College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 501A
Seminar in Engineering Analysis
Fall 2004 Number: 17472 Instructor: Larry Caretto
November 30 Homework Solutions
Hoffman, page 494, problem 64 – Solve problem 58 by the second order equilibrium
method.
We first convert the differential equation into a finite difference equation, using second-order
expressions for the derivatives: dy/dx = (yn+1 – yn-1)/(2h) and d2y/dx2 = (yn+1 + yn-1 – 2yn)/h2. This
gives the following finite difference equation.
y n 1  y n 1  2 y n
y  y n 1
 1  xn  n 1
 1  xn  y n  1
2
2h
h
Multiplying this equation by h2 and rearranging gives the following result.


1  xn 
1  xn 


2
2
1  h
 y n1   2  1  xn h y n  1  h
 y n1  h
2 
2 


This is a set of equations that we can use to solve for the y values. From the initial condition that
y(0) = 1 we have y0 = 1 so that the initial equation in the set becomes.
 2  1  x h y
2
1
1
1  x1 
1  x1 
1  x1 



2
2
 1  h
 y 2  h  1  h
 y 0  h  1  h

2 
2 
2 



We can use the second-order backwards difference equation1 to obtain a finite-difference
equivalent of the zero gradient boundary condition at x = 1. This is the last grid node,
corresponding to n = N.
dy
dx

x  xN
y N 2  4 y N 1  3 y N
0
2h

y N 2  4 y N 1  3 y N 2
With h = 0.125, the boundary node at x = 1 is N = 8. We have almost a tridiagonal system, but in
such a system the last equation can only have two unknowns, yN-1, and yN. We have two
equations that involve the last three unknowns. In addition to the zero gradient equation above,
we also have the general finite-difference equation for n = N – 1 = 7; this equation is shown
below.


1  x7 
1  x7 


2
2
1  h
 y6   2  1  x7 h y 7  1  h
 y8  h
2 
2 


1
This equation was presented in the class notes and is equation 5.101 on page 271 of Hoffman.
Engineering Building Room 1333
Email: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
November 30 homework solutions
ME501A, L. S. Caretto, Fall 2004
Page 2
We can combine the equation above with the zero gradient equation to obtain a new equation
that has only y7 and y8 as unknowns. To do this we write the zero gradient equation for N = 8 and
use the result to eliminate y6 from the two equations. Substituting the zero gradient equation for
y6 into the above equation gives.


1  x7 
1  x7 


2
2
1  h
4 y7  3 y8    2  1  x7 h y7  1  h
 y8  h
2
2




The algebraic manipulation below gives us an equation that we can use as the last equation in a
tridiagonal system of equations.
 2  1  x h
2
7

1  x7
1  x7 

2
 4  2h1  x7  y7  1  h
 3  3h
 y8  h
2
2 

2  1  x h
7


2

 2h y 7   2  2h1  x7 y8   h 2
We still have to use the equation 1  h


1  x7 
1  x7 

2
 y6   2  1  x7 h y7  1  h
 y8 = h2
2 
2 

as one of our equations. The purpose of the algebraic manipulations above was to convert one
pair of equations, both of which had three variables, into an equivalent pair of equations, one of
which has only two variables.
We now have a set of linear equations that forms a tridiagonal matrix with the general equation
Anyn-1 + Bnyn + Cnyn+1 = Dn. For all equations except the first (n = 1) and last (n = 8) equations,
these coefficients are defined as follows.
An  1  h
1  xn
2
Bn  2  1  xn h 2
Cn  1  h
1  xn
2
Dn  h 2
For the constant step size used here, x n = nh.
The first and last equations in the set were derived above. The first equation has no A coefficient
and the last equation has no C coefficient. The values of the other coefficients for the first and
last equations are
B1  2  1  x1 h 2
C1  1  h

A8  2  1  x7  h 2  2h

1  x1
2
D1  h 2  1  h
B8  2  2h1  x7 
1  x1
2
D8  h 2
We can use the Thomas algorithm for applying Gauss elimination to a tridiagonal system of
equations. In this algorithm we compute values of Fn and En for a back-substitution process using
the following initial equations: E1 = -C1/B1 and F1 = D1/B1. The remaining values of En and Fn are
found from the following equations.
En 
 Cn
Bn  An E n1
and
Fn 
Dn  An Fn1
Bn  An E n1
The last value of y, yN – in this problem, N = 8 – is found from the following equation.
November 30 homework solutions
ME501A, L. S. Caretto, Fall 2004
yN 
Page 3
DN  AN FN 1
BN  AN E N 1
The various coefficients and the solution by back substitution, yn = Fn + En yn+1 are shown in the
table below.
n
0
1
2
3
4
5
6
7
8
xn
0
0.125
0.25
0.375
0.5
0.625
0.75
0.875
1
An
Bn
Cn
Dn
En
Fn
0.921875
0.914063
0.90625
0.898438
0.890625
0.882813
1.560547
-1.98242
-1.98047
-1.97852
-1.97656
-1.97461
-1.97266
-1.9707
-1.53125
1.070313
1.078125
1.085938
1.09375
1.101563
1.109375
1.117188
-0.91406
0.015625
0.015625
0.015625
0.015625
0.015625
0.015625
0.015625
0.539901
0.727113
0.826506
0.891009
0.938224
0.97566
1.007039
0.461084
0.276134
0.180212
0.120315
0.078759
0.047949
0.024072
-0.54466
yn
1
0.616945
0.288685
0.017262
-0.19716
-0.35631
-0.46371
-0.52442
-0.54466
The two approaches are seen to give significant differences. To determine which method is more
accurate for this problem I ran the calculation using h = 0.005 and constructed the following table
which compares the two methods for different grid sizes. The two methods give approximately
the same results, to three significant figures, for N = 200; these presumably more accurate results
are closer to the shoot-and-try method than to the finite-difference approach. This confirms that
the shoot-and-try method is usually more accurate for boundary value problems in ordinary
differential equations.
N=8
x
0
0.125
0.25
0.375
0.5
0.625
0.75
0.875
1
Finite
differences
1
0.61695
0.28868
0.01726
-0.19716
-0.35631
-0.46371
-0.52442
-0.54466
Shootand-try
1
0.57307
0.20780
-0.09345
-0.33045
-0.50520
-0.62172
-0.68578
-0.70445
N = 200
Finite
Shootdifferences
and-try
1
1
0.59250
0.59242
0.22739
0.22724
-0.07474
-0.07494
-0.31352
-0.31377
-0.49075
-0.49103
-0.61019
-0.61049
-0.67736
-0.67766
-0.69910
-0.69941
No statement is implied in this table about the relative accuracy of the two methods for N = 200.
If we solved the equation using N = 20000 and compared both solutions for N = 200 to the N =
2000 grid, we would expect the shoot-and-try to be more accurate. Of course, this accuracy
would not be needed if we only wanted answers to three significant figures.
November 30 homework solutions
ME501A, L. S. Caretto, Fall 2004
Page 4
Hoffman, page 497, problem 106 – Consider the eigenproblem described by equation
8.178. The exact solution of this eigenproblem is = ±n (n = 1, 2, …). The finite-difference
equation corresponding to the eigenproblem is given by equation (8.185). The numerical
solution of the eigenproblem for x = 1/5 is presented in Table 8.27. Determine the
solution to the eigenproblem for (a) x = 1/6 and (b) x = 1/8. Compare the results of the
three solutions in normalized form; that is k/.
The finite-difference equations for this eigenproblem, as derived in section 8.9 of Hoffman, have
the following general form: –yn-1 + (2 – k2h2)yn – yn+1 = 0. For a grid numbered from zero to N, h =
1/N and the boundary conditions are y0 = yN = 0. The first equation is (2 – k2h2)y1 – y2 = 0 and the
last equation is (2 – k2h2)yN-1 – yN = 0. This set of equations has the conventional matrix
eigenvalue form, (A – I)y = 0. where  = k2h2, y is a column vector, and A is the following
tridiagonal matrix
 2 1 0 0  0 0 0 
 1 2  1 0  0 0 0 


 0 1 2 1  0 0 0 


0 0 1 2  0 0 0 

A
 0 0 0 1  0 0 0 




  



 0 0 0 0   1 2  1


 0 0 0 0  0  1 2 
The solution to the eigenvalues, Det(A – I) = 0 can be found by starting with a grid that x = ¼.
Such a grid has only three nodes (besides the two boundary nodes, and the eigenvalue problem
is found as follows. To simplify the algebra, we define Z = 2 – = 2 – k2h2.
Z 1 0
 1 Z  1  Z 3  0  0  0  Z  Z  Z 3  2Z
0 1 Z
For four nodes, we have to solve the equation below. We evaluate the derivative by expanding
by the first row. We see that the first minor determinant is the one that we evaluated previously
for three rows.
Z 1 0 0
Z 1 0
1 1 0
1 Z 1 0
 Z  1 Z  1  (1) 0 Z  1  Z ( Z 3  2Z )  ( Z 2  1)  Z 4  3Z 2  1
0 1 Z 1
0 1 Z
0 1 Z
0 0 1 Z
For five nodes we have the following result. Again we expand by the first row using previous
results to evaluate the minor derivatives. Again we find that the first minor is the previous
determinant for four rows. We evaluate the second minor by another minor expansion, this time
using the first column for the expansion. Here we see that the only minor determinant in this
expansion is the entire determinant for three rows.
November 30 homework solutions
ME501A, L. S. Caretto, Fall 2004
Z
1
0
0
0
1
Z
1
0
0
0
1
Z
1
0 Z
0
0
1
Z
1
0
0
0
1
Z
Z
1
0
0
1
Z
1
0
0
1
Z
1
0
0
1
Z
Page 5
1 1
0
0
0
Z
1
0
0
1
Z
1
0
0
1
Z
 (1)
 Z ( Z 4  3Z 2  1)  (1)( Z 3  2Z )  Z 5  4Z 3  3Z  0
The fifth order equation can be written as Z(Z4 – 4Z2 + 3) = 0 so that one root is Z = 0. The other
four roots are the roots of a quadratic equation for Y = Z 2. This quadratic equation can be written
as Y2 – 4Y + 3 = (Y – 3) (Y – 1) = 0 whose roots are Y = 3 and Y = 1. This gives the following
roots for Z = Y : 3 , 1, –1, and – 3 in addition to the previous root, Z = 0. Since we have
defined Z = 2 –  = 2 – k2h2, we can find k as follows, in this case where k = 1/6.
k
2Z
2Z

 6 2Z
1
h
6
The solution for x = 1/.8 requires the expansion of two more determinants. First we consider the
determinant for six nodes in the field. As usual we expand it using the first row. 2
Z 1 0 0 0 0
Z 1 0 0 0
1 1 0 0 0
1 Z 1 0 0 0
1 Z 1 0 0
0 Z 1 0 0
0 1 Z 1 0 0
 Z 0  1 Z  1 0  (1) 0  1 Z  1 0
0 0 1 Z 1 0
0 0 1 Z 1
0 0 1 Z 1
0 0 0 1 Z 1
0 0 0 1 Z
0 0 0 1 Z
0 0 0 0 1 Z




Z Z 5  4Z 3  3Z  (1) Z 4  3Z 2  1  Z 6  5Z 4  6Z 2  1  0
We can now obtain the determinant that we need for x = 1/8.
Z 1 0 0 0 0
1 Z 1 0 0 0
0 1 Z 1 0 0
0 0 1 Z 1 0
0 0 0 1 Z 1
0
0
0
0

0
0
0
0
0
Z 1 0 0 0 0
1 1 0 0 0 0
0
1 Z 1 0 0 0
0 Z 1 0 0 0
0
0 1 Z 1 0 0
0 1 Z 1 0 0
 (1)
0 Z
0 0 1 Z 1 0
0 0 1 Z 1 0
0
0 0 0 1 Z 1
0 0 0 1 Z 1
1 Z 1
0 0 0 0 1 Z
0 0 0 0 1 Z
0 1 Z


 

 Z Z 6  5Z 4  6 Z 2  1  (1) Z 5  4Z 3  3Z  Z Z 6  6Z 4  10Z 2  4  0
2
At this point we see a general recurrence relationship. If D N denotes the determinant for a grid
with N nodes (not counting boundary nodes) we can write D N = ZDN-1 – DN-2.
November 30 homework solutions
ME501A, L. S. Caretto, Fall 2004
Page 6
In this case, as with x = 1/6, we see that one root of the characteristic equation is Z = 0 and the
remaining roots can be found by the solution of a cubic equation in Y = Z 2: Y3 – 6Y2 + 10Y – 2 = 0. The
roots to this cubic were found using the goal seek method in Excel. One root was Y = 2 so that the cubic
equation could be divided by Y – 2 to yield the quadratic equation Y2 – 4Y + 2 = 0 whose roots are Y =
2  2 . This gives the values of Z = Y as  2 and
2  2 as well as the root Z = 0 previously
noted. As before we can determine the eigenvalue, k, from the values of Z using the following equation
and setting h = 1/8 in this case.
k
2Z
2Z

8 2Z
1
h
8
The eigenvalues found in this solution (and the ones in the text example for h = 0.2) are compared to the
exact eigenvalues of in the table below. For the values of h = 1/6 and 1/8 computed here, the values of Z
as well as the values of k/ are shown.
Exact
k/
1
2
3
4
5
6
7
h = .2
Error
k/
0.9837
1.63%
1.8710
6.45%
2.5752
14.16%
3.0273
24.32%
Z
1.7321
1
0
-1
-1.7321
h = 1/6
k/
0.9885
1.9099
2.7009
3.3080
3.6896
Error
1.15%
4.51%
9.97%
17.30%
26.21%
Z
1.8478
1.4142
0.7654
0
-0.7654
-1.4142
-1.8478
h = 0.125
k/
0.9935
1.9490
2.8295
3.6013
4.2347
4.7053
4.9951
Error
0.65%
2.55%
5.68%
9.97%
15.31%
21.58%
28.64%
As expected, the use of finer spacing gives more accurate values for the lower eigenvalues. Making the
grid finer provides estimates of higher eigenvalues; a grid with a spacing of 1/N gives N+1 eigenvalues.
However, the higher eigenvalues are always less accurate than the lower ones.
The MATLAB commands and output below shows how MATLAB can be used to determine the
eigenvalues of the ODE for any grid size. The number of grid nodes is specified by the variable m. Here
we set m = 7 to get the same results as in the finest grid for this problem. However, by setting m to
another value you could use the same commands to get a larger set of eigenvalues in which the first ones
would be more accurate than the ones found here. The commands uses the MATLAB funcion
diag(V,n) to construct the tridiagonal matrix; there is additional information regarding this function
following the list of commands and screen output.
>> %Solution of numerical ODE eigenvalue problem by MATLAB
>> format long
>> m = 7
%get more significant figures
%define size of system // output deleted here
>> h = 1/(m+1)
% Get step size // output deleted here
%Command below defines tridiagonal matrix; see explanation in text below
>> A=-diag(ones(m-1,1),1)-diag(ones(m-1,1),-1)+2*diag(ones(m,1))
A =
2
-1
-1
2
0
-1
0
0
0
0
0
0
0
0
November 30 homework solutions
0
0
0
0
0
-1
0
0
0
0
>> eig(A)
2
-1
0
0
0
-1
2
-1
0
0
ME501A, L. S. Caretto, Fall 2004
0
-1
2
-1
0
0
0
-1
2
-1
Page 7
0
0
0
-1
2
% Get eigenvalues of matrix
ans =
0.15224093497743
0.58578643762690
1.23463313526982
2.00000000000000
2.76536686473018
3.41421356237310
3.84775906502257
>> sqrt(ans)/h
% Get eigenvalues of ODE
ans =
3.12144515225805
6.12293491784144
8.88912372831364
11.31370849898476
13.30351379684072
14.78207252018059
15.69256448645169
The MATLAB function diag(V,n), used above, creates a matrix that is all zeros except for one diagonal,
specified by the argument, n. The elements in the diagonal are taken from the one dimensional matrix
specified in the first parameter, V. The size of the diagonal matrix is determined by the number of
elements in V and the diagonal on which the elements are placed. If n = 0 the elements of V are
distributed along the principal diagonal. If n is a nonzero (positive or negative) argument, the elements
are placed in the diagonal that is n diagonals above (for positive n) or n diagonals below (for negative n)
the principal diagonal.
In the diag(V,n) commands above, the vector, V, is created by using another MATLAB function, ones.
The command ones(m, n), used above, creates a matrix with m rows and n columns filled with ones.
The command ones(m-1, 1), used twice above, creates a column vector of length m – 1, filled with
ones. The tridiagonal matrix required for this problem is created as the sum of three individual matrices:
(1) one with 2 in the principal diagonal, (2) one with -1 in the diagonal above the principal diagonal, and
(3) one with -1 in the diagonal below the principal diagonal. Note that the size of the vector used when
the diagonals above or below the principal diagonal are filled is one less than the size required to fill the
principal diagonal.
The eig command is used to get the matrix eigenvalues. These eigenvalues are represented as a column
matrix in MATLAB. Here we have not set this matrix equal to any other symbol so MATLAB gives it the
default symbol ans. We can take the square root of the ans matrix and divide it by the scalar h to get the
desired result for the ODE eigenvalues. We could further divide this result by pi to get the normalized
eigenvalues.