394 Chapter 14Partial differential equations
(14.4)
We show that a solution of this equation can be written as
f(x, 1 y) 1 = 1 X(x) 1 × 1 Y(y) (14.5)
in which the solution, a function of the two variables xand y, is expressed as the
product of a function of xonly and a function of yonly. We have
becauseY(y)does not depend on x. Similarly,
becauseX(x)is a constant with respect to y. Substitution off 1 = 1 XYand its derivatives
into the differential equation (14.4) then gives
(14.6)
and, dividing throughout byf 1 = 1 X(x) 1 × 1 Y(y),
(14.7)
The first set of terms in square brackets on the left side of (14.7) depends only on
the variable x, the second set only on y. If the total is constant (zero here) it follows
that each of these sets of terms must be separately constant if xand yare independent
variables. Thus a change in the value of xcannot lead to a change in the value of the
terms that depend on yonly, and vice-versa. Therefore, if the first set of terms equals
the constantCthen the second set is equal to−C(for the total to be zero):
(14.8)
andCis called a separation constant. Then
(14.9)
(14.10)
dY
dy
=−CY
dX
dx
=CX
11
X
dX
dx
C
Y
dY
dy
=, =−C
11
0
Xx
dX x
dx Y y
dY y
() dy
()
()
()
=
Yy
dX x
dx
Xx
dY y
dy
()
()
()
()
+= 0
∂
∂
=
f
y
X
dY
dy
∂
∂
=
∂
∂
=
f
x
XY
x
Y
dX
dx
()
∂
∂
∂
∂
=
f
x
f
y
0