Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 217

5.7.3 Method of Continuation for Spherically-Symmetric Problems

We now apply the method of continuation to solve
⎧
⎨

r 2 ∂

u t
τ 0 +

A 2 1

r 2 ∂

u

u tt = + f ( r , t ) , 0 < r < + ∞ , 0 < t ,

∂ r ∂ r

(5.93)

u r ( 0 , t )= 0 ,

⎩

u ( r , 0 )= ϕ ( r ) , u t ( r , 0 )= ψ ( r ) .

By the solution structure theorem, we can first seek the solution of
⎧
⎨

u t τ 0 +

A 2 u tt = Δ u , 0 < r < + ∞ , 0 < t ,

u r ( 0 , t )= 0 , (5.94)

⎩

u ( r , 0 )= 0 , u t ( r , 0 )= ψ ( r ) .

should be an even
function of r . To obtain the solution of PDS (5.94), consider an auxiliary problem
⎧
⎨

Based on the given boundary condition u r ( 0 , t )= 0, the u ( r , t )

u t
τ 0 +

A 2 u tt = Δ u , − ∞ < r < + ∞ , 0 < t ,

(5.95)

ψ ( r ) , 0 ≤ r ,

⎩

u ( r , 0 )= 0 , u t ( r , 0 )= Ψ ( r )=

ψ ( − r ) , r < 0 ,

where Ψ ( r ) comes from an even prolongation of ψ ( r ) .

By function transformation v ( r , t )= ru ( r , t ) , PDS (5.95) is transformed to

⎧
⎨

v t
τ

A 2 v rr , − ∞ <

0 +

v tt = r < + ∞ , 0 < t ,

r

(5.96)

ψ ( r ) , 0 ≤ r ,

⎩

v ( r , 0 )= 0 , v t ( r , 0 )=

ψ ( −

) , < .

r r r 0

Its solution is available in Eq. (5.30). Thus the solution of PDS (5.94) is,

u ( r , t )= W ψ ( r , t )

t

I 0 b

2

τ 0 r + At
r 2 Ar e −

1

2

2

= ( At ) − ( r − ρ ) ρψ ( ρ ) d ρ , (5.97)

− At

where 0.

Finally, the solution of PDS (5.93) reads, by the solution structure theorem,

ψ ( ρ ) should be ψ ( − ρ ) if ρ <

1 W ϕ (

t

τ 0 + ∂

u = r , t )+ W ψ ( r , t )+ W f τ ( r , t − τ ) d τ , (5.98)

∂ t

0

1 2 a √ τ 0 ,

where f τ = f ( r , τ ) , b = ϕ ( − r )= ϕ ( r ) , f ( − r , t )= f ( r , t ) .