Method of Separation of Variables for Laplace Equation - CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


4.2 Laplace’s equation
The two dimensional Laplace’s equation is
u
xx
+ u
yy
= 0 (4.120)
We will show that a separation of variables also works for this equation
As an example consider the boundary conditions
u(x, 0) = 0, u(x, 1) = x¡ x
2
(4.121a)
u(0, y) = 0, u(1, 0) = 0. (4.121b)
If we assume separable solutions of the form
u(x, y) = X(x)Y(y), (4.122)
then substituting this into (4.120) gives
X
00
Y+ XY
00
= 0. (4.123)
Dividing by XY and expanding gives
X
00
X
+
Y
00
Y
= 0, (4.124)
and since each term is only a function of x or y, then each must be constant
giving
X
00
X
= l,
Y
00
Y
=¡l. (4.125)
From the ?rst of (4.121a) and both of (4.121b) we deduce the boundary
conditions
X(0) = 0, X(1) = 0, Y(0) = 0. (4.126)
Page 2


4.2 Laplace’s equation
The two dimensional Laplace’s equation is
u
xx
+ u
yy
= 0 (4.120)
We will show that a separation of variables also works for this equation
As an example consider the boundary conditions
u(x, 0) = 0, u(x, 1) = x¡ x
2
(4.121a)
u(0, y) = 0, u(1, 0) = 0. (4.121b)
If we assume separable solutions of the form
u(x, y) = X(x)Y(y), (4.122)
then substituting this into (4.120) gives
X
00
Y+ XY
00
= 0. (4.123)
Dividing by XY and expanding gives
X
00
X
+
Y
00
Y
= 0, (4.124)
and since each term is only a function of x or y, then each must be constant
giving
X
00
X
= l,
Y
00
Y
=¡l. (4.125)
From the ?rst of (4.121a) and both of (4.121b) we deduce the boundary
conditions
X(0) = 0, X(1) = 0, Y(0) = 0. (4.126)
The remaining boundary condition in (4.121a) will be used later. As seen
the in previous section, in order to solve the X equation in (4.125) subject
to the boundary conditions (4.126), it is necessary to set l = ¡k
2
. The X
equation (4.125) as the general solution
X = c
1
sin kx+ c
2
cos kx (4.127)
To satisfy the boundary conditions in (4.126) it is necessary to have c
2
= 0
and k = np, k2Z
+
so
X(x) = c
1
sin npx. (4.128)
From (4.125), we obtain the solution to the Y equation
Y(y) = c
3
sinh npy+ c
4
cosh npy (4.129)
Since Y(0) = 0 this implies c
4
= 0 so
X(x)Y(y) = a
n
sin npx sinh npy (4.130)
where we have chosen a
n
= c
1
c
3
. Therefore, we obtain the solution to
(4.120) subject to three of the four boundary conditions in (4.121)
u =
¥
å
n=1
a
n
sin npx sinh npy. (4.131)
The remaining boundary condition is (4.121a) now needs to be satis?ed,
thus
u(x, 1) = x¡ x
2
=
¥
å
n=1
a
n
sin npx sinh np. (4.132)
This looks like a Fourier sine series and if we let A
n
= a
n
sinh np, this
becomes
¥
å
n=1
A
n
sin npx = x¡ x
2
. (4.133)
which is precisely a Fourier sine series. The coef?cients A
n
are given by
A
n
=
2
1
Z
2
0
³
x¡ x
2
´
sin npx dx
=
16
n
3
p
3
(1¡ cos np), (4.134)
Page 3


4.2 Laplace’s equation
The two dimensional Laplace’s equation is
u
xx
+ u
yy
= 0 (4.120)
We will show that a separation of variables also works for this equation
As an example consider the boundary conditions
u(x, 0) = 0, u(x, 1) = x¡ x
2
(4.121a)
u(0, y) = 0, u(1, 0) = 0. (4.121b)
If we assume separable solutions of the form
u(x, y) = X(x)Y(y), (4.122)
then substituting this into (4.120) gives
X
00
Y+ XY
00
= 0. (4.123)
Dividing by XY and expanding gives
X
00
X
+
Y
00
Y
= 0, (4.124)
and since each term is only a function of x or y, then each must be constant
giving
X
00
X
= l,
Y
00
Y
=¡l. (4.125)
From the ?rst of (4.121a) and both of (4.121b) we deduce the boundary
conditions
X(0) = 0, X(1) = 0, Y(0) = 0. (4.126)
The remaining boundary condition in (4.121a) will be used later. As seen
the in previous section, in order to solve the X equation in (4.125) subject
to the boundary conditions (4.126), it is necessary to set l = ¡k
2
. The X
equation (4.125) as the general solution
X = c
1
sin kx+ c
2
cos kx (4.127)
To satisfy the boundary conditions in (4.126) it is necessary to have c
2
= 0
and k = np, k2Z
+
so
X(x) = c
1
sin npx. (4.128)
From (4.125), we obtain the solution to the Y equation
Y(y) = c
3
sinh npy+ c
4
cosh npy (4.129)
Since Y(0) = 0 this implies c
4
= 0 so
X(x)Y(y) = a
n
sin npx sinh npy (4.130)
where we have chosen a
n
= c
1
c
3
. Therefore, we obtain the solution to
(4.120) subject to three of the four boundary conditions in (4.121)
u =
¥
å
n=1
a
n
sin npx sinh npy. (4.131)
The remaining boundary condition is (4.121a) now needs to be satis?ed,
thus
u(x, 1) = x¡ x
2
=
¥
å
n=1
a
n
sin npx sinh np. (4.132)
This looks like a Fourier sine series and if we let A
n
= a
n
sinh np, this
becomes
¥
å
n=1
A
n
sin npx = x¡ x
2
. (4.133)
which is precisely a Fourier sine series. The coef?cients A
n
are given by
A
n
=
2
1
Z
2
0
³
x¡ x
2
´
sin npx dx
=
16
n
3
p
3
(1¡ cos np), (4.134)
and since A
n
= a
n
sinh np, this gives
a
n
=
4(1¡(¡1)
n
)
n
3
p
3
sinh np
. (4.135)
Thus, the solution to Laplace’s equation with the boundary conditions
given in (4.121) is
u(x, y) =
4
p
3
¥
å
n=1
1¡(¡1)
n
n
3
sin npx
sinh npy
sinh np
. (4.136)
Figure 11 show both a top view and a 3¡ D view of the solution.
0.001 u  = y x 0.2 u  = 0.1 u  = 0.05 u  = 0.025 u  = 1.0
0.5
0.0
1.0 0.0 0.5
0.0
0.1
0.2
0.5
u 1.0
y 1.0
x Figure 11. The solution (4.120) with the boundary conditions (4.121)
In general, using separation of variables, the solution of
u
xx
+ u
yy
= 0, 0 < x < L
x
, 0 < y < L
y
, (4.137)
subject to
u(x, 0) = 0, u(x, 1) = f(x),
u(0, y) = 0, u(1, y) = 0,
is
u =
¥
å
n=1
b
n
sin
npx
L
x
sinh
npy
L
y
sinh
np
L
y
, (4.139)
Page 4


4.2 Laplace’s equation
The two dimensional Laplace’s equation is
u
xx
+ u
yy
= 0 (4.120)
We will show that a separation of variables also works for this equation
As an example consider the boundary conditions
u(x, 0) = 0, u(x, 1) = x¡ x
2
(4.121a)
u(0, y) = 0, u(1, 0) = 0. (4.121b)
If we assume separable solutions of the form
u(x, y) = X(x)Y(y), (4.122)
then substituting this into (4.120) gives
X
00
Y+ XY
00
= 0. (4.123)
Dividing by XY and expanding gives
X
00
X
+
Y
00
Y
= 0, (4.124)
and since each term is only a function of x or y, then each must be constant
giving
X
00
X
= l,
Y
00
Y
=¡l. (4.125)
From the ?rst of (4.121a) and both of (4.121b) we deduce the boundary
conditions
X(0) = 0, X(1) = 0, Y(0) = 0. (4.126)
The remaining boundary condition in (4.121a) will be used later. As seen
the in previous section, in order to solve the X equation in (4.125) subject
to the boundary conditions (4.126), it is necessary to set l = ¡k
2
. The X
equation (4.125) as the general solution
X = c
1
sin kx+ c
2
cos kx (4.127)
To satisfy the boundary conditions in (4.126) it is necessary to have c
2
= 0
and k = np, k2Z
+
so
X(x) = c
1
sin npx. (4.128)
From (4.125), we obtain the solution to the Y equation
Y(y) = c
3
sinh npy+ c
4
cosh npy (4.129)
Since Y(0) = 0 this implies c
4
= 0 so
X(x)Y(y) = a
n
sin npx sinh npy (4.130)
where we have chosen a
n
= c
1
c
3
. Therefore, we obtain the solution to
(4.120) subject to three of the four boundary conditions in (4.121)
u =
¥
å
n=1
a
n
sin npx sinh npy. (4.131)
The remaining boundary condition is (4.121a) now needs to be satis?ed,
thus
u(x, 1) = x¡ x
2
=
¥
å
n=1
a
n
sin npx sinh np. (4.132)
This looks like a Fourier sine series and if we let A
n
= a
n
sinh np, this
becomes
¥
å
n=1
A
n
sin npx = x¡ x
2
. (4.133)
which is precisely a Fourier sine series. The coef?cients A
n
are given by
A
n
=
2
1
Z
2
0
³
x¡ x
2
´
sin npx dx
=
16
n
3
p
3
(1¡ cos np), (4.134)
and since A
n
= a
n
sinh np, this gives
a
n
=
4(1¡(¡1)
n
)
n
3
p
3
sinh np
. (4.135)
Thus, the solution to Laplace’s equation with the boundary conditions
given in (4.121) is
u(x, y) =
4
p
3
¥
å
n=1
1¡(¡1)
n
n
3
sin npx
sinh npy
sinh np
. (4.136)
Figure 11 show both a top view and a 3¡ D view of the solution.
0.001 u  = y x 0.2 u  = 0.1 u  = 0.05 u  = 0.025 u  = 1.0
0.5
0.0
1.0 0.0 0.5
0.0
0.1
0.2
0.5
u 1.0
y 1.0
x Figure 11. The solution (4.120) with the boundary conditions (4.121)
In general, using separation of variables, the solution of
u
xx
+ u
yy
= 0, 0 < x < L
x
, 0 < y < L
y
, (4.137)
subject to
u(x, 0) = 0, u(x, 1) = f(x),
u(0, y) = 0, u(1, y) = 0,
is
u =
¥
å
n=1
b
n
sin
npx
L
x
sinh
npy
L
y
sinh
np
L
y
, (4.139)
where
b
n
=
2
L
x
Z
L
x
0
f(x) sin
npx
L
x
dx. (4.140)
In the next three examples, we will construct solutions of Laplace’s equa-
tion when we have nonzero boundary condition on each of the remaining
three sides of the region 0 < 1, 0 < y < 1.
Example 8
Solve
u
xx
+ u
yy
= 0, (4.141)
subject to
u(x, 0) = 0, u(x, 1) = 0, (4.142a)
u(0, y) = 0, u(1, y) = y¡ y
2
. (4.142b)
Assume separable solutions of the form
u(x, y) = X(x)Y(y) (4.143)
Then substituting this into (4.141) gives
X
00
Y+ XY
00
= 0. (4.144)
Dividing by XY and expanding gives
X
00
X
+
Y
00
Y
= 0, (4.145)
from which we obtain
X
00
X
= l,
Y
00
Y
=¡l (4.146)
From (4.186) we deduce the boundary conditions
X(0) = 0, Y(0) = 0, Y(1) = 0. (4.147)
Page 5


4.2 Laplace’s equation
The two dimensional Laplace’s equation is
u
xx
+ u
yy
= 0 (4.120)
We will show that a separation of variables also works for this equation
As an example consider the boundary conditions
u(x, 0) = 0, u(x, 1) = x¡ x
2
(4.121a)
u(0, y) = 0, u(1, 0) = 0. (4.121b)
If we assume separable solutions of the form
u(x, y) = X(x)Y(y), (4.122)
then substituting this into (4.120) gives
X
00
Y+ XY
00
= 0. (4.123)
Dividing by XY and expanding gives
X
00
X
+
Y
00
Y
= 0, (4.124)
and since each term is only a function of x or y, then each must be constant
giving
X
00
X
= l,
Y
00
Y
=¡l. (4.125)
From the ?rst of (4.121a) and both of (4.121b) we deduce the boundary
conditions
X(0) = 0, X(1) = 0, Y(0) = 0. (4.126)
The remaining boundary condition in (4.121a) will be used later. As seen
the in previous section, in order to solve the X equation in (4.125) subject
to the boundary conditions (4.126), it is necessary to set l = ¡k
2
. The X
equation (4.125) as the general solution
X = c
1
sin kx+ c
2
cos kx (4.127)
To satisfy the boundary conditions in (4.126) it is necessary to have c
2
= 0
and k = np, k2Z
+
so
X(x) = c
1
sin npx. (4.128)
From (4.125), we obtain the solution to the Y equation
Y(y) = c
3
sinh npy+ c
4
cosh npy (4.129)
Since Y(0) = 0 this implies c
4
= 0 so
X(x)Y(y) = a
n
sin npx sinh npy (4.130)
where we have chosen a
n
= c
1
c
3
. Therefore, we obtain the solution to
(4.120) subject to three of the four boundary conditions in (4.121)
u =
¥
å
n=1
a
n
sin npx sinh npy. (4.131)
The remaining boundary condition is (4.121a) now needs to be satis?ed,
thus
u(x, 1) = x¡ x
2
=
¥
å
n=1
a
n
sin npx sinh np. (4.132)
This looks like a Fourier sine series and if we let A
n
= a
n
sinh np, this
becomes
¥
å
n=1
A
n
sin npx = x¡ x
2
. (4.133)
which is precisely a Fourier sine series. The coef?cients A
n
are given by
A
n
=
2
1
Z
2
0
³
x¡ x
2
´
sin npx dx
=
16
n
3
p
3
(1¡ cos np), (4.134)
and since A
n
= a
n
sinh np, this gives
a
n
=
4(1¡(¡1)
n
)
n
3
p
3
sinh np
. (4.135)
Thus, the solution to Laplace’s equation with the boundary conditions
given in (4.121) is
u(x, y) =
4
p
3
¥
å
n=1
1¡(¡1)
n
n
3
sin npx
sinh npy
sinh np
. (4.136)
Figure 11 show both a top view and a 3¡ D view of the solution.
0.001 u  = y x 0.2 u  = 0.1 u  = 0.05 u  = 0.025 u  = 1.0
0.5
0.0
1.0 0.0 0.5
0.0
0.1
0.2
0.5
u 1.0
y 1.0
x Figure 11. The solution (4.120) with the boundary conditions (4.121)
In general, using separation of variables, the solution of
u
xx
+ u
yy
= 0, 0 < x < L
x
, 0 < y < L
y
, (4.137)
subject to
u(x, 0) = 0, u(x, 1) = f(x),
u(0, y) = 0, u(1, y) = 0,
is
u =
¥
å
n=1
b
n
sin
npx
L
x
sinh
npy
L
y
sinh
np
L
y
, (4.139)
where
b
n
=
2
L
x
Z
L
x
0
f(x) sin
npx
L
x
dx. (4.140)
In the next three examples, we will construct solutions of Laplace’s equa-
tion when we have nonzero boundary condition on each of the remaining
three sides of the region 0 < 1, 0 < y < 1.
Example 8
Solve
u
xx
+ u
yy
= 0, (4.141)
subject to
u(x, 0) = 0, u(x, 1) = 0, (4.142a)
u(0, y) = 0, u(1, y) = y¡ y
2
. (4.142b)
Assume separable solutions of the form
u(x, y) = X(x)Y(y) (4.143)
Then substituting this into (4.141) gives
X
00
Y+ XY
00
= 0. (4.144)
Dividing by XY and expanding gives
X
00
X
+
Y
00
Y
= 0, (4.145)
from which we obtain
X
00
X
= l,
Y
00
Y
=¡l (4.146)
From (4.186) we deduce the boundary conditions
X(0) = 0, Y(0) = 0, Y(1) = 0. (4.147)
The remaining boundary condition in (4.186) will be used later. As seen
the in previous problem, in order to solve the Y equation in (4.146) subject
to the boundary conditions (4.147), it is necessary to set l = k
2
. The Y
equation (4.146) as the general solution
Y = c
1
sin ky+ c
2
cos ky (4.148)
To satisfy the boundary conditions in (4.147) it is necessary to have c
2
= 0
and k = np so
Y(y) = c
1
sin npy. (4.149)
From (4.146), we obtain the solution to the X equation
X(x) = c
3
sinh npx+ c
4
cosh npx (4.150)
Since X(0) = 0 this implies c
4
= 0. This gives
X(x)Y(y) = a
n
sinh npx sin npy (4.151)
where we have chosen a
n
= c
1
c
3
. Therefore, we obtain
u =
¥
å
n=1
a
n
sinh npx sin npy. (4.152)
The remaining boundary condition is (4.186) now needs to be satis?ed,
thus
u(1, y) = y¡ y
2
=
¥
å
n=1
a
n
sinh np sin npy. (4.153)
If we let A
n
= a
n
sinh np, this becomes
¥
å
n=1
A
n
sin npy = y¡ y
2
. (4.154)
Comparing with previous problem, we ?nd that interchanging x and y
interchanges the two problems and thus we con conclude that
A
n
=
16
n
3
p
3
(1¡ cos np), (4.155)