Selçuk J. Appl. Math. Selçuk Journal of Vol. 11. No.1. pp. 95-105 , 2010 Applied Mathematics
On the Solution of a Mathematical Model of Heat Distribution of Composite Medium
Hülya Y¬ld¬r¬m Ekinci
Ni¼gde University, Faculty of Education Department of Mathematics, 51200, Ni¼gde, Türkiye
e-mail: hulyaekinci@ nigde.edu.tr
Received Date: April 21, 2009 Accepted Date: February 8, 2010
Abstract. In this study, the eigenvalues and eigenfunctions of the spectral problem arising in …nding of the heat distribution of a composite medium con-sisting of two layers in contact are investigated and a formula for the heat distribution is given.
Key words: Heat distribution; eigenvalue; eigenfunction; composite medium. 2000 Mathematics Subject Classi…cation: 15A18, 15A39, 15A90.
1. Introduction
Heat transfer and temperature distribution in composite materials are subject to investigation in many industrial problems [1, 2]. For the heat distribution of a composite material which is consisting of so many parallel layers that are contacted to each other, a formula about the eigenvalue and eigenfunction of auxiliary spectral problem is given in Chapter 8 in Öz¬s¬k [7].
In this study, the existence of in…nite number of eigenvalues of auxiliary spectral problem in [4, 8] is proved. For some special cases we have found explicit forms of eigenvalues. We have, in general, found and asymptotic formula for the eigenvalues.
2. Temperature Distribution in Composite Materials
It will be useful to de…ne one-dimensional heat conduction formula related with the characteristics of the materials before giving the boundary conditions for calculation of the temperature distribution.
De…nition: Let U (x; t) denote the normalized heat distribution, where tis time and x is distance. The time-dependent one-dimensional heat distribution equation [4] is given in general by the formula
(1) @U (x; t)
@t = K
@2U (x; t)
@x2 + (1=c) R(x; t) h [U (x; t) U0]
Here, c is the heat composite of unit volume,R(x; t) is the heat generated in distance x and in time t; K is thermal di¤usivity, h is the positive surface heat transfer coe¢ cient and U0 is the ambient temperature in (1). All quantities
except U (x; t) and R(x; t) in (1) are assumed to be constant in general. Let us assume that two material (1) with thermal conductivity 1 and material (2) with thermal conductivity 2be constant at the boundary x = a, as shown in Fig.1.
Figure 1
At point x = a there may be a heat absorption or heat generation in general. In this case, using
lim
x!af (x) = f (a 0) and xlim!af (x) = f (a + 0)
x < a x > a
the following equations can be written
1U (a 0; t) = 2U (a + 0; t)
1Ux(a 0; t) = 2Ux(a + 0; t)
where 1, 2, 1and 2 are all positive quantities [7].
3. Separation of Variables Method Auxiliary Spectral Problem Separation of variables is a convenient method for solving partial di¤erential equations and used extensively in the literature. Theoretical investigations of
eigenvalues and eigenfunctions for the solution related with the separation of variables were studied under the concept of spectral theory.
Let’s write the following equations at x = a and assume that R(x; t) 0, h 0 in (1). Then (1) becomes, (2) Ut= K1Uxx(0 < x < a; t > 0) Ut= K2Uxx(a < x < b; t > 0) (3) U (0; t) = 0; U (b; t) = 0 (4) U (a 0; t) = U (a + 0; t) Ux(a 0; t) = Ux(a + 0; t) (5) U (x; 0) = f (x); 0 < x < b where K1; K2 and ; are all positive quantities [6].
Let’s seek the solution (2) by taking the conditions of (3) and (4) into consid-eration and regarding out the (5).
(6) U (x; t) = e ty(x); 0 < x < b;
where is a complex valued parameter and y(x) is a function of only x . If we put the solution of U (x; t) in (6) into (2), (3), (4) we obtain,
(7) y00(x) = 2 1y(x); 0 < x < a y00(x) = 2 2y(x); a < x < b (8) y(0) = y(b) = 0 (9) y(a 0) = y(a + 0) y0(a 0) = y0(a + 0) where 1= 1=pK1 and 2= 1= p K2 [3, 5].
De…nition: If there is a solution of y(x) to (7), (8) and (9) which is di¤erent from zero for a complex value of , then is called eigenvalue and y(x) is called eigenfunction corresponding to .
For the solution of (2), (3) and (4) in the form of (6), it is necessary and suf-…cient to …nd a “ ” as eigenvalue of (7), (8), (9) and y(x) is an eigenfuction corresponding to this eigenvalue. Spectrum of the problem is a group of eigen-values for (7), (8) and (9). Spectral problem is the investigation of eigeneigen-values [1, 2].
4. Study of Spectral Problem
De…nition: For complex , if there is a non-trivial and twice di¤erentiable functiony(x) satisfying (7) over (0; a) and (a; b),then “ ”is called an eigenvalue and y(x) is called an eigenfunction.
Lemma 1: Eigenvalues of problems (7), (8), (9) are real.
Proof: Let be an eigenvalue of (7), (8) and (9), and y(x) be an eigenfunction corresponding to . We multiply equation (7) by y and integrate from 0 to a then we obtain a Z 0 y00(x)y(x)dx = 21 a Z 0 jy(x)j2dx; (0 < x < a) and b Z a y00(x)y(x)dx = 22 b Z a jy(x)j2dx; (a < x < b):
By using these equations, we get
y0(a 0)y(a 0) a Z 0 y0(x) y0(x)dx = 2 1 a Z 0 jy0(x)j2dx y0(a + 0)y(a + 0) + b Z a y0(x)y0(x)dx = 2 2 b Z a jy (x)j2dx; respectively.
From equation (9), we obtain
(10) y0(a + 0) y (a + 0) a Z 0 jy0(x)j2dx = 21 a Z 0 jy (x)j2dx (11) y0(a + 0)y(a + 0) + b Z a jy0(x)j2dx = 22 b Z a jy (x)j2dx
Both sides of (11) is multiplied by and summed by (10), giving a Z 0 jy0(x)j2dx + b Z a jy0(x)j2dx = 21 a Z 0 jy (x)j2dx + 22 b Z a jy (x)j2dx
From the last equality the following expression can be written,
(12) = a Z 0 jy0(x)j2dx + b Z a jy0(x)j2dxt 2 1 a R 0 jy (x)j 2 dx + 2 2 b R a jy (x)j 2 dx :
The last equation implies that is real.
Lemma 2: The following orthogonal condition is satis…ed by eigenfunctions y1(x) and y2(x) with the corresponding eigenvalues 1 and 2 [7].
(13) 21 a Z 0 y1(x)y2(x)dx + 22 b Z a y1(x)y2(x)dx
Proof: Let’s consider the interval 0 < x < a …rst
(14) y001(x) = 1 21y1(x)
(15) y002(x) = 2 21y2(x)
multiplying both sides of equality (14) by y2(x) taking the conjugate of equality
(15) and then multiplying both sides by y1(x) give
y100(x)y2(x) = 1 21y1(x)y2(x)(x)
y1(x)y002(x) = 2 21y1(x)y2(x):
After subtracting these two equations, we can integrate over 0 < x < a and it gives a Z 0 [ y100y2(x) + y1(x)y200(x)dx = ( 1 2) 21 a Z 0 y1(x)y2(x)dx:
Solving the last equation by integration by parts and using conditions (8) and (9), we get (16) y10(a 0)y2(a 0) + y1(a 0)y20(a 0) = ( 1 2) 21 a Z 0 y1(x)y2(x)dx:
If we repeat the same process for (14) and (15) for the interval a < x < b and when we take y001(x) = 1 22y1(x) y002(x) = 2 22y2(x); we obtain, y001(x)y2(x) = 1 22y1(x)y2(x) y1(x)y002(x) = 2 22y1(x)y2(x):
These two equations are substituted each other and integrated over a < x < b so the following equality is obtained, by using (8) and (9)
(17) y0(a + 0) y(a + 0) + y(a + 0)y0(a + 0) = ( 1 2) 22 b
Z
a
y1(x)y2(x)dx
Using (9), (16) and (17) and 16= 2, we have
2 1 a Z 0 y1(x)y2(x)dx + 22 b Z a y1(x)y2(x)dx
Lemma 3: Eigenvalues of problems (7), (8), (9) has one layered. That is, for each eigenvalue there is only one linearly independent eigenfunction.
Proof: Suppose that linearly independent eigenfunctions y1(x) and y2(x)
cor-respond to eigenvalue . Then, y1(x) and y2(x) are linearly independent on
at least one of the intervals (0; a) and (a; b). Suppose the contrary that; there exist constants C1 and C2 such that y2(x) = C1y1(x)( 0 < x < a ), y2(x) =
C2y1(x)( a < x < b ). Substituting these into (4),
y1(a 0) = y1(a + 0); y2(a 0) = y2(a + 0)
y10(a 0) = y 0 1(a + 0); y 0 2(a 0) = y 0 21pt(a + 0) C1y1(a 0) = C2y1(a + 0) C1y 0 1(a 0) = C2y 0 1(a + 0):
Then, since y(x) 6= 0 we see that C1 = C2. Therefore, there is only one linear
independent eigenfunction corresponding to eigenvalue .
In order to investigate eigenvalues, let’s set the characteristic equation. Let '(x; ) (0 < x < a ) be the solution of the initial value problem
(18) y00(x) = 21y(x) (0 < x < a )
(19) '(0; ) = 0; '0(0; ) = 1;
and
(x; ) ( a < x < b ) the solution of the initial value problem
(20) y00(x) = 22y(x) ( a < x < b )
(21) (b; ) = 0; 0(b; ) = 1:
Lemma 4: Eigenvalues of equations (7), (8), (9) and the roots of the function ( ) are coincident where
( ) = '(a 0; ) (a + 0; ) '0(a 0; ) 0(a + 0; )
( ) = '(a 0; ) 0(a + 0; ) '0(a 0; ) (a + 0; )
Proof: It can easily be seen that the solution of (18) satisfying the condition of y(0) = 0 is
and solution of (20) satisfying the condition of y(b) = 0 y(x) = C2 (x; ) ( a < x < b ):
Then the following equations should be satis…ed for satisfaction of (9). C1'(a 0; ) = C2 (a + 0; )
C1'0(a 0; ) = C2 0(a + 0; )
At least one of the constants C1 and C2 should be di¤erent from zero for y(x)
to be non-trivial. Then we obtain,
( ) = '(a 0; ) (a + 0; ) '0(a 0; ) 0(a + 0; ) = 0:
In order to seek the presence of eigenvalues of (7), (8) and (9) characteristic equation which is satis…ed by eigenvalues should be obtained. For that reason, let’s write = s2 in (7). The general solution of (7) satisfying the conditions of (8) can be written as
y00(x) = s2 21y(x) (0 < x < a)
y00(x) = s2 2
2y(x) (a < x < b):
The solution of these equations
y(x) = C1cos s 1x + C2 sin s 1x ( 0 < x < a)
y(x) = C3 cos s 2x + C4sin s 2x ( a < x < b)
using y(0) = 0 in the …rst equation, we get C1= 0:
For C2= c1, we …nd
y(x) = c1sin s 1x (0 < x < a)
second equation becomes
0 = C3 cos s 2b + C4sin s 2b
Then,
C4 = C3
cos s 2b sin s 2b Let’s substitute this into
y(x) = C3cos s 2x C3
cos s 2b
y(x) = C3 sin s 2b
[ cos s 2b sin s 2x + sin s 2b cos s 2x] :
For C3 sin s 2b = c2, y(x) = c2sin s 2(b x) (a < x < b): As a result, we …nd (22) y(x) = c1sin s 1x (0 < x < a) y(x) = c2sin s 2(b x) (a < x < b) ; where c1; c2 2 R.
In this phase we can investigate the values of . For (22) to satisfy (9), we must have c1 sin s 1x = c2sin s 2(b x) s 1c1 cos s 1x = s 2 c2 cos s 2(b x) sin s 1a sin s 2(b x) 1cos s 1x 2cos s 2(b a) c1 c2 = 0 0 since c1 = 0 _ c2 = 0 , sin s 1a sin s 2(b x) 1cos s 1x 2cos s 2(b a) = 0:
( ) = 2sin(s 1a) cos(s 2(b a)) + 1sin(s 2(b x)) cos(s 1x) = 0 we take particularly 2 = 1, sin(s 2(b a) + s 1a) = 0: Then, we obtain s [ ( 1 2) a + 2b] = n sn = ( 1 2)a + 2b n; n = 0; 1; 2; ::: n= s2n= ( 1 2)a + 2b 2 n2; n = 0; 1; 2; ::: For. 2 6= 1,
2 1
sin(s 1a) cos 2(b a) + sin(s 2(b x)) cos(s 1x) = 0
2(b a) = k
1a = 1
A = 2
1
sin sk cos sl + A sin sl cos sk = 0:On the other hand, using trigonometric identities
sin cos = 1
2 [sin( + ) + sin( )] sin sk cos sl = 1
2(sin [s(k + 1)] + sin [s(k 1)] ) A sin sl cos sk = (A=2) (sin [s(k + 1)] sin [s(k 1)])
2 sin sk cos sl = sin [s(k + 1)] + sin [s(k 1)]
2 A sin sl cos sk = A sin [s(k + 1)] A sin [s(k 1)]
2 sin sk cos sl + 2A sin sl cos sk = (1+A) sin [s(k + 1)] +(1 A) sin [s(k 1)] and setting
1 + A = B ; 1 A = C k + 1 = d1 and k 1 = d2;
then we see that
(23) B sin sd1 + C sin sd2 = 0:
5. Temperature Distribution Formula
The solution of equations (2), (3), (4) and (5) isU (x; t) and it is written as,
U (x; t) =
1
X
n=1
Where the coe¢ cients Cnin the series solution can be obtained from the function
f (x) inside the initial condition described by (5)
f (x) = 1 X n=1 Cnyn(x): Then, we can …nd Cn= hf; yni hyn; yni = 2 1 a R 0 f (x)yn(x)dx + 22 b R a f (x)yn(x)dx 2 1 a R 0 y2 n(x)dx + 22 b R a y2 n(x)dx n=1,2,3,. . . 6. Conclusion
The heat distribution problem in composite medium given in (7), (8) and (9) was investigated by separating of variables. In the analysis of the spectral problem we have done, the reality of the eigenvalues, the orthogonality of the eigenfunctions, the eigenvalues and layered of the problem were investigated. Then the existence of the in…nite number of eigenvalues was stated and proved. The problem of the asymptoticness of the roots was not considered here but can be done for a future research.
The originality of this study is mentioned in Öz¬s¬k [7]. There, some basic for-mulas about layered composite medium are given except for a spectral analysis. In this study, all of the spectrum properties of the problem were analyzed and eigenvalues and eigenfunctions were investigated.
References
1. F. V. Atkinson, Discrete and Continuous Boundary Problems, Academic Press, New York, (1964).
2. R. E. Bellman, Stability Theory of Di¤erential Equations McGraw-Hill, New York, (1953).
3. P. E. Bulvain, and V. M. Kascheev, Solution of The Non-homogeneous Heat Con-duction Equation for Multilayered Bodies.Int Chem Engng 5 (1965), 112-115. 4. R. V. Churchill, Operational Mathematics McGraw-Hill Kogakusha Ltd Tokyo, (1972).
5. G. P. Mulholland, Cobblem H. Di¤usion Through Composite Media. Int J Heat Mass Transfer, 15 (1972), 147-160.
6. G. Oturanç, On the Investigation of Temperature Distribution of Composite Mater-ial. Sevennth National Matematical Symposium, Adana, Turkey (In Turkish), (1995). 7. N. Öz¬¸s¬k, Heat Conduction, John Willey & Sons, 2nd, (1980).
8. C. W. Title, Boundary Value Problem in Composite Media, J Appl Phys, 36 (1965), 1586-1488.
