \mathbf{a}_{0}^{I}\Big{)}\Big{|}+\ \sum_{i=1}^{n}\sum_{j=1}^{n}\Delta\mathbf{u}_{i,j} \Big{(}\mathbf{a}_{0}^{I}\Big{)}\odot\mathbf{\overline{\varepsilon}}_{0,j}\odot \mathbf{\overline{\varepsilon}}_{0,j}\end{split}  \tag{42}\]  
Then, the displacement interval of the \(r\)-th degree of freedom within the subinterval domain \(\mathbf{a}_{0}^{I}\) can be expressed as follows:  
\[u_{r,0}^{I}=[\underline{\mathbf{u}}_{0}(r),\overline{\mathbf{u}}_{0}(r)] \tag{43}\]  
where the subscript \(r\) of \(u_{r,0}^{I}\) denotes the \(r\)-th degree of freedom of the structure, the subscript \(0\) denotes the initial subinterval on the expansion route and \(\underline{\mathbf{u}}_{0}(r)\) and \(\overline{\mathbf{u}}_{0}(r)\) denote the \(r\)-th element of the vectors \(\underline{\mathbf{u}}_{0}\) and \(\overline{\mathbf{u}}_{0}\), respectively. To avoid solving for all the intervals, a natural idea is to find an expansion route such that the bounds of \(u_{r}^{I}\) can be predicted by solving the equilibrium equations on it. To predict the upper bound of the displacement response \(u_{r}^{I}\), the subinterval should move in the direction that makes it increase.