Applications to Science and Engineering

CHAPTER 10

Applications to Science and Engineering

10.1

Introduction

This chapter is devoted to describing some applications of functional equations to Science and Engineering. When designing a mathematical model to represent a physical reality it is a normal practice to choose functions that satisfy certain conditions. The aim of sections 10.3, 10.4 and 10.5 is to show how functional equations can be used to impose some constraints on the class of admissible functions. Each functional equation states a given property the mathematical model must satisfy. With the help of these examples, a new, powerful and rigorous methodology of model design is presented. We start by using a motivating example in Section 10.2. In Section 10.3 we derive the general structure of the laws of Science and show that an arbitrary selection of these laws can lead to inconsistencies. Section 10.4 is devoted to a statistical model for lifetime analysis, which is based on a compatibility condition and some known results from the theory of extreme value distributions. Section 10.5 considers an interesting example in which we explain in detail all the steps followed by a three member team to design a fatigue life model, starting with the separate proposals of the three members and their physical or empirical bases and ending with a consensus solution. Finally, in Section 10.6 functional networks are used to predict values of magnitudes satisfying differential and/or difference equations and to obtain the differential/difference equation associated with a set of data. As shown in that section, the estimation of its coefficients is done by simply solving systems of linear equations. 233

234

Chapter 10. Applications to Science and Engineering

z/ z/

u(x)

= "el

Figure 10.1: Graphical illustration of the tax function.

10.2

A motivating

example

We start by giving an example where the need for careful selection of models is illustrated. E x a m p l e 10.1 (Families o f t a x f u n c t i o n s ) . Let us assume that the person responsible for the tax policy of the European Community decides to establish a two-parameter family of tax functions given by

x2+x u(x) = Cx + D'

(10.1)

where x is the income, u(x) is the due tax and C and D are the two parameters (see Figure 10.1). Let us suppose, also, that this family of functions is to be utilized in all country members, using their corresponding monetary units. Because the Expression (10.1) depends on two parameters, C and D, the legislator has only two degrees of freedom, that is, he/she is free to fix two points of the tax function. Let us assume that for country A a tax amount of 10 monetary units for an income of 100 units and 200 monetary units for an income of 1000 units are chosen. This leads to the following system of equations, values of constants C and D and tax function 10= 100C+D 200-

10002 + 1000 1000C+D

/ / 799/ C=

180

D = 10190 18

u(z) = ls~

+ z)

799x + 101900'

(~0.2)

where x and u(x) are in monetary units of country A. If the model (10.1) is to be used in another country, B, whose monetary unit is such that r units are equivalent to 1 unit of country A, and we desire to fix the same points as above, then the model should be fitted by the conditions: 100r units of income should pay 10r units of tax and 1000r units of income

10.2. A motivatingexample

235

u(x) lO.O-r i r = 0.005 7.5 5.0 2.5 0.0 0

I

J

25

5

!

75

"!

100

x

Figure 10.2: Tax function for several values of r.

should pay 200r units of tax. Thus, now the system of equations, the C and D constants and the tax function become

lOr= lO02r2+lOOr lOOrC+ D 200r -

I

10002r 2 + 1000r 1000rC + D

~(y) =

I C = 80Or-1 180r

I

D - 10000r + 190 18

(~o.3)

180r(y 2 + Y) ( 8 0 0 r - 1)y + 100000r 2 + 1900r'

u(x),

where now, instead of x and we use y and in the m o n e t a r y units of country B.

v(y) because they are measured

In order to compare the tax functions in countries A and B, we now change y and into the monetary units of country and y = rx) to get

v(y)

A (v(y)= ru(x)

180(rx 2 + x) u (x) = (800r - 1) x + 100000r + 1900'

(10.4)

which is not only different from the tax function in (10.2) but also depends on r. This implies t h a t the citizens of countries A and B pay different taxes. Because of these facts, we say t h a t the family of models (10.1) is inconsistent. Figure 10.2 shows tax functions (10.4) for several values of r (1, 0.01 and 0.005). If, instead of (10.1), the following family of tax functions is selected x2

u(z) = c z + D'

(~o.5)

236

Chapter 10. Appfications to Science and Engineering w(x,r) ] Estimation ~ in c o u n ~

lc~

Estimation ....... ~ u n t r y A

-

tcou~

Figure 10.3: Two different ways of obtaining model A.

i0r i00r / / '0 /

then, using the same conditions, instead of (10.3) and (10.4), we get = 100rC + D

200r

10002r2 = ~ T D

C = --

=~

D-

9 5000r 9

9y 2 v(Y) = 40y + 5000r

(10.6)

and 9x 2

u(x) = 40x + 5000"

(10.7)

Note that in this case (10.7) can be obtained from (10.6) replacing y by xt and v(y) by u(x)t. Then, the family of models (10.5) is consistent. Thus, we realize that the selection of the u(x) family of tax functions, such as those in (10.1) and (10.5), cannot be arbitrary; on the contrary, it must satisfy some conditions in order to avoid the above problems. Figure 10.3 shows a diagram in which we illustrate the two different processes we can follow in order to get model (10.7). We want the diagram in Figure 10.3 to be commutative; that is, we want a family of functions u(x, C, D) such that the same model A can be obtained either proceeding directly, by estimating the model in country A, or through model B, i.e., estimating the model in country B and then changing the monetary units. Let the taxes associated with incomes xl and x2 in country A be Ul and u2, respectively. By proceeding through Model B, we first solve the system of equations in the two unknowns C(r) and D(r) :

ulr - v[xir, C(r),D(r)] } ( C(r) u2r = v[x2r, C(r), n(r)] =~ n(r)

=~ v(y) = v[y, C(r), n(r)]

(10.8)

and then we change the monetary units to obtain the tax function for country A 'I

u(x) = t-u[rx, C(r),D(r)]. r

(10.9)

If, on the contrary, we proceed directly to Model A, we solve the system of equations in the two unknowns C(1) and D(1) :

u, = u[xl, C(1),D(1)] } u2 = u[x2, C(1),D(1)] ~

{C(1) D(1)

~

u(x) = u[x, C(1),D(1))].

(10.10)

10.3.

237

Laws of science

For models (10.9) and (10.10) to coincide, we must have u[rx, C(r),D(r)]

= ru[x,

C(1), D(1)].

(10.11)

Equation (10.11) is a functional equation, which, by denoting w(x, r) - u[x, C(r),

can be written as

D(r)],

~ ( ~ , ~) = ~(~, 1).

(10.12)

We can give the following physical interpretation to functional Equation (10.12). Let w ( x , r ) , where x and w are in monetary units of country A, be the tax associated with an income x in country B. Then Equation (10.12), which is written in terms of monetary units of country B, says that a citizen of country B who earns x, in monetary units of A, that is, r x , in monetary units of B, pays the same taxes as a citizen of A earning x, in monetary units of A. Equation (10.12) is the equation for homogeneous functions (5.35) and then its general solution is (see Theorem 5.9) r

(10.13)

where r is an arbitrary function. Expression (10.13) indicates that any consistent model can be written as a function of one single genuine parameter, r, and one arbitrary function, r and that if the tax function r is known for country A, then the tax function for any other country B must be given by (10.13). |

10.3

Laws

of science

In a physical system there are some fundamental variables, such as length, time and space; from them, secondary or derived variables are obtained by certain, more or less complicated, formulas. In other cases, formulas relate to different variables, not necessarily fundamental. However, not every formula generates a valid variable, but only those satisfying some extra conditions (see Aczdl (1987b), pp.35-70). It is necessary that a change of location or/and scale of the independent variables keep the same formula structure up to a change of location or/and scale of the derived or dependent variable. In other words, the formula should remain invariant under location and scale changes. This condition can be written as the following functional equation: u ( r l X l ~- P l , r2x2 -{- P 2 , - - - , rnXn + Pn) = R(rl,r2,...,rn;pl,p2,...,pn)u(xl,x2,...,xn) +P(rl

, r2, . . . , rn; p l , P2, . . . , p n ) (ri, x, E ]R++; i = l , 2 , . . . , n ) ,

(10.14)

Chapter 10. Applications to Science and Engineering

238

where u ( x l , x 2 , . . . ,xn) is the formula that gives the derived variable as a function of the fundamental variables and P and R are functions associated with the location and scale changes of the derived variable, respectively. When a variable is allowed for location and scale changes we say that it has an interval scale. If a variable is allowed for scale changes only we say that it has a ratio scale. Examples of variables with interval scales are time, temperature and location. Examples of ratio scale variables are length, area, volume, speed and acceleration. Equation (10.14) shows the more general situation, in that it includes a maximum of restrictions. However, some simpler cases can occur, depending on whether or not the location or the scale changes exist for the fundamental or the derived variables or whether they are homogeneous or heterogeneous for all the variables. In this manner, we can deal with very many different situations. The following theorems and corollaries give the solution of Equation (10.14) for six different cases. T h e o r e m 10.1 ( G e n e r a l f o r m u l a for t h e laws of S c i e n c e I). The general forms of dependent real variables with interval scale non-constant and continuousat-a-point when all fundamental or independent variables have the same ratio scale, i.e.,the general solutions of the functional equation u(rx) = R(r)u(x) + P(r);

r, x, e JR++,

(i = 1, 2 , . . . , n),

(10.15)

where scalars such as r must be interpreted as (r, r,... ,r) when occupying a vector position, are

u(x) = f (__~2 ~3 Xl

Xl

U(X) -----x~f ( x2xl'Xl x3

xn)

+ clog(x1); R(r) : 1; P(r) : c log(r),

--

Xl

Xn)

--

Xl

+ b; R ( r ) = rC; P ( r ) = b [ 1 - rC].

(10.16) |

Setting x = 1 = ( 1 , 1 , . . . , 1) into (10.15) and subtracting this from Proof: (10.15) we get v(rx) -- R(r)v(x) + v(r), (~0.17) v(x) = u ( x ) - u ( 1 ) . We distinguish two cases: 9(a) R ( r ) - - 1. Then (10.17) becomes v(rx) -- v(x) + v(r). If x -

( s , s , . . . , s ) then

v(,-s) - v ( ~ ) + v(,-),

239

10.3. Laws of science

which is Equation (3.9). Thus, we have v(r) = c log(r),

and then

( lX)

v(x)=v

Xl

(

=v

X2 X3 Xl Xl

X~ /

1, m , _ _ , . . . ,

+ clog(x1),

thus, we finally obtain

~(x) = f (z2~,, ~1~3,..., z~) + clog(~,) ' where Xl

,

Xl

,...,

v

-

1,

-- , --..., Xl

R(~)

=

1,

P(r)

=

clog(r).

Xl

z ~ ) + u(1)

9(b) R ( r ) is not identically 1. If we set x = s into (10.17) and take into account the symmetry between r and s we deduce v(rs) - R ( r ) v ( s ) + v(r)

=~ v(rs) = R ( s ) v ( r ) + v(s).

(10.18)

Because of the above condition on R(r), there exists an r0 such that R(ro) ~ 1 and then v(ro)

~(~) = ~[R(~)- 1]; ~ = R(~o)- 1"

(10.19)

Since u, and thus v, is not constant, a ~ 0 and then the functional Equation (10.18) becomes a [ R ( r s ) - 1] = R ( r ) a [ R ( s ) -

1] + a [ R ( r ) - 1]

=~

R(rs)= R(r)R(s),

which is Cauchy's Equation (3.10) and then, taking into account (10.19), we get R(r) = r e r v(r) = a(r c - 1). (10.20) With this, Equation (10.17) transforms to

~(~x) = ~ v ( x ) + ~ ( ~ - 1) ~

~(~x) = ~ ( x ) ,

where w(x) = v(x) + a. Thus, we have

w(x)

= W(Xl lx) =x w(1 x2xlxlx3 ) =

~

Xl Xl

~

--a~

(10.21)

Chapter 10. Appfications to Science and Engineering

240 and then

:

Xl

,

Xl

,...,~

+b; P ( r ) = b ( 1 - r C ) .

Xl

C o r o l l a r y 10.1 ( R a t i o scale v a r i a b l e s ) . The general form of dependent real-

valued variables with ratio scale non-constant and continuous-at-a-point when all fundamental or independent variables have the same ratio scale, i.e., the general solution of the functional equation u(rx) = R(r)u(x) ; r, xi e lit++, (i = 1 , 2 , . . . , n )

(10.22)

(X2xl' x3 Xl '" "" ' X-~I)

(1o.23)

is U(X) "-- x~ f

' R ( r ) : rC"

Proof: Making P ( r ) = 0 in (10.16) we get either c = 0 or b = 0 and the resulting solution satisfies (10.22). Thus, (10.23) holds, n E x a m p l e 10.2 ( A r e a s o f t w o - p a r a m e t e r families o f s u r f a c e s ) . Let u(x, y) be the area of a family of figures depending on two length parameters, as, for example, a family of ellipses with semi-axes x and y, or a family of rectangles with sides x and y. Then, we have two independent variables with the same dimensions. If we apply the ratio scale r to both variables we find that, according to Corollary 10.1, the only possibility for u(x, y) is

u(~, y) = ~ I

~

,

but we know (see Example 3.1) that R(r) = r 2. Thus, the formula, given the area of a two-parameter family, must be of the form

\ X/ where f is an arbitrary non-negative function. ellipses and rectangles we have

Note that for the family of

f (x) = r x and f (x) = x, respectively.

I

T h e o r e m 10.2 ( G e n e r a l f o r m u l a for t h e laws o f S c i e n c e II). The general forms of dependent real-valued variables with interval scale non-constant and continuous-at-a-point when all fundamental or independent variables have ratio scale, i.e., the general solutions of the functional equation u(rx) = R ( r ) u ( x ) + P(r);

r , x E ]R~+

(10.24)

241

10.3. Laws of science are

n

u(x) = ~ ci log xi + b, i--1

(10.25)

R(r) = 1, n

P ( r ) = ~ ci log ri, i=l

and

n

u(x) = a IX x~' + b, i=1 n

ci

(~o.26)

R(r) = l-I r i , i=l

P ( r ) = b ~ - 11 ~?

9

i=l

Proof:

The proof is similar to that in Theorem 10.1.

C o r o l l a r y 10.2 ( R a t i o scale v a r i a b l e s ) . The general form of dependent real-valued variables with ratio scale non-constant and continuous-at-a-point when all fundamental or independent variables have ratio scale, i.e., the general solution of the functional equation u(rx) = R(r)u(x); is

n

r,x e ~++

n

u(x)=~

(10.27)

n

H ~ ,c,. , n(r)=

1-I ~ c, ,,

i=l

i--1

(10.28)

n

with a # O and

E

c2 i #0.

i--1

Proof: Making P ( r ) = 0 in (10.25) and (10.26) we get either ci = 0 (i = 1, 2 , . . . , n) or b -- 0 and then (10.28) holds. | T h e o r e m 10.3 ( G e n e r a l f o r m u l a for t h e laws of Science I I I ) . The general form of dependent real-valued variables with interval scale non-constant and continuous-at-a-point when all fundamental or independent variables have interval scale with the same value of r, i.e., the general solution of the functional equation u ( r x + p) = R ( r , p ) u ( x ) + P ( r , p ) ;

r, xi,pi e P~+, (i = 1 , 2 , . . . , n )

(10.29)

n

u(x) = ~ cixi + b, i--1

(lO.3O)

R(r, p) = r , n

P(r, p) = b(1 - r) + E cipi. i=1

Chapter I0. Applications to Science and Engineering

242

X

Threshold curve

No fatigue failure zone

Figure 10.4: Regression model.

Proof:

For the proof see Acz~l (1987b), pp. 57-61.

|

C o r o l l a r y 10.1 The only non-null solution of the functional equation

u(rx + p) = R(r, p)u(x); r, xi,p~ e JR+, (i = 1, 2 , . . . , n) is

u(x)-b;

R ( r , p ) = 1,

(10.31) (10.32)

where b is an arbitrary constant. P r o o f : Making P(r, p) = 0 in (10.30) we get b = 0 and c~ - 0 (i = 1, 2 , . . . , n). Thus, the only possible solution is the constant solution (10.32). i

10.4

A statistical m o d e l for lifetime analysis

In many practical engineering situations, the lifetime variable, T, appears as a random variable which depends on one regressor variable, X. This is, for example, the case of the fatigue life of wires, strands or tendons, the time up to breakdown in solid dielectrics or the time up to failure of marine breakwaters, which depend on the regressor variables stress range, voltage stress or wave height, respectively. As a consequence, a regression model as shown in Figure 10.4, could be a convenient approach to the problem. The model is completely established as soon as the cumulative conditional distribution function of lifetime, F(t, x), is defined for every value x of X. Two different ways in which the engineer can tackle the problem are: 1. Using standard linear regression models in order to fit the experimental data.

243

10.4. A statistical model for lifetime analysis

2. Creating adequate models not only to fit the experimental data but also to satisfy physical and theoretical considerations. By the first approach we mean the use of ready-made regression models, i.e., models not specially designed for the problem under consideration, but very well recognized by statisticians and experienced engineers. This is the most generally accepted approach because of its simplicity, its widespread use and the possibility of performing many standard and simple analyses, such as confidence limit analysis for example. In the cases in which the application of the first approach is not satisfactory, the engineer tries to develop tailor-made regression models. These models can either reflect his experience and feeling about the problem or be based on physical and theoretical considerations. In the following paragraphs, we derive a statistical model for lifetime analysis related to the weakest link principle with a wide applicability to engineering problems. 10.4.1

Derivation

of the fatigue model

Castillo et al. (1985) justify the following assumptions for the fatigue model: 1. Weakest link principle: This principle establishes that the fatigue lifetime of a longitudinal element is the minimum fatigue life of its constituent pieces. 2. Stability: The selected distribution function type must hold (be valid) for different specimen lengths. 3. Limit behavior: To include the extreme case of the size of the supposed pieces constituting the element going to zero, or the number of pieces going to infinity, it is convenient for the distribution function family to be an asymptotic family (see Galambos (1987) and Castillo (1988)). 4. L i m i t e d range: Experience shows that the lifetime T and the stress range X, have a finite lower end, which must coincide with the theoretical lower end of the selected cdf. This implies that the Weibull distribution is the only one satisfying requirements 1 to 4. 5. Compatibility: In the X - T field, the cumulative distribution function of the lifetime given stress range, F(t; x), should be compatible with the cumulative distribution function of the stress range given lifetime, E(x; t). These conditions lead to the following functional equation

F(t,x)

=

l-exp 1

exp

-[t-~(x)] ~'(~)} ~t(X) J

~(t)

J

(10.33)

= S(~;t),

244

Chapter 10. Appfications to Science and Engineering

X

p = 0.95 p = 0.90 p = 0.50

J

p=0.10

t

Lifetime T

Figure 10.5: WShler field of model 1.

where ~/t(x), c~t(x) and ~t(x) are the location, scale and shape parameters of the Weibull laws for given x, and "~z(t), ax(t) and ~x(t) are the location, scale and shape parameters of the Weibull laws for given t. Expression (10.33) is equivalent to the functional equation

[t-~/t(x)] ~'(~)

~(~)

=

[x-~/~(t) ~x(t)

~(t)

"

(10.34)

G6mez-Bay6n (1984) and Castillo and Galambos (1987b) have shown that the only feasible solutions of Equation (10.33) axe the three models: 9M o d e l 1 (see Figure 10.5):

F(t,x) = 1 -

(10.35)

e -[(t-A)(x-B)/C+D]~.

9M o d e l 2 (see Figure 10.6):

F(t,x) = 1 - e-[C(t-A)E(~-B)~

(10.36)

9M o d e l 3:

F(t, x) = 1 - e

- [ C ( t - A ) E ( x - B)DeFl~

t - A ) l o g ( x - B)] (10.37)

where A,B, C, E > O, D and F are arbitrary constants.

10.5

Statistical models tudinal elements

for f a t i g u e life o f l o n g i -

One of the most important problems when dealing with the statistical analysis of the fatigue life of longitudinal elements is the size effect; that is, the influence of length on the survivor function.

10.5. Statistical models for fatigue life of longitudinal elements

245

X A = 0.95 =0.90

..... pp= 0.50 =0.10

I

I

Lifetime T

Figure 10.6: WShler field of model 2. S

,

L,_

J J

'1

"1 sI

!

9

It_

q

'I S2

'1 Sn

Figure 10.7: Illustration of the hypothesis of independence.

By longitudinal element we understand an element satisfying the following two conditions: 9only one dimension is important in the behavior of the element and 9if the element is longitudinally divided into imaginary pieces (see Figure 10.7) all pieces are subject to the same external action (stress, force, etc.) Several models have been given in the past to solve this problem, but, unfortunately, most of them are based on the assumption of independence of the fatigue life of non-overlapping pieces. This assumption states that if an element of length s, such as that shown in Figure 10.7, is hypothetically divided into several pieces of lengths sl,s2, ..., sn, then the survivor function of the element S(s, z) must satisfy the equation n

s(,, z) = I-[ s(,,, z). i=1

Here we shall abandon the independence assumption and, making use of the functional equations theory, we shall state the problem in a very different way. We shall assume here that a team of three members is required to design a consensus model for the analysis of the fatigue life of longitudinal elements.

246

Chapter 10. Applications to Science and Engineering

However, they are required to give separate proposals before joining together and reaching a consensus. The three proposals associated with the three members will be denoted by models 1, 2 and 3, respectively. Model 1 For the sake of simplicity we assume n = 2, that is, the element of length x + y is divided into two non-overlapping pieces of lengths x and y. We also assume that there exists a function S(x, z) that gives the survivor function of a piece of length x and that the survivor function of the element can be calculated in terms of that of the two pieces. In other words, S(x, z) must satisfy the following functional equation

S(x + y,z) = H[S(x,z),S(y,z)],

(10.38)

where the function H indicates how the survivor function of the element can be obtained from those of the pieces. It is worthwhile mentioning that Equation (10.38) implies the associativity and commutativity character of the H function and the dependence of the survivor function S on the total length of the element. In fact we can write

S(x + y + z,t)

= = = =

H[S(x + y,t),S(z,t)] H[H[S(x,t),S(y,t)],S(z,t)] H [ S ( x , t ) , S ( y + z,t)] H[S(x, t), H[S(y, t), S(z, t)]]

and

S(x + y, t) = H[S(x, t), S(y, t)] = S(y + x, t) = H[S(y, t), S(x, t)]. Thus, the survivor function of an element of length s is independent of the number and size of the sub-elements into which it is divided in order to calculate it, using (10.38). In the following paragraphs we solve functional Equation (10.38) in two different forms. The functional Equation (10.38) is a particular case of the functional equation S[G(x,y),z] = g [ M ( x , z ) , N ( y , z ) ] , (10.39) with M = N = S and G(x,y) = x + y. In this case we can easily satisfy all regularity conditions in Theorem 7.10, because S(x, 0) = 1 and we can choose families of survivor functions such that Sl(x,c) r 0 and functions H such that H1 ~- 0 and //2 ~- 0. The general solution of (10.39) is (see Theorem 7.10 and Expression (7.80))

s(~, z) e(~,y)

= =

t[f(z)g -~(~) + ~(z) + Z(z)], g{h(~)+ k(y)],

H(x,y) M(x,z) N(x,z)

= = =

l[m(x) + n(y)], m - l [ f ( z ) h ( x ) + a(z)] rt-l[f(z)k(x) + f](z)]

(10.40)

247

10.5. Statistical models for fatigue life of longitudinal elements

where g, h, k, l, m and n are arbitrary strictly monotonic continuously ditterentiable functions and f, a and/3 are arbitrary continuously differentiable functions. Thus, for Equation (10.38) we have

s(~, z)

=

z[f(z)g-~(~) + ~(z) + ~(z)]

g[h(x)+k(y)]

= = =

m - l [ f ( z ) h ( x ) + a(z)] n - l [ f ( z ) k ( x ) + fl(z)] x+y,

(10.41)

from which g - l ( x + y) = h(x) + k(y),

which is Pexider's equation I with the general continuous-at-a-point solution (see Theorem 4.1) g - l ( x ) = A x + B + C; h(x) = A x + B; k(x) = A x + C.

With this, expressions (10.41) become S(x, z)

l [ f ( z ) ( A x + S + C) + a(z) + fl(z)] m - l [ f ( z ) [ A x + B] + a(z)] n - l [ f ( z ) [ A x + C] + fl(z)],

= = =

and making A f ( z ) x = u we obtain

s(~, z)

= = =

Z[u + (B + C ) f ( z ) + ~(z) + ~(z)] m - l [ u + B f ( z ) + ~(z)] ~-~[~ + e l ( z ) + Z(z)],

(~o.42)

and, using Lemma 6.1, we get u

(B + C ) f ( z ) + a(z) + 3(z)

=

cu+a

=~ c = l ,

a=0,

=

m-l(x

=

B f ( z ) + a(z) + b =~ fl(z) = b - C f ( z ) .

- b)

Then, (10.42) becomes S ( x , z ) = m - l [ u + B f ( z ) + a(z)] = n - l ( u + b),

which implies u

=

ClU + a l

==~

=

Clb+bl=b+bl

Cl = 1, al = 0,

m-l(x) By(z) + ~(z)

and finally we get the desired solution

=~ a ( z ) = b + b l - B f ( z ) ,

248

Chapter 10. Appfications to Science and Engineering

Model

1: The general solution of (10.38) is:

S ( x , z ) = w[f(z)x]; H ( x , y ) = w[w-l(x) + w-l(y)], where we have made W--I(x) = [//(X)- b] A " Due to the weakest link principle and because S(x, z) is a survivor function, it must be non-increasing in z and x. Then, in addition, we must have

S(x,O)= l

I f ( o ) = o ; ~ ( o ) = 1], or If(o) = ~ ; ~(~)= 1], o , .

==~ w[f(O)x]= l

If(o) = - ~

; ~(-~)=

1]

{ [f(~) = o; ~(o) = o], or S(x, oo) = 0 =r w[f(oo)x] = 0

=~

[f(oo)

= cx:) ; w ( o o ) = 0],

[f(~)

= -~

; ~(-~)

or.

= 0]

If w(x) = exp(Dx) we get the model of independence. The structure of the function H reveals its above mentioned associative and commutative character. We can solve (10.38) in a much easier way if we observe that the variable z plays the role of one parameter, i.e., for any fixed value of z, Equation (10.38) can be written in the form

S(x + y) = HIS(x), S ( y ) ] , and due to the associative character of H we can write (see Theorem 6.6)

H ( x , y ) = w[w-l(x) + w-l(y)], and its substitution into (10.38) leads to

S(~ + v, z)

=

~{~-~[s(~,z)]+~-~[s(y,z)]}

v(= + v, ~) = C(=, z) + G(v, z), with G(x,z) = w - l S ( x , z ) , which, for z held constant, is Cauchy's Equation (3.7) and then (see Theorem 3.3):

G(x, z) = f ( z ) x ==~ S(x, z) = w[f(z)x]. Model 2 Member 2 in the team wants to start from the following result : Bogdanoff and Kozin (1987) based on some experimental results of Picciotto (1970), suggest the following model for the survivor function

s(z, z) = S(y, z) ~ ' ~ ,

(~o.43)

10.5. Statistical models for fatigue life of longitudinal elements

249

1.0 0.B 0.6 0.4 0.2 0.0

9

0

|

1O0

-

i

-

i

9

,

-

200 300 400 Element length

,

500

9

,

600

Figure 10.8: Experimental and theoretical survivor functions for lengths 30, 60 and 90 cm. (from Bogdanoff and Kozin (1987)).

where S(x, z) and S(y,z) are the survivor functions associated with two elements of lengths x and y, respectively, and N(y, x) is an unknown function (see Example 6.2). Figure 10.8 shows the experimental survivor functions and those obtained using model (10.43) (see Bogdanoff and Kozin (1987)). Note that (10.43) is an implicit function of S(x, z), or in other words, it is a functional equation. Thus, it must be solved to know what the Bogdanoff and Kozin proposal is. Castillo et al. (1990a) showed that the only compatible functions for N(y,x) are those of the form Y ( y , x ) = q(x) q(y) From Example 6.2, we get Model2.

S(x,z)=p(z)q(z); N ( y , x ) = q(x) q(y)"

(10.44)

For S(x, z) to be a survivor function it must be non-increasing in z and we must have s ( ~ , 0 ) = 1 ~ p(0)~(~) = 1 ~ p ( 0 ) = 1, S(x, oo)=O =~ p(co) q(z)=O :~ p ( c o ) = O . If q(x) = x we get the model of independence. The hazard function associated with S(x, z) is

-p'(z) h(x,z) = ~ q ( x ) p(z)

= -[logp(z)]'q(x) = s(z)q(x),

which shows that Model 2 is the Cox-proportional hazards model (see Cox (1972)). Thus, functional Equation (10.43) characterizes the proportional hazards Cox-model.

Chapter 10. Applications to Science and Engineering

250 Model 3

Member 3, based on expression (10.43), assumes that the survivor function of one element of length x can be obtained from the survivor function of one element of length y and a given, but unknown, function of x and y. In other words he assumes that the survivor function must satisfy the functional equation

S(x, z) = K[S(y, z), N ( x , y)],

(10.45)

which is a particular case of (6.36)

G(x, y) = K [ M ( x , z), N(y, z)], with S(x, y) = C(y, x) = M(y, x). If we choose F and N to be invertible with respect to their first argument and K invertible with respect to its first argument for a fixed value of the second, then the regularity conditions in Theorem 6.5 hold and the general solution of the last equation is (see Expression (6.37)):

G(x,y) M(x,y)

= =

f-l~o(x) +q(y)]; l - l ~ ( x ) + r(y)];

g(x,y) N(x,y)

= =

f - l [ l ( x ) + n(y)]; n-'[q(x) - r(y)],

(10.46)

and then, for (10.45) we must have

s(~,z) K(x,y) N(y,z)

= /-~[p(z)+q(~)] =

z-~[p(z)+

=

f - i l l ( x ) + n(y)],

r(~)],

=

n-l[q(y) - r(z)],

(10.47)

which, by Lemma 6.1, implies

p(z) y-~(~)

= =

c p ( z ) + a =~ c = l , 1-a(x-a-b)

q(x)

=

cr(z) + b = r(z) + b,

a=0,

and then, from Expression (10.47), model 3 becomes Model 3- S(x,z)

K(x,y) N(x,y)

= =

=

l - l ~ ( z ) + r(x)], l-l[l(x) -]- m(y)], m - l [ r ( x ) - r(y)],

(~o.48)

where we have made re(x) = n(x) - b. For S(x, z) to be a survivor function it must be non-increasing in z and we must have [/(1) = p(0) = -c~], or s ( ~ , 0) = z - ~ o ( 0 ) + r(~)] = 1 ~ [Z(X) = p(0) = ~ ] , or [r (~) = z(1) - p(0)]

S(x, oo) =/-X[p(c~) + r(x)] = 0 =r

[/(0) = p(c~) = c~], or [/(0) = p(oo) = -c~], or.

[r(~) = t(0) - p ( ~ ) ]

10.5. Statistical models for fatigue life of longitudinal elements

251

I f / - I ( x ) -----exp[Dexp(Cx)] we get the model of independence. One important aspect to point out here is that Equation (10.45) is more than a simple generalization of Equation (10.43). In fact, it includes some extra compatibility conditions, in the sense that no arbitrary N ( y , x ) is admissible in Model 3, even though, initially, the function N ( y , x ) seems to be arbitrary. In order to prove this, we show that the only admissible N functions are those appearing in Model 2 (Expression (10.44)). Let us assume that function K is that implied from Equation (10.48), that is K ( x , y) = x u = l-X[l(x) + m(y)] =~ l(x u) = l(x) + m(y), which, by making the change of variable u = log(x), can be written

l[exp(uy)] =/[exp(u)] + m(y). This is a Pexider functional equation with solution

l(x) = c log[a log(x)]; re(y) = c log(y). Thus, we finally get

N(x,y)

=

m-a[r(x) - r(y)] = exp [r(x) - r(Y)] = q(x)

q(y)'

which is model 2. Reaching a

consensus

In the second and final step the team is required to join and reach a consensus. Normally, a consensus solution is understood as a linear combination of the quantitative judgments of several individuals. However, in many cases the consensual solution reached does not satisfy many of the properties that were satisfied by the solutions in the proposals given by the different individuals (see Genest and Zidek (1986)). This fact is irrelevant when one tries to use the consensus model to make some evaluations, such as to calculate some probabilities, for example, but becomes a very serious inconvenience when one tries to model a physical system. In fact, the functional equations (10.38), (10.43) and (10.45) state some properties, which the different members understand the physical system must satisfy. Thus, any member would not accept models violating his/her associated functional equation. Thus, in the following, we shall understand consensus as the intersection of the three families of models, if it exists, i.e., as models satisfying all the requirements. We start by analyzing the common part of Models 1 and 2 (see the corresponding equations).

s(~, z)

=

~ { / ( z ) : ] = p(z)~(:),

Chapter 10. Applications to Science and Engineering

252 which implies

log[S(x, z)] = log {w[f(z)x]} = q(x)log[p(z)], and making the change of variable u = f ( z ) we get

log{w[ux]} = q(x) log{p[f-l(u)]}, which is Pexider's functional Equation (4.4) with the general continuous-at-apoint solution (see Theorem 4.4) log{[w(x)]} = A B x c, q ( x ) = Ax C, logIp[f-l(x)]} = B x C. Thus,

S(x,z) p(z)

= =

exp {AB[f(z)x] C} ; w(x) exp[BfC(z)]; q(x)

=

=

exp(ABxC)

Axe,

'

9

(10.49)

which shows that Models 1 and 2 are not coincident but they share the common model

S ( x , z ) = exp { A S [ f ( z ) x ] C} =_ S ( x , z ) = exp[f(z)x] c = ~(z) xc,

(10.50)

where/3(z) is an arbitrary positive function. Model (10.50) for C = 1 becomes the model of independence. The hazard function for this model is

dZ(z)

h(x z) =

dz

'

~(zi

d{log[/3(z)]}xC = s(z)xe"

xc = -

dz

If now we look for the common part of Models 1 and 3 we get the functional equation s ( ~ , z ) = t-lip(z)+ ~(~)] = ~[f(z)~], which, by making the change of variable u = f(z), becomes Pexider's Equation (4.3)

t[~(~)]

: p [ f - ' ( ~ ) ] + r(~),

with the general continuous-at-a-point solution (see Theorem 4.3)

l[w(x)] = Alog(BCx);

p[f-l(x)] = Alog(Bx);

r(x) = Alog(Cx),

and then we finally get

w(x) = l-l[Alog(BCx)];

f - l ( x ) - p-l[Alog(Bx)];

r(x) = Alog(Cx),

which shows that Model 1 is a particular case of Model 3. Finally, we compare Models 2 and 3. For the coincidence we must have

s ( ~ , z) = L-1 [p~ (z) + r(~)] = p ( z ) qr

10.5. Statistical models for fatigue life of longitudinal elements

253

Figure 10.9: Illustration of separate and consensus proposals.

and taking logarithms we get

q(x) logp(z) = log{l-1 [pl(z) + r(x)]} which implies

l{exp[q(x) logp(z)]} = Pl (z) + r(x), which is Pexider's Equation (4.3). Thus, we have

and then

/[exp(x)]

=

r[q-l(x)]

=

pl {p-l[exp(x)]}

=

l-1(~) q(x) p(z)

= =

Alog(BCx), Alog(Bx), Alog(Cx),

exp [exp(x/A)/(BC)], exp (r(x)/A)/B, exp [exp (pl (z)/A)/C],

which shows that Model 2 is a particular case of Model 3. Thus, we can conclude that a consensus model could be model (10.50), which is the family of models common to all three members of the team. Figure 10.9 shows the required separate and consensus proposals as well as the common proposals associated with all three groups of only two members. As a final conclusion, we can add that functional equations can prove themselves to be a very powerful tool to be used in model design. As a matter of fact, the engineer can state all the conditions to be satisfied by the desired model in terms of functional equations. Then, by first solving the resulting system and then in terms of its general solution, one can make the selection by playing with the remaining degrees of freedom.

254

Chapter 10. Applications to Science and Engineering z(t+u Ul ~

0{I

X2

z(t+u2)

z(t)

X8

Figure 10.10: Functional network associated with functional Equation (7.2).

10.6

Differential, functional and difference equations

In this section we use the functional network methodology described in Chapter 9 for predicting values of magnitudes satisfying differential, functional and/or difference equations, and for obtaining the difference and differential equation associated with a set of data. As we shall show, the estimation of the differential or difference equation coefficients is carried out simply solving systems of linear equations, in the cases of equally or unequally spaced or missing data points. 10.6.1

A motivating

example

In this section we use the example in Section 7.2. Equation (7.2) can be represented by the network in Figure 10.10, where I is used to refer to the identity function. Similarly, Equation (7.4) can be represented by the network in Figure 10.11. Both cases correspond to functional networks, such as those introduced in Chapter 9. Therefore, we proceed by applying the functional networks formalism, as shown in Section 9.5. S t e p 1 ( S t a t e m e n t of t h e problem): Understanding of the problem to be solved, which was done in Chapter 7. S t e p 2 (Initial topology): Based on the knowledge of the problem, the topology of the initial functional network is selected. For example, the vibrating functional, (7.2), and difference, (7.4), equations of the mass problem have led to the two functional networks in Figures 10.10 and 10.11, respectively. S t e p 3 (Simplification): In this step, the initial functional network is simplified using functional equations. It must be noted that when there are coincident

10.6. Differential, functional and difference equations

255

Xl

t

z(t+2u)

z(t)

z(t+u) Figure 10.11: Functional network associated with difference Equation (7.4).

neural outputs, they must coincide in values, and this leads to a functional equation which allows the initial topology of the functional network to be simplified. This is not the case of the functional networks in Figures 10.10 and 10.11. Thus, no simplification is possible here. For some illustrative examples On this step, the reader is referred to Section 9.6. S t e p 4 ( U n i q u e n e s s of r e p r e s e n t a t i o n ) : In this step, uniqueness conditions for the neural functions to be unique must be found. For example, in the case of the network in Figure 10.10, we can consider the possibility of the existence of two sets of functions {a0, a l , 5} and {a~, a~, 5* } such that

z(t + U~) = -

-

a o ( u ~ , u 2 ) z ( t ) + ~ ( u ~ , u 2 ) z ( t + ~ x ) +~(t;u~,~2) C~(Ul, ~t2)Z(t)-~ Oi (Ul, ~t2)Z(t -~-Ul)-~- 5"($; ~tl, U2);

(10.51)

that is, about the existence of two different functional networks with the same structure leading to the same outputs for the same inputs. S t e p 5 ( D a t a collection): For the learning to be possible we need some data. In this step, we consider two different cases: (a) equally spaced data, and (b) unequally spaced data. For the sake of clarity, the following steps of the functional network approach will be described for each of the two cases. Equally spaced data We start by analyzing the case with equally spaced data. Let us assume that we have available the data given in Table 10.1, which consists of the vibrating mass displacements z corresponding to different times t. In the case of equally spaced data (constant u), we use Equation (7.4), where co(u) and al(U) for constant u are constants, and function 5(t) can be approximated by a linear combination of a set of linearly independent functions {r = 1 , . . . , m}. If z(tj) for j = 0 , . . . , n are the observed data for equally spaced times tj, the solution of a differential equation of order k with constant coefficients in

256

Chapter 10. Applications to Science and Engineering

Table 10.1: Observed displacements z of system in Figure 7.2 for different times t. t ! z 0.0 0.100 0 . 2 0.238 0.4 -0.441 0.6 -0.760 0.8 0.151 1.0 0.924 1.2 0.450 1.4 -0.145 1.6 0.077 1.8 0.098 2.0 -0.632 2.2 -0.764 2.4 0.258 2.6 0.876 2.8 0.287 3.0 -0.189 3.2 0.105 3.4 0.006 3.6 -0.737 3.8 -0.671 4.0 0.417

t l z 0.177 0.04 0.155 0.24 0.44 -0.590 0.64 -0.660 0.84 0.381 0.921 1.04 0.284 1.24 1.44 -0.156 0.144 1.64 1.84 -0.012 2.04 -O.758 2.24 -0.623 0.474 2.44 2.64 0.833 2.84 0.130 3.04 -0.170 0.158 3.24 3.44 -0.126 3.64 -0.834 3.84 -0.494

t 0.08 0.28 0.48 0.68 0.88 1.08 1.28 1.48 1.68 1.88 2.08 2.28 2.48 2.68 2.88 3.O8 3.28 3.48 3.68 3.88

z It 0.12 0.241 0.32 0.036 -0.708 0.52 -0.507 0.72 0.92 0.586 1.12 0.861 1.32 0.130 -0.130 1.52 1.72 0.188 -0.153 1.92 -0.843 2.12 -0.435 2.32 2.52 0.654 2.72 0.740 -0.005 2.92 -0.119 3.12 3.32 0.180 -0.282 3.52 -0.880 3.72 -0.280 3.92

z [ t 0.277 q -0.113 I -0.781 ' -0.31 ~ 0.752 ' 0.754 0.001 -0.074 0.199 -0.313 -0.876 -0.214 0.786 0.608 -0.107 -0.047 0.163 -0.446 -0.870 -0.045

[

z 0.278 -0.277 -0.800 -0.086 0.867 0.612 -0.093 -0.000 0.170 -0.478 -0.850 0.023 0.861 0.452 -0.169 0.032 0.105 -0.603 -0.799 0.194

10.6.

257

Differential, functional and difference equations

k+m

g(j) = Z c i (I)ji-k X i=k+ 1

Zj+k-1

/

Zj+k+2

+~)--"'~

Zj+k

k f(zj ..... Zj+k-1)= Z CiZj+i-1 i=l

zj Figure 10.12- Functional network associated with Equation (10.52).

z(t)

can be approximated using the model k

k+m

Z

zj+k=~cizj+i-l+ i=1

cir J

(~o.52)

J=O,'",n-k,

i=k+l

where u = tj+l - tj for j = 0 , . . . , n - 1, zj = z(to + j u ) , r = r Cl , . . . , Ck+m are constant coefficients.

+ j u ) and

The functional network associated with (10.52) is given in Figure 10.12. Step 6 (Learning)" At this point, the neural functions are estimated (learned) by using some minimization methods. In our example, the e r r o r ej+k a t the point t j + k = to + ( j + k ) u using the approximation given by (10.52) becomes k e j + k = zj+~ - ~

k+rn

~, z j + , _ ~ -

i--1

~

~{_~;

j = o , . . . , ~ - k.

(lO.53)

i=k+l

Thus, the parameters Cl,..., Ck+m can be estimated by minimizing

=

ej+k = j--O

zj+k"=

c~zj+~_~ "=

c~r i--k+l

k

.

(~o.54)

258

C h a p t e r 10.

A p p l i c a t i o n s to Science and Engineering

The minimum is obtained for

2 Oar

=

j =o

zj+~ -

~=1

c~z3+~_] -

z=k+ 1

c~r

zj+~_l = O,

r = 1,...,k, 2 (~Cr

=

5= 0

i=1

i=k+l

i-k

k

(lO.55) =o.

r=k+l,...,k+m.

(A11 A12)(c1) (bi)

This leads to the system of linear equations with k + m unknowns Ac = b r

--A21

+ [

. . A22

.

.

. c2

.

.

.

(10.56)

b2

From (10.55) we can write the expressions for each element a~s of A and br of b: n-k ars

=

if

r-1,...,k,

s--1,...,k,

if

r-1,...,k,

s=k+l,...,k+m,

if

r-k+l,...,k+m,

s-1,...,k,

s-k V/r-k

if

r=k+l,...,k+m,

s=k+l,...,k+m,

Zj+kZjTr-1

if

r = 1,...,k,

zj+k4rT~_k "

if

r = k + 1 , . . . , k + m.

~_~ Z j + s _ l Z j + r _

1

j=0

n-k

a~

--

~

r s_kZjTr_l

5=0

n-k

a.

=

Z ~§

j=O

n--k

j=0

n-k

br j=o

n-k

br

=

Z

j=o

(10.57)

Finally, from (10.56) we get c = A-lb,

(10.58)

which gives the solution. Returning to the system in Figure 7.2, if we use the functions {r (t), r

r

r (t), r

= { 1, sin(t), cos(t), sin(2t), cos(2t)}.

and the equally spaced observed displacements for different times shown in Table 10.1 or Figure 10.13, we get:

10.6. Differential, functional and difference equations

259

m

0.75

9

0.5

9 , "

0.25

.

.~

.

mmm 9

9

9

-0.25

"

9

9

m m,

".."

1

92

",,~"

9

4

-0.5 ,,.

-0.75

Figure 10.13:

A

9

%-

.

Observed data z for the displacement of system in Figure 7.2.

24.40 23.60 -3.67 5.08 3.56 -4.46 -1.63

23.60 24.40 -3.58 4.80 3.52 -4.12 -2.10

-3.67 -3.58 99.00 41.70 -19.10 13.80 11.90

5.08 4.80 41.70 43.0 6.91 -6.85 -20.10

3.56 3.52 -19.10 6.91 55.50 21.60 -12.0

b =

21.30 23.70 -3.34 4.35 3.25 -3.52 -2.62

,

-4.46 -4.12 13.80 -6.85 21.60 50.40 6.18

-1.63 ~ -2.10 11.90 -20.10 -12.20 6.18 48.60 j

(10.59)

and

(10.60)

which leads to

c =

-1.0130 ~ 1.9442 -0.0214 0.0335 -0.0155 0.0154 0.0091

(10.61)

S t e p 7 ( M o d e l v a l i d a t i o n ) : Finally, using (10.52) with the values in (10.61) we can predict displacements which are visually indistinguishable from those in

Chapter 10. Applications to Science and Engineering

260

Table 10.2: Observed displacements z of system in Figure 7.2 for different rand o m times t.

t

Z

0.000 0.258 0.413 0.639 0.741 0.831 1.220 1.590 1.780 1.950 2.120 2.340 2.490 2.740 3.000 3.110 3.310 3.520 3.750 3.850

0.100 0.105 -0.491 -0.664 -0.195 0.333 0.377 0.054 0.135 -0.437 -0.876 -0.099 0.682 0.545 -0.189 -0.O65 0.169 -0.437 -0.823 -0.454

t 0.001 0.308 0.494 0.640 0.781 1.050 1.370 1.610 1.790 1.960 2.140 2.410 2.600 2.760 3.001 3.200 3.340 3.520 3.750 3.880

Z

0.102 -0.067 -0.739 -0.661 0.038 0.915 -0.109 0.094 0.125 -0.487 -0.871 0.315 0.877 0.432 -0.189 0.103 0.133 -0.453 -0.821 -0.294

t 0.009 0.321 0.534 0.660 0.781 1.100 1.450 1.610 1.820 1.970 2.220 2.420 2.610 2.820 3.040 3.220 3.420 3.630 3.770 3.940

Z

0.119 t 0.028 -0.117 0.327 -0.794 0.561 -0.588 0.666 0.784 0.039 1.110 0.826 -0.154 1.580 1.620 0.097 1.820 0.057 1.980 -0.520 -0.707 2.230 2.430 0.385 2.670 0.867 2.830 0.223 -0.173 3.090 3.260 0.136 -0.047 3.480 -0.817 3.690 -0.769 3.770 3.950 0.057

Figure 10.13. In fact, we get a m a x i m u m and a m e d i u m absolute prediction error of To test the possibility of over-fitting, m e a n squared error) for the training d a t a obtaining the following results

Z

0.155 -0.140 -0.800 -0.565 0.056 0.786 0.036 0.110 0.052 -0.542 -0.646 0.434 0.767 0.164 -0.109 0.173 -0.300 -0.884 -0.758 0.127

t O.040 0.331 0.588 0.695 0.813 1.200 1.580 1.660 1.890 2.040 2.330 2.450 2.710 2.840 3.110 3.310 3.500 3.700 3.780 3.970

Z

0.177 -0.155 -0.777 -0.436 0.230 0.459 0.048 0.166 -0.188 -0.747 -0.178 0.517 0.640 0.115 -0.066 0.172 -0.345 -0.883 -0.750 0.238

absolute prediction error of 0.0334 0.0132. we have obtained the R M S E (root and a set of 1000 test d a t a points,

RMSEtraininz = 0.018; RMSEtestm9 = 0.042, which shows t h a t the error increase is not very high. S t e p 8 ( U s e o f t h e m o d e l ) : At this step, the model is ready to be used. Unequally

spaced data

Table 10.2 shows 100 observed displacements of the system in Figure 7.2 for r a n d o m times. In this case we use the Expression (10.58) to predict t h e behavior of t h e system using these observed displacements and two different models a p p r o x i m a t i n g the functions a0, c~1 and 5 involved in it. Note t h a t this approach is also valid for the case of missing data.

10.6.

Differential, functional and difference equations

z(t)

I

I

1

I

I

..+ ~+

I

1

o +

%

0.5

261 I

Observed Predicted

o Q

g

+

0

+

~

8

o

+

-0.5

o$ I

-1

0.5

0

1

I

I

1

1.5

2

O_ +

I

'

t

2.5

3

3.5

4

Figure 10.14: Observed and predicted displacements.

M o d e l 1: S t e p 6 ( L e a r n i n g ) " Suppose that functions c~0, c~: and (f are approximated by c~0(u:,u2)

--

al + a2u: + a3u2,

(10.62) (10.63)

a:(u:,u2)

=

bl + b2u: + b3u2,

5(t, u:, u2)

=

c: + c2 sin(t) + c3 cos(t) +c4 sin(2t) + c5 cos(2t),

(10.64)

where ai, b~ and ci are parameters to be estimated. To this aim, we define the function 100

F ( a , b, c) = Z

[z* (ti) - z(ti)] 2,

(10.65)

i----3

where z* (t~) is the predicted displacement for time t, using Expression (10.58). The minimum of this function is attained at: a: bl ca c4

= = =

1.603, -0.591, -0.007, 0.014,

a2 --

--23.663,

b2 =

26.331,

c2 = c5 =

0.030, -0.015.

a3 = b3 = c3 =

15.514, -18.058, -0.013,

S t e p 7 ( M o d e l validation)" Using these parameters, the medium absolute prediction error is E - 0.064. Figure 10.14 shows the observed and predicted displacements of the system in Figure 7.2.

262

Chapter 10. Applications to Science and Engineering

To test the possibility of over-fitting, we have obtained the RMSE (root mean squared error) for the training data and a set of 1000 test data points, obtaining the following results

RMSEtrainin9 = 0.11; RMSEtestin9 = 0.15, which shows that the error increase is small. M o d e l 2: S t e p 6 ( L e a r n i n g ) : Suppose that functions c~0, ~1 and 5 are approximated by c~0(ul,u2)

=

al + a2ul + a3u2,

(10.66)

~l(ul,u2)

=

bl + b 2 u l +b3u2,

(10.67)

5(t, ul,u2)

=

cl(ul,u2) +c2(ul,u2)sin(t)

(10.68)

-{-c3(Ul, u 2 ) c o s ( t ) + c 4 ( u l , u 2 ) s i n ( 2 t )

(10.69)

+c5(ul, u2) cos(2t),

(10.70)

where, c4(ul,u2) = c41 + c42ul + C / 3 U 2 , f o r i = 1 , . . . , 5 , and ai,bi and c4j are parameters to be estimated. To this purpose, we define the function 100

f ( a , b, c) = ~

[z*(ti) - z(ti)] 2 ,

(10.71)

i--3

where z*(ti) is the predicted displacement for time ti using Expression (10.58). The minimum of this function is attained for: al = bl = cll c14 = C22--

c25 = C33 - -

1.749, -0.694, 0.042, -0.077, 0.006, -0.049, 0.014,

a2 = b2 = c12 = ClS = C23 =

c31 = C34 =

-27.282, 30.703, 1.706, -2.620, 0.302, -0.189, -1.870,

a3 = b3 = c13 = c21 = c24 = c3~ = C3S =

16.893, -20.628, -1.999, 3.415, -0.717, 1.196, 0.929.

S t e p 7 ( M o d e l v a l i d a t i o n ) : Using these parameters, the medium absolute prediction error is E - 0.060. This shows that it is not worthwhile including non-constant c4(ul, u2), i = 1,2,3, 4,5 functions.

Exercises 10.1 To solve the problem in Section 10.5 a fourth member proposes a model of the type S(x, u + v) = H[S(x, u), S(x, v)]. Obtain its general solution, discuss the new consensus solution and give it a physical interpretation.

10.6.

263

Differential, f u n c t i o n a l a n d difference e q u a t i o n s

10.2 Obtain a general equation for hydraulic problems knowing that in the most general case you can have the following magnitudes: 9Four geometric length parameters: length (a), width (b), depth (c) and roughness (r) 9One cinematic parameter: velocity (v) 9One parameter related to internal forces: pressure (p) 9Density (d) 9Specific weight (g) 9Viscosity (m) 9Surface stress (s) 9Elasticity (E) 10.3 Consider the physical relations: e = log(vt).

and e = C l -Jr

c2t.

where e is the space, v the velocity and t the time. Are they valid physical relations? H i n t : perform the change of variables e ---, ree + eo; v --, rvV + Vo; t --, rtt + to

and see what you get. 10.4 Consider the formula VT -- ~ U ~ ( S 1 / 2 / K ) ' r - l ( q s

- q).

where * VT evaporation under turbulent conditions [LT -1]

- q humidity difference (saturation deficit) where q is the specific humidity of air and qs is the specific humidity of saturation at the temperature of the water surface (both are dimensionless)

9 qs

9 K mean turbulent diffusion coefficient of water vapor in air [L2T -1] 9 Swater surface area [L2] 9 Uwind speed (mean value) [ L T -1] 9/3 and ~/are dimensionless constants. Is it a valid physical relation?

264

Chapter 10. Applications to Science and Engineering

10.5 Build a fatigue model with a Weibull type survival function and check which of the three models proposed in Section 10.5 it belongs to. 10.6 Add a new proposal for a model of the statistical analysis of the fatigue life of longitudinal elements (size effect) described in Section 10.5 such that the resulting model after consensus with the three other proposals becomes more restrictive than the model in (10.50). 10.7 Design a functional network to solve the differential equation

z"(~) + (~ + b)z'(~) + ~bz(~) = o. H i n t : Use the results in Example 7.12. 10.8 Design two functional networks to solve the differential equations xf'(~)

- kf(~)

k f(x) f (x) - x

= 0

x2

and f'(~)

- f(x)

= 0.

discussed in Examples 7.14, 7.15 and 7.16, respectively.