The approach to solving the problem of complex relaxation spectra is presented. Presentation for the XI International Conference on Defect interaction and anelastic phenomena in solids. Tula, 2007.

1. Identification of theIdentification of the
Mathematical Models ofMathematical Models of
Complex Relaxation ProcessesComplex Relaxation Processes
in Solidsin Solids
Bakhrushin V.E.Bakhrushin V.E.
University of HumanitiesUniversity of Humanities
“ZISMG”, Zaporozhye, Ukraine“ZISMG”, Zaporozhye, Ukraine

2. Relaxation processes:
- internal friction;
- dispersion of modulus;
- stress relaxation;
- elastic aftereffect.
Parameters:
- interstitial concentrations for different states;
- interstitial solubility;
- local diffusion coefficients;
- activation energies for jumps.

3. Identification tasks
1. To choose the type of mathematical model: ideal
Debay peak (model of the standard linear body); the
sum of ideal peaks (processes); enhanced Debay
peak; the sum of enhanced peaks; the sum of peaks +
background.
2. To determine the quantity of relaxation processes
3. To determine the parameters of relaxation processes

4. ( )
n
1 1 1 1 i
0 0i
i 1 0i
E E 1 1
Q T Q exp Q cosh
RT R T T
− − − −
=
   
= − + −  ÷ ÷
    
∑
0i
i 0i
kT
E RT ln
hf
=
Model of spectrum at Snoeck relaxation area:
1
0Q ,E−
– background intensity and activation energy;
1
0i 0iQ ,T−
– i-th peak height and temperature
– i-th peak activation energy
f – sample vibration frequency

5. Parameters, which must be determined are:
1
0i 0in,Q ,T−
or
1
0i in,Q ,E−
( )0,14139 0,003245
0T 12,89967 2,706674f 0,04547 0,04929f E,= + + − +
An error for 300 – 800 К interval at f = 20 – 60 Hz
is not more, then 1 %
From Wert & Marx formula such approximation may
be obtained:
( )( )
m 2
1 1
j j
j 1
S Q Q T min− −
=
= − →∑
( )( )
2
1 1
m
j j
1 2
j 1 j
Q Q T
S min
− −
=
−
= →
σ
∑

6. ( )
( )
2
exp z/
z
 − β
 ψ =
β π
( )mz ln /= τ τ z2β = σ
Peak enhance may be taken into account by the model
of log-normal distribution of relaxation time:

7. max
max id
δ
ρ =
δ
2
0,0853 0,197 0,970ρ = ϕ + ϕ +
n
1 1 1 i
0i
i 1 0i
E 1 1
Q Q cosh
R T T
− − −
=
  
= −  ÷
ρ  
∑
In this case:
( )mlnϕ = ωτ
For ρ value from N.P. Kushnareva & V.S. Petchersky
data such approximation may be obtained:

8. Graphic decomposition
2 peaks without error
-8
-4
0
4
1,4 1,7 2 2,3
arcch(Q-1
m/Q-1
)
103
/T

9. 4 peaks + error
-6
-3
0
3
6
1,2 1,5 1,8 2,1 2,4
arcch(Q-1
m/Q-1
)

10. Linear least-squares method:
( )
n
1
j i i j i
i 1
x A cosh [B (x c )]−
=
ϕ = −∑
1
i 0i j j i 0iA Q , x 1/T , c 1/T−
= = =
i
i i
1 k
B ln
c hfc
=

11. ( )
( )
( )
( )
( )
( )
n
j
j j
j 1 1
n
j
j j
j 1 2
n
j
j j
j 1 n
x
y x 0;
A
x
y x 0;
A
x
y x 0,
A
=
=
=
∂ϕ
 − ϕ =  ∂
∂ϕ
 − ϕ =  ∂
∂ϕ
 − ϕ =  ∂
∑
∑
∑
LLLLLLLLLLLL

12. ( )j 1
i j i ij
i
x
cosh [B (x c )] F
A
−
∂ϕ
= − =
∂
n n n n
2
1 1j 2 1j 2 j n 1j nj j 1j
j 1 j 1 j 1 j 1
n n n n
2
1 2 j 1j 2 2 j n 2 j nj j 2 j
j 1 j 1 j 1 j 1
1 nj 1j 2 nj
A F A F F A F F y F ;
A F F A F A F F y F ;
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A F F A F
= = = =
= = = =
+ + + =
+ + + =
+
∑ ∑ ∑ ∑
∑ ∑ ∑ ∑
L
L
n n n n
2
2 j n nj j nj
j 1 j 1 j 1 j 1
F A F y F
= = = =
+ + =∑ ∑ ∑ ∑L

13. 1 11 2 12 n 1n 1
1 21 2 22 n 2n 21
1 n1 2 n2 n nn n
A W A W ... A W Z ;
A W A W ... A W Z ;
. . . . . . . . . . . . . . . . . . . . . . . . . .
A W A W ... A W Z ,
+ + + =
 + + + =


 + + + =
n
1 1
ik i k
j 1 j i j k
1 1 1 1
W cosh B cosh B
x c x c
− −
=
       
 = − − ÷ ÷  ÷  ÷  ÷ ÷        
∑
n
1
i i i
j 1 j i
1 1
Z y cosh B
x c
−
=
  
= −  ÷ ÷
   
∑

14. The advantages of linear least-squares
method:
- simple realization;
- sufficient accuracy (up to 10 % for the main peaks
heights)
Method disadvantages:
- linearization error;
- necessity of peak temperatures preliminary definition;
- possibility to obtain an ill-condition system;
- supposition of uniformly precise of the data;
- supposition of absolute accuracy of temperature
measurements;
- possibility of obtaining the negative values of peak
heights.

15. The method of gradient descent
(linearized least square method)
Main differences:
- an expression of ideal peak is linearized by Taylor
series expansion in the neighborhood of some point
(initial estimate) with abandonment of only linear
terms;
- the possibility to choose the different types of
objective function (cancellation of supposition that the
data have the same errors)
[L. Crer et. al, 1969; M.S. Ahmad et. al, 1971;
O.N. Razumov et. al., 1974; A.I. Efimov et. al., 1982.]

16. ( )( ) ( )
1 1 1m
j j
2
j 1 j k
Q Q T Q T
0, k 1,2,...q
a
− − −
=
− ∂
= =
σ ∂
∑
{ }1 1 1
0 01 0n 1 n 01 0na E,Q ,Q ,...,Q ,E ,...,E ,T ,...,T− − −
=
q 3n 2= +
( )( )
2
1 1
m
j j
1 2
j 1 j
Q Q T
S min
− −
=
−
= →
σ
∑
From
we can obtain:
at general case, q 3n= without background,
q 2n= for ideal Debay peaks.

17. After linearization we obtain:
( ) ( )
m
0 1
k k k mkm
1
a a a M Z ,−
=
∆ = − = ∑l
( ) ( )1 1m
k 2
j 1 j k
Q T Q T1
M ,
a a
− −
=
 ∂ ∂
=  ÷ ÷σ ∂ ∂ 
∑l
l
( )( ) ( )
1 1 1m
j j
2
j 1 j
Q Q T Q T
Z ,
a
− − −
=
− ∂
=
σ ∂
∑l
l
where:
derivatives are determined in 0
ka

18. Adjusted values:
0
k k ka a a , 0 1.= + γ∆ < γ ≤
From the definition of ka∆ follows, that it corresponds
with the general formula of gradient search methods:
0
k k 1a a gradS .= −β
Gradient methods realize an iteration procedure, in
which such stopping conditions may be used:
p p 1
k ka a ;−
− < ε p p 1
1 1S S ;−
− < δ 1gradS ;< ξ , , 0.ε δ ξ >

19. Problems and disadvantages:
- poor convergence at the case of large number of
peaks;
- possibility of iteration stopping at the critical point,
which is not the point of minimum;
- possibility of getting into a loop, when the objective
functional S is ravine;
- absence of realization at standard libraries of the most
popular software packages;
- М matrix must be positively defined at the every step
of iterations

20. ( ) ( )(k 1) (k) 1 (k) (k)
H ,+ −
= −X X X G X
Quasi-Newton algorithm
{ }1 1 1
0 01 0n 1 n 01 0na E,Q ,Q ,...,Q ,E ,...,E ,T ,...,T− − −
=
2 2 2
2
1 1 2 1 q
2 2 2
2
2 1 2 2 q
2 2 2
2
q 1 q 2 q
S S S
...
a a a a a
S S S
...
a a a a aH
... ... ... ...
S S S
...
a a a a a
 ∂ ∂ ∂
 ÷
∂ ∂ ∂ ∂ ∂ ÷
 ÷∂ ∂ ∂
 ÷
∂ ∂ ∂ ∂ ∂=  ÷
 ÷
 ÷
 ÷∂ ∂ ∂
 ÷
∂ ∂ ∂ ∂ ∂ 
1
2
q
S
a
S
aG
...
S
a
∂ 
 ÷∂
 ÷
∂ ÷
 ÷∂=
 ÷
 ÷
 ÷∂
 ÷
 ÷∂ 

21. ij
ij
h , i j;
c
0, i j.
 =
= 
≠
n
1 T
1 k k k
k 1
P z ,
−
=
= ∑ v v
° 1 1 1
1H C PC ,
− − −
=
zk are eigenvalues and vk are eigenvectors of matrix:
Grinshtadt technique:
1 1
P C HC ,− −
=
It is necessary to provide the positive definiteness of
Hesse matrix or to find an approximation of Н-1

22. and
F is a Fisher criterion value for the corresponding
numbers of degrees of freedom and significance level,
∆2
- sum of errors squares (relative errors) of
experimental points.
Adequacy criteria for spectrum models:
2
S
F≤
∆
2
F
S
∆
≤
( )0,052
S
F F 2,0...2,5> ≈ ⇒
∆
number of model parameters must be increased;
2
F
S
∆
> ⇒
number of model parameters must be decreased.

23. Quasi-unimodelity (an absence of physically
different minimums) of objective functional, that is all
minimums of objective functional correspond to the
same physical model of a spectrum.
Deviation from quasi-unimodality may be caused
with:
- the presence of excess peaks in the model;
- absence of some essential peak in the model;
- presence at the real spectrum of some collateral
peak, which height is close to measurement error.

24. Absence of model residuals serial correlation
(Darbin & Watson criterion):
( )
m 2
j j 1
j 2
m
2
t
j 1
e e
d ,
e
−
=
=
−
=
∑
∑
( )1 1
j j je Q Q T− −
= − - model residuals.
d 2≈ - serial correlation is absent;
d 0→
d 4≈
- positive serial correlation;
- negative serial correlation
(there are excess peaks).

25. 0
5
10
500 600 700 800
Q-1
·103
T, K
Given data (4 peaks + error)

26. Initial approach (4 peaks) σ=0,1

27. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
The result of decomposition
S=0,30
F=0,74

28. -0,25
-0,1
0,05
0,2
500 600 700 800
∆Q-1
·103
T, K
Residuals
d=2,18

29. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
Initial approach 1 (3 peaks) σ=0,1

30. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
The result of decomposition 1 (3 peaks)
S=20,52
F=50,31

31. -2
-1
0
1
500 600 700 800
∆Q-1
·103
T, K
Residuals
d=0,33

32. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
Initial approach 2 (3 peaks) σ=0,1

33. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
The result of decomposition 2 (3 peaks)
S=13,58
F=33,28

34. -2
-1
0
1
500 600 700 800
∆Q-1
·103
T, K
Residuals
d=0,54

35. 0
4
8
12
500 600 700 800
Q-1
·103
T, K
Initial approach 1 (5 peaks) σ=0,1

36. Initial approach 2 (5 peaks)
0
4
8
12
500 600 700 800
Q-1
·103
T, K
σ=0,1

37. Given 3_1 3_2 4 5_1 5_2
T1 570 571,3 578,6 569,4 569,4 569,4
T2 620 632,5 619,8 619,8 619,8
T3 690 698,1 685,5 690,2 690,2 690,2
T4 750 742,4 749,2 749,2 749,2
T5 499,6 718,2
Q1 6 6,3 7,4 5,9 5,9 5,9
Q2 3 3,5 3,1 3,1 3,1
Q3 12 13,0 12,2 12,0 12,0 12,0
Q4 3 3,6 2,9 2,9 2,9
Q5 0,0 0,0
σ=0,1

38. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
σ=0,3Initial approach 1 (4 peaks)

39. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
Initial approach 2 (4 peaks) σ=0,3

40. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
Initial approach 3 (4 peaks) σ=0,3

41. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
The result of decomposition
s=3,01
F=0,76

42. -0,5
-0,2
0,1
0,4
500 600 700 800
∆Q-1
·103
T, K
d=1,20
Residuals

43. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
Initial approach 1 (3 peaks) σ=0,3

44. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
Initial approach 2 (3 peaks) σ=0,3

45. 0
4
8
12
500 600 700 800 T, K
Q-1
·103
Initial approach 3 (3 peaks) σ=0,3

46. The result of decomposition
0
4
8
12
500 600 700 800 T, K
Q-1
·103
s=15,87
F=3,99

47. -2,5
-1
0,5
500 600 700 800
∆Q-1
·103
T, K
Residuals
d=0,57

48. 0
4
8
12
500 600 700 800
Q-1
·103
T, K
Initial approach 1 (5 peaks) σ=0,3

49. 0
4
8
12
500 600 700 800
Q-1
·103
T, K
The result of decomposition
s=2,92
F=0,74

50. -0,5
-0,2
0,1
0,4
0,7
500 600 700 800
∆Q-1
·103
T, K
Residuals
d=1,21

51. 0
4
8
12
500 600 700 800
Q-1
·103
T, K
Initial approach 2 (5 peaks) σ=0,3

52. 0
4
8
12
500 600 700 800
Q-1
·103
T, K
The result of decomposition
s=1,41
F=0,35

53. -0,5
-0,2
0,1
0,4
0,7
500 600 700 800
∆Q-1
·103
T, K
Residuals
d=2,47

54. Given 3 4 5_1 5_2
T1 570 578,7 569,8 571,2 569,7
T2 620 621,8 622,8 621,3
T3 690 686,5 691,2 691,2 690,6
T4 750 749,2 757,1 757,2 751,7
T5 537,5 851,0
Q1 6 7,2 5,9 5,7 5,8
Q2 3 3,0 2,9 2,9
Q3 12 12,4 12,2 12,2 12,0
Q4 3 3,6 2,9 2,9 2,9
Q5 0,3 0,6
σ=0,3

55. 0
10
20
450 550 650 750
Q-1
·103
T, K
0,18
0,36
0,42
0,60
[N], at.%:
Nb – 2 at.% W – N (3 peaks, 1 result)

56. -3,5
-2,5
-1,5
-0,5
0,5
1,5
2,5
450 550 650 750
∆Q-1
·103
T, K
d: 0,69; 0,52
1,94; 1,14
Residuals

57. 0
10
20
450 550 650 750
Q-1
·103
T, K
0,18
0,36
0,42
0,60
[N], at.%:
Nb – 2 at.% W – N (3 peaks, 2 result)

58. -3,5
-2,5
-1,5
-0,5
0,5
1,5
2,5
450 550 650 750
∆Q
-1
·103
T, K
d: 0,85; 1,33
1,88; 1,27
Residuals

59. 0
10
20
450 550 650 750
Q-1
·103
T, K
[N], at.%: 0,18
0,36
0,42
0,60
Nb – 2 at.% W – N (4 peaks)

60. -2,2
-1,2
-0,2
0,8
1,8
2,8
450 550 650 750
∆Q-1
·103 d: 1,83; 2,10
2,84; 1,67
Residuals

61. 0
10
20
450 550 650 750 T, K
Q-1
·103
[N], at.%: 0,18
0,36
0,42
0,60
Nb – 2 at.% W – N (5 peaks, 1 result)

62. Residuals
-2
-1
0
1
2
450 550 650 750
∆Q
-1
·103
T, K
d: 1,60; 2,36
3,09; 1,83

63. 0
10
20
450 550 650 750 T, K
Q-1
·103
[N], at.%: 0,18
0,36
0,42
0,60
Nb – 2 at.% W – N (5 peaks, 2 result)

64. -2
-1
0
1
2
450 550 650 750
∆Q-1
·103
T, K
d: 2,54; 2,36
2,96; 2,25
Residuals

65. T1 T2 T3 T4 Т5 S F
3_1 539,1 657,2 685,2 70,1 2,0
3_2 647,1 673,6 748,7 86,5 1,6
4 537,9 650,9 674,7 749,0 43,3 3,3
5_1 528,8 656,9 676,4 748,0 593,0 35,7 4,0
5_2 535,5 641,4 674,4 745,5 665,5 34,1 4,2
Q1 Q2 Q3 Q4 Q5
3_1 2,2 18,2 13,4
3_2 9,3 21,3 2,7
4 2,1 10,8 19,5 2,6
5_1 1,5 13,6 15,9 2,6 2,1
5_2 2,0 7,0 17,9 2,6 5,8

66. Nb – 12 at.% W Nb – 6
at.%
W
4
peaks
5 peaks 4
peaks1 set 2 set
E1, kJ/mol 102,2 86
E2, kJ/mol 109,5 111,5 110,1 109,8
E3, kJ/mol 116,3 116,9 116,5 116,5
E4, kJ/mol 128,9 129,6 129,1 128,3
E5, kJ/mol 1456 145,9 145,9 145,9

67. 0
0,3
0,6
0,9
500 600 700 800
Nb – 2 at.% Hf – 0,32 at.% N (3 peaks):
E = 1,47; 1,61; 1,76 kJ/mol;
d = 0,90; F = 1,73.

68. 0
0,3
0,6
0,9
500 600 700 800
Nb – 2 at.% Hf – 0,32 at.% N (4 peaks):
E = 1,29; 1,48; 1,62; 1,79 kJ/mol;
d = 1,26; F = 2,64.

69. 0
0,3
0,6
0,9
500 600 700 800
Nb – 2 at.% Hf – 0,32 at.% N (5 peaks, 1 result):
E = 1,44; 1,54; 1,63; 1,77; 1,91 kJ/mol;
d = 1,06; F = 3,21.

70. 0
0,3
0,6
0,9
500 600 700 800
Nb – 2 at.% Hf – 0,32 at.% N (5 peaks, 2 result):
E = 1,26; 1,46; 1,57; 1,64; 1,80 kJ/mol;
d = 1,55; F = 3,58.

71. n
N i
i 1
0i
i 2 2 2
i
i
0i
i
M(T) M M (T);
M
M (T) ;
1 4 f
E 1 1
exp
R T T
.
2 f
=


 = − ∆


∆
∆ =
+ π τ
   
 −  ÷
   τ = π
∑
The temperature dependence of dynamic elastic
modulus in a case of n processes, which satisfy the
model of standard linear body, may be determined
from a system:
MN – non-relaxed modulus.

72. Model parameters, which must be identified, are:
0i 0iM ,T .∆
( ) ( )
m 2
1 exp j j
j 1
S M T M T min,
=
 = − → ∑
We are to solve such minimization problem:
( )exp jM Twhere are experimental data for modulus at Tj.
(*)

73. Functional (*) has a great number of minimums,
so the result of minimization strongly depends on initial
assumption.
Adequate model may be obtained by using as T0i
initial values the results of relaxation spectrum
decomposition and setting initial values as0iM∆
1
0i N 0iM 2M Q .−
∆ =
T0i values after minimization are very close with the
initial ones, and values change essentially. But
there is a correlation (r = 0,90 – 0,97) between partial
Snoek peaks heights and results for :
0iM∆
0i
1
0i H
M
2,00 0,15.
Q M−
∆
= ±
0iM∆

74. Nb – 12 at.% W – N
[N], at.%:  - 0,11;
■ – 0,16;
▲ – 0,22;
♦ - 0,31