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3.7 The Two Dimensional Heat Equation
top surface
y
b
bottom surface
a
x
0
Figure 3.7-1 A thin rectangular plate with insulated top and bottom surfaces
We will solve the two dimensional heat equation for a thin rectangular plate shown in Figure 3.71. The top and bottom surfaces are insulated and the edges are kept at zero temperature.
  2u  2u 
u
= c2  2  2 
t
y 
 x
(3.7-1)
The boundary conditions are
and
u(0,y,t) = u(a,y,t) = 0
0<y<b
u(x,0,t) = u(x,b,t) = 0
0<x<a
The initial temperature distribution is u(x,y,0) = f(x,y) for 0 < x < a, 0 < y < b.
We assume that u(x,t) can be separated into X(x), a function of x alone, Y(y), a function of y
alone, and T(t), a function of t alone.
u(x,t) = X(x) Y(y) T(t)
From the following relations
dT
u
=X
t
dt
dX
u
=T
x
dx

d2X
 2u
=
T
x 2
dx 2
u
dY
=T
y
dy

 2u
d 2Y
=
T
y 2
dy 2
Eq. (3.7-1) becomes
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XY
 d2X
dT
d 2Y
= c2T  Y
X 2
2
dt
dy
 dx



Divide the above equation by c2XYT to obtain
1 d2X
1 dT
1 d 2Y
=
+
=  k2 = constant
X dx 2
c 2T dt
Y dy 2
Since the RHS depends on x and y only, and the LHS depends on t only, they must equal to a
constant  k2. The constant must be negative for non-trivial solution. The dependence on X and Y
can also be separated into a dependence on X only and a dependence on Y only.
1 d2X
1 d 2Y
=

 k2 =  2 = constant
2
2
X dx
Y dy
The equation containing Y(y) is rearranged to
1 d 2Y
+ (k2  2 ) = 0
2
Y dy
Let 2 = k2  2 or k2 = 2 + 2, then
1 d 2Y
=  2
Y dy 2
The ODE with respect to x is
d2X
=  2X  X = C1cos(x) + C2sin(x)
dx 2
The constant C1 can be determined from the boundary condition
x = 0, X = 0 = C1(1) + C2(0)  C1 = 0
At x = a, X = 0 = C2sin(a)
To avoid the trivial solution, C2  0 and
sin(a) = 0  a = m  m =
m
a
The ODE with respect to y is
d 2Y
=  2Y  Y = d1cos(x) + d2sin(x)
2
dy
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The constant d1 can be determined from the boundary condition
y = 0, Y = 0 = d1(1) + d2(0)  d1 = 0
At y = b, Y = 0 = d2sin(b)
To avoid the trivial solution, d2  0 and
sin(b) = 0  b = n  n =
n
b
2
k mn
=  m2 +  n2
Therefore
The ODE with respect to t is
1 dT
2
2
c 2t 
=  k mn
 T = Amn’exp  k mn
2
c T dt
Let 
2
mn
 m2 n2 
 m 2 2 n 2 2 
= k c =  2  2  c2  mn = c  2  2 
b 
b 
a
 a
2
mn
1/ 2
2
The solution umn(x,y,t) is then
umn(x,y,t) = C2sin(
m
n
x) d2sin(
y) Amn’exp  2mn t 
a
b
The general solution is a linear combination of all the particular solutions

u(x,y,t) =


Amnsin(
n 1 m 1
m
n
x) sin(
y) exp  2mn t  , where Amn = C2 d2 Amn’
a
b
The constants Amn can be obtained from the initial condition

u(x,y,0) = f(x,y) =


Amnsin(
n 1 m 1
Multiply both sides of the equation by sin(
b
a
0
0

m
n
x) sin(
y)
a
b
m' 
n' 
x) sin(
y) and integrate over the area ab
a
b
m' 
n' 
x) sin(
y)dxdy =
a
b
 
b a
m' 
m
n
n' 
Amn   [sin(
x) sin(
y)] [sin(
x) sin(
y)]dxdy


0 0
a
b
a
b
n 1 m 1
f ( x, y ) sin(
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m' 
m
n
n' 
x) sin(
y)] [sin(
x) sin(
y)]dxdy is nonzero only for the
a
b
a
b
values of m’ = m and n’ = n.
The integral
b
a
0
0

[sin(
b
a
0
0

[sin(
m' 
ab
m
n
n' 
x) sin(
y)] [sin(
x) sin(
y)]dxdy =
a
b
a
b
4
Therefore
Amn =
4
ab
b
a
0
0

f ( x, y ) sin(
m
n
x) sin(
y)dxdy
a
b
For a numerical example, let a = b = 1, c =

u(x,y,t) =


Amnsin(
n 1 m 1
1

, and f(x,y) = 100, we have
m
n
x) sin(
y) exp  2mn t 
a
b
where
 m2 n2 
mn = c  2  2 
b 
a
1 1
Amn = 400 

0 0
Amn =
sin(
1/ 2
= (m2 + n2)1/2
m
n
x) sin(
y)dxdy
a
b
400 [1  ( 1) m ][1  ( 1) n ]
2
mn
Amn = 0 if m = even or n = even, otherwise Amn = 4. Let m = 2l + 1, n = 2k + 1

u(x,y,t) =


l 0 k 0
m
n
1
sin(
x) sin(
y) exp  2mn t 
a
b
( 2l  1)( 2k  1)
A Matlab program is listed in Table 3.7-1 to plot u(x,y,t) at t = 1. The plot only provides a
relative temperature distribution as shown in Figure 3.7-2. The actual values on the coordinated
should be ignored.
__________ Table 3.7-1 Matlab program to plot u(x,y,t) ___________
% Two dimensional heat problem
%
x=[0:20]/20;y=x';n1=length(x);
X=ones(n1,1)*x;Y=y*ones(1,n1);
uxy=zeros(n1,n1);
t=1;
for n=0:4
for m=0:4
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687320022
np=2*n+1;mp=2*m+1;
lamdas=np*np+mp*mp;
uxy=sin(np*pi*X).*sin(mp*pi*Y)*exp(-lamdas*t)/(np*mp)+uxy;
end
end
mesh(uxy)
Figure 3.7-2 Temperature distribution over a rectangular plate
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