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ENGR 516 Computational Methods for Graduate Students
Catholic University of America
Assignment #7
Elliptic PDE
ENGR516 Assignment #5: Two Dimensional PDEs
Problem #1: Alternating Direction Implicit (ADI) Method
Research and write a brief explanation of the ADI Method and solve:
Two Dimensional LaPlace Equation: Electric Potential Over a Flat Plate with Point
Charge
!
!
!
∇ 2u(x, y) = f (x, y) for -1 ≤ x ≤ 1, -1 ≤ y ≤ 1
boundary conditions: u(x,y) = 0 for all boundaries
f (0.5, 0.5) = −1
f (−0.5, −0.5) = 1
elsewhere : f (x, y) = 0
Two Dimensional Temperature Diffusion:
⎛ ∂ 2u(x, y,t) ∂ 2u(x, y,t) ⎞ ∂u(x, y,t)
10 −4 ⎜
+
⎟⎠ =
∂x 2
∂y 2
∂t
⎝
!
!
!
for 0 ≤ x ≤ 4, 0 ≤ y ≤ 4 0 ≤ t ≤ 5000
u(x, y, 0) = 0
u(x, y,t) = e y cos x − e x cos y for x = 0, x = 4, y = 0, y = 4
!
!
Present results for t= 5000
Problem #2: Crank-Nicolson Problem
Solve the Two-Dimensional Temperature Problem above using Crank-Nicolson Method.
ENGR 516 Computational Methods for Graduate Students
Catholic University of America
Assignment #6
EigenValue Problems
1.) The moment of inertia, Ix, Iy, and the product of inertia Ixy of the cross sectional area
shown in the figure are: 1
y
inertia I xy of the
20 mm
4 mm
O
4
mm
s of the matrix
3 mm
2 mm
24 mm
on of the eigen-
solving the charprincipal axes of
).
x
4.23 The moment of inertia I x , I y , and the product of inertia I xy of the
cross-sectional area shown in the figure are:
I x = 5286 mm4, I y = 4331 mm4, and
I xy = 2914 mm4
The principal moments of inertia are the eigenvalues of the matrix
y
20 mm
O
24 mm
5286 2914
, and the principal axes are in the direction of the eigen2914 4331
vectors. Determine the principal moments of inertia by solving the characteristic equation. Determine the orientation of the principal axes of
inertia (unit vectors in the directions of the eigenvectors).
2 mm
Solution
2 mm
The eigenvalues of the matrix
5286 2914
are the principal moments of inertia. They are dete
2914 4331
the roots of the characteristic equation:
1
( 5286 – λ ) ( 4331 – λ ) – ( 2914 ) ( 2914 ) = 0
ipal moments of inertia.
are determined
2.) They
The structure
of anfrom
acetylene
molecule may be idealized as four masses1 connected by
which simplifies to 14402270 – 9617λ + λ 2 = 0 .
4.50 The structure of the C H (acetylene) molecule may be ide-
( 2914 ) ( 2914 ) = 0
2
two
springs
shown
in2 the
figure
below.
Applying
thequadratic
equation
of motion; we can
The
equation
is solved
by using
formula:
alized
as fouras
masses
connected
two
springs
(see
discussion
in the
4.50 The
structure of by
the C
2H2 (acetylene) molecule may be idegenerate
system
model
forequation
the amplitudes
of
vibration
of each
atom:
Problem a4.25).
By
applying
the
motion,
thediscussion
following
2
alized
as four masses
connected
byoftwo
springs (see
in
or λ 1 = 7761.36 and
genvalue must satisfy:
(i)
= λi
9617
( 9617 ) – 4 ( 1 ) ( 14402270 )
1
Problem
By applying
equation
ofλmotion,
the
following
--------------------------------------------------------------------------------------- or λ 1 = 7761.36
and λ 2 = 1855.64
system of equations
can4.25).
be written
forthethe
amplitudes
vibration
1 2 = of
2
system of equations can be written for the amplitudes of vibration
of each atom: of each atom:
4.50 Thekstructure ofkthe C2H2 (acetylene) molecule may be idek-------k CH
CH
CH
CH
By
- – ω2
– -------0 eigenvectors corresponding to each eigenvalue must satisfy:
2
--------- – ωalized
0 definition,
0 the
as– -------four
by0two springs
(see discussion in
m - masses connected
mH
mH
m HH
A 1 the 0following
k CC
( k CH
+ k CC )
k CH By
4.25).
applying
the equation
of motion,
(i)
(i)
– -------- ---------------------------- – ω2
– -------0 A
λ 2k = Problem
1855.64
5286 2914 u 1 = λ u 1
kwritten
( k CH +ofk CC
m)C
mfor
mcan
1A 2
00vibration
C
C
CH system
CC
equations
be
the
amplitudes
of
2
=
– --------- ----------------------------- – ω
– -------0
i
0
k CC
k CH + k CC )
k CH A A 3
(i)
m C of eachmatom:
m C(---------------------------2914 4331 u ( i )
C
0
– -------- – ω 2 – -------- 2 = 0
u2
2
0
mC
mC
mC
A4
0
A3
k CH + k CC ) k2
k CH
CC
k CH(---------------------------k CH –2k-------kthe
CHwith
CH
2
Starting
eigenvalue:
0
-–ω
- - – ωfirst
0
–0-------- – ----------------------- – ω
- 0
mH
0
m C – -------mmCH 0A 4
mC
mH
mH
2
2
2 are the restoring
( 1 )force spring ( 1 )
(1)
frequency,
k CHkCH =k5.92
k× 10 kg/s22 and kCC A= 115.8 × 100 kg/s
( k CHω+iskthe
k CH where
u1
u1
CC )
CC CH
5286 2914 u
0-----------------------------------C–H
- – and
---------bonds,
– ω respectively,
– -------- constants
- – ω 2–the
-------0
representing
C–C
and m H = 1amu and m C = 112amu=areλ the
=
7761.36
1
mH
mC
m C m H–27
A2
mC
(1)
(1)
= 0 2914 4331 u ( 1 )
masses of the atoms ( 1amu = 1.6605 × 10 kg).
u2
u2
2
2
2
0
(
+
k
)
A
k
k
k
2πc- (where
2 andCH
2 are the
CC
CH × 10
CC(frequencies),
3corresponding
(a)
Determine
the
eigenvalues
ω
and
the
wavelengths
λ
=
-------2
where ω is the
frequency,
kg/s
k
=
5.92
k
=
15.8
×
10
kg/s
restoring
force
spring
0
– -------- CH
----------------------------- – ω
– ---------CC
(ω
1)
(1)
0
mC 8
m C areA 4redundant
m C two equations
These
and yield u = 1.1772u .
u1
0
(i)
u2
m/s isand
the speed
light). respectively, and m H = 1amu and 1m C = 12amu 2are the
c = the
3 × 10C–H
constants representing
C–Cofbonds,
(1)
u1
(1)
(1)
= 7761.36
u2
(1)
u2
(1)
772u 2
1.
u1
.
k – 27
k to the eigenvalues found in part (a). From the eigenvectors,
(b) Determine the eigenvectors corresponding
CH
CH
2
(1) 2
0the relative
-------- – ω 2 moving toward, or away from
= 1.6605
× –10
masses of the0atomsdeduce
( 1amu
kg).
motion
of-------the- atoms
(i.e.,
Since
the
eigenvector
a unit vector: ( u (11 ) )each
+ other?).
( u2 ) = 1 .
m
m Hare theyis
H
Solution
(a) Determine the
eigenvalues ω (frequencies),
and2 the corresponding2 wavelengths
λ = 2πc
--------- (where
2
2
ω force spring
where ω is the frequency, k CH = 5.92 × 10 kg/s and k CC = 15.8 × 10 kg/s are the restoring
8 The following script file solves this problem:
representing
theof
C–H
and C–C bonds, respectively, and m H = 1amu and m C = 12amu are the
is the speed
light).
c = constants
3 × 10 m/s
– 27
(b) Determine
eigenvectors
corresponding
= 1.6605 × 10to the
masses the
of the
atoms ( 1amu
kg).eigenvalues found in part (a). From the eigenvectors,
clear, motion
clc
deduce
the
relative
of
the
atoms
(i.e.,
are
theyand
moving
toward, or away
from eachλother?).
(a) Determine the eigenvalues ω (frequencies),
the corresponding
wavelengths
= 2πc
--------- (where
Excerpts from this work may be reproduced by instructors
for distribution on a not-forω
for testing
of light).or instructional purposes only to students enrolled in courses for which t
c = 3 × 10 m/s is the speed
M(1,1)=kCH/mH; M(1,2)=-M(1,1);
(b) Determine
the
eigenvectors
corresponding
to the
eigenvalues
found in part (a).or
From
the eigenvectors,
has
been
adopted.
Any
other reproduction
translation
of this work beyond that p
The following
script
file
solves
this
problem:
M(2,1)=-kCH/mC; M(2,2)=(kCH+kCC)/mC; M(2,3)=-kCC/mC;
deduce
the relative motion
of the atoms
(i.e.,
are they
toward,
or awayStates
from each
other?). Act without the permis
Sections
107
or 108
ofmoving
the 1976
United
Copyright
M(3,2)=M(2,3); M(3,3)=M(2,2);
M(3,4)=M(2,1);
copyright owner is unlawful.
Solution M(4,3)=M(1,2); M(4,4)=-M(4,3);
Solution
kCH=5.92e2; kCC=15.8e2; mH=1.6605e-27; mC=12*1.6605e-27;
c=3e8;8 M=zeros(4);
clcfollowing script file solves this problem:
The
ructors for distributionclear,
on a not-for-profit
basis
kCC=15.8e2; mH=1.6605e-27; mC=12*1.6605e-27;
nts enrolled in courseskCH=5.92e2;
for which the
textbook
Excerpts
from this work may be reproduced by instructors for distribution on a not-for-profit basis
c=3e8; M=zeros(4);
for testing or instructional purposes only to students enrolled in courses for which the textbook
slation of this work beyondclear,
that permitted
by Any other reproduction or translation of this work beyond that permitted by
has
adopted.
clcbeen
M(1,1)=kCH/mH;
M(1,2)=-M(1,1);
Sections of
107 the
or 108 of the 1976 United States Copyright Act without the permission of the
Copyright Act without the permission