Transcript hat function

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: