## 3 Adaptive Subinterval Finite Element Method Based on Dynamic Sensitivity Analysis
### Existing Subinterval Finite Element Method  
However, the solution accuracy of the improved interval finite element method is affected by the level of uncertainty. The higher the level of uncertainty, the lower the accuracy of the solution. The subinterval finite element method can be employed when solving the response of structures with large uncertainty. The uncertainty parameter \(a_{i}^{I}(i=1,2,...,n)\) is divided into \(m_{i}\) subintervals and can be expressed as follows:  
\[a_{i,i}^{I}=[\underline{a}_{i}+2(t-1)\Delta a_{i}/m_{i},\underline{a}_{i}+2t \Delta a_{i}/m_{i}],(t_{i}=1,2,...,m_{i}) \tag{32}\]  
where \(a_{i,i}^{I}\),is the \(t_{i}\)-th subinterval of the parameter \(a_{i}^{I}\). The number of subintervals for different parameters can be determined based on its level of uncertainty. Now, there are a total of \(m_{1}m_{2}...m_{n}\) subinterval combinations, and each combination can be expressed as follows:  
\[\mathbf{a}_{t_{1}\cdots t_{i}\cdots t_{n}}^{I}=\Big{(}a_{t_{1},1}^{I},...,a_{ t_{i},i}^{I},...,a_{n,n}^{I}\Big{)},(t_{i}=1,2,...,m_{i}) \tag{33}\]  
The level of uncertainty for each subinterval is as follows:  
\[\mu_{t_{i},i}=\frac{1}{m_{i}}\frac{a_{i}^{R}}{\underline{a}_{i}+(2t-1)a_{i}^{ R}/m_{i}}\approx\frac{1}{m_{i}}\frac{a_{i}^{R}}{a_{i}^{C}}=\frac{1}{m_{i}}\mu_{i} \tag{34}\]  
Dividing the interval into more subintervals results in lower uncertainty levels for each subinterval. According to interval error estimation theory [40], lower uncertainty levels lead to higher computational accuracy. Determining the upper and lower bounds of the response for a combination of \(m_{1}m_{2}...m_{n}\) subintervals is a non-deterministic polynomial (NP)-hard problem. The enumeration method is used to solve this problem. For the subinterval combination \(\mathbf{a}_{t_{1}t_{2}\cdots t_{n}}^{I}\), subinterval static equilibrium equations can be written as follows:  
\[\mathbf{K}\Big{(}\mathbf{a}_{t_{1}\cdots t_{i}\cdots t_{n}}^{I}\Big{)} \mathbf{u}\Big{(}\mathbf{a}_{t_{1}\cdots t_{i}\cdots t_{n}}^{I}\Big{)}=\mathbf{ f}\Big{(}\mathbf{a}_{t_{1}\cdots t_{i}\cdots t_{n}}^{I}\Big{)},(i=1,2, \cdots n;t_{i}=1,2,\cdots,m_{i}) \tag{35}\]Applying the interval finite element method proposed in this paper to each subinterval, the range of displacement responses for each subinterval can be found. For a combination of two adjacent subintervals, the following equation always holds:  
\[\mathbf{K}\left(\mathbf{a}_{i_{1}\ldots i_{l}\ldots t_{n}}^{I}\right)\cap\mathbf{ K}\left(\mathbf{a}_{i_{1}\ldots i_{l}+1\ldots t_{n}}^{I}\right)=\mathbf{K} \left(\mathbf{a}_{i_{1}}^{I}\ldots\overline{\mathbf{a}}_{i_{l}}=\underline{ \mathbf{a}}_{i_{l}+1}\ldots\mathbf{a}_{i_{n}}^{I}\right) \tag{36}\]  
\[\mathbf{f}\left(\mathbf{a}_{i_{1}\ldots i_{l}\ldots t_{n}}^{I}\right)\cap \mathbf{f}\left(\mathbf{a}_{i_{1}\ldots i_{l}+1\ldots t_{n}}^{I}\right)= \mathbf{f}\left(\mathbf{a}_{i_{1}}^{I}\ldots\overline{\mathbf{a}}_{i_{l}}= \underline{\mathbf{a}}_{i_{l}+1}\ldots\mathbf{a}_{i_{n}}^{I}\right) \tag{37}\]  
For two adjacent subintervals, both the stiffness matrix and load vector have overlapping parts. As a result, the intersection of the interval displacement solutions is not the empty set. \[\mathbf{u}\left(\mathbf{a}_{i_{1}\ldots i_{l}\ldots t_{n}}^{I}\right)\cap \mathbf{u}\left(\mathbf{a}_{i_{1}\ldots i_{l}+1\ldots t_{n}}^{I}\right)\
eq\varnothing \tag{38}\]  
The above equation shows that the displacement solutions of the subintervals can form a single connected region.