2.10 Semi-Infinite Rod 189
differential equation can be separated into two ordinary equations as usual:
φ′′(x)
φ(x)=T′(t)
kT(t)=const. (5)There is just one boundary condition onu, which requires thatφ( 0 )=0. The
boundedness condition also requires thatφ(x)remain finite asx→∞.Itis
easy to check (see Exercises) that a positive separation constant produces func-
tionsφ(x)that cannot fulfill both the boundary and boundedness conditions
without being identically 0. Thus, we must choose a negative separation con-
stant,−λ^2. The differential equation, together with the boundary and bound-
edness conditions, forms asingulareigenvalue problem (singular because of
the semi-infinite interval),
φ′′+λ^2 φ= 0 , 0 <x, (6)
φ( 0 )= 0 ,φ(x) bounded asx→∞. (7)The general solution of the differential equation is
φ(x)=c 1 cos(λx)+c 2 sin(λx),which is bounded for any choice of the constants and for any value ofλ.The
boundedness condition told us to use a negative constant in Eq. (5) and now
contributes nothing further.
Applying the boundary condition atx=0showsthatc 1 =0, leavingφ(x)=
c 2 sin(λx). In this singular eigenvalue problem, there are no “special” values
ofλ:Anyvalue produces a nonzero solution of the differential equation that
also satisfies the boundary and boundedness conditions. (But negative values
ofλproduce no new solutions.) Recalling that any constant multiple of a so-
lution of a homogeneous problem is still a solution, we choosec 2 =1and
summarize the solution of the singular eigenvalue problem as
φ(x;λ)=sin(λx), λ > 0. (8)The solution of Eq. (5) forT(t), with constant−λ^2 ,is
T(t)=exp(
−λ^2 kt)
.
For any value ofλ^2 ,thefunction
u(x,t;λ)=sin(λx)exp(
−λ^2 kt)
satisfies Eqs. (1)–(3). Equation (1) and the boundary condition Eq. (2) are ho-
mogeneous, and Eq. (3) is homogeneous in effect; therefore any linear combi-
nation of solutions is a solution. Since the parameterλmay take on any value,