Download College of Engineering and Computer Science Mechanical

College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 501A
Seminar in Engineering Analysis
Fall 2004 Number: 17472 Instructor: Larry Caretto
October 19 Homework Solutions
Page 204, problem 5 – Find the power series solution of the differential equation y’’ – 3y’ +
2y = 0.
In the power series solution we postulate a solution of the following form

y ( x )   a n ( x - x0 ) n
[1]
n 0
Here the an are unknown coefficients. We can differentiate this series twice to obtain.
dy 
  nan ( x - x0 ) n1
dx n0
and
d2y 
  n(n  1)an ( x - x0 ) n2
2
dx
n 0
[2]
Substituting equations [1] and [2] into our original differential equation gives the following result.



n 0
n 0
n 0
 n(n  1)an ( x - x0 ) n2  3 nan ( x - x0 ) n1  2 an ( x - x0 ) n  0
[3]
We want to rewrite the first two sums in this equation so that the have the same form as the final
sum. We consider each of these separately. In the first sum of equation [3], the first two terms
(for n = 0 and n = 1) are zero. Thus we can write this sum as follows, where we introduce a new
index, m = n-2.



n 0
n2
m 0
 n(n  1)an ( x - x0 ) n2  0  0   n(n  1)an ( x - x0 ) n2   (m  1)(m  2)am2 ( x - x0 ) m [4]
In a similar fashion, we note that the first (n = 0) term in the second sum is zero and we rewrite
this sum using the new index m = n – 1 to give



n 0
n 1
m 0
 nan ( x - x0 ) n1  0   nan ( x - x0 ) n1   (m  1)am1 ( x - x0 ) m
[5]
We now want to substitute equations [4] and [5] back into equation [3]. Before doing this we note
that the summation index, n, in the final sum is a dummy index and re can replace this by m = n.
This gives the following result in place of equation [3].

 (m  1)(m  2)a
m 0

m 2

( x - x0 )  3 (m  1)a m 1 ( x - x0 )  2 a m ( x - x0 ) m  0
Engineering Building Room 1333
Email: [email protected]
m
m
m 0
Mail Code
8348
[6]
m 0
Phone: 818.677.6448
Fax: 818.677.7062
October 19 homework solutions
ME501A, L. S. Caretto, Fall 2004
Page 2
Since each sum in this equation has the same lower and upper limits we can combine them as
follows.

 (m  1)(m  2)a
m 0
m 2
 3(m  1)a m 1  2a m ( x - x0 ) m  0
[7]
In order for this series sum to equal zero for any value of x, the coefficient of each power of x
must equal zero. Thus we must satisfy the following equation for each value of m.
(m  1)(m  2)am 2  3(m  1)am1  2am  0
[8]
We can solve this equation for am+2.
am2 
3(m  1)a m 1  2a m
(m  1)( m  2)
[9]
We can use this equation to find any value of am+2 in terms of previous coefficients am+1 and am.
Since we cannot solve this equation for a0 and a1, we will leave these constants to be used to
satisfy boundary conditions on the differential equation. Setting m = 0 in equation [9], gives the
value of a2 as follows.
a 2  a0 2 
3(0  1)a01  2a0 3a1  2a 0

(0  1)(0  2)
2
[10]
Setting m = 1 in equation [9] gives us the value of a3 in terms of a1 and a2. We can use equation
[10] to obtain an equation that then gives a3 in terms of a0 and a1.
a3  a1 2
 3a  2a 0
6 1
3(1  1)a11  2a1 6a 2  2a1
2


 
(1  1)(1  2)
6
6

  2a1
7 a  6a 0

 1
6
[11]
We can continue in this fashion to try to deduce a general formula for am. Next we have m = 2.
a 4  a 2 2
 7a  6a 0   3a1  2a 0 
9 1
  2

3(2  1)a 21  2a 2 9a3  2a 2
6
2



  15a1  14a 0 [12]



(2  1)( 2  2)
12
12
24
We do not appear to have a simple solution yet. The solution that we have so far, with terms up
to and including (x – x0)4, can be written as follows.
y  a0  a1 ( x  x0 ) 
3a1  2a0
7a  6a0
15a1  14a0
( x  x0 ) 2  1
( x  x0 ) 3 
( x  x0 ) 4  
2
6
24
We see that the values of a0 and a1 are given by the initial conditions: a0 = y(x0) and a1 = y’(x0).
Although we know that the differential equation has a simple solution in terms of exponentials,
general power series solutions will not have simple expressions and we can just leave the
solution in the form above, with the knowledge that we can get the coefficients for additional
terms in the series by the recursion equation [9]. In a numerical application of a power series
October 19 homework solutions
ME501A, L. S. Caretto, Fall 2004
Page 3
solution, we would not even need to reduce the individual coefficients by algebra to terms
involving a0 and a1. Instead, we could use equation [9] to obtain numerical results for the
coefficients.

Page 205, problem 23 – Shift the index in the power series

s 2
s ( s  1) s 1
x so that the
s2 1
power under the summation sign is x m. Verify your result by writing the first few terms for
each form of the series. Determine the radius of convergence for the series.
To accomplish the shift we define a new index m = s – 1 to get the term xm in the summation sign.
To do this we have to replace s by m + 1 everywhere in the summation. At the lower limit of s =
2, we have m = 2 – 1 = 1. Thus, the summation becomes
s( s  1) s 1  (m  1)(m  1  1) m11  (m  1)(m  2) m
x 
x

x

2
2
(m  1) 2  1
s 2 s  1
m 1
m 1 (m  1)  1

The first four terms in the original series (for all values of s from 2 to 5) are
s( s  1) s 1 6 x 12 x 2 20 x 3 30 x 4
x 





2
5
10
17
26
s 2 s  1

The first four terms in the modified series (for all values of m from 1 to 4) are
(m  1)( m  2) m 6 x 12 x 2 20 x 3 30 x 4
x 





2
5
10
17
26
m 1 ( m  1)  1

The first four terms are the same in both series.
We see that the terms multiplying xm (or xs-1) increase in each term. Thus the series will diverge
unless the x terms decrease. However, the x terms will decrease only if |x|<1. Thus this the
radius of convergence for this series is 1.
Page 216, problem 9 – Find a basis of solutions for the differential equation (x + 2)2y’’+ (x +
2)y’ - y = 0, using the Frobenius method.
To simplify this task we will use z = x + 2 as the independent variable. For this substitution we
will still have d2y/dz2 = d2y/dx2 and dy/dz = dy/dx. Thus we seek the solution to z2y’’ + zy - y = 0.
or y’’ + y’/z - y/z2 = 0.
We write the solution using Frobenius method in the following form.


n 0
n 0
y( z )  z r  an z n   an z nr
[1]
Here the an are unknown coefficients and the value of r is also unknown. We can differentiate
this series twice to obtain.
dy 
  (n  r )an z n r 1
dz n0
and
d2y 
  (n  r )( n  r  1)an z n r 2
2
dz
n 0
[2]
October 19 homework solutions
ME501A, L. S. Caretto, Fall 2004
Page 4
We can substitute equations [1] and [2] into our differential equation: y’’ + y’/z - y/z2 to obtain the
following equation.

 (n  r )(n  r  1)a
n 0
n
z
n r 2
1 
1  

n  r 1 
   (n  r )a n z
  2   an z nr   0
z  n 0
 z  n 0

[3]
If we combine the z2 and z terms outside the sums with the z terms inside the sums, each series
sum has a term in zn+r-2, and we can combine these into a single series as follows.
 (n  r )(n  r  1)  (n  r )  1an z nr 2   (n  r ) 2  1an z nr 2  0


n 0
n 0
[4]
The first (n = 0) term in this series is (r2 – 1aoxr-2). The indicial equation required to set the
coefficient of xr-2 = 0, without setting a0 = 0 is r2 = 1. So the roots of the indicial equationare r =
1. This is the case where the two roots differ by an integer. We proceed by examining the first
solution where r = 1. From equation [4] we see that we have to satisfy the following equation for
the coefficient of each power of x to vanish.
(n  r )
2



 1 a n  (n  1) 2  1 a n  n(n  2)a n  0
[5]
The only way to satisfy this equation for values of n > 0 is for a n = 0 for n > 0. Thus, the power
series solution in equation [1] has only a single term.


n 0
n 0
y ( z )   a n z n  r   a n z n 1  a 0 z
[6]
In the case where the two roots of the indicial equation differ by an integer, the second solution is
given by the following equation, where r2 is the second root of the indicial equation and k may be
zero.


n 0
n 0
y 2 ( z )  ky1 ( z ) ln( z )  z r2  An z n  ky1 ( z ) ln( z )   An z n  r2
[7]
If we differentiate this soltuion two times we obtain the following results.
dy 2 ( z )
dy ( z )
y ( z) 
 k 1 ln( z )  k 1
  An (n  r2 ) z n  r2 1
dz
dz
z
n 0
[8]
dy 22 ( z )
dy12 ( z )
2k dy1 ( z ) ky1 ( z ) 
k
ln( z ) 

  An (n  r2  1)(n  r2 ) z n r2 2
2
dz
dz
z dz
z
n 0
[9]
If we substitute this result into our differential equation : y’’ + y/z – y/z2 = 0, we obtain
October 19 homework solutions
ME501A, L. S. Caretto, Fall 2004
Page 5
dy12 ( z )
2k dy1 ( z ) ky1 ( z ) 
k
ln( z ) 

  An (n  r2  1)( n  r2 ) z n  r2 2 
2
dz
z dz
z
n 0

ky1 ( z ) 
1  dy1 ( z )
1 

n  r2 1 
k
ln(
z
)


A
(
n

r
)
z

ky
(
z
)
ln(
z
)

An z n  r2   0


n
2
1


2 
z
dz
z
n 0
n 0
 z 

[10]
We can rearrange this equation, collecting all the terms multiplying ln(z) as follows.
 dy 2 ( z ) 1 dy1 ( z ) 1

2k dy1 ( z ) ky1 ( z )
k 1

 2 y1 ( z ) ln( z ) 


z dz
z dz
z
z2
 dz


1  y1 ( z ) 
 1
n  r2  2
An (n  r2  1)( n  r2 ) z
 k
  An (n  r2 ) z n  r2 1   2

z
z
n 0
n 0
 z
[11]

A z
n 0
n
n  r2
0
The first term in brackets that multiplies kln(z) is simply the original differential equation. Since y1
is a solution to this equation, the term in brackets is zero. In addition there are two ky1/z2 terms
with opposite signs that cancel. Eliminating these terms and combining the common summation
terms gives the following equation.
2k dy1 ( z ) 
  (n  r2  1)( n  r2 )  (n  r2 )  1An z n  r2  2  0
z dz
n 0
[12]
Since y1 = a0z, dy1/dz = a0. Substituting dy1/dz = a0.and r2 = -1 into equation 12 and simplifying
the summation term gives.




2k
2k
a 0   (n  r2 ) 2  1 An z n  r2  2 
a 0   n(n  2) An z n 3  0
z
z
n 0
n 0
[13]
Writing the first few terms in this equation gives the following result.

0  A1 z  2  2ka0 z 1  3 A3  8 A4 x   n(n  2) An z n 3  0
[14]
n 0
We must have A1 = k = A3 = A4 = 0 to set the coefficients of the initial terms to zero. In addition,
all subseuqent values of An for n > 4 must be zero. Thus, the only possible nonzero values of A n
are A0 and A2. Setting k = 0 and r2 = -1 in equation

y 2 ( z )  ky1 ( z ) ln( z )   An z n  r2  A0 z 1  A2 z
[15]
n 0
We see that this second solution contains the first solution, y = a 0z. Thus we can use any linear
combination of z and z-1 as a basis for the solutions to the original differential equation. Since we
made an initial choice to solve our equation for z = x + 2 instead of x, we see that we can write
our solutions in terms of the original x variable as y1 = x + 2 and y2 = 1/(x + 2).
October 19 homework solutions
ME501A, L. S. Caretto, Fall 2004
Page 226, problem 23 – Show that
x

Page 6
J 1 ( x)dx   x  J ( x)  c .
Equation (25) on page 223 of Kreyszig states that


d 
x J ( x)   x  J 1 ( x) . If we multiply
dx
this equation by –dx and take the indefinite integral of each side we obtain the result that




  d x  J ( x)   x  J 1 ( x)dx  const . Since  d x  J ( x)  x  J ( x) , and the
constant can be placed on any side of the equation, we have obtained the desired result.
Page 232, problem 7 – Using the substitutions y = u x and z = kx2/2, show that the
differential equation y’’ + k2x2y = 0 reduces to Bessel’s equation. Find a general solution
to the original equation in terms of Bessel functions.
From the definition of y we have dy/dx = xi/2du/dx + ux-1/2/2 and d2y/dx2 = x1/2d2u/dx2 - ux-3/2/4 +
x-1/2du/dx. Substutituing this result into our original differential equation gives.
x
1
2
3 u
1
d 2u
 1 du
x 2
 x 2  k 2 x 2 ux 2  0
2
dx
4
dx
[1]
Multiplying this equation by x3/2 gives
x2
d 2u
du u
x
  k 2 x 2 ux 2  0
2
dx 4
dx
[2]
If we shift the independent variable from x to z = kx 2/2, so that x = (2z/k)1/2 and dx/dz =
(2/k)1/2z-1/2/2 =(2kz)-1/2, we have
1 du
du dz du

 (2kz) 2
dx dx dz
dz
[3]
1 du 
1 d 
1 du 
d 2 u d  du  dz d  du  dz d 
  
(2kz) 2
 (2kz) 2
(2kz) 2
 
2



dx  dx  dx dz  dx  dx dz 
dz 
dz 
dz 
dx
1
2

1 d u
1 z 2 du 
d 2u
du
2
2

 (2kz) (2kz)

(
2
k
)

2
kz
k
2
2
2 dz 
dz
dz
dz


1
[4]
2
Substituting x = (2z/k)1/2 and the results of equations [3] and [4] into equation [2] gives.
2z 
d 2u
du   2 z 
 2kz 2  k
 
k 
dz   k 
dz
 4z 2
1
2
2kz 2 du  u  4 z 2 u
1
dz
4
2
d u
du
du u
 2z
 2z
  4z 2u  0
2
dz
dz 4
dz
Combining the first derivative terms and dividing the equation by 4 gives
[5]
October 19 homework solutions
ME501A, L. S. Caretto, Fall 2004
z2
Page 7
d 2u
du  2 1 
z
  z  u  0
2
dz 
16 
dz
2
We see that this equation has the form of Bessel’s equation, z
[6]


d 2u
du
z
 z 2  2 u  0 ,
2
dz
dz
with  = ¼. Since this is a non-integer , we can write a general solution as u = AJ 1/4(z) +
BJ-1/4(z). We are actually interested in a solution to our original equation with x as the
independent variable and y as the dependent variable. Using the definitions of u and z we can
obtain the desired final result as y = x1/2[AJ1/4(kx2/2) + BJ-1/4(kx2/2).
Page 232, problem 14 – The function I(x) = i-J(ix), where i2 = -1, is called the modified
Bessel function of the first kind of order . Show that I(x) is a solution of x2y’’ + xy’ – (x2 +
2) = 0.
We can test this solution by plugging it into the differential equation. Because i is a constant, we
can readily obtain the necessary derivatives of I(x) to obtain the following result.
x2



 
2 
J (ix )
d i  J (ix )
d2y
dy
2
2
2 d i

x

x


y

x

x
 x 2   2 i  J (ix )  0
2
2
dx
dx
dx
dx



If we define a new variable, z = ix, so that z2 = -x2, and dx = d(z/i), we can substitute this variable
and bring the constant i- outside the derivative operators to obtain.
 z i
2

d 2 J ( z ) z  d J ( z )
 i
  z 2   2 i  J ( z )  0
2
i
d ( z / i)
d z




Cancelling common factors of i, and –1 in the second and first term and the common factor of i-
in all remaining terms gives.
z2
d 2 J ( z )
d J ( z )

z
 z 2   2 J ( z )  0
2
dz
dz


The resulting equation is simply Bessel’s equation for which J (z) is a solution. Thus, the
proposed solution to the original differential equation is correct.
To obtain the power series solution for I(x) we use the the power series equation for J(x). We
can mainpulate the result using definition of i2 = -1 to eliminate the (-1)m term in the numerator of
the sum.
(1) m ix 
I ( x)  i J (ix )  i ix   2 m 

m! (m    1)
m 0 2




2m


(1) m i   x 
(1) m  1 x 2 m
x 2 m 
x  2 m 
 x  2 m 
  2 m 
m! (m    1)
m! (m    1) m 0 2
m! (m    1)
m 0 2
m 0 2


2m
2m
m
The final result is the series for I(x) given in the problem statement.