Download chapter 84 an introduction to partial differential equations

CHAPTER 84 AN INTRODUCTION TO PARTIAL
DIFFERENTIAL EQUATIONS
EXERCISE 318 Page 887
1. Determine the general solution of
Since
∂u
= 4ty then
∂y
∂u
= 4ty
∂y
u = ∫ 4t y d y = (4t) ∫ y d y
= ( 4t )
u = 2ty 2 + f (t )
i.e.
2. Solve
Since
y2
+ f (t ) = 2ty 2 + f (t )
2
∂u
= 2t cos θ given that u = 2t when θ = 0
∂t
∂u
= 2t cos θ
∂t
then
u = ∫ 2t cos θ d t = (2 cos θ) ∫ t d t
= ( 2 cos θ )
u = 2t when θ = 0, hence,
t2
+ f (θ )= t 2 cos θ + f (θ )
2
2t = t 2 + f (θ )
f (θ =
) 2t − t 2
from which,
Hence,
u = t 2 cos θ + 2t − t 2
or
u = t 2 ( cos θ − 1) + 2t
3. Verify that u(θ, t) = θ 2 + θ t is a solution of
∂u
= 2θ+ t
∂θ
Hence,
and
∂u
∂u
−2
=
t
∂θ
∂t
∂u
= 0 + θ =θ
∂t
∂u
∂u
∂u
∂u
−2
=
t
− 2 = (2θ + t ) − 2(θ )= 2θ + t − 2θ = t which verifies that
∂θ
∂t
∂θ
∂t
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4. Verify that u = e − y cos x is a solution of
∂ 2u ∂ 2u
+
=
0
∂x 2 ∂y 2
∂ 2u
= − e − y cos x
2
∂x
Since u = e − y cos x then
∂u
= e − y (− sin x)
∂x
Also,
∂u
= − e − y cos x
∂y
Hence,
∂ 2u ∂ 2u
+
=
− e − y cos x + e − y cos x =
0
∂x 2 ∂y 2
5. Solve
Since
and
∂ 2u
= e − y cos x
∂x 2
∂ 2u
∂u
π
= 8e y sin 2 x given that at y = 0,
= sin x, and at x = , u = 2 y 2
∂x∂y
∂x
2
∂ 2u
= 8e y sin 2 x then integrating partially with respect to y gives:
∂x∂y
∂u
=
∂x
∫ 8e
y
sin
2 x d y 8e y sin 2 x + f ( x)
=
From the boundary conditions,
sin x = 8e 0 sin 2x + f(x)
i.e.
and
∂u
= sin x when y = 0, hence
∂x
from which, f(x) = sin x – 8 sin 2x
∂u
= 8e y sin 2 x + sin x − 8sin 2 x
∂x
Integrating partially with respect to x gives:
u = ∫ [8e y sin 2 x + sin x − 8sin 2 x]d x =
−4 e y cos 2 x − cos x + 4 cos 2 x + f ( y )
From the boundary conditions, u = 2 y 2 when x =
π
2
, hence
2 y 2 = –4e y cos π – 0 + 4 cos π + f(y)
= 4e y – 4 + f(y) from which, f(y) = 2 y 2 – 4e y + 4
Hence, the solution of
∂ 2u
= 8e y sin 2 x is given by:
∂x∂y
u = −4 e y cos 2 x − cos x + 4 cos 2 x + 2 y 2 – 4e y + 4
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∂u
∂ 2u
6. Solve = y ( 4 x 2 − 1) given that at x = 0, u = sin y and
= cos 2y
2
∂x
∂x
∂ 2u
Since = y ( 4 x 2 − 1)
∂x 2
x = 0 when
∂u
=
∂x
then
∂u
= cos 2y, hence, cos 2y = 0 + f(y)
∂x
∂u
 4 x3

= y
− x  + cos 2 y
∂x
 3

Hence,
and
u=
∫
  4 x3


 x4 x2 
x y  −  + x cos 2 y + F ( y )
 y  3 − x  + cos 2 y  d =

 3 2 
 

x = 0 when u = sin y, hence,
Thus,
 4 x3

2 −1 d
y
4
x
=
x
y
− x  + f ( y)
(
)

∫
 3

sin y = F(y)
 x4 x2 
u = y  −  + x cos 2 y + sin y
 3 2 
∂u
∂ 2u
7. Solve = sin ( x + t ) given that
= 1 when t = 0, and when u = 2t when x = 0
∂x
∂x∂t
∂ 2u
Since = sin ( x + t ) then integrating partially with respect to t gives:
∂x∂t
∂u
=sin ( x + t ) d t =
− cos ( x + t ) + f ( x)
∂x ∫
From the boundary conditions,
∂u
= 1 when t = 0, hence
∂x
1 = – cos x + f(x) from which, f(x) = 1 + cos x
i.e.
∂u
=− cos ( x + t ) + 1 + cos x
∂x
Integrating partially with respect to x gives:
u = ∫ [− cos ( x + t ) + 1 + cos x]d x =
− sin ( x + t ) + x + sin x + f (t )
From the boundary conditions, u = 2t when x = 0, hence
2t = – sin t + 0 + sin 0 + f(t) = – sin t + f(t) from which, f(t) = 2t + sin t
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∂ 2u
Hence, the solution of = sin ( x + t ) is given by:
∂x∂t
u =− sin ( x + t ) + x + sin x + 2t + sin t
8. Show that u(x, y) = xy +
Since, u = xy +
x
∂ 2u
∂ 2u
+y
=
2x
is a solution of 2 x
y
∂x∂y
∂y 2
x
∂u
1
= y+
then
y
∂x
y
and
∂u
x
=x −
=x − xy −2
∂y
y2
∂ 2u 2 x
=
∂y 2 y 3
and
∂ 2u
∂ 
x 
1
= x −  =
1−
∂x∂y ∂x 
y2 
y2
Also,
Hence,
L.H.S. = 2 x

 2x 
∂ 2u
∂ 2u
1 
+y
= 2 x 1 −  + y  
∂x∂y
∂y 2
y2 

 y3 
= 2x −
2x 2x
+
=
2 x = R.H.S.
y2 y2
9. Find the particular solution of the differential equation
conditions that when y = π,
Since
∂2 y
= cos x cos y given the initial
∂x∂y
∂u
= x, and when x = π, u = 2 cos y.
∂x
∂2 y
= cos x cos y then integrating partially with respect to y gives:
∂x∂y
∂u
=
∂x
cos y d y
∫ cos x=
From the boundary conditions,
x = cos x sin π + f(x)
cos x sin y + f ( x)
∂u
= x when y = π, hence
∂x
from which, f(x) = x
∂u
i.e.= cos x sin y + x
∂x
Integrating partially with respect to x gives:
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u = ∫ [cos x sin y + x=
]d x sin x sin y +
x2
+ f ( y)
2
From the boundary conditions, u = 2 cos y when x = π , hence
2 cos y = sin π sin y +
=
π2
2
π2
2
+ f(y)
+ f(y) from which, f(y) = 2 cos y –
Hence, the solution of
π2
2
∂2 y
= cos x cos y is given by:
∂x∂y
u = sin x sin y +
π2
x2
+ 2 cos y −
2
2
10. Verify that φ=
( x, y ) x cos y + e x sin y satisfies the differential equation
∂ 2φ ∂ 2φ
+
+ x cos y =
0
∂x 2 ∂y 2
∂φ
Since φ = x cos y + e x sin y then = cos y + e x sin y
∂x
∂φ
=
− x sin y + e x cos y
∂y
and
Hence,
L.H.S. =
and
∂ 2φ
= e x sin y
∂x 2
and
∂ 2φ
=
− x cos y − e x sin y
∂y 2
∂ 2φ ∂ 2φ
+
+ x cos
=
y e x sin y + ( − x cos y − e x sin y ) + x cos y
∂x 2 ∂y 2
= e x sin y − x cos y − e x sin y + x cos y =
0=
R.H .S .
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© 2014, John Bird
EXERCISE 319 Page 888
1. Solve T′′ = c 2 µT
Since T′′= c2µT
given c = 3 and µ = 1
and c = 3 and µ = 1, then T′′– (3)2(1)T = 0
i.e.
T′′ – 9T = 0
If T′′ – 9T = 0 then the auxiliary equation is:
m2 − 9 =
0 i.e. m 2 = 9
Thus, the general solution is:
from which, m =
9 = ±3
T = A e 3t + B e − 3t
2. Solve T′′ – c 2 µT = 0 given c = 3 and µ = –1
Since T′′– c2µT = 0 and c = 3 and µ = –1, then T′′– (3)2(–1)T = 0
i.e.
T′′ + 9T = 0
If T′′ + 9T = 0 then the auxiliary equation is:
from which, m = −9 =± j 3 or 0 ± j3
m2 + 9 =
0 i.e. m 2 = −9
Thus, the general solution is:
T = e0 { A cos 3t + B sin 3t}
= A cos 3t + B sin 3t
3. Solve X′′ = µX
given µ = 1
Since X′′ = µX and µ = 1, then X′′ – X = 0
If X′′ – X = 0 then the auxiliary equation is:
m 2 − 1 =0
i.e.
Thus, the general solution is:
m2 = 1
from which,
m = 1 or m = –1
X = A e x + B e− x
4. Solve X′′ – µX = 0 given µ = –1
Since X′′ – µX = 0 and µ = –1, then
X ′′ + X = 0
If X′′ + X = 0 then the auxiliary equation is:
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© 2014, John Bird
m 2 + 1 =0
i.e.
Thus, the general solution is:
m 2 = −1
from which,
m = −1 =± j
X = e0 { A cos x + B sin x}
= A cos x + B sin x
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EXERCISE 320 Page 892
1. An elastic string is stretched between two points 40 cm apart. Its centre point is displaced 1.5 cm
from its position of rest at right-angles to the original direction of the string and then released
with zero velocity. Determine the subsequent motion u(x, t) by applying the wave equation
∂ 2u 1 ∂ 2u
with c 2 = 9
=
∂x 2 c 2 ∂t 2
The elastic string is shown in the diagram below
1. The boundary and initial conditions given are:
u (0, t ) = 0 

u (40, t ) = 0 
u(x, 0) = f(x) =
1.5
x
20
=−
0 ≤ x ≤ 20
1.5
60 − 1.5 x
x + 3=
20
20
20 ≤ x ≤ 40
 ∂u 
 ∂t  = 0 i.e. zero initial velocity
t =0
2. Assuming a solution u = XT, where X is a function of x only, and T is a function of t only,
then
∂u
= X 'T
∂x
and
∂ 2u
= X '' T
∂x 2
and
Substituting into the partial differential equation,
gives:
X '' T =
∂u
= XT '
∂y
and
∂ 2u
= XT ''
∂y 2
∂ 2u 1 ∂ 2u
=
∂x 2 c 2 ∂t 2
1
XT ''
c2
i.e.
X '' T =
1
XT '' since c 2 = 9
9
X '' T ''
=
9T
X
3. Separating the variables gives:
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© 2014, John Bird
Let constant, µ =
X '' T ''
then
=
9T
X
µ=
X ''
X
X′′ – µX = 0
from which,
µ=
and
T ''
9T
T′′ – 9µT = 0
and
4. Letting µ = – p 2 to give an oscillatory solution gives
X′′ + p 2 X = 0
and
and the auxiliary equation is: m 2 + p 2 = 0 from which, m =− p 2 =
± jp
T′′ + 9 p 2 T = 0
and the auxiliary equation is:
m 2 + 9 p 2 = 0 from which, m =
−9 p 2 =
± j3 p
5. Solving each equation gives:
X = A cos px + B sin px
and
T = C cos 3pt + D sin 3pt
Thus, u(x, t) = {A cos px + B sin px}{C cos 3pt + D sin 3pt}
6. Applying the boundary conditions to determine constants A and B gives:
(i) u(0, t) = 0, hence 0 = A{C cos 3pt + D sin 3pt} from which we conclude that A = 0
Therefore,
u(x, t) = B sin px {C cos 3pt + D sin 3pt}
(1)
(ii) u(40, t) = 0, hence 0 = B sin 40p{C cos 3pt + D sin 3pt}
B ≠ 0 hence sin 40p = 0 from which, 40p = nπ and p =
7. Substituting in equation (1) gives: u(x, t) = B sin
or, more generally,=
u n ( x, t )
∞
nπ x
40
∑ sin
n =1
nπ
40
3nπ t
3nπ t 

+ D sin
C cos

40
40 

nπ x 
3nπ t
3nπ t 
+ Bn sin
 An cos

40 
40
40 
(2)
where An = BC and Bn = BD
8. From equation (8), page 890 of textbook,
An =
=
2 L
nπ x
f ( x) sin
dx
∫
L 0
L
40  60 − 1.5 x 
2  20  1.5 
nπ x
nπ x 
x  sin
dx+∫ 
d x

 sin

∫
0
20
40   20 
40
40
 20 

Each integral is determined using integration by parts (see Chapter 68, page 739) with the result:
An =
(8)(1.5)
nπ
12
nπ
sin
sin
=
2
2
2
2
nπ
2 nπ
2
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© 2014, John Bird
From equation (9), page 890 of textbook,
Bn =
2
cnπ
∫
L
0
g ( x) sin
nπ x
dx
L
 ∂u 
 ∂t  = 0 = g(x) thus, Bn = 0
t =0
Substituting into equation
(2) gives: un ( x, t )
=
∞
∑ sin
n =1
∞
=
∑ sin
n =1
Hence,
nπ x 
3nπ t
3nπ t 
+ Bn sin
 An cos

40 
40
40 
nπ x  12
nπ
3nπ t
nπ t 
+ (0) sin
cos
 2 2 sin

40  n π
2
40
50 
u(x, t) =
12
π2
∞
1
∑n
n =1
2
sin
nπ
nπ x
3nπ t
sin
cos
2
40
40
2. The centre point of an elastic string between two points P and Q, 80 cm apart, is deflected a
distance of 1 cm from its position of rest perpendicular to PQ and released initially with zero
velocity. Apply the wave equation
∂ 2u 1 ∂ 2u
where c = 8, to determine the motion of a point
=
∂x 2 c 2 ∂t 2
distance x from P at time t.
The elastic string is shown in the diagram below.
The boundary and initial conditions given are:
u (0, t ) = 0 

u (80, t ) = 0 
u(x, 0) = f(x) =
1
x
40
=−
1
x+2
40
0 ≤ x ≤ 40
40 ≤ x ≤ 80
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 ∂u 
 ∂t  = 0 i.e. zero initial velocity
t =0
Assuming a solution u = XT, where X is a function of x only, and T is a function of t only,
then
∂u
= X 'T
∂x
and
∂ 2u
= X '' T
∂x 2
∂u
= XT '
∂y
and
Substituting into the partial differential equation,
X '' T =
gives:
1
XT ''
c2
X '' T =
i.e.
1
XT '' since c = 8
64
X '' T ''
=
64T
X
X '' T ''
then
=
X
64T
µ=
X ''
X
X′′ – µX = 0
from which,
∂ 2u
= XT ''
∂y 2
∂ 2u 1 ∂ 2u
=
∂x 2 c 2 ∂t 2
Separating the variables gives:
Let constant, µ =
and
µ=
and
T ''
64T
T′′ – 64µT = 0
and
Letting µ = – p 2 to give an oscillatory solution gives
X′′ + p 2 X = 0
and
and the auxiliary equation is: m 2 + p 2 = 0 from which, m =− p 2 =
± jp
T′′ + 64 p 2 T = 0
and the auxiliary equation is:
−64 p 2 =
± j8 p
m 2 + 64 p 2 = 0 from which, m =
Solving each equation gives:
X = A cos px + B sin px
and
T = C cos 8pt + D sin 8pt
Thus, u(x, t) = {A cos px + B sin px}{C cos 8pt + D sin 8pt}
Applying the boundary conditions to determine constants A and B gives:
(i) u(0, t) = 0, hence 0 = A{C cos 8pt + D sin 8pt} from which we conclude that A = 0
Therefore,
u(x, t) = B sin px {C cos 8pt + D sin 8pt}
(1)
(ii) u(80, t) = 0, hence 0 = B sin 80p{C cos 8pt + D sin 8pt}
B ≠ 0 hence sin 80p = 0 from which, 80p = nπ and p =
Substituting in equation (1) gives: u(x, t) = B sin
or, more generally, =
u n ( x, t )
∞
8nπ t
8nπ t 
nπ x 
+ D sin
C cos

80
80 
80 
∑ sin
n =1
nπ
80
nπ x 
nπ t
nπ t 
+ Bn sin
 An cos

80 
10
10 
1322
(2)
© 2014, John Bird
An =
(8)(1)
8
nπ
nπ
sin
sin
=
2
2
2
2
2 nπ
2
nπ
Bn =
2
cnπ
∫
L
0
g ( x) sin
nπ x
dx
L
 ∂u 
 ∂t  = 0 = g(x) thus, Bn = 0
t =0
∞
∑ sin
Substituting into equation
(2) gives: un ( x, t )
=
n =1
∞
=
∑ sin
n =1
Hence,
u(x, t) =
nπ x 
nπ t
nπ t 
+ Bn sin
 An cos

80 
10
10 
8
π
2
∞
nπ x  8
nπ
nπ t
nπ t 
cos
+ (0) sin
 2 2 sin

80  n π
2
10
10 
1
∑n
n =1
1323
2
sin
nπ
nπ x
nπ t
sin
cos
2
80
10
© 2014, John Bird
EXERCISE 321 Page 894
1. A metal bar, insulated along its sides, is 4 m long. It is initially at a temperature of 10°C and at
time t = 0, the ends are placed into ice at 0°C. Find an expression for the temperature at a point P
at a distance x m from one end at any time t seconds after t = 0
The temperature u along the length of the bar is shown in the diagram below
The heat conduction equation is
∂ 2u 1 ∂u
and the given boundary conditions are:
=
∂x 2 c 2 ∂t
u(0, t) = 0, u(4, t) = 0 and
u(x, 0) = 10
Assuming a solution of the form u = XT, then, X = A cos px + B sin px
T = k e − p 2 c 2t
and
Thus, the general solution is given by: u(x, t) = {P cos px + Q sin px} e − p2c2t
u(0, t) = 0 thus 0 = P e − p2c2t from which, P = 0 and u(x, t) = {Q sin px} e − p2c2t
Also, u(4, t) = 0 thus 0 = {Q sin 4p} e − p2c2t
Since Q ≠ 0, sin 4p = 0 from which, 4p = nπ where n = 1, 2, 3, … and p =
∞
Hence,
u(x, t) =

∑ Q
n
e − p2c2t sin
n =1
nπ
4
nπ x 

4 
The final initial condition given was that at t = 0, u = 10, i.e. u(x, 0) = f(x) = 10
∞
Hence,
10 =

∑ Q
n
n =1
sin
nπ x 

4 
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© 2014, John Bird
where, from Fourier coefficients, Qn = 2 × mean value of 10 sin
nπ x
from x = 0 to x = 4
4
nπ x 

cos
4

2
nπ x
4  = − 20 cos 4nπ − cos 0  = 20 (1 − cos nπ )
d x = 5 −
Qn = ∫ 10sin
 nπ
0
nπ 
nπ 
4
4
4



4

0
4
i.e.
= 0 (when n is even) and
∞
Hence, the required solution is: u(x, t) =

∑ Q
n
e − p2c2t sin
n =1
40
(when n is odd)
nπ
nπ x 

4 
nπ x
1 − n2π 2c2t
=
e 16 sin
∑
π n ( odd ) =1 n
4
40
∞
2. An insulated uniform metal bar, 8 m long, has the temperature of its ends maintained at 0°C, and
at time t = 0 the temperature distribution f(x) along the bar is defined by f(x) = x(8 – x). If c 2 = 1,
solve the heat conduction equation
∂ 2u 1 ∂u
to determine the temperature u at any point in
=
∂x 2 c 2 ∂t
the bar at time t.
The temperature u along the length of bar is shown in the diagram below
The heat conduction equation is
∂ 2u 1 ∂u
and the given boundary conditions are:
=
∂x 2 c 2 ∂t
u(0, t) = 0, u(8, t) = 0 and
u(x, 0) = 0
Assuming a solution of the form u = XT, then, X = A cos px + B sin px
and
T = k e − p 2 c 2t
Thus, the general solution is given by: u(x, t) = {P cos px + Q sin px} e − p2c2t
u(0, t) = 0 thus 0 = P e − p2c2t from which, P = 0
and u(x, t) = {Q sin px} e − p2c2t
Also, u(8, t) = 0 thus 0 = {Q sin 8p} e − p2c2t
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© 2014, John Bird
Since Q ≠ 0, sin 8p = 0 from which, 8p = nπ where n = 1, 2, 3, … and p =
∞
Hence,
u(x, t) =

∑ Q
n
e − p2c2t sin
n =1
nπ
8
nπ x 

8 
The final initial condition given was that at t = 0, u = 0, i.e. u(x, 0) = f(x) = 0
∞
Hence,
0=

∑ Q
n
sin
n =1
nπ x 

8 
where, from Fourier coefficients, Qn = 2 × mean value of x(8 – x) sin
i.e.
Qn =
nπ x
from x = 0 to x = 8,
8
8
2 8
nπ x
1 8
nπ x
nπ x 
x
(8
−
x
)
sin
=
d
x
8
x
sin
d
x
−
x 2 sin
d x

∫
∫
∫
0
0
0
8
8
4
8
8

Using integration by parts,
 (16)(8)

1 8
nπ x
(16)(8)

 − ( 0)
8
x
sin
d
x
=
−
cos
n
π
+
sin
n
π
2


4 ∫0
8
nπ
n
π
(
)


and
 (2)(8) 2
  (4)(8) 2 
1 8 2
nπ x
(4)(8) 2
(4)(8) 2

−−

−
=
π
−
π
−
π
x
x
n
n
n
sin
d
cos
sin
cos
2
3
 nπ
  ( nπ )3 
4 ∫0
8
n
π
n
π
(
)
(
)

 

3
8
= 0 (when n is even) and   (when n is odd)
π 
nπ x 

Hence, the required solution is: u(x, t) = ∑ Qn e − p2c2t sin
=
8 
n =1 
∞
8
= 
π 
3
 nπ 

−

Q
e
 n 8
∑
n =1 

∞
2
(1) t
sin
nπ x 

8 
nπ x
1 − n2π 2t
e 64 sin
∑
3
8
n ( odd ) =1 n
∞
3. The ends of an insulated rod PQ, 20 units long, are maintained at 0°C. At time t = 0, the
temperature within the rod rises uniformly from each end reaching 4°C at the mid-point of PQ.
Find an expression for the temperature u(x, t) at any point in the rod, distant x from P at any time
t after t = 0. Assume the heat conduction equation to be
∂ 2u 1 ∂u
and take c 2 = 1
=
∂x 2 c 2 ∂t
The temperature along the length of the rod is shown in the diagram below
1326
© 2014, John Bird
The heat conduction equation is
∂ 2u 1 ∂u
and the given boundary conditions are:
=
∂x 2 c 2 ∂t
u(0, t) = 0, u(20, t) = 0 and
u(x, 0) = 0
Assuming a solution of the form u = XT, then, X = A cos px + B sin px
T = k e − p 2 c 2t
and
Thus, the general solution is given by: u(x, t) = {P cos px + Q sin px} e − p2c2t
u(0, t) = 0 thus 0 = P e − p2c2t from which, P = 0
and u(x, t) = {Q sin px} e − p2c2t
Also, u(20, t) = 0 thus 0 = {Q sin 20p} e − p2c2t
Since Q ≠ 0, sin 20p = 0 from which, 20p = nπ where n = 1, 2, 3, … and p =
∞
Hence,
u(x, t) =

∑ Q
n
e − p2c2t sin
n =1
nπ
20
nπ x 

20 
where, from Fourier coefficients, 2 × the mean value from x = 0 to x = 20
Qn =
20 
2  10  2 
nπ x
2
nπ x 

x  sin
d x + ∫  − x + 8  sin


∫
10
20  0  5 
20
20 
 5

(see above diagram)
10
20
 2
  2 
 
n
x
n
x
π
π


n
x
n
x
n
x
2
2
π
π
π
  − x cos
   x  cos
 
sin
sin
8cos
1    5 
20
5
20


5
20
5
20
20
 
 
=
+
−
−

2
2
10  
 nπ 
 nπ 
 nπ   
 nπ   
 nπ 





 








 20 
 20 
 20   10 
20
20






0


=
 
  −4 cos nπ 4sin nπ
1  
2 +
2
 
2
π
n
10 


 nπ 



   20 
 20 

 

 

 − ( 0 )  +  8cos nπ + 0 − 8cos nπ
   nπ 

 nπ 
  




 20 

   20 
1327
 
nπ
nπ
nπ
4sin
8cos
  4 cos
2 −
2 −
2
−
2

π
π
n
n





 nπ 


   20 



 20 
  
 20 
 
 
  

  
  
© 2014, John Bird

nπ
nπ
nπ
nπ
nπ 
 −4 cos

4sin
4 cos
4sin
8cos
1 
2 +
2 + 8cos nπ − 8cos nπ −
2 +
2 +
2 
=


2
2
10   nπ 
 nπ 
 nπ 
 nπ 
 nπ  
 nπ 
 nπ 












  20 
 20 
 20 
 20 
 20  
 20 
 20 

nπ
nπ
nπ
 −8cos
8sin
8cos
1 
2 +
2 +
2
=

2
10   nπ 
 nπ 
 nπ 






  20 
 20 
 20 


nπ

 8sin
1


2
=

2
 10   nπ 

  20 






= 0 when n is even
2
nπ  320
nπ
8  20 
=
when n is odd
sin

 = 2 2 sin

10  nπ 
2  n π
2
nπ − π 2 n2 2(1) t
nπ x 
 320
sin
e 20 sin


∑
2 2
2
20 
n =1  n π
∞
Hence, the required solution is: u(x, t) =
=
320
π2
π t
1
nπ
nπ x −  n400


sin
sin
e
∑
2
2
20
n ( odd ) =1 n
∞
1328
2
2
© 2014, John Bird
EXERCISE 322 Page 897
1. A rectangular plate is bounded by the lines x = 0, y = 0, x = 1 and y = 3. Apply the Laplace
equation
∂ 2u ∂ 2u
+
=
0 to determine the potential distribution u(x, y) over the plate, subject to
∂x 2 ∂y 2
the following boundary conditions: u = 0 when x = 0
0 ≤ y ≤ 2,
u = 0 when x = 1
0 ≤ y ≤ 2,
u = 0 when y = 2
0 ≤ x ≤ 1,
u = 5 when y = 3
0≤x≤1
Initially a solution of the form u(x, y) = X(x)Y(y) is assumed, where X is a function of x only, and
Y is a function of y only. Simplifying to u = XY, determining partial derivatives, and substituting
into
∂ 2u ∂ 2u
+
=
0 gives:
∂x 2 ∂y 2
X′′Y + XY′′ = 0
X ''
Y ''
= −
X
Y
Separating the variables gives:
Letting each side equal a constant, – p 2 gives the two equations:
X′′ + p 2 X = 0
from which,
and
Y′′ – p 2 Y = 0
X = A cos px + B sin px
and Y = C e py + D e − py or
Y = C cosh py + D sinh py or
Y = E sinh p(y + φ)
Hence
u(x, y) = XY = {A cos px + B sin px }{E sinh p(y + φ)}
or
u(x, y) = {P cos px + Q sin px }{sinh p(y + φ)}
where P = AE and Q = BE
The first boundary condition is: u(0, y) = 0, hence 0 = P sinh p(y + φ) from which, P = 0
u(x, y) = Q sin px sinh p(y + φ)
Hence,
The second boundary condition is: u(1, y) = 0, hence 0 = Q sin p(1) sinh p(y + φ)
from which,
sin p = 0, hence,
p = nπ
for n = 1, 2, 3, …
The third boundary condition is: u(x, 2) = 0, hence, 0 = Q sin px sinh p(2 + φ)
sinh p(2 + φ) = 0
from which,
Hence,
and
φ = –2
u(x, y) = Q sin px sinh p(y – 2)
Since there are many solutions for integer values of n
∞
u(x, y) =
∑ Qn sin px sinh p( y − 2) =
n =1
∞
∑Q
n
sin nπ x sinh nπ ( y − 2)
(a)
n =1
1329
© 2014, John Bird
∞
The fourth boundary condition is: u(x, 3) = 5 = f(x), hence, f(x) =
∑Q
n
sin nπ x sinh nπ (3 − 2)
n =1
From Fourier series coefficients,
Qn sinh nπ = 2 × the mean value of f(x) sin nπ x from x = 0 to x = 1
2 1
10
10
 cos nπ x 
=−
5sin nπ x d=
x 10  −
( cos nπ − cos 0 ) = (1 − cos nπ )
∫

0
nπ
nπ
1
nπ  0

1
i.e.
=
= 0 (for even values of n), =
Hence,
=
Qn
20
(for odd values of n)
nπ
20
20
cosech nπ
=
nπ (sinh nπ ) nπ
∞
Hence, from equation (a),
u(x, y) =
∑Q
n
sin nπ x sinh nπ ( y − 2)
n =1
20
=
π
∞
∑
n ( odd ) ) =1
1
cos ech nπ sin nπ x sinh nπ ( y − 2)
n
2. A rectangular plate is bounded by the lines x = 0, y = 0, x = 3, y = 2. Determine the potential
distribution u(x, y) over the rectangle using the Laplace equation
following boundary conditions:
u (0, y) = 0
0 ≤ y ≤ 2,
∂ 2u ∂ 2u
+
=
0 , subject to the
∂x 2 ∂y 2
u (3, y) = 0
u (x, 2) = 0 0 ≤ x ≤ 3,
0 ≤ y ≤ 2,
u (x, 0) = x(3 – x) 0 ≤ x ≤ 3
Initially a solution of the form u(x, y) = X(x)Y(y) is assumed, where X is a function of x only, and
Y is a function of y only. Simplifying to u = XY, determining partial derivatives, and substituting
into
∂ 2u ∂ 2u
+
=
0 gives:
∂x 2 ∂y 2
X′′Y + XY′′ = 0
X ''
Y ''
= −
X
Y
Separating the variables gives:
Letting each side equal a constant, – p 2 gives the two equations:
X′′ + p 2 X = 0
from which,
and
Y′′ – p 2 Y = 0
X = A cos px + B sin px
and Y = C e py + D e − py or
Y = C cosh py + D sinh py or
Y = E sinh p(y + φ)
Hence
u(x, y) = XY = {A cos px + B sin px}{E sinh p(y + φ)}
or
u(x, y) = {P cos px + Q sin px }{sinh p(y + φ)}
1330
© 2014, John Bird
The first boundary condition is: u(0, y) = 0, hence 0 = P sinh p(y + φ) from which, P = 0
u(x, y) = Q sin px sinh p(y + φ)
Hence,
The second boundary condition is: u(3, y) = 0, hence 0 = Q sin 3p sinh p(y + φ)
from which,
sin 3p = 0, hence,
3p = nπ
i.e. p =
nπ
3
for n = 1, 2, 3, …
The third boundary condition is: u(x, 2) = 0, hence, 0 = Q sin px sinh p(2 + φ)
sinh p(2 + φ) = 0
from which,
Hence,
and
φ = –2
u(x, y) = Q sin px sinh p(y – 2)
Since there are many solutions for integer values of n,
∞
u(x, y) =
∑ Qn sin px sinh p( y − 2) =
n =1
∞
∑Q
n
sin
n =1
nπ
nπ
x sinh
( y − 2)
3
3
(a)
The fourth boundary condition is: u(x, 0) = x(3 – x) = 3x – x 2 = f(x),
∞
hence,
f(x) =
∑Q
n
sin
n =1
nπ
nπ
x sinh
(−2)
3
3
From Fourier series coefficients,
Qn sinh
−2nπ
nπ
= 2 × the mean value of f(x) sin
x from x = 0 to x = 3
3
3
=
2 3
nπ
( 3x − x 2 ) sin x d x
∫
0
1
3
3
3

 
 
n
x
n
x
n
x
n
x
n
x
π
π
π
π
π










2
  −3 x cos 
 3sin 
   − x cos 
 2 x sin 
 2 cos 
 

3 
3  
3 
3 
3  





= 2
+
−
+
+

2
2
3



 
n
π
n
π




n
n
π
π




 nπ 





 
 






 
 3 
 3 
3
3
3








0
0

by integration by parts (see Chapter 68)




2 
 −9 cos nπ 9 cos nπ 2 cos nπ
 27

= 2
2 − 2 cos nπ ) 
+
−
+
= 2
(
3
3
3
3
 nπ 
n π

 nπ 
 nπ  
  nπ 






  3 
 3 
 3 
 3  
=
54
( 2 − 2 cos nπ )
n3π 3
= 0 (for even values of n),
=
216
(for odd values of n)
n3π 3
1331
© 2014, John Bird
Hence, Qn
=
216
216
−2nπ
=
cosech
3
3
−2nπ  n π
3

n3π 3  sinh

3 

Hence, from equation (a),
∞
∞
nπ
nπ
216
−2nπ
nπ x
nπ
u(x,
y) = ∑ Qn sin
=
x sinh
( y − 2) ∑
cosech
sin
sinh
( y − 2)
3 3
3
3
3
3
3
n 1=
n 1 n π
=
216
π
3
∞
∑
n ( odd ) ) =1
1332
nπ x
nπ
1
2nπ
cosech
sin
sinh
(2 − y )
3
n
3
3
3
© 2014, John Bird