Mixed Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 151

Solution.

1. u 3 ( x , t ) .Tofind u 3 ( x , t ) , we first develop the Green function G satisfying

⎧
⎨

a 2 G xx ,

G t + τ 0 G tt = 0 < x < l , 0 < τ < t < + ∞ ,

G x ( 0 , t )= G x ( l , t )+ hG ( l , t )= 0 , (4.21)

⎩

− ξ ) τ 0

G | t = τ = 0 , G t | t = τ = δ ( x .

By taking the boundary conditions into account, we use the eigenfunctions in
Row 6 in Table 2.1 to expand G so that

+ ∞
m = 1 T m ( t ) cos β m x ,

β m = μ m l G = ,

x
lh . Substituting it into

where the μ m are the positive zero-points of f ( x )= cot x −

the equation of (4.21) yields

τ 0 T m + T m +( β m a

2 T m =

) 0 .

Its general solution is

t − τ

2τ 0

e −

T m ( t )= [ a m cos γ m ( t − τ )+ b m sin γ m ( t − τ )] ,

where a m and b m are undetermined constants, and

1

1
2

1
2

1
2

2

r 1 , 2 = −

τ 0 ±

− 4 τ 0 ( β m a ) = −

τ 0 ± γ m i

.

τ 0

Applying the initial condition G | t = τ = 0 yields a m = 0. To satisfy the initial

− ξ ) τ 0 we have condition G t | t = τ = δ ( x

+ ∞

m = 1 b m γ m cos β m x = δ ( x − ξ )

,

τ 0

which requires

1
γ m M m

l

δ ( x − ξ )

τ 0

1
τ 0 γ m M m cos

b m = cos β m x d x = β m ξ . (4.22)

0

Finally, we have the Green function

+ ∞
m = 1

1
τ 0 γ m M m cos

t − τ

2τ 0

e − G = β m ξ cos β m x sin γ m ( t − τ ) .