Finite Element Analysis. Barna Szabó

      The first term on the right is , the second term is zero on account of Theorem 1.3, the third term is positive for any in . Therefore is minimum.

      Theorem 1.4 states that the error depends on the exact solution of the problem and the definition of the trial space .

      The finite element method is a flexible and powerful method for constructing trial spaces. The basic algorithmic structure of the finite element method is outlined in the following sections.

1.3.1 The standard polynomial space

      The standard polynomial space of degree p, denoted by , is spanned by the monomials defined on the standard element

      (1.49)

      The choice of basis functions is guided by considerations of implementation, keeping the condition number of the coefficient matrices small, and personal preferences. For the symmetric positive‐definite matrices considered here the condition number C is the largest eigenvalue divided by the smallest. The number of digits lost in solving a linear problem is roughly equal to . Characterizing the condition number as being large or small should be understood in this context. In the finite element method the condition number depends on the choice of the basis functions and the mesh.

The standard polynomial basis functions, called shape functions, can be defined in various ways. We will consider shape functions based on Lagrange polynomials and Legendre¹² polynomials. We will use the same notation for both types of shape function.

      Lagrange shape functions

      Lagrange shape functions of degree p are constructed by partitioning into p sub‐intervals. The length of the sub‐intervals is typically but the lengths may vary. The node points are , and . The ith shape function is unity in the ith node point and is zero in the other node points:

      (1.50)

      These shape functions have the following important properties:

      (1.51)

For example, for the equally spaced node points are , , . The corresponding Lagrange shape functions are illustrated in Fig. 1.3.

      Exercise 1.5 Sketch the Lagrange shape functions for .

      Legendre shape functions

      For we have

      (1.52)

      For we define the shape functions as follows:

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