1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

302 Chapter 5 Higher Dimensions and Other Coordinates

Because differentiation with respect tox,y,ortgives the same result inside
or outside the integral with respect toz, and because of the boundary condi-
tion (12), we find that

1

c

∫c

0

(

∂^2 u

∂x^2 +

∂^2 u

∂y^2 +

∂^2 u

∂z^2

)

dz=∂

(^2) v
∂x^2 +
∂^2 v
∂y^2 ,
andvsatisfies the two-dimensional heat equation,
∂^2 v
∂x^2 +
∂^2 v
∂y^2 =

1

k

∂v

∂t,^0 <x<a,^0 <y<b,^0 <t.

(See the exercises for details and for boundary and initial conditions.)
Ifz-variation cannot be ignored, we could try to get rid of they-variation
by introducing an average in that direction,

w(x,z,t)=

1

b

∫b

0

u(x,y,z,t)dy.

From the boundary condition (13) we find that
∫b

0

∂^2 u

∂y^2

dy=∂u

∂y

(x,b,z,t)−∂u

∂y

(x, 0 ,z,t)

=

(h

κ

)[(

T 2 −u(x,b,z,t)

)

+

(

T 2 −u(x, 0 ,z,t)

)]

.

Ifbis small — the parallelepiped is more like a plate — we may accept the ap-
proximationu(x,b,z,t)+u(x, 0 ,z,t)≡ 2 w(x,z,t),whichwouldmake,from
the preceding expression,

1

b

∫b

0

∂^2 u

∂y^2

dy≡

( 2 h

bκ

)(

T 2 −w(x,z,t)

)

.

After applying the averaging process to Eqs. (10), (11), (12), and (14) we ob-
tain the following two-dimensional problem forw:

∂^2 w

∂x^2 +

∂^2 w

∂z^2 +

2 h

bκ(T^2 −w)=

1

k

∂w

∂t,^0 <x<a,^0 <z<c,^0 <t,

( 15 )

w( 0 ,z,t)=T 0 ,w(a,z,t)=T 1 , 0 <z<c, 0 <t,( 16 )

∂w

∂z(x,^0 ,t)=^0 ,

∂w

∂z(x,c,t)=^0 ,^0 <x<a,^0 <t,(^17 )

w(x,z, 0 )=^1

b

∫b

0

f(x,y,z)dy, 0 <x<a, 0 <z<c.( 18 )