Problem 1
Solving problem 6.17 in the textbook: V 1
¥ 4 Given the set of points:
x -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
f(x) -3.632 -0.3935 1 0.6487 1.282 -4.518 -8.611 -12.82 15.91 -15.88 9.402 9.017
0 Writing a Matlab mction that calculates the
Lecture 1: Numerical IntegrationThe Trapezoidal and Simpsons Rule
A problem
The probability of a normally distributed (mean and standard deviation ) event
occurring between the values
a and b is
B
P ( a x b) =
where A =
2
1
e t 2 dt
2
A
(2)
a
b-
and B =
Lecture 2: Numerical integrationGauss Quadrature
f(x)
h
x1
x2
x3
For the general case shown above
The area for a single strip using the Trapezoidal rule is
x2
h
h
f ( x) dx 2 f ( x ) + 2 f ( x )
1
2
x1
The area for the pair of strips via Simpsons is
x3
h
Two-dimensional Numerical Integration over a rectangle
We would like to develop numerical schemes for evaluation integrals of the
form
d f ( x , y) dy dx
ac
b
Where the plan form a x b, and c y d is a rectangle.
This problem is asking us to find the volu
Ce 3101 Due Wed April 4 BEFORE 11.55pm
0.2
S(t)
Following the use of appropriate dimensionless parameters the infiltration of a moisture front s (t ) into
a particular heterogeneous layered soil under a constant ponded dimensionless height h0 = 0.2 is
giv
CE3101 Homework 4: Due Friday March 9 11:55pm
The solution of a diffusion transport equation (see second of semester) results in the
following set of equations in 9 (nine) unknowns x = ( x1; x2 ;.; x9 )
Ax = b
(1)
Where the only non-zero elements of A and
3101 Homework 1 Due on Tuesday 7th (no grade reduction if a handed in by 5pm on Feb 8th)
The sediment unit flux (volume per length per time) at a point x in the long profile of the Earth surface
can be described by an equation of the form
x
q ( x, ) = K (
Quiz 4/19 CE: 3101, Name: =
Using 3 nodes (numbered 1, 2 and 3) and a spacing of = 0.5
Numerically solve the following BVP
d 2u
= 2, u (0) = 1,
dx 2
du
= 1
dx x =1
Using the notes can write as
U1= 1
U1-2U2+U3= (.5X.5)X2 = .5
U2-U3=2X0.125+1*.5 = .75
Or
-2
Quiz5
Write the 3rd order IVP
d 3 y d 2 y dy
d2y
2+
= sin( y ), 2
dx
dx 3 dx
dx
=
x=0
dy
= y (0) = 1
dx x
As a first order system
Solution:
Let S=dy/dx and C = dS/dx
Then equation can be written as
(dC/dx)-C+S=sin(y)
And the following first order system
In class Quiz, Feb 23
Name:Solution
The two cubics
y = x3
And
y = 5 x3 6 x 2 + 2
Meet at the point x = 1, y = 1
Why do these two cubics, NOT represent cubic splines meeting at the Knot point x = 1, y = 1 ?
Cubics meet at point x=1
y = x3 = 1 : y = 5x3 6x