Finite Element Analysis. Barna Szabó

any perturbation of u by will increase . Therefore is minimum at and hence

      (1.39)

      Therefore we have

(1.40)

      where the last two terms are zero because . Integrating the first term by parts,

and, substituting this into eq. (1.40), we get

      (1.41)

      Since this holds for all , the bracketed expression must be zero. In other words, the solution of the differential equation

      (1.42)

minimizes the potential energy defined by eq. (1.38). This is the strong form of the problem.

Remark 1.2 The procedure in Example 1.2 is used in the calculus of variations for identifying the differential equation, known as the Euler⁹‐Lagrange¹⁰ equation, the solution of which maximizes or minimizes a functional. In this example the solution minimizes the potential energy on the space .

      Remark 1.3 Whereas the strain energy is always positive, the potential energy may be positive, negative or zero.

1.3 Approximate solutions

      The trial and test spaces defined in the preceding section are infinite‐dimensional, that is, they span infinitely many linearly independent functions. To find an approximate solution, we construct finite‐dimensional subspaces denoted, respectively, by , and seek the function that satisfies for all . Let us return to the introductory example described in Section 1.1 and define

where ϕi () are basis functions. Using the definitions of and given in eq. (1.12), we write the bilinear form as

(1.43)

      Similarly,

      (1.44)

      where ri is defined in eq. (1.12). Therefore we can write

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