ANALYTICAL SOLUTION OF CATTANEO EQUATION
Maria Lupa
Institute of Mathematics and Computer Science, Czestochowa University of Technology, Poland email: lupa@imi.pcz.pl
Abstract. The Cattaneo equation supplemented by adequate boundary and initial conditions is considered. The solution of this equation is found in analytic way. In the final part of the paper the computations basing on the solution presented are shown.
1. Cattaneo equation
The following equation is considered
2 2
2 2
u τ u u
t t a x
∂ + ∂ = ∂
∂ ∂ ∂ (1)
where u = u (x, t) is an unknown function, a, τ are the positive constants. The equa- tion (1) is supplemented by the following boundary conditions
( ) ( )
0, 0 : 0, 0
0, : , 0
t x u t
t x L u L t
> = =
> = = (2)
and initial ones
( )
00
0 , 0 : , 0 0
0 , 0 : 0
t
x L t u x u
x L t u
t
=< < = = >
< < = ∂ =
∂
(3)
2. Analytical solution
The equation (1) can be written in the form
2 2
2 2
τ 1
u u u 0
x a t a t
∂ − ∂ − ∂ =
∂ ∂ ∂ (4)
To solve the problem (4), (2), (3), the Fourier method [1] is applied. So, one assumes that
( ) ( )
0
,
n,
n
u x t u x t
∞
=
= ∑ (5)
where
( ) , ( ) ( )
n n n
u x t = X x T t (6)
and then
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
2
/ //
2
2
/ //
2
, ,
,
, ,
,
n n
n n n n
n n
n n n n
u x t u x t
X x T t X x T t
t t
u x t u x t
X x T t X x T t
x x
∂ ∂
= =
∂ ∂
∂ ∂
= =
∂ ∂
(7)
Putting (7) into (4) one obtains
( ) ( ) ( ) ( ) ( ) ( )
//
τ
//1
/n n n n n n
0
X x T t X x T t X x T t
a a
− − = (8)
this means
( ) ( ) ( )
( ) ( )
( )
// // /
τ 1
n n n
0
n n n
X x T t T t
X x − a T t − a T t = (9)
or
( ) ( )
( ) ( ) ( )
// /
//
τ 1
n n
n
n n
T t T t
X x a a
X x T t
= + (10)
It is assumed that
( ) ( )
( ) ( ) ( )
// /
//
2
τ 1
λ
n n
n
n
n n
T t T t
X x a a
X x T t
= + = − (11)
where λ
n≠ 0 are the constants.
If the function X
n(x), T
n(t) will be the solutions of equations
( ) ( )
( ) ( )
( )
// /
//
2 2
τ 1
λ , λ
n n
n
n n
n n
T t T t
X x a a
X x T t
= − + = − (12)
then these functions will fulfil the equation (9). The equations (12) can be written in the form
( ) ( )
//
λ
20
n n n
X x + X x = (13)
and
( ) ( ) ( )
// / 2
τ T
nt + T
nt + a λ
nT t
n= 0 (14)
The solution of equation (13) is following
( ) cos λ sin λ
n n n n n
X x = A x + B x (15)
Taking into account the boundary conditions (2) one has
( ) 0 0, ( ) sin λ 0 λ π
n n n n n n
X A X L B L n
= = = = → = L (16)
this means
( ) sin π
n n
X x B n x
= L (17)
The equation (14) can be written in the form
( ) ( )
2 2( )
// /
2
τ
n nn π a
n0
T t T t T t
+ + L = (18)
Under the assumption that
2
4
2π
2τ 0
L − n a > (19)
one obtains the following solution of equation (18)
( ) exp exp ( α ) exp α ( )
n
2τ
n n n nT t t C t D t
= − − +
(20)
where
2 2 2
4 π τ
α
n
2 τ
L n a
L
= − (21)
Finally, the function (5) has a form
( ) ( ) ( )
1
, exp sin π exp α exp α
2τ
n n n n nt n x
u x t E t F t
L
∞
=
= − ∑ − + (22)
where E
n= B
nC
n, F
n= B
nD
nare the constants.
Because the equation (4) is linear, so the function (22) fulfils the equation (4) and boundary conditions (2).
Now, the initial conditions (3) should be taken into account, this means
( ) ( )
01
, 0
n nsin π
n
u x E F n x u
L
∞
=
= ∑ + = (23)
and
( )
0 1
, 1 1 π
α + α sin 0
2τ
n n2τ
n nt n
u x t n x
E F
t L
∞
= =
∂
= − − + − =
∂ ∑ (24)
For the arguments x ∈ [ − L, L] function u (x, 0) can be extended on the uneven func- tion
( ) ( )
( )
0
0
0,
, , 0
, 0 0, 0
, 0,
0,
x L
u x L
u x x
u x L
x L
= −
− ∈ −
= =
∈
=
(25)
Taking into account the expansion of this function into a Fourier series one obtains
( )
0 0
0
2 π 2
sin d 1 1
π
L
n
n n
u n x u
E F x
L L n
+ = ∫ = − − (26)
From condition (24) results that the zero function is expanded into the Fourier series. In such case
1 1
α + α 0
2τ
nE
n2τ
nF
n
− − + − =
(27)
The solution of the system of equations (26), (27) is following
( )
0
1 2α τ
1 1
π 2α τ
n n
n
n
E u n
− +
= − − (28)
and
( )
0
1 2α τ
1 1
π 2α τ
n n
n
n
F u n
+
= − − (29)
Finally, one obtains
( ) ( )
( ) ( )
0
1
1 1 π
, exp sin
2 π τ 2 τ
1 2α τ 1 2 α τ
exp α exp α
α α
n
n
n n
n n
n n
u t n x
u x t
n L
t t
∞
=
− −
= −
− + − + +
∑
(30)
or
( ) ( )
( ) ( )
0
1
2 1 2 1
2 1 2 1
2 1 2 1
2 1 π
, exp 1 sin
π τ 2 τ 2 1
1 2α τ 1 2 α τ
exp α exp α
α α
k
k k
k k
k k
u t k x
u x t
k L
t t
∞
=
− −
− −
− −
−
= −
−
− + − + +
∑
(31)
3. Example of computations
The heat transfer in biological tissue [2] of thickness L = 0.01 m is considered.
The initial temperature of tissue equals u (x, 0) = 37
oC. On the surfaces x = 0 and x = L the Dirichlet condition u (0, t) = u (L, t) = 0 is accepted. The following values of parameters are assumed: thermal diffusivity a = 2.67⋅10
−7m
2/s and relaxation time τ = 0.05 s. The values of L, a, τ fulfil the assumed condition (19).
The temperature field in the domain analyzed is described by equation (1), the boundary conditions (2) and the initial conditions (3). The solution of the problem formulated is given by equation (31).
Fig. 1. Temperature distribution for times 10, 20, 30, 40 and 50 seconds
Fig. 2. Cooling curves at the points 0.002 m, 0.003 m and 0.005 m
In Figure 1 the temperature distribution for times 1 - 10, 2 - 20, 3 - 30, 4 - 40, 5 - 50 second is shown. Figure 2 presents the cooling curves at the points 1 - x = 0.002 m, 2 - x = 0.003 m and 3 - x = 0.005 m.
References
[1] Kącki E., Partial differential equations, WNT, Warsaw 1995 (in Polish).
[2] Majchrzak E., Mochnacki B., Sensitivity analysis and inverse problems in bio-heat transfer modelling, CAMES 2006, 13, 85-108.