Hassani S. Mathematical Methods: For Students of Physics and Related Fields
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596 Laplace’s Equation: Cartesian Coordinates
Multiplying the second equation by e
√
α
2
b
and using B = −A,weobtain
A
*
e
2
√
α
2
b
− 1
+
=0 ⇒ A =0 ore
2
√
α
2
b
=1.
The first choice (A =0)andA = −B yields a trivial solution again. Therefore, weBoundary
conditions force
α
2
to be
imaginary.
have to assume that the second choice holds. However, even with the second choice,
if we restrict ourselves to the real numbers, the only solution of e
2
√
α
2
b
=1would
be α
2
= 0 which is a contradiction because we are dealing precisely with the case
of α
2
= 0. It follows that
√
α
2
must be a complex number. In fact, recalling that
e
2inπ
=1foranyintegern, we immediately get
2
√
α
2
b =2inπ ⇒
√
α
2
b = inπ ⇒ α
2
= −
nπ
b
2
,n= ±1, ±2,....
Note that n = 0 is excluded because this choice would make α
2
=0.
We now turn to the X equation. Since α
1
+ α
2
=0,weobtain
α
1
= −α
2
=
nπ
b
2
,n= ±1, ±2,...,
and
d
2
X
dx
2
−
nπ
b
2
X =0 ⇒ X(x)=Ce
nπx/b
+ De
−nπx/b
.
To be physically meaningful, the potential must remain finite as x → +∞.It
follows from the last equation that either n is negative and D =0,orn is positive
and C = 0. Either choice will lead to the same final result as the reader may verify.
Choosing positive values of n with C = 0, the potential can be written as
Φ
n
(x, y)=ADe
−nπx/b
*
e
inπy/b
− e
−inπy/b
+
= A
n
e
−nπx/b
sin
nπy
b
,
whereweusedA = −B and introduced a new constant A
n
. We also subscripted the
potential because for every n, we get a different function for Φ. All such functions
are solutions of Laplace’s equation and therefore, so is their sum. In fact, it is only
the sum that is general enough to result in the final solution. We thus write
Φ(x, y)=
∞
n=1
Φ
n
(x, y)=
∞
n=1
A
n
e
−nπx/b
sin
nπy
b
. (25.5)
This is a Fourier series in y with x dependent coefficients. The potential will be
completely determined if the constants A
n
can be determined. This is where the
last unused information comes in: The potential at x =0isV . Substituting this
information in Equation (25.5) yields
V =Φ(0,y)=
∞
n=1
A
n
sin
nπy
b
from which A
n
can be determined using the Fourier series techniques. We leave it
for the reader to show that A
n
=2V [1 − (−1)
n
]/(nπ) (see Problem 25.1), or
A
n
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
4V
nπ
if n is odd,
0ifn is even.
25.2 Cartesian Coordinates 597
By writing n =2k +1with k =0, 1, 2,..., the potential in the region of interest
becomes
Φ(x, y)=
4V
π
∞
k=0
e
−(2k+1)πx/b
2k +1
sin
(2k +1)πy
b
(25.6)
=
4V
π
e
−πx/b
sin
πy
b
+
e
−3πx/b
3
sin
3πy
b
+
e
−5πx/b
5
sin
5πy
b
+ ···
.
Because of the exponential factor, the series converges very rapidly, and for large
values of x the potential very quickly drops to zero. Figure 25.3 shows the potential
function (in arbitrary units) as a function of x and y.
Example 25.2.1 illustrates the general feature of solving Laplace’s equation
by the separation of variables in Cartesian coordinates. This feature works in
other coordinate systems as well. The separation of variables results in some
ODEs which involve parameters (in the case above, the α’s) to be determined
by some of the BCs. All values of these parameters—which in all cases of
interest to us will turn out to be integers—consistent with the used boundary
conditions are allowed and must be taken into account, i.e., an infinite sum
(with as yet undetermined coefficients) over such parameters is to be formed
as the most general solution of Laplace’s equation. By applying the remaining
BCs, the undetermined coefficients can be evaluated, resulting in the unique
solution appropriate for the geometry of the problem. If the geometry extends
to infinity in a certain direction, then such an infinity is to be considered as a
BC. It is extremely useful to take into account any symmetry of the problem
as such symmetries will simplify the solution considerably. The symmetry in
the z-direction of Example 25.2.1 saved us the trouble of solving one (out of
three) complete ODE.
Figure 25.3: The potential function inside the semi-infinite box of Figure 25.2 when
only 20 terms of the infinite series are kept. Note how quickly the potential drops to
zero along the x-axis due to the exponential factor.
598 Laplace’s Equation: Cartesian Coordinates
Example 25.2.2. Steady-state heat conduction problems also obey Laplace’s
equation. So, let us consider a rectangular medium infinite in the z direction en-infinitely long
rectangular heat
conductor
closed by two pairs of parallel slabs of width a and b as shown in Figure 25.4. The
temperatures of the slabs of width a—assumed parallel to the x-axis—are zero. The
temperatures of the other two slabs are T
1
and T
2
. Wewanttofindthetemperature
at all points in the region enclosed after the equilibrium is reached.
As in Example 25.2.1, we can ignore the z-dependence and write T (x, y)=
X(x)Y (y)whereX and Y satisfy Equation (25.3) with α
1
= −α
2
. For exactly the
same reason as in Example 25.2.1, α
2
cannot be zero and Y can only be of the form
Y (y)=A
n
sin
nπy
b
,n=1, 2,...,
where the subscript on A
n
reminds us that different constants can be chosen to
multiply different sine functions. The solution for X will, however, be different. We
still have
X(x)=C
n
e
nπx/b
+ D
n
e
−nπx/b
,n=1, 2,...,
but neither C
n
nor D
n
is zero this time. Multiplying the two functions and redefining
the constants, we can write
T
n
(x, y)=
A
n
e
nπx/b
+ B
n
e
−nπx/b
sin
nπy
b
and the most general infinite series solution becomes
T (x, y)=
∞
n=1
A
n
e
nπx/b
+ B
n
e
−nπx/b
sin
nπy
b
. (25.7)
So far, we have used only two of the four BCs. The remaining two will deter-
mine the unknowns A
n
and B
n
. Substituting these BCs yields the following two
equations:
T
1
= T (0,y)=
∞
n=1
(A
n
+ B
n
)
≡E
n
sin
nπy
b
=
∞
n=1
E
n
sin
nπy
b
,
T
2
= T (a, y)=
∞
n=1
A
n
e
nπa/b
+ B
n
e
−nπa/b
≡F
n
sin
nπy
b
=
∞
n=1
F
n
sin
nπy
b
,
y
x
Τ
= 0
Τ
= 0
b
Τ
= T
1
Τ
= T
2
a
Figure 25.4: The cross section of the two pairs of parallel slabs maintained at different
temperatures.
25.2 Cartesian Coordinates 599
where we have redefined the constants multiplying the sine functions. As in Example
25.2.1 and Problem 25.1, we have
E
n
=
2T
1
nπ
[1 − (−1)
n
],F
n
=
2T
2
nπ
[1 − (−1)
n
].
These relations show that only odd terms of the infinite series are of relevance, and
they are given by
E
2k+1
≡ A
2k+1
+ B
2k+1
=
4T
1
π(2k +1)
, (25.8)
F
2k+1
≡ A
2k+1
e
(2k+1)πa/b
+ B
2k+1
e
−(2k+1)πa/b
=
4T
2
π(2k +1)
.
These are two equations in two unknowns which can be solved to get
A
2k+1
=
2(T
2
− T
1
e
−(2k+1)πa/b
)
π(2k + 1) sinh[(2k +1)πa/b]
,
B
2k+1
=
2(T
1
e
(2k+1)πa/b
− T
2
)
π(2k + 1) sinh[(2k +1)πa/b]
.
Substituting in Equation (25.7)—with n replaced by 2k + 1—and rearranging terms
yields
T (x, y)=
4
π
∞
k=0
T
1
sinh
*
(2k+1)π(a−x)
b
+
+ T
2
sinh
*
(2k+1)πx
b
+
(2k +1)sinh
*
(2k+1)πa
b
+
sin
(2k +1)πy
b
. (25.9)
The reader is urged to verify that when T
2
=0anda →∞, we recover the result
of Example 25.2.1—with V replaced by T
1
—as we should. Figure 25.5 shows the
potential function (in arbitrary units) as a function of x and y.
The examples treated so far may give the impression that α
1
or α
2
is
never zero. This has to do with the specific BCs imposed on Φ (or T ). In
Figure 25.5: The potential function inside the box of Figure 25.4 for the special case
of a = b and T
1
= T
2
when only 20 terms of the infinite series are kept.
600 Laplace’s Equation: Cartesian Coordinates
both examples, Y vanishes at both y =0andy = b. Such a BC excludes
α
2
= 0 because the corresponding Y ,namelyY = Ay + B, cannot satisfy
those conditions unless Y = 0 identically.
Example 25.2.3.
To see how the α
1
= −α
2
= 0 terms can enter in the game, let
us modify the temperatures of the plates and strips of Example 25.2.2 so that the
bottom plate and the left strip are held at T = 0 while the top plate is held at T
1
and the right strip at T
2
.
Let us write the most general solution of Laplace’s equation obtained from sep-
arating variables including the α
1
=0=−α
2
term. Since the nonzero α
1
and
α
2
are of opposite signs, one of them will be positive and will have real square
roots and the other pure imaginary roots. Let us assume that α
1
is positive.
Then X will be of exponential type and Y of imaginary exponential or trigono-
metric type. It follows that the most general solution of Laplace’s equation can be
written as
T (x, y)=(A
0
x + B
0
)(C
0
y + D
0
) (25.10)
+
∞
α
A
α
e
√
αx
+ B
α
e
−
√
αx
.
C
α
sin(
√
αy)+D
α
cos(
√
αy)
/
,
where we have used α for α
1
= −α
2
. Itisconvenienttoimposethey BCs first. So,
since T (x, 0) = 0, we have
0=(A
0
x + B
0
) D
0
+
∞
α
A
α
e
√
αx
+ B
α
e
−
√
αx
D
α
which should hold for arbitrary values of x. This can happen only if D
0
= D
α
=0.
So, absorbing the multiplicative constant C
0
and C
α
into the A’s and B’s, we get a
new expression for the temperature:
T (x, y)=(A
0
x + B
0
) y +
∞
α
A
α
e
√
αx
+ B
α
e
−
√
αx
sin(
√
αy).
The other y BC gives
T
1
=(A
0
x + B
0
) b +
∞
α
A
α
e
√
αx
+ B
α
e
−
√
αx
sin(
√
αb).
For this to hold for arbitrary x, we need to haveTheimportanceof
α
1
=0term is
displayed here by
the relation
between B
0
and
T
1
.
A
0
=0,B
0
b = T
1
⇒ B
0
=
T
1
b
, sin(
√
αb)=0 ⇒ α =
nπ
b
2
.
The temperature function reduces to
T (x, y)=
T
1
b
y +
∞
n=1
A
n
e
nπx/b
+ B
n
e
−nπx/b
sin
nπy
b
.
We now impose the other two BCs. These will give us the following two equations:
0=T (0,y)=
T
1
b
y +
∞
n=1
(A
n
+ B
n
)sin
nπy
b
,
T
2
= T (a, y)=
T
1
b
y +
∞
n=1
A
n
e
nπa/b
+ B
n
e
−nπa/b
sin
nπy
b
.
25.2 Cartesian Coordinates 601
Multiplying both sides of these equations by sin(mπy/b) and integrating from 0 to
b yields the following two equations for A
m
and B
m
:
A
m
+ B
m
= −
2T
1
b
2
#
b
0
y sin
mπy
b
dy =
2T
1
mπ
(−1)
m
,
A
m
e
mπa/b
+ B
m
e
−mπa/b
=
2
b
#
b
0
T
2
−
2T
1
b
y
sin
mπy
b
dy (25.11)
=
2
mπ
[T
2
+(−1)
m
(T
1
− T
2
)] .
These two equations can be solved to obtain the remaining unknown coefficients A
m
and B
m
.
A couple of remarks are in order. The preceding example illustrated clearly
the importance of the α
1
= 0 term: Had we not included it in the expansion
of T , we would not have obtained the answer. This is overlooked in most
elementary treatments of Laplace’s equation. It is worthwhile to emphasize
this point.
Box 25.2.1. Always start with the most general solution of Laplace’s
equation, including the term corresponding to the case in which the con-
stants of the separation of variables are zero, as given in Equation (25.10).
Then apply the BCs, keeping in mind that there may be a preferred order
for such an application.
In Example 25.2.3, the order in which the y BCs were applied first was the
preferred choice.
The second remark has to do with the choice of the functional form of
X and Y . In Example 25.2.3, we chose X to be exponential and Y to be
trigonometric. We could just as well have chosen Y to be exponential and X
to be trigonometric. The appearance of the series would have changed, but
the value of T at any point in the region of interest would have been the same
for both series. This is due to the uniqueness of the solution of Laplace’s
equation.
4
Example 25.2.4. The examples treated so far have been exclusively in two di-
mensions. We now consider a three-dimensional problem. Although this particular a three-
dimensional
example of the
application of
Laplace’s equation
problem can be solved more quickly by relying on our intuition (as we did in Exam-
ple 25.2.1, for example), we shall start from the most general solution, as prescribed
by Box 25.2.1.
Suppose that the four lateral sides of widths a and b of a semi-infinite rectangular
conducting tube are grounded and the closed base is held at potential V .Thecross
section of this tube is shown in Figure 25.4 where it is assumed that the tube starts
at z = 0 and extends to infinity in the positive z-direction. We are interested in
finding the potential inside this tube.
4
The representation of the same function by different series should be familiar to the
reader from calculus where f(x) can be written as a Taylor expansion about any point in
its domain of definition. Although such expansions look different, they all represent the
same function.
602 Laplace’s Equation: Cartesian Coordinates
We start with Equation (25.3) which holds for all solutions of Laplace’s equationthe four
alternatives for
constants α
1
and
α
2
in Cartesian coordinates. There are four different cases to consider:
1. α
1
=0=α
2
:Inthiscase,X(x) is of the generic form Ax + B,andwithy or
z replacing x, this is also the generic form of Y and Z. Let us denote these
solutions as X
0
, Y
0
,andZ
0
.
2. α
1
=0,α
2
=0: Inthiscase,X(x) is of the generic form Ax + B.ButY and
Z are either exponential or trigonometric. Let us denote these solutions as
X
0
, Y
α
2
,andZ
α
2
.
3. α
1
=0,α
2
=0: Inthiscase,Y (y) is of the generic form Ay + B.ButX and
Z are either exponential or trigonometric. Let us denote these solutions as
Y
0
, X
α
1
,andZ
α
1
.
4. α
1
=0,α
2
=0: InthiscaseX, Y ,andZ are either exponential or trigono-
metric. Let us denote these solutions as X
α
1
, Y
α
2
,andZ
α
1
+α
2
.
The most general solution for the potential, encompassing all values of α
1
and α
2
,is
Φ(x, y, z)=X
0
(x)Y
0
(y)Z
0
(z)+X
0
(x)
α
2
Y
α
2
(y)Z
α
2
(z) (25.12)
+ Y
0
(y)
α
1
X
α
1
(x)Z
α
1
(z)+
α
1
=0
α
2
=0
X
α
1
(x)Y
α
2
(y)Z
α
1
+α
2
(z).
We now apply the BCs. Since Φ(0,y,z)=Φ(a, y, z) = 0 for arbitrary y and z,Boundary
conditions severely
restrict the terms
of the infinite
sums in (25.12).
and since each term in Equation (25.12) is independent of all others, we conclude
that X
0
(0) = 0 = X
0
(a)andX
0
(0) = 0 = X
0
(a). It follows that A and B are both
zero for X
0
and X
0
.So,X
0
(x)=0=X
0
(x). Similarly, Y
0
(y) = 0, and Φ is reduced
to the last term (the double sum) of (25.12). Furthermore, since both X
α
1
and Y
α
2
vanish at the two ends of their respective ranges, we expect them to be periodic,
i.e., of trigonometric type. So, the most general solution is now
Φ(x, y, z)=
α
1
,α
2
[A
α
1
cos(
√
α
1
x)+B
α
1
sin(
√
α
1
x)]
· [C
α
2
cos(
√
α
2
y)+D
α
2
sin(
√
α
2
y)] Z
α
1
+α
2
(z).
If this is to vanish at x = 0 for arbitrary y and z,thenA
α
1
must be zero; and if
Φ(a, y, z) = 0 for all y and z, then all coefficients of the product of the y and z
functions in the sum must be zero. These coefficients—after setting A
α
1
equal to
zero—are of the form sin(
√
α
1
a). It follows that
√
α
1
a = mπ ⇒ α
1
=
mπ
a
2
,m=1, 2,...,
where we have excluded the negative values of m as in the previous examples. An
entirely analogous reasoning leads to C
α
2
=0and
√
α
2
b = nπ ⇒ α
2
=
nπ
b
2
,n=1, 2,....
The z-dependence is exponential, and since the potential cannot diverge at large
values of z, the positive exponent will be absent. Absorbing all multiplicative con-
stants into (a single doubly indexed) one, we can now write
Φ(x, y, z)=
∞
m,n=1
A
mn
sin
mπx
a
sin
nπy
b
e
−π
√
m
2
/a
2
+n
2
/b
2
z
. (25.13)
25.3 Problems 603
The unknown constants A
mn
are determined by using the last BC: atwo-dimensional
Fourier series
V =Φ(x, y, 0) =
∞
m,n=1
A
mn
sin
mπx
a
sin
nπy
b
. (25.14)
This is a double Fourier series.
Theorem 25.2.5. The coefficients of the double Fourier series (25.14) can be
calculated by multiplying both sides by sin
jπx
a
sin
kπy
b
and integrating the result
from 0 to a in the x variable and from 0 to b in y:
A
jk
=
4
ab
#
a
0
#
b
0
Φ(x, y, 0) sin
jπx
a
sin
kπy
b
dx dy
It now follows that
A
jk
=
4V
ab
#
a
0
#
b
0
sin
jπx
a
sin
kπy
b
dx dy =
4V
ab
#
a
0
sin
jπx
a
dx
#
b
0
sin
kπy
b
dy
=
4V
ab
a
πj
[1 − (−1)
j
]
b
πk
[1 − (−1)
k
]
or
A
jk
=
4V
π
2
1 − (−1)
j
j
1 − (−1)
k
k
.
It is clear that only the odd terms of the double sum will contribute. Thus, the final
answer for the potential inside [Equation (25.13)] is
Φ(x, y, z)=
16V
π
2
∞
m,n=1
sin[(2m +1)πx/a]
2m +1
sin[(2n +1)πy/b]
2n +1
· e
−π
√
(2m+1)
2
/a
2
+(2n+1)
2
/b
2
z
.
By its very construction, this function satisfies Laplace’s equation as well as all the
BCs. Therefore, by the uniqueness theorem it must represent the unique potential
for the region of interest.
25.3 Problems
25.1. Given that V =
∞
n=1
A
n
sin(nπy/b)whereV is a constant in the
interval (0,b), show that A
n
=2V [1 − (−1)
n
]/(nπ).
25.2. A long hollow cylinder with square cross section of side a has three sides
grounded and the fourth side maintained at potential V
0
(see Figure 25.6).
Find the potential at all points inside.
25.3. Example 25.2.1 treated the case in which the plate at x =0washeldat
the constant potential V . Now suppose that it is held at a potential that varies
with y. Use Equation (25.5) to find the potential as a function of x and y when
604 Laplace’s Equation: Cartesian Coordinates
y
x
Φ
= 0
Φ
= 0
a
Φ
= 0
Φ
= V
0
Figure 25.6: The cross section of the conducting cylinder extended along the z-axis.
(a) Φ(0,y)=
V
0
b
2
y(y − b)
(b) Φ(0,y)=
V
0
b
y
(c) Φ(0,y)=V
0
sin
πy
b
.
25.4. In Example 25.2.2, we assumed constant temperatures for the left and
right plates. Now suppose that the top and bottom plates are as before, but
the left plate is held at a varying temperature given by
T (0,y)=
T
0
b
2
y(y − b).
Use Equation (25.7) to find the temperature as a function of x and y when
(a) T (a, y)=0;
(b) T (a, y)=
T
0
b
2
y(y −b);
(c) T (a, y)=T
0
;
(d) T (a, y)=
T
0
b
y;
(e) T (a, y)=T
0
sin
πy
b
.
25.5. Suppose that the top and bottom plates of Example 25.2.2 are as before,
but the left plate is held at a varying temperature given by
T (0,y)=T
0
sin
2πy
b
.
Use Equation (25.7) to find the temperature as a function of x and y when
(a) T (a, y)=0;
(b) T (a, y)=
T
0
b
2
y(y −b);
(c) T (a, y)=T
0
;
25.3 Problems 605
(d) T (a, y)=
T
0
b
y;
(e) T (a, y)=T
0
sin
πy
b
.
25.6. Solve Equation (25.8) for A
2k+1
and B
2k+1
and substitute in (25.7) to
obtain (25.9).
25.7. Verify that when T
2
=0anda →∞, Equation (25.9) approaches the
result of Example 25.2.1—with V replaced by T
1
.
25.8. Derive Equation (25.11). Assume that T
1
= T
2
= T
0
and solve for A
m
and B
m
.
25.9. Obtain the expression for A
jk
in Theorem 25.2.5.
25.10. Find the potential inside a cube with sides of length a when the top
side is held at a constant potential V
0
with all other sides grounded (zero
potential).
25.11. Find the electrostatic potential inside a cube with sides of length a if
all faces are grounded except the top, which is held at a potential given by:
(a)
V
0
a
x, 0 ≤ x ≤ a. (b)
V
0
a
y, 0 ≤ y ≤ a.
(c)
V
0
a
2
xy, 0 ≤ x, y ≤ a. (d) V
0
sin
π
a
x
, 0 ≤ x ≤ a.