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Finite Element Method History Application • Consider the two point boundary value problem 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} 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 (VM): where VM is the span of the set of hat functions {Φ1, Φ2,…, ΦM} Procedure in Solving the Problem Numerically 3. From the discrete variational formulation obtained previously, obtain the matrix equation and Procedure in Solving the Problem Numerically 4. Solve the matrix equation A = b. If then the approximate solution 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 Variational Formulation • Suppose u is a solution of (D). Then • Take any . Variational Formulation • Integrating the left hand side, we get Variational Formulation • The given boundary conditions lead to Variational Formulation • Since v is an arbitrary element of V, we conclude that any solution u of (D) is also a solution of v. Variational Formulation Variational Formulation • The equation can be written as Variational Formulation • Let us prove the reverse. Suppose that u is a solution of (V). Then Variational Formulation • So, Variational Formulation • is continuous and bounded in the open interval (0,1) Variational Formulation • Since Variational Formulation • If are continuous in (0,1), then is also continuous in (0,1). So, u is also a solution of (D). Uniqueness of Solution • If then for any are two solutions of (V), , Uniqueness of Solution • Subtracting the two equations, we get Uniqueness of Solution • Since it is true for any So, it is true for Uniqueness of Solution • So • Moreover, . in (0,1), where f is continuous on [0,1]. Uniqueness of Solution • But • So 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 The Hat Functions • Including the end points 0 and 1, we consider the node points where The Hat Functions • For j = 1,…,M, we define the hat function to be linear in the intervals and with but for . The Hat Functions • The hat function is also defined to be zero outside the open interval The Subspace of ss • Define the subset of to be the collection of all functions in such that is linear on each subinterval The Subspace • Consider the nodes Let of ss 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 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 . The Subspace • But space • So of ss is also contained in the vector 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 Discretization of the Variational Formulation • The equation holds if so for • Then is the hat function we have 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: Discretization of the Variational Formulation which can also be written as • In matrix form, we write Discretization of the Variational Formulation • The stiffness matrix A has entries and the load vector b has components 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 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, The Existence of A-1 The Existence of A-1 The Existence of A-1 • 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. 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]. 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. Example Problem • Consider the following problem: