Finite Element Analysis. Barna Szabó

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Finite Element Analysis - Barna Szabó

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satisfies equation (1.5) is called the strong solution. The generalized formulation has the following important properties:

      1 The exact solution, denoted by , exists for all data that satisfy the conditions where α and β are real numbers, and f is such that satisfies the definitive properties of linear forms listed in Section A.1.2 for all . Note that κ, c and f can be discontinuous functions.

      2 The exact solution is unique in the energy space, see Theorem 1.1.

      3 If the data are sufficiently smooth for the strong solution to exist then the strong and weak solutions are the same.

      4 This formulation makes it possible to find approximations to with arbitrary accuracy. This will be addressed in detail in subsequent sections.

      Exercise 1.2 Assume that

and
are given. State the generalized formulation.

converges to u left-parenthesis x right-parenthesis equals x plus b in the space upper E left-parenthesis upper I right-parenthesis as n right-arrow infinity. For the definition of convergence refer to Section A.2 in the appendix.

      This exercise illustrates that restriction imposed on u prime (or higher derivatives of u) at the boundaries will not impose a restriction on upper E left-parenthesis upper I right-parenthesis. Therefore natural boundary conditions cannot be enforced by restriction. Whereas all functions in upper E left-parenthesis upper I right-parenthesis are continuous and bounded, the derivatives do not have to be continuous or bounded.

Geometric representation of exercise 1.3: The function un(x).
.

      (1.35)integral Subscript 0 Superscript script l Baseline f v d x less-than infinity for all v element-of upper E left-parenthesis upper I right-parenthesis period

      1.2.2 The principle of minimum potential energy

      (1.36)pi left-parenthesis u right-parenthesis equals Overscript def Endscripts one half upper B left-parenthesis u comma u right-parenthesis minus upper F left-parenthesis u right-parenthesis

      on the space ModifyingAbove upper E With tilde left-parenthesis upper I right-parenthesis.

      Proof: For any v element-of upper E Superscript 0 Baseline left-parenthesis upper I right-parenthesis, double-vertical-bar v double-vertical-bar Subscript upper E Baseline not-equals 0 we have:

      (1.37)StartLayout 1st Row 1st Column pi left-parenthesis u plus v right-parenthesis equals 2nd Column one half upper B 
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