College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 501B
Seminar in Engineering Analysis
Spring 2009 Class: 14443 Instructor: Larry Caretto
February 23 Homework Solutions
1.
Obtain the solution for Laplace’s equation in a rectangle for the boundary conditions
shown below.
 2u  2u

 0 0  x  L, 0  y  H
x 2 y 2
u
u
u(0, y )  u( L, y )  0
0
 g N ( x)
y y  0
y y  H
In the notes on the solution of Laplace’s equation, we obtained the solution by separation of
variables for the case where the value of u, rather than the gradient, was specified at y = 0 and y
= H. Equation [8] in those notes gives the separation of variables solution shown below and we
showed that we must have B = 0 to satisfy the boundary condition that u = 0 at x = 0.
u( x, y)  Asin( x)  B cos(x)C sinh( y)  D cosh( y)
Setting B = 0, taking the y derivative of the solution, and setting y = 0 to satisfy the boundary
condition that u/y = 0 at y = 0 gives
u
( x, y)  A sin( x)C cosh( y)  D sinh( y)
y
u
( x,0)  0  A sin( x)C cosh(  0)  D sinh(  0)  AC sin( x)
y
The condition of zero gradient at y = 0 can only be satisfied for all x if C = 0. We set C = 0 and
apply the resulting equation to x = L to consider the boundary condition that u = 0 at x = L.
u ( L, y )  A sin( L) D cosh( y )  0
This equation can be satisfied if λL = nπ. As usual, this gives an infinite series of solutions that
satisfies the differential equation and three of the four boundary conditions.

u ( x, y )   Cn sin( n x) cosh( n y )
n 1
with
n 
n
L
This leaves the boundary condition that u/y = gN(x) at y = H. Taking the y derivative of the
previous equation and setting y = H gives
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February 23 homework solutions
ME501B, L. S. Caretto, Spring 2009

u
( x, y )  g N ( x)   nCn sin( n x) sinh( n H )
y
n 1
with
Page 2
n 
n
L
As usual we obtain an orthogonal eigenfunction expansion by multiplying the equation above by
sin(λmx)dx = sin(mx/L)dx and integrating from 0 to L. This gives the result below.
L

L
 mx 
 mx  n
 nx 
 nH 
g N ( x) sin 
Cn sin 
dx  sin 

 sinh 
dx
L 
L  n 1 L
L 
L 




0
0




L
L
n
m
 nH 
 mx   nx 
 mH 
2  mx 

Cn sinh 
Cm sinh 
 sin 
 sin 
dx 
 sin 
dx
L
L
L
L
L
L
L










n 1
0
0


L
Using the result that
 sin
0
2

L
 mx 

dx  and solving for Cm gives.
2
 L 
Cm 
2
m
L
 mx 
( x) sin 
dx
 L 
 mH 
sinh 

 L 
g
0
N
Substituting this result into the solution for u(x,y) gives the following general solution to this
problem.
 nx 
 ny  2
 mx 
sin 
g N ( x) sin 
 cosh 

dx


 L 
 L  n 0
 L 
u ( x, y )  
 nH 
n 1
sinh 

 L 
L
This expression for Cm is similar to the one where the boundary conditions were u = 0 at y = 0
and u = uN(x) at y = H.
2.
What is the solution if gN(x) is a constant, GN?
Setting gN(x) = GN gives the following result for Cm.
L
2G N  L
2 L
 mx 
 mx 
G N sin 

cos
dx



m 0
m  m  L  0
 L 
Cm 

 mH 
 mH 
sinh 
sinh 


 L 
 L 
February 23 homework solutions
ME501B, L. S. Caretto, Spring 2009
1   1
1  cosm
2
m2

m

2 LGN
Cm 
2 LGN
 mH 
sinh 

 L 
 mH 
sinh 

 L 
m

 4 LGN
 m2


 0


Page 3
m odd
m even
This equation for Cm gives the following result for u(x,y).
 nx 
 ny 
sin
cosh




4 LGN 
L 
L 


u ( x, y ) 
 2  nH 
 2 n1
n sinh 

 L 
3. Solve Laplace’s equation for the potential, u(x,y), in a rectangular region with 0  x  L
and 0  y  H, with zero gradient boundary conditions at x = 0 and x = L (for all y
values) as shown below. This is similar to the first problem solved in the notes on
solving Laplace’s equation except that the side boundary conditions that u(0,y) =
u(L,y) = 0 are replaced by zero gradient boundary conditions. Parts (a) to (c) below
outline the steps in the solution with reference to the notes.
 2u  2u

 0 0  x  L, 0  y  H
x 2 y 2
u
x
x 0
u

x
 u ( x,0)  0
xL
[1]
u ( x, H )  u N ( x )
(a) You should obtain a general solution similar to the one given by equation [13] in
the notes on solving Laplace’s equation. (This problem is similar to problem 9 on
page 561 of Kreyszig that was assigned for the February 9 homework on the
diffusion equation.)
Use separation of variables.
u(x,t) = X(x)Y(y)
[2]
Since X(x) is a function of x only and Y(y) is a function of y only, we obtain the following result
when we substitute equation [2] into equation [1].
 2u  2u  2 X ( x)Y ( y)  2 X ( x)Y ( y)
 2 X ( x)
 2Y ( y)



 Y ( y)
 X ( x)
0
x 2 y 2
x 2
y 2
x 2
y 2
If we divide the final equation through by the product X(x)Y(y), and move the y derivative to the
other side of the equal sign, we obtain the following result.
[3]
February 23 homework solutions
ME501B, L. S. Caretto, Spring 2009
Page 4
1  2 X ( x)
1  2Y ( y)


X ( x) x 2
Y ( y) y 2
[4]
The left hand side of equation [4] is a function of x only; the right hand side is a function of y only.
The only way that this can be correct is if both sides equal a constant. This also shows that the
separation of variables solution works. In order to simply the solution, we choose the constant to
be equal to2. This gives us two ordinary differential equations to solve.
1 d 2 X ( x)
 2
2
X ( x) dx

1 d 2Y ( y )
 2
2
Y ( y ) dy
[5]
Equation [5] shows that we have two separate differential equations, each of which has a known
general solution. These equations and their general solutions are shown below.
d 2 X ( x)
 2 X ( x)  0
2
dx
d 2Y ( y )
 2Y ( y )  0
2
dy


X ( x)  A sin( x)  B cos(x)
[6]
Y ( y )  C sinh( y )  D cosh( y )
[7]
From the solutions in equations [6] and [7], we can write the general solution for u(x,y) = X(x)Y(y)
as follows and its partial derivative with respect to x as follows.
u( x, y )  A sin( x )  B cos( x )C sinh( y )  D cosh( y )
u
( x, y )   A cos( x )  B sin( x )C sinh( y )  D cosh( y )
x
[8]
We now apply the boundary conditions shown with the original equation [1] to evaluate the
constants A, B, C, and D. If we substitute the boundary condition at x = 0 into equation [8], get
the following result.
u
(0, y )  0   A cos(  0)  B sin(  0)C sinh( y )  D cosh( y )
x
[9]
Because sin(0) = 0 and cos(0) = 1, equation [9] will be satisfied for all y only if A = 0. Thus, we
set A = 0. Next we apply the solution in equation [8] (with A = 0) to the boundary condition at y =
0.
u( x,0)  0  B cos(x)C sinh( 0)  D cosh( 0)  BD cos(x)
[10]
This boundary condition will be satisfied for all x only if D = 0. The third boundary condition that
the normal gradient is zero at x = L may be applied to the second equation in [8] (after using the
results from the first two boundary conditions) as follows.
u
( L, y )  0    B sin( L)C sinh( y )
x
[11]
February 23 homework solutions
ME501B, L. S. Caretto, Spring 2009
Page 5
Equation [11] can only be satisfied if the sine term is zero. This will be true only if L is an
integral times . If n denotes an integer, we must have
L  n
or

n
L
[12]
Since the cosine of zero is one, we must include n = 0 as one of the possible eigenvalues in this
solution. This contrasts with the sine where inclusion of n = 0 in the eigenvalues does not give us
another eigenfunction. However, if n = 0 then  = 0 and our differential equation for Y(y) in
equation [7] becomes
d 2Y ( y )
 0 , whose solution is Y(y) = Cy + D. To meet the boundary
dy 2
condition that u(x,0) = 0 we must have Y(0) = C(0) + D = 0 or D = 0, leaving Y0(y) = Cy.
Since any integral value of n gives a solution to the original differential equations, and the
boundary conditions, the most general solution is one that is a sum of all possible solutions, each
multiplied by a different constant. In the general solution for one value of n, which we can now
write for n > 0 as BCcos(nx)sinh(ny) with n = n/L.
For m = 0, giving 0 = 0, the solution is BCcos(0x)y = BCcos(0)y = BCy = C0y, . (As usual, we
write the product of two constants, BC, as the single constant, C n, which may be different for each
value of n.) The general solution which is a sum of all solutions with different values of n is
written as follows

u( x, y )  C0 y   Cn cos( n x ) sinh( n y )
n 1
n 
n
L
[13]
Here we have separated out the initial term for n = 0 because it leads to a different solution for
Y(y).
(b) Obtain the equation equivalent to equation [16] in the notes giving the coefficients
in the eigenfunction expansion for u(x,H) = uN(x).
We can apply equation [13] to this boundary condition by setting y = H and u(x,H) = u N(x). This
gives the following result.

 nH   nx 
u( x, H )  uN ( x )  C0 H   Cn sinh 
 cos

 L   L 
n 1
[14]
We can simplify this equation by using the orthogonality relationships for integrals of the cosine.
If we multiply both sides by cos(mx/L), where m is another integer, which may be zero, and
integrate from a lower limit of zero to an upper limit of L, we get the following result.

 mx 
 mx 
 mx   nx 
 nH 
u
(
x
)
cos
dx

C
H
cos
dx

Cn cos




 cos
 tanh 
dx

0
0
0


 L 
 L 
 L   L 
 L 
0
0 n 1
L
L
L
 nH 
2  mx 
 C0 HL m 0  (1   m 0 ) tanh 
  cos 
dx
 L 0
 L 
L
[15]
February 23 homework solutions
ME501B, L. S. Caretto, Spring 2009
Page 6
L
 L
 mx 
 mx 
Since  cos
sin 
dx  
 = 0, unless m = 0 in which case the integral equals L,
L
m

L



 0

0
L
the term multiplying C0 is zero unless m = 0. (We expect this from the orthogonality of the cosine
terms that include the cosine of zero.) We write the C0 term on the right with the Kronecker delta
to show that it is present only if m = 0. We multiply the general term by (1 – m0) to show that it is
present only if m is not equal to zero. As usual, we have used the orthogonality relationships to
drop all terms in the summation except for the term where m = n. Solving [15] gives separate
results for m = 0 and m ≠ 0. For m ≠ 0, we have the following result.
 mx 
 mH 
2  mx 
0 uN cos L dx Cm sinh  L 0 cos  L dx
L
L
L
L
 mH   x
 2mx 
 mH  L
 Cm sinh 
sin 
 
  Cm sinh 

 L   2 4m  L  0
 L 2
[16]
This leads to the following expression for Cm
 1 L

u N ( x)dx
 HL 0
L

Cm   u ( x) cos mx dx
N

 L 
0
L
 mH 

sinh 


2
 L 


m0
[18]
m0
(c) Obtain the equation equivalent to equation [20] in the notes giving the solution for
the boundary condition that u(x,H) = uN(x) = U, a constant.
Applying equation [17] for Cm to the case where uN(x) = U, a constant gives the following result.
 1 L
U
Udx 


H
0
 HL
L

m x 
Cm    U cos
dx
L


 0
0

 m H  L

 sinh 
 L 2

m0
[17]
m0
The result that Cm = 0 unless m = 0 comes from the previous result that
L
 L
 mx 
 mx 
0 cos L dx   m sin  L 0 = 0, unless m = 0. Applying the results of equation [17] to
L
the general solution in equation [13] gives the following simple solution for u(x,y) for the boundary
conditions of this problem.
February 23 homework solutions
ME501B, L. S. Caretto, Spring 2009
u ( x, y ) 
Page 7
yU
H
[18]
This solution satisfies the differential equation and the boundary conditions in equation [1]. After
some thought, it even makes sense physically. If the problem is one of heat transfer and both the
x = 0 and the x = L side have a zero gradient, there is no heat flow in through these sides. Since
the top and bottom are uniform (the top temperature is U and the bottom is zero) there is only
one-dimensional heat transfer, giving the linear result of equation [18].
4.
Obtain the solution to Laplace’s equation for 0 ≤ x ≤ L and 0 ≤ y ≤ H where u(x,0) =
uS(x), u(x,H) = uN(x), and ∂u/∂x = 0 for all y at x = 0 and x = L. What is the solution if
uS(x) = U1 and uN(x) = U2? (This is the same as problem 3, with the boundary
conditions at y = 0 and y = H replaced by the following: u(x,0) = U 1 and u(x,H) = U2. Use
the results from that solution to get the solution here. If your solution takes more than
one page you are on the wrong track.)
We can find the solution to this problem, u(x,y) as the superposition of two solutions to Laplace’s
equation: u(x,y) = u1(x,y) + u2(x,y) where u1 and u2 both solve Laplace’s equation and they have
the following boundary conditions.

u1(x,0) = uS(x), u1(x,H) = 0, and ∂u1/∂x = 0 for all y at x = 0 and x = L, and

u2(x,0) = 0, u2(x,H) = uN(x), and ∂u2/∂x = 0 for all y at x = 0 and x = L.
Equations [13] and [18] in the solution to problem 3, above, give the solution for u2 as follows.

u2 ( x, y )  C0 y   Cn cos(n x) sinh( n y )
n 1
 1 L

0 uN ( x )dx
HL
L

mx 
Cm    u N ( x ) cos
dx
 L 
0

 m H  L

 t sinh 
 L 2

n 
n
L
m0
m0
We can obtain the solution for u1 from the solution for u2 by replacing y in the solution for u2 by H
– y and replacing uN by uS. This gives the following equations for the solution u1.

u1 ( x, y )  D0 ( H  y )   Dn cos(n x) sinh n ( H  y ) 
n 1
n 
n
L
February 23 homework solutions
ME501B, L. S. Caretto, Spring 2009
 1 L

0 uN ( x)dx
HL

L
mx 
Dm    uS ( x) cos
dx
 L 
0

 mH  L

 sinh 
 L 2

Page 8
m0
m0
For any given pair of boundary conditions, uN(x) and uS(x), we can evaluate the coefficients Cm
and Dm, substitute the results into the equations for u1 and u2, respectively, and add the results to
get the desired solution, u(x,y) = u1(x,y) + u2(x,y). For the case where uS(x) = U1 and uN(x) = U2,
we have the following results.
 1 L
U
U 2 dx  2
m0


H
 HL 0
L
mx 
Cm    U 2 cos
dx
 L 
0
0 m0

 mH  L

 sinh 
 L 2

 1 L
U
U1dx  1
m0


H
 HL 0
L
mx 
Dm    U1 cos
dx
 L 
0
0 m0

 mH  L

 sinh 
 L 2

With these coefficients the solutions for u1 and u2 become simple linear solutions: u1 = (U2y/H)
and u1 = (U1/H)(H – y). Adding these two solutions gives the desired result.
u(x,y) = u1(x,y) + u2(x,y) = (U1/H)(H – y) + u1 + (U2y/H) = U1 + (U2 – U1)y/H
5.
Find the temperature u(x,y) in a thin metallic square plate of side a = 12 with insulated
faces if the left, lower, and right sides are kept at 0oC and the upper side at the
temperature sin(πx/12) really sin(πx/a).
From the notes on the solution to Laplace’s equation we know that the general solution for the
case where the lower, left and right sides are zero; it is given by equation [13] in those notes (with
L = H = a here).

u( x, y )   Cn sin( n x ) sinh( n y )
with
n 
n 1
n
a
The general solution can be written as follows when the upper face at y = a is equal to sin(πx/a).

u( x, a )   Cn sin( n x ) sinh( n a )
n 1
with
n 
n
a
We can use the orthogonality relationships for integrals of the sine. If we multiply both sides by
sin(mx/a), where m is another integer, and integrate from a lower limit of zero to an upper limit of
a, we get the following result.
February 23 homework solutions
ME501B, L. S. Caretto, Spring 2009
Page 9

L m1
 m x    x 
 mx   nx 
 na 
sin
sin
dx


Cn sin 




 sin 
 sinh 
dx

0  a   a 

2
 a   a 
 a 
0 n 1
a
a

LCm sinh m 
 mx   nx 
2  mx 
   Cn sin 
 sin 
 sinh n dx  Cm sinh m  sin 
dx 
2
 a   a 
 a 
n 1 0
0
a
a
a
Here we note that the first integral,
L
 mx   x 
 sin  dx  m1 is zero unless m = 1, in
a  a
2
 sin 
0
which case the integral is L/2. The right hand side of this equation follows the usual development
of an expansion in orthogonal eigenfunctions. The net result for C m is
 1
 sinh   m  1

Cm  
0
m 1


When we substitute this expression for Cm into the general solution we obtain the following result
(after also substituting λn = nπ/a).
 x 
 y 
sin   sinh  
a
 a 
u ( x, y )   
sinh(  )
6.
Sketch or plot some of the isotherms in problem 4. Give physical reasons for the
general shape of these curves.
The overall shape will be that of half a sine wave in the x direction, damped by the factor
sinh(πy/a) / sinh(π) in the y direction. At the top, where y = a, the value of u will be sin(x/a),
which will be zero at x = 0 and x = a with a maximum value of 1 at x = a/2. The isotherm plot will
be similar to the contours we have seen for a boundary condition where the top value is a
constant, but there will be larger regions of lower temperatures because the top temperatures are
close to zero near the ends.
February 23 homework solutions
ME501B, L. S. Caretto, Spring 2009
Page 10
Isotherms
for 648
Problem
5
Solutions to Problem
39 on Page
of Kreyszig
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
y/a
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
0.2
0.4
0.6
x/a
0.8
1
0