Download Set 9

EGR 599
Advanced Engineering Math II
_____________________
LAST NAME, FIRST
Problem set #9
1. Use the Galerkin Finite-Element Method to approximate the solution to the boundary-value
problem
2
2

x,
for 0  x  1 with y(0) = y(1) = 0
4
4
16
using x0 = 0, x1 = 0.3, x2 = 0.7, x3 = 1 and compare the results to the actual solution
2
1
1



y(x) =  cos x 
sin x + cos x.
2
2
4
3
3
6
Ans:
y” +
u1
u2
y=
cos
Galerkin
Exact
0.076894
0.074469
0.079885
0.077129
2. Use the formula  =
 Ldx
 r dx
2
and the trial function x(1  x) to estimate the smallest
eigenvalue in equation xy” + y = 0 subject to y(0) = y(1) = 0.
Ans: 4
3. Solve the one-dimensional heat conduction equation
d 2T
=  f(x)
dx 2
for a 10-cm rod with boundary conditions of T(0, t) = 50 and T(10, t) = 100 and a uniform heat
source of f(x) = 20. Use the trial function T = ax2 + bx + c
Ans:
T =  10x2 + 105x + 50
4. a) Use the Rayleigh-Ritz method to approximate the solution of
y” = 3x + 1,
y(0) = 0,
y(1) = 0,
using a quadratic in x as the approximating function.
b) Solve the problem by collocation, setting the residual to zero at x = 0.5.
c) Solve the problem by Galerkin’s method.
Ans:
u(x) = 1.25x(x  1) for (a), (b), and (c)
5. Develop the elements equations for a 10-cm rod with boundary conditions of T(0, t) = 40 and
T(10, t) = 100 and a uniform heat source of f(x) = 20. Employ four equal-size elements of length
= 2.5 cm. Compute the temperature distribution for the entire rod.
Ans:
x
T
0
40
2.5
242.5
5
320
6 Use Galerkin’s method to develop an element equation for D
0 D
d 2c
dc
U
 kc
2
dx
dx
R D
~
~
d 2c
dc
~
U
 kc
2
dx
dx

x1

x2
U

k
dc
d 2c
 U - kc = 0.
2
dx
dx
~
~
 d 2c
dc
~  N dx

U
 kc
D
 i
2
dx
 dx

x2
D
7.5
272.5
x1
~
d 2c
N i ( x )dx
dx 2
(1)
~
dc
N i ( x)dx
dx
(2)
x2
x1

x2
x1
~ N ( x ) dx
c
i
Term (1):
(3)
10
100
D

c  c2 
 dc

( x1 )  1

~
d 2c
 dx

x 2  x1 
N i ( x )dx  D
c2  c1 
dc
dx 2


( x2 ) 

x 2  x1 
 dx

x2
x1
Term (2):

x2

x2
~
dc
N i ( x)dx 
dx
x1
N i ( x )dx 
x1



x2
x1
c2  c1
N i ( x)dx
x 2  x1
x 2  x1
2
~
c  c1
dc
N i ( x)dx  2
dx
2
x2
x1
U

x2
x1
 c2
~

dc
N i ( x )dx  U 
c
dx
 2

 c1 
2 
 c1 

2 
Term (3):
k

x2
x1
~N ( x )dx   k ( x 2  x1 )
c
i
2
 c1 
 
c2 
Total element equation [(1) + (2) + (3)]
a11
a
 11
a11   c1  b1 
  
a11  c2  b2 
where
a11 
D
U k
   x2  x1 
x2  x1 2 2
a 22  
a12 
D
U

x 2  x1 2
D
U k
   x2  x1 
x2  x1 2 2
b1   D
dc
( x1 )
dx
b2  D
dc
( x2 )
dx
a 21 
D
U

x 2  x1 2