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(1.99) can be solved for π∞ to obtain an estimate for the exact value of the potential energy.

      The relative error in energy norm corresponding to the ith finite element solution in the sequence is estimated from

(1.100)

      Usually the percent relative error is reported. This estimator has been tested against the known exact solution of many problems of various smoothness. The results have shown that it works well for a wide range of problems, including most problems of practical interest; however, it cannot be guaranteed to work well for all conceivable problems. For example, this method would fail if the exact solution would happen to be energy‐orthogonal to all basis functions associated with (say) odd values of i.

      Remark 1.12 From equation (1.92) we get

(1.101)

On plotting vs. N on log‐log scale a straight line with the slope will be seen for sufficiently large N. The estimated value of β, corresponding to the ith solution in the sequence, is denoted by βi. It is computed from eq. (1.98):

      (1.102)

      Examples

      The properties of the finite element solution with reference to a family of model problems is discussed in the following. The problems are stated as follows: Find such that

(1.103)

      where κ and c are constants and is defined such that the exact solution is:

(1.104)

      As explained in Section 1.5.1, when α is not an integer, the case considered in the following, then this solution lies in the space . Therefore the asymptotic rate of h‐convergence on uniform meshes, predicted by eq. (1.92), is and the asymptotic rate of p‐convergence on a fixed mesh is .

      We selected this problem because it is representative of the singular part of the exact solutions of two‐and three‐dimensional elliptic boundary value problems.

      Referring to Theorem 1.3, we have for all therefore . Consequently for the kth element the load vector in the local numbering convention is:

(1.105)

      where by definition .

When then the first derivative of is infinity in the point . To avoid having in the integrand, the first term in eq. (1.105) is integrated by parts:

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