Potential Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 352

By the principle of superposition, its solution is a superposition of two parts: u 1
from v 0 πδ ( ω )

a 2 f

and u 2 from − ( t ) . The former satisfies

⎧
⎨

u

a 2

2

p n )

u

+ ( ω + = 0 ,

sin p n b
p n

u

⎩ | t = 0 = v 0 πδ ( ω ) .

sin p n b
p n

e − a 2

2

+ p n ) t . Its inverse transformation

Its solution is u

( ω

( ω , p n , t )= v 0 πδ ( ω )

leads to

p n + σ

2

sin p n b cos p n y
p n

e − a 2

2

p n )

2 v 0 πδ ( ω ) n

( ω + t

u 1 ( ω , y , t )= .

p n + σ

2

b ( )+ σ

In order to find u 2 , by the solution structure theorem, we consider the PDS
⎧
⎨

a 2 v yy +( ω

2 v

v t − a ) = 0 , ( 0 , b ) × ( 0 , + ∞ ) ,

y = 0 =

v y = b =

∂

∂ v v

0 ,

y + σ

0 ,

⎩

∂ y ∂

a 2 f v | t = τ = − ( τ ) .

Its solution is

p n + σ

2

sin p n b cos p n y
p n

e − a 2

2

p n )(

2 a 2 f ( τ ) ∑ n

( ω + t − τ )

v = − .

p n + σ

2

b ( )+ σ

Therefore,

t

u 2 ( ω , y , t )= v d τ

0

2 a 2 t
0

p n + σ

2

sin p n b cos p n y
p n

e − a 2

2

+ p n )( t − τ ) d

( τ ) ∑ n

( ω

= − f τ .

b ( p n + σ

2

)+ σ

Thus the solution of PDS (7.41) is

u ( ω , y , t )= u 1 ( ω , y , t )+ u 2 ( ω , y , t ) .

Its inverse transformation yields the solution of PDS (7.40)

p n + σ

2

sin p n b cos p n y
p n

2 t

n

e − ( ap n ) u ( x , y , t )= 2 v 0

p n + σ

2

b ( )+ σ

t

p n + σ

2

2 a
√ π

( τ )
√ t

f sin p n b cos p n y

p n

2

− τ ) e − x 2 4 a 2

− τ n

e − ( ap n )

( t / ( t − τ ) d

− τ

p n + σ

2

b ( )+ σ 0

2 v 0 −

t

4 a 4 p n τ − x 2 ( t − τ )

2 a
√ π

( τ )
√ t

f

= e

4 a 2

d τ

( t − τ )

− τ 0

p n + σ

2

sin p n b cos p n y
p n

2 t

· n e − ( ap n ) ,

p n + σ

2

b ( )+ σ