1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2.9 Generalities on the Heat Conduction Problem 187
We now begin to assemble the solution. For eachn= 1 , 2 , 3 ,...,wn(x,t)
=φn(x)Tn(t)satisfies Eqs. (9)–(11). As these are all linear homogeneous equa-
tions, any linear combination of solutions is again a solution. Thus the tran-
sient temperature has the form

w(x,t)=

∑∞

n= 1

anφn(x)exp

(

−λ^2 nkt

)

.

The initial condition Eq. (12) will be satisfied if we choose theanso that

w(x, 0 )=

∑∞

n= 1

anφn(x)=g(x), l<x<r.

The convergence theorem tells us that the equality will hold, except possibly at
a finite number of points, iff(x)— and thereforeg(x)— is sectionally smooth.
Thusw(x,t)is the solution of its problem, if we choose

an=

∫r

l∫gr(x)φn(x)p(x)dx

lφ^2 n(x)p(x)dx

.

Finally, we can write the complete solution of Eqs. (1)–(4) in the form

u(x,t)=v(x)+

∑∞

n= 1

anφn(x)exp

(

−λ^2 nkt

)

. (16)

Working from the representation Eq. (16) we can draw some conclusions
about the solution of Eqs. (1)–(4).

1.Since all theλ^2 nare positive,u(x,t)does tend tov(x)ast→∞.

2.For anyt 1 >0, the series foru(x,t 1 )converges uniformly inl≤x≤r

because of the exponential factors; thereforeu(x,t 1 )is a continuous func-

tion ofx. Any discontinuity in the initial condition is immediately elimi-

nated.

3.For large enough values oft, we can approximateu(x,t)by

v(x)+a 1 φ 1 (x)exp

(

−λ^21 kt

)

.

(To judge how largetmight be, we need to know something about thean

and theλn.) Becauseφ 1 (x)is of one sign on the intervall<x<r(that

is,φ 1 (x)>0orφ 1 (x)<0 for allxbetweenlandr), the graph of our

approximation will lie either above or below the graph ofv(x)but will

not cross it (provided thata 1 =0).