158 Chapter 2 The Heat Equation
and separate these equalities into two ordinary differential equations linked by
the common parameterλ^2 :
φ′′+λ^2 φ= 0 , 0 <x<a, (4)
T′+λ^2 kT= 0 , 0 <t. (5)The boundary conditions onucan be translated into conditions onφ,be-
cause they are homogeneous conditions. The boundary conditions in product
form are
∂u
∂x( 0 ,t)=φ′( 0 )T(t)= 0 , 0 <t,
∂u
∂x(a,t)=φ′(a)T(t)= 0 , 0 <t.To satisfy these equations, we must have the functionT(t)always zero (which
would makeu(x,t)≡0), or else
φ′( 0 )= 0 ,φ′(a)= 0.The second alternative avoids the trivial solution.
We now have a homogeneous differential equation forφtogether with ho-
mogeneous boundary conditions:
φ′′+λ^2 φ= 0 , 0 <x<a, (6)
φ′( 0 )= 0 ,φ′(a)= 0. (7)A problem of this kind is called aneigenvalue problem. We are looking for those
values of the parameterλ^2 for which nonzero solutions of Eqs. (6) and (7)
may exist. Those values are calledeigenvalues, and the corresponding solutions
are calledeigenfunctions. Note that the significant parameter isλ^2 ,notλ.The
square is used only for convenience. It is worth mentioning that we already
saw an eigenvalue problem in Section 3 and in the Euler buckling problem of
Chapter 0.
The general solution of the differential equation in Eq. (6) is
φ(x)=c 1 cos(λx)+c 2 sin(λx).Applying the boundary condition atx=0, we see thatφ′( 0 )=c 2 λ=0, giv-
ingc 2 =0orλ=0. We put aside the caseλ=0andassumec 2 =0, so
φ(x)=c 1 cos(λx). Then the second boundary condition requires thatφ′(a)=
−c 1 λsin(λa)=0. Once again, we may havec 1 =0orsin(λa)=0. Butc 1 = 0
makesφ(x)≡0. We choose therefore to make sin(λa)=0byrestrictingλto