1. Finite Element Method

2. History

3. Application

4. Consider the two point boundary value problem

5. Procedure in Solving the Problem Numerically 1. Obtain the Variational Formulation where V = {v:v continuous on [0,1], v’ piecewise continuous, bounded on [0,1], v(0)=v(1)=0}

6. Procedure in Solving the Problem Numerically 2. Discretize the variational formulation. This means that for a chosen M € N, we subdivide [0,1] into M +1 subintervals each of length h = 1/(M+1) and get the formulation (V M ): where V M is the span of the set of hat functions {Φ 1, Φ 2,…, Φ M }

7. Procedure in Solving the Problem Numerically 3. From the discrete variational formulation obtained previously, obtain the matrix equation and

8. Procedure in Solving the Problem Numerically 4. Solve the matrix equation A = b. If then the approximate solution

9. Steps Variational Formulation Uniqueness of Solution Hat Functions Discretization of the Variational Formulation Existence of A -1 Convergence of the Approximate Solution uM to the Exact Solution u

10. Variational Formulation Suppose u is a solution of (D). Then Take any.

11. Variational Formulation Integrating the left hand side, we get

12. Variational Formulation The given boundary conditions lead to

13. Variational Formulation Since v is an arbitrary element of V, we conclude that any solution u of (D) is also a solution of v.

14. Variational Formulation

15. The equation can be written as

16. Variational Formulation Let us prove the reverse. Suppose that u is a solution of (V). Then

17. Variational Formulation So,

18. Variational Formulation is continuous and bounded in the open interval (0,1)

19. Variational Formulation Since

20. Variational Formulation If are continuous in (0,1), then is also continuous in (0,1). So, u is also a solution of (D).

21. Uniqueness of Solution If are two solutions of (V), then for any,

22. Uniqueness of Solution Subtracting the two equations, we get

23. Uniqueness of Solution Since it is true for any it is true for So,

24. Uniqueness of Solution So. Moreover, in (0,1), where f is continuous on [0,1].

25. Uniqueness of Solution But So

26. The Hat Functions Consider the interval [0,1]. For a chosen, we subdivide [0,1] into M +1 subintervals. Choose the subintervals to be of length

27. The Hat Functions Including the end points 0 and 1, we consider the node points where

28. The Hat Functions For j = 1,…,M, we define the hat function to be linear in the intervals and with but for.

29. The Hat Functions The hat function is also defined to be zero outside the open interval

30. The Subspace of ss Define the subset of to be the collection of all functions in such that is linear on each subinterval

31. The Subspace of ss Consider the nodes Let

32. The Subspace of ss So, any function is uniquely determined by its values at the nodes Similarly, any is a unique linear combination of the hat functions

33. The Subspace of ss Consider the hat functions Recall the span of HM to be the set of all possible linear combinations of hat functions in HM.

34. The Subspace of ss But is also contained in the vector space So

35. Discretization of the Variational Formulation To solve the variational problem numerically is to solve its discretized form: Now, we have shown earlier that for some vector

36. Discretization of the Variational Formulation The equation holds if is the hat function so for we have Then

37. Discretization of the Variational Formulation which can be written as This yields a system of M linear equations with M unknowns The are precisely the values of at the nodes. The system is as follows:

38. Discretization of the Variational Formulation which can also be written as In matrix form, we write

39. Discretization of the Variational Formulation The stiffness matrix A has entries and the load vector b has components

40. The Existence of A -1 Note that A is a symmetric matrix since To show that A is nonsingular, we will show that A is positive definite. In other words, we will show that for every nonzero vector in

41. The Existence of A -1 Let where the zero vector in It is possible for to have some components that are zero but not all. Then,

42. The Existence of A -1

44. Thus for any nonzero vector we have to be strictly positive to prove that A is positive definite. So we proceed further by noticing that some component – of is nonzero. So We have shown that A is positive definite, hence A is nonsingular.

45. Convergence of the approximate solution to the exact solution Theorem: If is an approximate solution of then for every we have where is the minimum value of over the whole closed interval [0,1].

46. Convergence of the approximate solution to the exact solution Note that exists since is continuous on [0,1] so that the Extreme-Value Theorem applies. So as M grows bigger, we can expect the error to shrink to zero.

47. Example Problem Consider the following problem: