Unformatted text preview: ouble
Fourier sine series of h. As in the case of ordinary Fourier sine series, the bmn ’s
can be determined by an explicit formula. To determine it, we multiply both
sides of (5.12) by (2/a) sin(pπx/a), where p is a positive integer, and integrate
with respect to x to obtain
a h(x, y ) sin(pπx/a)dx
0 ∞ = bmn
m,n=1 a 2
a sin(pπx/a) sin(mπx/a)dx sin(nπy/b).
0 The expression within brackets is one if p = m and otherwise zero, so
∞ a 2
a h(x, y ) sin(pπx/a)dx =
0 bpn sin(ny/b).
n=1 Next we multiply by (2/b) sin(qπy/b), where q is a positive integer, and integrate
with respect to y to obtain
ab a b h(x, y ) sin(pπx/a) sin(qπy/b)dxdy
∞ = bpn
b b sin(nπy/b) sin(qπy/b)dy .
0 The expression within brackets is one if q = n and otherwise zero, so we ﬁnally
bpq = 22
ab a b h(x, y ) sin(pπx/a) sin(qπy/b)dxdy.
0 0 Suppose, for example, that c = 1, a = b = π and
h(x, y ) = sin x sin y + 3 sin 2x sin y + 7 sin 3x sin 2y.
126 (5.13) In this case, we do not need to carry out the integration indicated in (5.13)
View Full Document
This document was uploaded on 01/12/2014.
- Winter '14