196 Chapter 2 The Heat Equation
substitute the formulas forA(λ)andB(λ)into Eq. (4):
u(x,t)=1
π∫∞
0[∫∞
−∞f(x′)cos(λx′)dx′cos(λx)+
∫∞
−∞f(x′)sin(λx′)dx′sin(λx)]
exp(
−λ^2 kt)
dλ.Combining terms we find
u(x,t)=1
π∫∞
0∫∞
−∞f(x′)[
cos(λx′)cos(λx)+sin(λx′)sin(λx)]
dx′×exp(
−λ^2 kt)
dλ=^1
π∫∞
0∫∞
−∞f(x′)cos(
λ(x′−x))
dx′exp(
−λ^2 kt)
dλ.If the order of integration may be reversed, we may write
u(x,t)=^1
π∫∞
−∞f(x′)∫∞
0cos(
λ(x′−x))
exp(
−λ^2 kt)
dλdx′.The inner integral can be computed by complex methods of integration. It
is known to be (Miscellaneous Exercises 32, Chapter 1)
∫∞0cos(
λ(x′−x))
exp(
−λ^2 kt)
dλ=√
π
4 ktexp[
−(x′−x)^2
4 kt]
, t> 0.This gives us, finally, a new form for the temperature distribution:
u(x,t)=√^1
4 kπt∫∞
−∞f(x′)exp[
−(x′−x)^2
4 kt]
dx′. (7)Using this form, we find the solution of the example problem solved earlier
(seeEq.(6))tobe
u(x,t)=√T^0
4 πkt∫a−aexp[
−(x′−x)^2
4 kt]
dx′. (8)Of the two formulas, Eqs. (4) and (7), for the solutionu(x,t), each has its
advantages. For simple problems we may be able to evaluate the coefficients
A(λ)andB(λ)in Eq. (5). However, it is a rare case indeed when the integral
in Eq. (4) can be evaluated analytically. The same is true for the integral in
Eq.(7).Thus,ifthevalueofuat a specificxandtis needed, either integral
would be calculated numerically. For large values ofkt, the exponential factor