Separation of Variables for Higher Dimensional Heat Equation

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

Objective: Let Ω be a planar region with boundary curve Γ. Consider heat conductionin Ω with fixed boundary temperature on Γ:

(PDE) ut − k(uxx + uyy) = 0 (x, y) in Ω, t > 0,

(BC) u(x, y, t) = 0 (x, y) on Γ, t > 0,

(IC) u(x, y, 0) = f(x, y) (x, y) in Ω.

(IDEA: Separation of Variables) We look for solutions of (PDE)-(BC) whose spatialstructure of the temprature distribution remains invariant with time. In these solutions onlythe amplitudes of the temprature distribution may change with time. Mathematically, we tryto classify all nontrivial solutions of the following form:

u(x, y, t) = T (t)Φ(x, y) 6≡ 0.

(THE EQUATIONS FOR PRODUCT SOLUTIONS) Plugging u = T (t)Φ(x, y) in(PDE)-(BC) we see: There is a constant α such that

(∗) T ′(t) − kαT (t) = 0 t > 0;

(∗∗) Φxx + Φyy − αΦ = 0 (x, y) in Ω,

(∗ ∗ ∗) Φ = 0 (x, y) on Γ.

We look for solutions of (*)-(**)-(***) that are 6≡ 0. Here, (*) and (**) are deduced from(PDE), and (***) follows from (BC). Equation (∗) is easy to solve. The main issue now is tosolve the eigenvalue problem (∗∗)-(∗ ∗ ∗).

(EIGENVALUES AND EIGENFUNCTIONS) Now focus on the boundary valueproblem (**)-(***) for Φ(x, y). The solution structure of this problem depends on the parametervalue α. It can be shown that for most choices of α, (**)-(***) only has the trivial solutionΦ(x, y) ≡ 0. The special values of α admiting nontrivial solutions are called eigenvalues of(**)-(***), and in that case, the corresponding nontrivial solutions Φ(x, y) 6≡ 0 are calledeigenfunctions.

Unfortunately, for a general domain Ω, it’s impossible to write down the eigenvalues andeigenfunctions of (∗∗)-(∗ ∗ ∗) in explicit form. We do, however, have a mathematical theoremabout this eigenvalue problem:

THEOREM. (i) The eigenvalues of (∗∗)-(∗∗∗) form a sequence of negative numbers αn∞

n=1

such that

0 > α1 > α2 ≥ α3 ≥ α4 ≥ · · · , limn→∞

αn = −∞.

(ii) The corresponding eigenfunctions Φn(x, y) are smooth functions in Ω.

(iii) Φ1 > 0 does not change sign in Ω.

(iv)∫∫

ΩΦm(x, y)Φn(x, y)dxdy = 0 for m 6= n.

Let α = αn be one of the eigenvalues. We now solve equation (*), which becomes

T ′(t) − kαnT (t) = 0 t > 0.

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The solutions areT (t) = Constant · ekαnt.

(CLASSIFICATION OF PRODUCT SOLUTIONS) Thus, a product solution of(PDE)-(BC) must be a constant multiple of

un(x, t) = ekαntΦn(x, y), n = 1, 2, 3, · · · .

(FROM PRODUCT SOLUTIONS TO GENERAL SOLUTIONS) The generalsolutions of (PDE)-(BC) are linear combinations of the product solutions:

u(x, t) =

∞∑

n=1

anekαntΦn(x, y),

where an 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 an. Let t = 0:

(FOURIER) f(x, y) =∞

∑

n=1

anΦn(x, y).

The coefficients an can be expressed in terms of f(x, y). The derivation is similar to the 1-Dcase. Details follow: Multiply both sides of (FOURIER) by Φm(x, y) and integrate over Ω:

∫∫

Ω

f(x, y)Φm(x, y)dxdy =

∞∑

n=1

an

∫∫

Ω

Φn(x, y)Φm(x, y)dxdy

= am

∫∫

Ω

Φm(x, y)2dxdy,

by Property (iv) of the above theorem.This finally gives the solution of the initial-boundary value problem (PDE)-(BC)-(IC):

u(x, t) =∞

∑

n=1

anekαntΦn(x, y),

where

an =

∫∫

Ωf(x, y)Φn(x, y)dxdy∫∫

ΩΦn(x, y)2dxdy

n = 1, 2, 3, · · · .

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2. The Case of a Rectangular Ω

Objective: Solve the initial-boundary value problem (PDE)-(BC)-(IC), for the case whereΩ is the rectangle 0 < x < a, 0 < y < b:

(PDE) ut − k(uxx + uyy) = 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 separationof variables: Thanks to the special symmetry of the rectangular domain, we can obtain explicitexpressions for the eigenvalues and eigenfunctions. As matter of fact, the eigenfunctions of theproduct form:

Φ(x, y) = X(x)Y (y)

will produce all eigenvalues. Although, in general there might be eigenfunctions of non-productforms, those non-product eigenfunctions are linear comibinations of product eigenfunctions. Inthis sense, we only need to solve (∗∗)-(∗ ∗ ∗) for product functions Φ(x, y) = X(x)Y (y). Thiswill separate the 2-D eigenvalue problem (∗∗)-(∗ ∗ ∗) to two 1-D eigenvalue problems which weknow 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) 6≡ 0.

Plugging u = T (t)X(x)Y (y) in (PDE)-(BC) we see: there are constants λ and µ such that

(∗) T ′(t) + k(λ + µ)T (t) = 0 t > 0;

(♥)x X ′′(x) + λX(x) = 0 (0 < x < a), X(0) = X(a) = 0;

(♥)y Y ′′(y) + µY (y) = 0 (0 < y < b), Y (0) = Y (b) = 0.

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

λm = (mπ/a)2, m = 1, 2 · · · ,

problem (♥)x has nontrivial solutions which are constant multiples of

Xm(x) = sin(mπx/a).

Another 1-D eigenvalue problem (♥)y is solved similarly: If and only if µ is one of the followingvalues

µn = (nπ/b)2, n = 1, 2 · · · ,

problem (♥)y has nontrivial solutions which are constant multiples of

Yn(y) = sin(nπy/b).

For λ = λm and µ = µn, we now solve equation (*), which becomes

T ′(t) + k(λm + µn)T (t) = 0 t > 0.

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The solutions are

T (t) = Constant · e−k(λm+µn)t = Constant · e−k[(mπ/a)2+(nπ/b)2]t.

(CLASSIFICATION OF PRODUCT SOLUTIONS) Thus, a (triple-factor) productsolution of (PDE)-(BC) must be a constant multiple of

um,n(x, y, t) = sin(mπx/a) sin(nπy/b)e−k[(mπ/a)2+(nπ/b)2]t, m, n = 1, 2, 3, · · · .

(SOLUTION FORMULA FOR THE INIT-BDRY VAL PROBLEM) The solutionof the initial-boundary value problem (PDE)-(BC)-(IC) is given by:

u(x, t) =

∞∑

m=1

∞∑

n=1

am,n sin(mπx/a) sin(nπy/b)e−k[(mπ/a)2+(nπ/b)2]t,

where

am,n =

∫ a

x=0

∫ b

y=0f(x, y) sin(mπx/a) sin(nπy/b)dxdy

∫ a

x=0

∫ b

y=0sin(mπx/a) sin(nπy/b)2dxdy

=4

ab

∫ a

x=0

∫ b

y=0

f(x, y) sin(mπx

a

)

sin(nπy

b

)

dxdy.

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EXERCISES

[1] Consider the 2-D heat equation in a rectangle, with top and bottom sides insulated, andleft and right boundary temperature fixed at 0:

(1) ut − uxx − uyy = 0 0 < x < a, 0 < y < b, t > 0,

(2i) uy(x, 0, t) = 0, uy(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 dataf(x, y).

(c) Find the Fourier series formula in the case where

f(x, y) = xy.

[2] Consider the 3-D heat equation in a 3-D box with insulated boundary conditions:

(4) ut − uxx − uyy − uzz = 0 0 < x < a, 0 < y < b, 0 < z < c, t > 0,

(5) ux(0, y, z, t) = ux(a, y, z, t) = uy(x, 0, z, t) = uy(x, b, z, t)

= uz(x, y, 0, t) = uz(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 dataf(x, y, z).

ANSWERS:

[1] (a) u(x, y, t) = C sin(mπx

a

)

cos(nπy

b

)

e−[(mπ/a)2+(nπ/b)2]t, where m is any positive in-

teger, n is any nonnegative integer, and C is any nonzero constant.

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(b) u(x, y, t) =

∞∑

m=1

∞∑

n=0

am,n sin(mπx

a

)

cos(nπy

b

)

e−[(mπ/a)2+(nπ/b)2]t,

where

am,0 =2

ab

∫ a

x=0

∫ b

y=0

f(x, y) sin(mπx

a

)

dxdy (m ≥ 1)

am,n =4

ab

∫ a

x=0

∫ b

y=0

f(x, y) sin(mπx

a

)

cos(nπy

b

)

dxdy (m ≥ 1, n ≥ 1)

(c) u(x, y, t) =∞

∑

m=1

ab(−1)m+1

mπsin

(mπx

a

)

e−(mπ/a)2t

+

∞∑

m=1

∞∑

n=1

4ab(−1)m1 − (−1)n

mn2π3sin

(mπx

a

)

cos(nπy

b

)

e−[(mπ/a)2+(nπ/b)2]t

[2] (a) u(x, y, z, t) = C cos(mπx

a

)

cos(nπy

b

)

cos

(

kπz

c

)

e−[(mπ/a)2+(nπ/b)2+(kπ/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

am,n,k cos(mπx

a

)

cos(nπy

b

)

cos

(

kπz

c

)

e−[(mπ/a)2+(nπ/b)2+(kπ/c)2]t,

where

a0,0,0 =1

abc

∫ a

x=0

∫ b

y=0

∫ c

z=0

f(x, y, z)dxdydz

am,0,0 =2

abc

∫ a

x=0

∫ b

y=0

∫ c

z=0

f(x, y, z) cos(mπx

a

)

dxdydz (m ≥ 1)

a0,n,0 =2

abc

∫ a

x=0

∫ b

y=0

∫ c

z=0

f(x, y, z) cos(nπy

b

)

dxdydz (n ≥ 1)

a0,0,k =2

abc

∫ a

x=0

∫ b

y=0

∫ c

z=0

f(x, y, z) cos

(

kπz

c

)

dxdydz (k ≥ 1)

am,n,0 =4

abc

∫ a

x=0

∫ b

y=0

∫ c

z=0

f(x, y, z) cos(mπx

a

)

cos(nπy

b

)

dxdydz (m ≥ 1, n ≥ 1)

a0,n,k =4

abc

∫ a

x=0

∫ b

y=0

∫ c

z=0

f(x, y, z) cos(nπy

b

)

cos

(

kπz

c

)

dxdydz (n ≥ 1, k ≥ 1)

am,0,k =4

abc

∫ a

x=0

∫ b

y=0

∫ c

z=0

f(x, y, z) cos(mπx

a

)

cos

(

kπz

c

)

dxdydz (m ≥ 1, k ≥ 1)

am,n,k =8

abc

∫ a

x=0

∫ b

y=0

∫ c

z=0

f(x, y, z) cos(mπx

a

)

cos(nπy

b

)

cos

(

kπz

c

)

dxdydz

(m ≥ 1, n ≥ 1, k ≥ 1)

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