Environmental Engineering Reference
In-Depth Information
3.2.3 Three-Dimensional Mixed Problems
Cuboid Domain
Consider finding the solution of
a 2
=
+
(
,
,
,
) ,
Ω × (
, + ) ,
u t
Δ
u
f
x
y
z
t
0
∂Ω =
(
,
,
,
)
,
(3.12)
L
u
u x
u y
u z
0
u
(
x
,
y
,
z
,
0
)= ϕ (
x
,
y
,
z
) ,
where
Ω
stands for the cuboid domain : 0
<
x
<
a 1
,
0
<
y
<
b 1
,
0
<
z
<
c 1 ,and
∂Ω
represents the six boundary surfaces of
. The boundary conditions can be different
on each of the six surfaces. We can readily find the solutions for all combinations
of different boundary conditions by following the procedure in Section 2.6, using
Table 2.1 and the solution structure theorem.
To demonstrate, we consider
Ω
u
(
0
,
y
,
z
,
t
)=
u x (
a 1 ,
y
,
z
,
t
)=
0
,
u y (
x
,
0
,
z
,
t
)=
u y (
x
,
b 1 ,
z
,
t
)+
h 2 u
(
x
,
b 1 ,
z
,
t
)=
0
,
(3.13)
u z (
x
,
y
,
0
,
t
)
h 1 u
(
x
,
y
,
0
,
t
)=
u z (
x
,
y
,
c 1 ,
t
)=
0
.
By the solution structure theorem, we first develop W ϕ (
x
,
y
,
z
,
t
)
, the solution for
the case f
0. Based on the given boundary conditions (3.13), we should use the
eigenfunctions in Rows 2, 6 and 8 in Table 2.1 to expand the solution so that
=
sin μ l z
c 1 + ϕ l
+
sin (
2 m
+
1
) π
x
cos μ n y
b 1
=
T mnl (
)
.
u
t
(3.14)
2 a 1
m
=
0
,
n
,
l
=
1
Substituting Eq. (3.14) into the equation of PDS (3.12) leads to
T mnl (
2
mnl T mnl (
t
)+ ω
t
)=
0
with the general solution
mnl t
C mnl e ω
T mnl (
t
)=
,
 
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