Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 128

3.4.2 Fundamental Solution of the One-Dimensional
Heat-Conduction Equation

+ ∞

Rewrite Eq. (3.27) into u = ϕ ( ξ ) V ( x , ξ , t ) d ξ ,where

− ∞

t e − ( x − ξ ) 2

1
2 a √ π

V ( x , ξ , t )= , t > 0 . (3.29)

4 a 2 t

Consider a unit point source at t = 0and x = x 0 such that ϕ ( x )= δ ( x − x 0 ) and

Eq. (3.27) becomes

+ ∞

u = δ ( ξ − x 0 ) V ( x , ξ , t ) d ξ = V ( x , x 0 , t ) ,

− ∞

which is the temperature distribution due to the initial unit point source at x = x 0 .

Therefore, V ( x , ξ , t ) in Eq. (3.29) must satisfy

V t =

a 2 V xx , − ∞ < x < + ∞ , 0 < t ,

(3.30)

V | t = 0 = δ ( x − ξ ) ,

and is the temperature distribution due to ϕ ( x )= δ ( x − ξ ) , an initial unit point

source at x = ξ .

defined by Eq. (3.29) is called the fundamental solution of the one-
dimensional heat-conduction equation .Byusing V , we may rewrite the solution of
PDS (3.25) by

The V ( x , ξ , t )

+ ∞ t + ∞

=

u ϕ ( ξ ) V ( x , ξ , t ) d ξ + d τ f ( ξ , τ ) V ( x , ξ , t − τ ) d ξ . (3.31)

− ∞ 0 − ∞

Therefore, V ( x , ξ , t ) in Eq. (3.29) can also be regarded as the fundamental solution

of PDS (3.25). Here

e − ( x − ξ ) 2

1
2 a π (

V ( x , ξ , t − τ )=

4 a 2

, t > τ (3.32)

( t − τ )

t − τ )

satisfies
V t =

a 2 V xx , − ∞ <

x < + ∞ , 0 < τ < t < + ∞ ,

(3.33)

V | t = τ = δ ( x − ξ )

or

V t

a 2 V xx = + δ ( x − ξ , t − τ ) , − ∞ < x < + ∞ , 0 < τ < t < + ∞ ,

(3.34)

| t = τ = .

V 0