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# Separation of Variables for Higher Dimensional Heat Equation

## 1. Heat Equation and Eigenfunctions of the Laplacian: An 2-D Example

Objective: Let be a planar region with boundary curve . Consider heat conduction
in with xed boundary temperature on :
(PDE) u
t
k(u
xx
+ u
yy
) = 0 (x, y) in , t > 0,
(BC) u(x, y, t) = 0 (x, y) on , t > 0,
(IC) u(x, y, 0) = f(x, y) (x, y) in .
(IDEA: Separation of Variables) We look for solutions of (PDE)-(BC) whose spatial
structure of the temprature distribution remains invariant with time. In these solutions only
the amplitudes of the temprature distribution may change with time. Mathematically, we try
to classify all nontrivial solutions of the following form:
u(x, y, t) = T(t)(x, y) 0.
(THE EQUATIONS FOR PRODUCT SOLUTIONS) Plugging u = T(t)(x, y) in
(PDE)-(BC) we see: There is a constant such that
() T

## (t) kT(t) = 0 t > 0;

()
xx
+
yy
= 0 (x, y) in ,
( ) = 0 (x, y) on .
We look for solutions of (*)-(**)-(***) that are 0. Here, (*) and (**) are deduced from
(PDE), and (***) follows from (BC). Equation () is easy to solve. The main issue now is to
solve the eigenvalue problem ()-( ).
(EIGENVALUES AND EIGENFUNCTIONS) Now focus on the boundary value
problem (**)-(***) for (x, y). The solution structure of this problem depends on the parameter
value . It can be shown that for most choices of , (**)-(***) only has the trivial solution
(x, y) 0. The special values of admiting nontrivial solutions are called eigenvalues of
(**)-(***), and in that case, the corresponding nontrivial solutions (x, y) 0 are called
eigenfunctions.
Unfortunately, for a general domain , its impossible to write down the eigenvalues and
eigenfunctions of ()-( ) in explicit form. We do, however, have a mathematical theorem
THEOREM. (i) The eigenvalues of ()-() form a sequence of negative numbers {
n
}

n=1
such that
0 >
1
>
2

3

4
, lim
n

n
= .
(ii) The corresponding eigenfunctions
n
(x, y) are smooth functions in .
(iii)
1
(iv)

m
(x, y)
n
(x, y)dxdy = 0 for m = n.
Let =
n
be one of the eigenvalues. We now solve equation (*), which becomes
T

(t) k
n
T(t) = 0 t > 0.
1
The solutions are
T(t) = Constant e
knt
.
(CLASSIFICATION OF PRODUCT SOLUTIONS) Thus, a product solution of
(PDE)-(BC) must be a constant multiple of
u
n
(x, t) = e
knt

n
(x, y), n = 1, 2, 3, .
(FROM PRODUCT SOLUTIONS TO GENERAL SOLUTIONS) The general
solutions of (PDE)-(BC) are linear combinations of the product solutions:
u(x, t) =

n=1
a
n
e
knt

n
(x, y),
where a
n
are constants.
(SOLUTION FORMULA FOR THE INIT-BDRY VAL PROBLEM) To obtain
the solution of the initial-boundary value problem (PDE)-(BC)-(IC), we now need to use (IC)
to determine the values of a
n
. Let t = 0:
(FOURIER) f(x, y) =

n=1
a
n

n
(x, y).
The coecients a
n
can be expressed in terms of f(x, y). The derivation is similar to the 1-D
case. Details follow: Multiply both sides of (FOURIER) by
m
(x, y) and integrate over :

f(x, y)
m
(x, y)dxdy =

n=1
a
n

n
(x, y)
m
(x, y)dxdy
= a
m

m
(x, y)
2
dxdy,
by Property (iv) of the above theorem.
This nally gives the solution of the initial-boundary value problem (PDE)-(BC)-(IC):
u(x, t) =

n=1
a
n
e
knt

n
(x, y),
where
a
n
=

f(x, y)
n
(x, y)dxdy

n
(x, y)
2
dxdy
n = 1, 2, 3, .
2
2. The Case of a Rectangular
Objective: Solve the initial-boundary value problem (PDE)-(BC)-(IC), for the case where
is the rectangle {0 < x < a, 0 < y < b}:
(PDE) u
t
k(u
xx
+ u
yy
) = 0 0 < x < a, 0 < y < b, t > 0,
(BC)

## u(x, 0, t) = 0, u(x, b, t) = 0 0 < x < a, t > 0,

u(0, y, t) = 0, u(a, y, t) = 0 0 < y < b, t > 0,
(IC) u(x, y, 0) = f(x, y) 0 < x < a, 0 < y < b.
THE EIGENVALUE PROBLEM ()-() can be solved by further separation
of variables: Thanks to the special symmetry of the rectangular domain, we can obtain explicit
expressions for the eigenvalues and eigenfunctions. As matter of fact, the eigenfunctions of the
product form:
(x, y) = X(x)Y (y)
will produce all eigenvalues. Although, in general there might be eigenfunctions of non-product
forms, those non-product eigenfunctions are linear comibinations of product eigenfunctions. In
this sense, we only need to solve ()-( ) for product functions (x, y) = X(x)Y (y). This
will separate the 2-D eigenvalue problem ()-( ) to two 1-D eigenvalue problems which we
know how to solve.
An ALTERNATIVE APPROACH is to start from the beginnning with product solu-
tions of the form:
u(x, y, t) = T(t)X(x)Y (y) 0.
Plugging u = T(t)X(x)Y (y) in (PDE)-(BC) we see: there are constants and such that
() T

()
x
X

()
y
Y

## (y) + Y (y) = 0 (0 < y < b), Y (0) = Y (b) = 0.

We know how to solve 1-D eigenvalue problem ()
x
: If and only if is one of the following
values

m
= (m/a)
2
, m = 1, 2 ,
problem ()
x
has nontrivial solutions which are constant multiples of
X
m
(x) = sin(mx/a).
Another 1-D eigenvalue problem ()
y
is solved similarly: If and only if is one of the following
values

n
= (n/b)
2
, n = 1, 2 ,
problem ()
y
has nontrivial solutions which are constant multiples of
Y
n
(y) = sin(ny/b).
For =
m
and =
n
, we now solve equation (*), which becomes
T

(t) + k(
m
+
n
)T(t) = 0 t > 0.
3
The solutions are
T(t) = Constant e
k(m+n)t
= Constant e
k[(m/a)
2
+(n/b)
2
]t
.
(CLASSIFICATION OF PRODUCT SOLUTIONS) Thus, a (triple-factor) product
solution of (PDE)-(BC) must be a constant multiple of
u
m,n
(x, y, t) = sin(mx/a) sin(ny/b)e
k[(m/a)
2
+(n/b)
2
]t
, m, n = 1, 2, 3, .
(SOLUTION FORMULA FOR THE INIT-BDRYVAL PROBLEM) The solution
of the initial-boundary value problem (PDE)-(BC)-(IC) is given by:
u(x, t) =

m=1

n=1
a
m,n
sin(mx/a) sin(ny/b)e
k[(m/a)
2
+(n/b)
2
]t
,
where
a
m,n
=

a
x=0

b
y=0
f(x, y) sin(mx/a) sin(ny/b)dxdy

a
x=0

b
y=0
{sin(mx/a) sin(ny/b)}
2
dxdy
=
4
ab

a
x=0

b
y=0
f(x, y) sin

mx
a

sin

ny
b

dxdy.
4
EXERCISES
 Consider the 2-D heat equation in a rectangle, with top and bottom sides insulated, and
left and right boundary temperature xed at 0:
(1) u
t
u
xx
u
yy
= 0 0 < x < a, 0 < y < b, t > 0,
(2i) u
y
(x, 0, t) = 0, u
y
(x, b, t) = 0 0 x a, 0 y b, t > 0,
(2ii) u(0, y, t) = 0, u(a, y, t) = 0 0 y b, t > 0,
(3) u(x, y, 0) = f(x, y) 0 < x < a, 0 < y < b.
(a) Find all nontrivial solutions u(x, y, t) to (1)-(2) of the form
u(x, y, t) = X(x)Y (y)T(t).
(b) Find the Fourier series solution formula for (1)-(2)-(3) with the general initial data
f(x, y).
(c) Find the Fourier series formula in the case where
f(x, y) = xy.
 Consider the 3-D heat equation in a 3-D box with insulated boundary conditions:
(4) u
t
u
xx
u
yy
u
zz
= 0 0 < x < a, 0 < y < b, 0 < z < c, t > 0,
(5) u
x
(0, y, z, t) = u
x
(a, y, z, t) = u
y
(x, 0, z, t) = u
y
(x, b, z, t)
= u
z
(x, y, 0, t) = u
z
(x, y, c, t) = 0 0 < x < a, 0 < y < b, 0 < z < c, t > 0,
(6) u(x, y, z, 0) = f(x, y, z) 0 < x < a, 0 < y < b, 0 < z < c.
(a) Find all nontrivial solutions u(x, y, z, t) to (4)-(5) of the form
u(x, y, z, t) = X(x)Y (y)Z(z)T(t).
(b) Find the Fourier series solution formula for (4)-(5)-(6) with the general initial data
f(x, y, z).
 (a) u(x, y, t) = C sin

mx
a

cos

ny
b

e
[(m/a)
2
+(n/b)
2
]t
, where m is any positive in-
teger, n is any nonnegative integer, and C is any nonzero constant.
5
(b) u(x, y, t) =

m=1

n=0
a
m,n
sin

mx
a

cos

ny
b

e
[(m/a)
2
+(n/b)
2
]t
,
where
a
m,0
=
2
ab

a
x=0

b
y=0
f(x, y) sin

mx
a

dxdy (m 1)
a
m,n
=
4
ab

a
x=0

b
y=0
f(x, y) sin

mx
a

cos

ny
b

dxdy (m 1, n 1)
(c) u(x, y, t) =

m=1
ab(1)
m+1
m
sin

mx
a

e
(m/a)
2
t
+

m=1

n=1
4ab(1)
m
{1 (1)
n
}
mn
2

3
sin

mx
a

cos

ny
b

e
[(m/a)
2
+(n/b)
2
]t
 (a) u(x, y, z, t) = C cos

mx
a

cos

ny
b

cos

kz
c

e
[(m/a)
2
+(n/b)
2
+(k/c)
2
]t
, where
m, n, k are any nonnegative integers and C is any nonzero constant.
(b) u(x, y, z, t) =

m=0

n=0

k=0
a
m,n,k
cos

mx
a

cos

ny
b

cos

kz
c

e
[(m/a)
2
+(n/b)
2
+(k/c)
2
]t
,
where
a
0,0,0
=
1
abc

a
x=0

b
y=0

c
z=0
f(x, y, z)dxdydz
a
m,0,0
=
2
abc

a
x=0

b
y=0

c
z=0
f(x, y, z) cos

mx
a

dxdydz (m 1)
a
0,n,0
=
2
abc

a
x=0

b
y=0

c
z=0
f(x, y, z) cos

ny
b

dxdydz (n 1)
a
0,0,k
=
2
abc

a
x=0

b
y=0

c
z=0
f(x, y, z) cos

kz
c

dxdydz (k 1)
a
m,n,0
=
4
abc

a
x=0

b
y=0

c
z=0
f(x, y, z) cos

mx
a

cos

ny
b

dxdydz (m 1, n 1)
a
0,n,k
=
4
abc

a
x=0

b
y=0

c
z=0
f(x, y, z) cos

ny
b

cos

kz
c

dxdydz (n 1, k 1)
a
m,0,k
=
4
abc

a
x=0

b
y=0

c
z=0
f(x, y, z) cos

mx
a

cos

kz
c

dxdydz (m 1, k 1)
a
m,n,k
=
8
abc

a
x=0

b
y=0

c
z=0
f(x, y, z) cos

mx
a

cos

ny
b

cos

kz
c

dxdydz
(m 1, n 1, k 1)
6
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