Biomedical Engineering Reference
In-Depth Information
into (7.83) results in:
d 2 T
d t 2 ¼
c 2
X
d 2 X
d x 2 þ
c 2
Y
d 2 Y
1
T
d y 2 ¼ u 2
:
(7.85)
Separating and solving the temporal term:
d 2 T
d t 2 þ
u 2 T
¼
0
(7.86)
results in
T
ð
t
Þ¼
E sin ut
þ
f cos ut
:
(7.87)
The spatial terms become:
d 2 X
d x þ
u 2
c 2 ¼
d 2 Y
d y 2 ¼ k 2
1
X
1
Y
:
(7.88)
Solving
u 2
c 2
k 2 X
d 2 X
d x 2 þ
¼
0
(7.89)
results in
q
ð
q
ð
2
2
k 2
k 2
X
ð
x
Þ¼
A sin
u
=
c
Þ
þ
B cos
u
=
c
Þ
:
(7.90)
Solving
d 2 Y
d y 2 þ
k 2 Y
¼
0
(7.91)
results in
Y
ð
y
Þ¼
C sin ky
þ
D cos ky
:
(7.92)
Applying the boundary conditions of z
¼
0at x
¼
0 and x
¼
L leads to B
¼
0 and
q
ð
2
k 2 L
A sin
u
=
c
Þ
¼
0
:
(7.93)
Thus
A sin mpx
L
X
ð
x
Þ¼
(7.94)
q
ð
2
k 2 L
where
u
=
c
Þ
¼
mp and m
¼
1, 2,
. Applying the boundary conditions of z
¼
0at y
¼
0
.
and y
¼
W leads to D
¼
0 and
C sin kW
¼
:
0
(7.95)
Thus
C sin mpy
W
Y
ð
y
Þ¼
:
(7.96)
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