1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

152 Chapter 2 The Heat Equation

Ifφhas the form given in the preceding, the boundary conditions require that
φ( 0 )=c 1 =0, leavingφ(x)=c 2 sin(λx).Thenφ(a)=c 2 sin(λa)=0.
We now have two choices: eitherc 2 =0, makingφ(x)≡0 for all values ofx,
or sin(λa)=0. We reject the first possibility, for it leads to the trivial solution
w(x,t)≡0. In order for the second possibility to hold, we must haveλ=
nπ/a,wheren=± 1 ,± 2 ,± 3 ,....Thenegativevaluesofndo not give any
new functions, because sin(−θ)=−sin(θ ).Henceweallown= 1 , 2 , 3 ,...
only. We shall setλn=nπ/a.
Incidentally, because the differential equations (13) and the boundary con-
ditions (14) forφ(x)are homogeneous, any constant multiple of a solution is
still a solution. We shall therefore remember this fact and drop the constantc 2
inφ(x). Likewise, we delete thecinT(t).
To review our position, we have, for eachn= 1 , 2 , 3 ,...,afunctionφn(x)=
sin(λnx)and an associated functionTn(t)=exp(−λ^2 nkt).Theproductwn(x,t)
=sin(λnx)exp(−λ^2 nkt)has these properties:

1. ∂∂^2 xw 2 n=−λ^2 nwn;∂w∂tn=−λ^2 nkwn; and thereforewnsatisfies the heat equa-
   tion.
   2.wn( 0 ,t)=sin( 0 )e−λ^2 nkt=0 for anynandt; and thereforewnsatisfies the
   boundary condition atx=0.
   3.wn(a,t)=sin(λna)e−λ^2 nkt=0 for anynandtbecauseλna=nπand
   sin(nπ)=0. Thereforewnsatisfies the boundary condition atx=a.
   Now we call on the Principle of Superposition in order to continue.

Principle of Superposition.
Ifu 1 ,u 2 ,...are solutions of the same linear, homogeneous equations, then so
is

u=c 1 u 1 +c 2 u 2 +···. 

In fact, we have infinitely many solutions, so we need an infinite series to
combine them all:

w(x,t)=

∑∞

n= 1

bnsin(λnx)exp

(

−λ^2 nkt

)

. (15)

Using an infinite series brings up questions about convergence that we are go-
ing to ignore. However, it is easy to verify that the function defined by the
series does satisfy the boundary conditions: Atx=0andatx=a,eachterm
is 0, so the sum is 0 as well. To check the partial differential equation, we have
to differentiatew(x,t)by differentiating each term of the series. This done, it
is easy to see that terms match and the heat equation is satisfied.