15CCM211A & 6CCM211B outline solutions 4 Part II
Setting t = 0 in eqn (0.3) of the Assignment sheet we obtain
(x) =n=1
An sin(npix
a
)Using question [1] we hence obtain the stated formula for
An.
Differentiating eqn (0.3) on the Assignment w.r.t. t and setting
t = 0 the
other boundary condition gives
(x) =n=1
Bnnpic
asin(npix
a
)and once again question [1] implies the remaining formula.
Consider now the specific example given, in which
(x) = 0 and (x) =
{310x, if 0 x 1
3320
(1 x), if 13 x 1 (1)
Since = 0 we have Bn = 0, while since L = 1 the An are given
by
An = 2
10
(x) sin (npix) dx
=3
5
1/30
x sin (npix) dx+3
10
11/3
(1 x) sin (npix) dx
= cos(npi/3)5npi
+3
5
sin(npi/3)
5n2pi2+
cos(npi/3)
5npi+
3
10
sin(npi/3)
5n2pi2
=9
10pi2sin(npi/3)
n2.
Substituting these values for An and Bn the asserted formula
results.
