Piecewise cubic interpolation:
x0
=a
x1
xN
=L
Piecewise cubic Hermite polynomials:
DOF: 4N 2(N 1) = 2(N + 1) prescribe f (xi ) and f (xi ) at i = 0, . . . , N .
Presentation of f in terms of Hermite b
Interpolation and approximation
0.1
0.1.1
Interpolation and approximation
Approximating functions
A geometric view of representing arbitrary functions in terms of some basis functions:
Vector Algebra:
2. For the Chebyshev density (x) =
1
1x2
the associated potential is:
(z) = log 
z
z2 1

2
From the limiting values (0) = log 21 and (1) = log 21 we observe that
2N
p(z) = eN N
0.2
throughout [1
Numerical Integration
Basic Idea: Integrate polynomial interpolants to approximate integrals.
fN
f1
f2
f0
h
x0 = a
x1
xN = b
x2
f (x) = pn (x) +
b
b
f (x) dx =
a
a
f (N +1) ()
(N + 1)!
N
N
j=0
(x xj )
Method 2: Lagrange interpolation of degree n.
Dene the following polynomial basis functions
k (xj )
= kj =
k (x)
each of degree n such that
0 k=j
1 k=j
(x x0 ) . . . (x xk1 )(x xk+1 ) . . . (x xn )
=
y0 x0
et
x
y
=
1
0
0 100
x = et x0
y = e100t y0
x
y
e100t
FE: If we were to use the FE method in the useful regime we would requre 2 < hk < 0
1 = 1 h < 2
2 = 100 h < 1/50
We do not particularly care
Truncation Error: The truncation error (T.E.) is the remainder you get when you substitute the
exact solution to Duxx + bu = f () into the dierence equation.
Th =
i.e.:
2
ui + Bi ui Fi = O(h2 ).
h2
A
Integrating Functions on Innite Intervals:
Eg.
I=
f (x) dx
0
If f (x) xp as x then
x1p

1p a
xp dx =
a
exists only if p > 1.
Truncate the Innite Interval:
c
f (x) dx +
I =
a
f (x) dx
c
= I1 + I2
Us
0.3.2 Integrating functions with singularities
1
I=
0
1
I=
0
dx
= 2x1/2
x1/2
1
=2
0
ex
dx
x2/3
We cannot just use the trapezoidal rule in this case as f0 . In
this case we use what are called open int
Numerical Solution of PDE
2. Introduction to PDE
2.1 Classication of PDE
1st order PDE
F (x, y, ux , uy ) = 0
Eg:
uux + ut = 1
shock waves in trac ow and uid mechanics
Solving 1st order PDE using the
Method 3.1: Special case of equally spaced
meshpoints:
Assume
ba
,
N
DIFFERENCE OPERATORS
xk = a + kh
where h =
k = 0, . . . , N
Forward:
fn = fn+1 fn
2 fn = fn+1 fn = fn+2 2fn+1 + fn
Backward:
2
fn =
Using the integral form of y = f (x, y(x)
xn+1
y(xn+1 ) = y(xn ) +
f (x, y(x)dx
xn
h
= y(xn ) + [f (xn , y (xn ) + f (xn+1 , y (xn+1 )] + O(h3 )
2
()
There are a number of dierent ways we can choose t