526 Higher Engineering Mathematics
the potential at any point within the rectangle OPQR.
The boundary conditions are:
u = 0 when x = 0 i.e. u(0, y) = 0 for 0≤ y ≤ b
u = 0 when x = a i.e. u(a, y) =0 for 0≤ y ≤ b
u = 0 when y = b i.e. u(x, b) =0 for 0≤ x ≤ a
u = f (x) when y = 0 i.e. u(x, 0) = f (x)
for 0 ≤ x ≤ a
As with previous partial differential equations, 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 = X Y , determining partial
derivatives, and substituting into
∂ 2 u
∂x 2 +
∂ 2 u
∂ y 2 = 0 gives:
X Y + X Y = 0
Separating the variables gives:
X
X
= −
Y
Y
Letting each side equal a constant, − p 2 , gives the two
equations:
X
+ p
2 X = 0 and Y
− p
2 Y = 0
from which, X = A cos px + B sin px and
Y = C e py + D e − py or Y = C cosh py + D sinh py (see
Problem 5, page 480 for this conversion).
This latter form can also be expressed as:
Y = E sinh p( y + φ) by using compound angles.
Hence u(x, y) = X Y
= {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 = B E.
The first boundary condition is: u(0, y) = 0, hence
0 = P sinh p(y + φ) from which, P = 0. Hence,
u(x, y) = Q sin px sinh p(y + φ).
The second boundary condition is: u(a, y) = 0,
hence
0 = Q sin pa sinh p(y + φ)
from which,
sin pa = 0, hence, pa = nπ or p =
nπ
a
for
n = 1, 2, 3, ...
The third boundary condition is: u(x, b) = 0,
hence, 0 = Q sin px sinh p(b + φ) from which,
sinh p(b + φ) = 0 and φ =−b.
Hence, u(x, y) = Q sin px sinh p(y − b) =
Q 1 sin px sinh p(b − y) where Q 1 =−Q.
Since there are many solutions for integer values of n,
u(x, y) =
∞
n=1
Q n sin px sinh p(b − y)
=
∞
n=1
Q n sin
nπ x
a
sinh
nπ
a
(b − y)
The fourth boundary condition is: u(x, 0) = f (x),
hence, f (x) =
∞
n=1
Q n sin
nπ x
a
sinh
nπb
a
i.e.
f (x) =
∞
n=1
Q n sinh
nπb
a
sin
nπ x
a
From Fourier series coefficients,
Q n sinh
nπb
a
= 2 × the mean value of
f (x) sin
nπ x
a
from x = 0 to x = a
i.e.
=
a
0
f (x) sin
nπ x
a
dx from which,
Q n may be determined.
This is demonstrated in the following worked
problem.
Problem 7. A square plate is bounded by the
lines x = 0, y = 0, x = 1 and y = 1. Apply the
Laplace equation
∂ 2 u
∂x 2 +
∂ 2 u
∂ y 2 = 0 to determine the
potential distribution u(x, y) over the plate, subject
to the following boundary conditions:
u = 0 when x = 0 0≤ y ≤ 1,
u = 0 when x = 1 0≤ y ≤1,
u = 0 when y = 0 0≤ x ≤ 1,
u = 4 when y = 1 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 = X Y , determining
partial derivatives, and substituting into
∂ 2 u
∂x 2 +
∂ 2 u
∂ y 2 = 0
gives:
X
Y + X Y
= 0
Separating the variables gives:
X
X
=−
Y
Y
Letting each side equal a constant, − p 2 , gives the two
equations:
X
+ p
2 X = 0 and Y
− p
2 Y = 0