1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

306 Chapter 5 Higher Dimensions and Other Coordinates

and the corresponding functionTis

Tmn=exp

(

−λ^2 mnkt

)

.

We now begin to assemble the solution. For each pair of indicesm,n(m=
1 , 2 , 3 ,...,n= 1 , 2 , 3 ,...) there is a function

umn(x,y,t)=φmn(x,y)Tmn(t)

=sin

(mπx

a

)

sin

(nπy

b

)

exp

(

−λ^2 mnkt

)

that satisfies the partial differential equation (1) and the boundary conditions
Eqs. (2) and (3). We may form linear combinations of these solutions to get
other solutions. The most general linear combination would be the double
series

u(x,y,t)=

∑∞

m= 1

∑∞

n= 1

amnφmn(x,y)Tmn(t), (14)

and any such combination should satisfy Eqs. (1)–(3). There remains the ini-
tial condition Eq. (4) to be satisfied. Ifuhas the form given in Eq. (14), then
the initial condition becomes

∑∞

m= 1

∑∞

n= 1

amnφmn(x,y)=f(x,y), 0 <x<a, 0 <y<b. (15)

The idea of orthogonality is once again applicable to the problem of selecting
the coefficientsamn. One can show by direct computation that

∫b

0

∫a

0

φmn(x,y)φpq(x,y)dx dy=







ab

4 ifm=pandn=q,

0 , otherwise.

(16)

Thus, the appropriate formula for the coefficientsamnis

amn=^4

ab

∫b

0

∫a

0

f(x,y)sin

(mπx

a

)

sin

(nπy

b

)

dx dy. (17)

Iffis a sufficiently regular function, the series in Eq. (15) will converge and
equalf(x,y)in the rectangular region 0<x<a,0<y<b.Wemaythensay
that the problem is solved. It is reassuring to notice that each term in the series
of Eq. (14) contains a decaying exponential, and thus, astincreases,u(x,y,t)
tends to zero, as expected.

Example.
Let us take the specific initial condition

f(x,y)=xy, 0 <x<a, 0 <y<b.