In numerical analysis, the interval finite element method (interval FEM) is a finite element method that uses interval parameters. Interval FEM can be applied in situations where it is not possible to get reliable probabilistic characteristics of the structure. This is important in concrete structures, wood structures, geomechanics, composite structures, biomechanics and in many other areas. The goal of the Interval Finite Element is to find upper and lower bounds of different characteristics of the model (e.g. stress, displacements, yield surface etc.) and use these results in the design process. This is so called worst case design, which is closely related to the limit state design. Worst case design requires less information than probabilistic design however the results are more conservative [Köylüoglu and Elishakoff 1998].
Applications of the interval parameters to the modeling of uncertainty Consider the following equation:
a x = b {\displaystyle ax=b}
where a and b are real numbers, and x = b a {\displaystyle x={\frac {b}{a}}} . Very often, the exact values of the parameters a and b are unknown. Let's assume that a ∈ [ 1 , 2 ] = a {\displaystyle a\in [1,2]=\mathbf {a} } and b ∈ [ 1 , 4 ] = b {\displaystyle b\in [1,4]=\mathbf {b} } . In this case, it is necessary to solve the following equation
[ 1 , 2 ] x = [ 1 , 4 ] {\displaystyle [1,2]x=[1,4]}
There are several definitions of the solution set of this equation with interval parameters.
United solution set In this approach the solution is the following set
x = { x : a x = b , a ∈ a , b ∈ b } = b a = [ 1 , 4 ] [ 1 , 2 ] = [ 0.5 , 4 ] {\displaystyle \mathbf {x} =\left\{x:ax=b,a\in \mathbf {a} ,b\in \mathbf {b} \right\}={\frac {\mathbf {b} }{\mathbf {a} }}={\frac {[1,4]}{[1,2]}}=[0.5,4]}
This is the most popular solution set of the interval equation and this solution set will be applied in this article. In the multidimensional case the united solutions set is much more complicated. The solution set of the following system of linear interval equations
[ [ − 4 , − 3 ] [ − 2 , 2 ] [ − 2 , 2 ] [ − 4 , − 3 ] ] [ x 1 x 2 ] = [ [ − 8 , 8 ] [ − 8 , 8 ] ] {\displaystyle {\begin{bmatrix}{[-4,-3]}&{[-2,2]}\\{[-2,2]}&{[-4,-3]}\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}={\begin{bmatrix}{[-8,8]}\\{[-8,8]}\end{bmatrix}}}
is shown on the following picture
∑ ∃ ∃ ( A , b ) = { x : A x = b , A ∈ A , b ∈ b } {\displaystyle \sum {_{\exists \exists }}(\mathbf {A} ,\mathbf {b} )=\{x:Ax=b,A\in \mathbf {A} ,b\in \mathbf {b} \}}
The exact solution set is very complicated, thus it is necessary to find the smallest interval which contains the exact solution set
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