1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2.3 Example: Fixed End Temperatures 151

allxin the interval 0<x0, the common value of the two
sides must be a constant, varying neither withxnort:

φ′′(x)

φ(x)

=p, T

′(t)

kT(t)

=p.

Now we have two ordinary differential equations for the two factor functions:

φ′′−pφ= 0 , T′−pkT= 0. (13)

The two boundary conditions onwmay also be stated in the product form:

w( 0 ,t)=φ( 0 )T(t)= 0 ,w(a,t)=φ(a)T(t)= 0.

There are two ways these equations can be satisfied for allt>0. Either the
functionT(t)≡0 for allt, which is forbidden, or the other factors must be
zero. Therefore, we have

φ( 0 )= 0 ,φ(a)= 0. (14)
Our job now is to solve Eqs. (13) and satisfy the boundary conditions (14)
while avoiding the trivial solution.
Case 1:Ifp>0, the solutions of Eqs. (13) are
φ(x)=c 1 cosh

(√

px

)

+c 2 sinh

(√

px

)

, T(t)=cepkt.

Now we apply the boundary conditions:

φ( 0 )=0: c 1 = 0 ,

φ(a)=0: c 2 sinh

(√

pa

)

= 0.

Because the sinh function is 0 only when its argument is 0 — clearly not true
of√pa—wehavec 1 =c 2 =0andφ(x)≡0, which is not acceptable.
Case 2:Ifwetakep=0, the solutions of the differential equations (13) are
φ(x)=c 1 +c 2 x,T(t)=c. The boundary conditions require

φ( 0 )=0: c 1 = 0 ,

φ(a)=0: c 2 a= 0.

Again we haveφ(x)≡0.
Case 3: We now try a negative constant. Replacingpby−λ^2 in Eqs. (13)
gives us the two equations

φ′′+λ^2 φ= 0 , T′+λ^2 kT= 0 ,

whose solutions are

φ(x)=c 1 cos(λx)+c 2 sin(λx), T(t)=cexp

(

−λ^2 kt

)

.