Mathematical Modelling for Energy Systems (Slides)

Mathematical Modeling

for

Energy Systems

O.Caligaris

Scuola politecnica di Ingegneria - Università di Genova Polo di Savona

of the most interesting topics in Differential Calculus; they concern with the problem to find a regular function y which satisfies a given relation, involving y itself and its derivatives, and some initial data.

The simplest example for this kind of problems is the free fall of an object

point P of mass m which falls under the sole influence of gravity and let us ignore air resistance; we have F = mg.

Experimental evidence shows that P falls down; we can use as reference system the straight line traveled by the material point assuming that the origin coincides with the ground and considering the line oriented upward.

of the material point P at time t, P moves down with velocity v (t) = ˙x (t)

and acceleration

a(t) = ¨x (t)

Since P falls under the sole influence of gravity, only the force F = mg acts on P and, by the first Newton Law we have

ma(t) = `mg Sothat

¨

into account that t0 = 0 and x (t0) = h. We have ˙ x (t) = `gt + c1 (2) and x (t) = `1 2gt 2_{+ c1t + c0 (3)}

for c0 e c1.

We can do this using initial values of velocity and position of the material point P

In fact, we can easily show, using 2 and 3, that:

v0 = ˙x (0) = c1 h0= x (0) = c0 (4)

and this seems natural as to determine the motion of the material point P is necessary to know also its initial position and velocity. If we take account of these data we can say that P moves along the x axis according to the law

x (t) = `1 2gt

Conservation of Energy Principle At time t the potential energy of P is

U(t) = mgx (t) while its kinetic energy is

1 2m ˙x

2_(t)

and its total energy

E (t) = 1 2m ˙x

2_{(t) + mgx (t)}

remains constant during the motion. 1

2m ˙x

2_{(t) + mgx (t) = mk (6)}

Using initial data v0 ed h0 we can compute

k = 1

2mv

2

(₁ 2m ˙x 2_{(t) + mgx (t) = mk} 1 2mv 2 0 + mgh0= k

can be used to describe the position x (t) of the material point P , however it is more difficult to derive x explicitly; moreover there is a problem about uniqueness of the solution.

6 can be written as 1 2x˙

2_{(t) = k ` gx (t) (7)}

and this equality shows that

k ` gx (t) must be positive, so that x (t) » k_g. We also remark that

take the square root of both sides ˙

x (t) = ˚ q

2k ` 2gx (t) (8)

divide by the second member ˙ x (t) p

2k ` 2gx (t)

= ˚1 (9)

multiply both sides by g

g ˙x (t) p 2k ` 2gx (t) = ˚g (10) integrate from t0 = 0 to t, Z t 0 g ˙x (s ) p 2k ` 2gx (s ) ds = ˚gt (11)

u = x (s ) ; du = ˙x (s )ds whence, Z x (t) h0 gdu p 2k ` 2gu = ˚gt (12) Since v0 = ˚ p

2k ` 2gh0, we can assert that

q 2k ` 2gx (t) `p2k ` 2gh0 = ˇgt (13) q 2k ` 2gx (t) = ˇgt ˚ v0 (14) 2k ` 2gx (t) = (ˇgt ˚ v0)2 (15) x (t) = k g ` 1 2g(ˇgt ˚ v0) 2 (16)

however we must observe that the choice of the sign holds true as long as ˙x (t), i.e. velocity, doesn’t vanish.

This possibility never occurs if v0 < 0 while it happens at t0= v_g0 if v0> 0.

In this case we have

˙

x (t0) = 0

obtain from 8 that x (t) = k_g and it must be x (t) » k_g, we deduce that x cannot

increase so that ˙

x (t) = ` q

2k ` 2gx (t) (17) Let us finally remark that, since we used the

Conservation of Energy Principle, we also found a constant solution x (t) = k_g; however this solution is meaningless for the motion

of the falling body x

y

O .

Population Growth Models.

In the years between 1798 and 1826 Thomas Robert Malthus published his book "An Essay on the Principle of Population" in which he claimed that the population growth would be

unsustainable for the earth’s resources and argued that the growth of a population could be controlled by diseases, famines and wars on the one hand and abortion, control birth, prostitution, late marriage and celibacy on the other hand.

Malthus’ exponential model assumes that the growth of a population is proportional to its consistency;

Let

n(t)

the number of individuals of a population at time t, let moreovert

c its growth rate, then n satisfies the following Cauchy problem:

( ˙

n(t) = cn(t) n(t0) = n0 dove c è il tasso di crescita

If we solve the equation we obtain

In order to apply the model we need to determine c using experimental data

To this end we only need

n0= n(t0) n1 = n(t1) for some t0, t1 Substituting t = t1 in 18 we get n1= n0ec (t1`t0) and c = ln(n1 n0) t1` t0

If n1 = 2n0 t1` t0 is the doubling time and for n1= 1₂n0 t = t1 is the half-life.

the growth rate c is given by

c = n(t)˙ n(t)

the following table contains data about the Total Midyear World Population: from http://www.census.gov/population/international/data/worldpop/table_population.php 1950 2,557,628,654 1951 2,594,919,657 1952 2,636,732,631 1953 2,681,994,386 1954 2,730,149,884 1955 2,782,001,154 1956 2,835,182,293 1957 2,891,211,793 1958 2,947,979,287 1959 3,000,544,325 1960 3,042,828,380 1961 3,083,799,968 1962 3,139,919,051 1963 3,209,631,895 1964 3,280,981,862 1965 3,350,186,115 1966 3,420,416,498 1967 3,490,051,163 1968 3,562,007,503 1969 3,636,825,800 1970 3,712,338,708 1971 3,789,941,225 1972 3,866,158,404 1973 3,941,664,971 1974 4,016,159,586 1975 4,088,621,062 1976 4,159,718,199 1977 4,231,619,236 1978 4,303,647,736 1979 4,378,565,589 1980 4,450,929,761 1981 4,533,928,518 1982 4,614,015,853 1983 4,695,112,999 1984 4,773,874,962 1985 4,855,692,131 1986 4,939,715,093 1987 5,026,262,667 1988 5,113,554,741 1989 5,200,376,354 1990 5,287,869,228 1991 5,370,833,520 1992 5,455,716,183 1993 5,538,201,967 1994 5,618,942,438 1995 5,699,768,392 1996 5,780,312,511 1997 5,859,124,817 1998 5,936,610,692 1999 6,013,679,354 2000 6,090,319,399 2001 6,167,064,399 2002 6,243,867,851 2003 6,320,371,175 2004 6,397,322,922 2005 6,474,229,144 2006 6,552,104,498 2007 6,630,764,007 2008 6,709,620,605 2009 6,788,203,578 2010 6,866,054,281 2011 6,943,437,438 2012 7,020,760,225 2013 7,098,495,231 2014 7,176,023,055 2015 7,253,260,112 2016 7,330,119,639

For t0= 1960, t1 = 1970 we have c = ln(n1 n0) t1` t0) ≈ :0199 and n(t) = 3:0428 ´ 109´ e:0199(t`1960) For t0= 1970, t1 = 1980 we have c = ln(n1 n0) t1` t0) ≈ :0181 and n(t) = 3:7123 ´ 109´ e:0181(t`1970)

The graph compares the population predicted with the Malthus’ Model with real population data.

The error in Malthus’ model rapidly increases with time. World population is overestimated.

Although in this case the exponential model is not completely appropriate to describe world population growth, it is able to represent a large number of physical systems.

As an example we can consider the radioactive decay; A radioactive substance decays proportionally to its mass

Thorium has an half-life of 1:65 ˆ 1010 years;

Therefore thorium decays according to the following law q = q0ect

where c can be calculated as

c =

ln(1₂)

1:65 ˆ 1010 = `1:33 ˆ 10

Malthus’ model assumes that the growth rate is constant, but an analysis of data shows that it is decreasing.

This fact leads to the hypothesis that the growth is conditioned by the number of individuals:

the greater the number of individuals, the lower the growth rate Verhulst (1846) developed a model which takes in account that the growth rate linearly decreases with the number of individuals.

Since

c (¯n) = a ` b ¯n = 0 () n =¯ a

b

so when n(t) = ¯n the growth rate vanishes and growth stops.

¯

n is a steady solution for the growth equation and represents a kind of equilibrium.

Verhulst model is defined by the Cauchy’s problem (

˙

n(t) = an(t) ` bn2<sub>(t) = n(t)(a ` bn(t))</sub> n(t0) = n0

The differential equation is known as logistic equation and the solution as logistic function.

The Logistic Equation

1 _{has two constant solutions}

n(t) = 0 n(t) = a

b = ¯n

which represent equilibrium solutions: when n(t) = 0 or n(t) = ¯n no further evolution is possible;

2 _since an ` bn2 > 0 for 0 < n < a b every solution is increasing when n0 < a<sub>b</sub> decreasing when n0 > a<sub>b</sub> 3 <sub>since</sub> ¨ n(t) = (a ` 2bn(t)) ˙n(t) every solution is

convex ( ˙n is increasing), when n(t) < _2ba

We can draw the directions field

and we may infer that n ! ¯n

increasing when n0< ¯n decreasing when n0> ¯n while it is constant when n0= ¯n Moreover when n(t) < a 2b = ¯ n 2

the rate of change increases and the solution is convex, and when

n(t) > a

2b =

¯ n 2

the rate of change decreases and the solution is concave,

The picture shows that if the initial number of individuals is greater than ¯n then the

population decreases and asymptotically approaches ¯n. On the contrary if the initial number of individuals is smaller than ¯n then the population increases and asymptotically approaches ¯n.

Moreover the growth rate is increasing as long as the number of individuals is smaller than ¯

n=2 and decreases from then on. ¯

n is, in essence, the maximum number of individuals that the environment can sustain and is usually called carrying capacity.

We can solve the Cauchy’s problem by separation of variables: It is easy to verify, using standard results, that the solution exists and that it is unique,

( ˙ n(t) = an(t) ` bn2<sub>(t) = n(t)(a ` bn(t))</sub> n(t0) = n0 moreover n(t) = 0 ; n(t) = a b are constant solutions of the equation.

Separating variables we get ˙ n(t) n(t)(a ` bn(t)) = 1 and Z t t0 ˙ n(s ) n(s )(a ` bn(s ))ds = t ` t0 Z n(t) n(t0) dx x (a ` bx ) = t ` t0

We have: 1 x (a ` bx ) = 1 a „ 1 x + b a ` bx « whence Z n(t) n0 dx x (a ` bx )dx = 1 a “ ln jx j ` ln ja ` bx j” n(t) n0 = = 1 aln „ <sub>n(t)</sub> a ` bn(t) « „ a ` bn0 n0 «

Let us remark that we can calculate the integral only when both n(t) and n0 _{belong to same interval in R n f0; a=bg so that a`bn(t)}n(t)

and c = n0

a`bn0 are both positive or negative. We get

t ` t0= 1 aln c „ <sub>n(t)</sub> a ` bn(t) « whence n(t) = a b + ce`a(t`t0) = 1 ˛ + ‚e`a(t`t0) (19)

We can determine a; ˛ e ‚ as soon as we know the population data at times t0` h, t0, t0+ h, n(t0` h) = p0 n(t0) = p1 n(t0+ h) = p2 Indeed we have 1 n(t) = ˛ + ‚e `a(t`t0) and 8 > < > : 1 p0 = ˛ + ‚e ah 1 p1 = ˛ + ‚ 1 p2 = ˛ + ‚e `ah

Let H = e`ah<sub>, we get</sub> 8 > < > : 1 p0 = ˛ + ‚ 1 H 1 p1 = ˛ + ‚ 1 p2 = ˛ + ‚H (20) whence (<sub>1</sub> p0 ` 1 p1 = ‚ `1 H ` 1´ = ‚ 1`H H 1 p1 ` 1 p2 = ‚(1 ` H) (21) 1 p0 ` 1 p1 1 p1 ` 1 p2 = 1 H

We finally obtain that: 1 H = p1p2(p1` p0) p0p1(p2` p1) H = p0p1(p2` p1) p1p2(p1` p0) and we can find ‚ e ˛ using 20,21.

‚ = 1 p1 ` 1 p2 1 ` H ; ˛ = 1 p1 ` ‚ Moreover a = 1 hln 1 H

For t0= 1970 h = 10 we have n(t) = 10 9 :0875 + 0:1818 ´ e`:0282(t`1970) For t0= 1960 h = 10 we have n(t) = 10 9 `0:8727 + 1:2013e`:0051(t`1960)

The graphs show how the choice of the initial data for the Cauchy problem is critical and affects the goodness of approximation.

Let us consider a population growing in an environment with fixed carrying capacity and let us assume that a fixed quantity of

individuals is harvested at constant rate

As an example we can consider a fish population subject to fishing or the population control for an isolated species living in a

Let us consider a population whose growth rate is a ` bn which is harvested at constant rate c .

We have ( ˙ n(t) = an(t) ` bn2<sub>(t) ` c</sub> n(t0) = n0 a; b; c > 0

Existence and uniqueness of the solution of this Cauchy problem can be verified by standard results and the differential equation admit constant solutions if the algebraic equation

`bx2

+ ax ` c = 0 has real solutions.

Therefore it is natural to consider three cases

When

a2` 4bc < 0 the algebraic equation

`bx2<sub>+ ax ` c = 0</sub>

has no real solutions.

˙ n(t) an(t) ` bn2<sub>(t) ` c</sub> = 1 Z t t0 ˙ n(s ) an(s ) ` bn2<sub>(s ) ` c</sub>ds = t ` t0 Z n(t) n(t0) dx ax ` bx2<sub>` c</sub> = t ` t0

Using partial fraction decomposition we get 1 ax ` bx2<sub>` c</sub> = ` 1 b 1 “ x `_2ba ”2 ` a2 4b2+ c b = = `1 b 1 “ x ` a 2b ”2 + ‹ = `1 b‹ 1 “ 1 p ‹ “ x ` a 2b ””2 + 1 = where ‹ =4bc ` a 2 4b2 Therefore Z dx ax ` bx2<sub>` c</sub> = ` 1 bp‹ arctanp1 ‹ „ x ` a 2b « + cost: and ` 1 bp‹ arctanp1 ‹ „ n(t) ` a 2b « + + 1 bp‹ arctanp1 ‹ „ n0` a 2b « = t ` t0 Moreover n(t) = a 2b+ p ‹ tan „ t0+ 1 bp‹ arctanp1 ‹ „ n0` a 2b « ` t «

Logistic Growth with Harvesting a = 2 b = 1 c = 4 a2` 4bc = `12 < 0

When

a2_{` 4bc = 0}

the algebraic equation

`bx2<sub>+ ax ` c = 0</sub>

has one real solutions with multiplicity 2.

We have `bx2<sub>+ ax ` c = 0 for x =</sub> a

2b and

n(t) = ¸ = a

2b

is a constant solution of the differential equation. Moreover we have 1 ax ` bx2<sub>` c</sub> = ` 1 b 1 “ x ` a 2b ”2 whence Zn(t) n0 dx ax ` bx2` c = 1 b 1 “ n(t) `<sub>2b</sub>a” `1 b 1 “ n0` a 2b ”= t ` t0 and n(t) = 1 2b a + 2bn0` a 1 + (n0b `a 2)(t ` t0) !

Logistic Growth with Harvesting

a = 2 b = 1 c = 1

When

a2_{` 4bc > 0} the algebraic equation

`bx2<sub>+ ax ` c = 0</sub>

has two real distinct solutions:

¸ = a 2b + s a2<sub>` 4bc</sub> 4b2 ; ˛ = a 2b ` s a2_{` 4bc} 4b2

There are two constant solutions for the differential equations and we have

`bx2

+ ax ` c = `b(x ` ¸)(x ` ˛) Using partial fraction decomposition we get

1 `bx2<sub>+ ax ` c</sub> = 1 b(˛ ` ¸) „ 1 x ` ¸ ` 1 x ` ˛ «

Therefore Z 1 `bx2<sub>+ ax ` c</sub>dx = 1 b(˛ ` ¸)(ln jx ` ¸j ` ln jx ` ˛j) + ’(x ) where ’ is piecewise constant

We have Z n(t) n0 1 `bx2<sub>+ ax ` c</sub>dx = = 1 b(˛ ` ¸)ln »„ n(t) ` ¸ n(t) ` ˛ « „ n0` ˛ n0` ¸ «– = t ` t0 whence n(t) = ¸ ` ˛n0`¸ n0`˛e `b(˛`¸)(t`t0) 1 ` n0`¸ n0`˛e `b(˛`¸)(t`t0)

Logistic Growth with Harvesting

a = 5 b = 1 c = 4

Looking at the graphs of the solutions we obtain that:

1 _{when a}2_{` 4bc < 0 the population dies out in a finite time}

2 _{when a}2_{` 4bc = 0}

the population dies out in a finite time if n0< _2ba

the population approaches the value a

2b asymptotically if

n0 > _2ba

3 _{when a}2_{` 4bc > 0}

the population dies out in a finite time if n0< ˛

the population approaches the value ˛ asymptotically if n0 > ˛

Therefore the population doesn’t dies out as long as

c » a

2 4b

and the initial number of individuals is sufficiently large. The value c = _4ba2 is the greatest harvesting rate which doesn’t cause the extinction of the population.

When the population doesn’t die out, it approaches asymptotically to the value

¸ = a 2b + s a2_{` 4bc} 4b2 < a b where a

Predator-Prey Model

The predator-Prey Model deals with the interactions between two species one of which feeds the other.

The model was independently developed by Vito Volterra and Alfred Lotka around 1925 .

The issue that aroused the interest in this problem deals with certain data concerning the type of fish caught in the Adriatic Sea in the years between 1920 and 1930

The biologist Umberto D’Ancona, examining data relating to the fish sold in the ports of Rijeka (Fiume), Trieste and Venice

observed an increase in the percentage of Selachii (predators such as sharks and rays) in the years between 1914 and 1918

The following table shows the data studied by D’Ancona Anno 1914 1915 1916 1917 1918 1919 1920 1921 1922 1923 % of Selaci 11.9 21.4 22.1 21.1 36.4 27.3 16.0 15.9 14.8 10.7 1914 1916 1918 19201922 19151917 1919 1921 1923 60 70 80 90 10 20 30 40

The data show an increase in the percentage of Selachii and a corresponding reduction in other fish

The military operations in the Adriatic Sea had certainly caused the reduction of fishing activities but this could explain a decrease in the fish caught, not an increase in selachi and a reduction in other fish.

D’Ancona, at first tried to explain the phenomenon with biological arguments, but later asked Vito Volterra to study the problem Volterra proposed the model that now bears its name and which provides an explanation of the observed data.

The same type of model was independently introduced by Alfred James Lotka in his work and was published in the book "Elements of Mathematical Biology" of 1924.

The problem can be stated as follows: x (t) is the number of prey at time t y (t) is the number of predators at time t

in the absence of predators the prey grows following an exponential malthusian model

˙

x (t) = ax (t)

the growth rate,a, of prey decreases in proportion to the number of predators

a ` ¸y (t)

In the absence of prey predators die out following a model Malthusian

˙

y (t) = `by (t)

The growth rate, b, of predators increases proportionally to the number of prey

The dynamics of the two populations can be described by the following differential system

(Lotka-Volterra Equations) ( ˙ x (t) = ax (t) ` ¸y (t)x (t) ˙ y (t) = `by (t) + ˛x (t)y (t) (22)

with initial values

(

x (t0) = x0 y (t0) = y0

This differential system to is simple but can not be integrated explicitly.

Nevertheless we can prove that

1 _{The differential system has two steady-state solutions (}b

˛; a ¸), (0; 0)

2 _{The orbits of the differential system lie in the first quadrant}

and are closed curves

3 _{The solutions of the differential system are periodic}

4 _{There is a conservation law : the quantity}

a ln(y ) ` ¸y + b ln(x ) ` ˛x

remains constant along a orbit, moreover the average population is constant for every orbit.

1 T Z T 0 x (s )ds = b ˛ ; 1 T Z T 0 y (s )ds = a ¸

We can find steady-solutions of the differential system (x (t); y (t)) = (‰; ”) solving ( 0 = a‰ ` ¸”‰ 0 = `b” + ˛”‰ and we get ‰ = 0; ” = 0 and ‰ = b ˛; ” = a ¸

To study the orbits ( x = x (t) y = y (t) of the system ( ˙ x (t) = ax (t) ` ¸y (t)x (t) = ffi(x (t); y (t)) ˙ y (t) = `by (t) + ˛x (t)y (t) = (x (t); y (t)) (23)

we can observe that, every non steady orbit can be locally represented as a function y = y (x ); indeed we can assume that x (t) is invertible and we substitute in y (t) and we get

(

x (t) = x

Deriving we obtain

˙

y (t) = y0(x (t)) ˙x (t)

where ˙ indicates derivation with respect to t and 0 derivation with respect to x . Therefore we get y0(x ) = (x ; y (x )) ffi(x ; y (x )) = `by (x ) + ˛xy (x ) ax ` ¸xy (x ) Separating variables y0(x ) „ a y (x ) ` ¸ « =„ `b x + ˛ «

and integrating with initial data (

x (t0) = ¯x

We obtain a ln y (x ) ` ¸y (x ) ` a ln ¯y + ¸ ¯y = `b ln x + ˛x + b ln ¯x ` ˛ ¯x (24) Now if we set g(y ) = a ln y ` ¸y f (x ) = `b ln x + ˛x G (y ) = g(y ) ` g( ¯y ) F (x ) = f (x ) ` f (¯x ) we can write 24 as g(y ) = f (x ) + g( ¯y ) ` f (¯x ) (25) that is G (y ) = F (x )

and if G`1 is the inverse of a suitable restriction of G we can locally write

The picture shows the graph of the function G (y ) = g(y ) ` g( ¯y ) y0= a ¸ g0= G (y0) = a ln a ¸` ¸ a ¸ = = a „ ln a ¸ ` 1 « > 0

The picture shows the graph of the function F (x ) = f (x ) ` f (¯x ) x0 = b ˛ f0= f (x0) = `b ln b ˛ + ˛ b ˛ = = `b „ ln b ˛ ` 1 « > 0

The equality

G (y ) = g(y ) = g( ¯y ) ` g(y ) = f (x ) ` f (¯x = F (x ) is true only when

[f0; g0] = (`1; g0] \ [f0_{; +1) 6= ∅}

We can draw an orbit of the system G`1(F (x )) using the graphical construction which is shown in the picture.

We have that:

F is decreasing for x 2 [xm; x0] F is increasing for x 2 [x0; xM] F reaches all values in [f0; g0],

G is increasing and invertible for [ym; y0] G is decreasing and invertible for [y0; yM] G reaches all values in [f0; g0]

Therefore

G`1 is defined in [f0; g0] with values in [ym; y0] or in [y0; yM] G`1(F (´)) is defined and decreasing in [xm; x0]

More orbits are shown below

The function

U(x ; y ) = a ln y ` ¸y + b ln x ` ˛x

remains constant on every orbit and can be used to state a conservation law.

The above considerations allow us to state that:

The orbits of the differential system are closed curves. (x0; y0) , which is the orbit of the non-zero steady solution, lies inside every orbit

The orbits entirely lies within the first quadrant, i.e. xm; xM; ym; yM > 0

Now we want to prove the periodicity of the solutions. To this end we observe that:

if (x (t); y (t)) is a solution of the differential system then also (x (t + T ); y (t + T )) is a solution indeed ( ˙ x (t + T ) = ax (t + T ) ` ¸y (t + T )x (t + T ) ˙ y (t + T ) = `by (t + T ) + ˛x (t + T )y (t + T )

Every (¯x ; ¯y ) lie on one and only one orbit of the differential system. Indeed if (¯x ; ¯y ) lie on the orbits defined by (x (t); y (t)) and (‰(t); ”(t)) then ( x (t0) = ¯x y (t0) = ¯y ( ‰(fi0) = ¯x ”(fi0) = ¯y for some t0; fi0. If we define

(

‰1(t) = x (t ` fi0+ t0) ”1(t) = y (t ` fi0+ t0)

then (‰1; ”1) is a solution of the differential system and ( ‰1(fi0) = ¯x ”1(fi0) = ¯y Therefore ( ‰(t) = ‰1(t) = x (t ` fi0+ t0) ”(t) = ”1(t) = y (t ` fi0+ t0)

We can also prove that the solutions of the differential system are periodic.

Every solution defines an orbit which can be represented as the union of the graphs of two regular functions defined on a compact interval. Therefore its length ‘ if finite.

Since ffi; vanish simultaneously only in (0; 0) and (˜x ; ˜y ), bearing in mind that these two points are not part of any of the orbits, except for the two stationary solutions, we have

˙ x2(t) + ˙y2(t) = ffi2(ˆx ; ˆy ) + 2(ˆx ; ˆy ) – – min (x ` x0)2+ (y ` y0)2> › xm» x » xM; ym» y » yM fffi2_{(x ; y ) +} 2_{(x ; y )g – m > 0}

It follows that Z +1 0 q ˙ x2_{(t) + ˙y}2_{(t)dt = +1}

and there is T > 0 such that

Z T

0 q

˙

x2_{(t) + ˙y}2_{(t)dt = ‘}

Since (x (t); y (t)) moves along its orbit with strictly positive

velocity, while the orbit has finite length, we can assert that after a finite time (x (t); y (t)) reaches the initial point again.

Since every point lies on one and only one orbit, (x (t); y (t)) must remain in the previous orbit and the solution is periodic.

Average values mantain constant along an orbit, that is 1 T Z T 0 x (s )ds = b ˛ ; 1 T Z T 0 y (s )ds = a ¸ (26)

Indeed dividing the first equation by x and the second one by y , we get (x (t)˙ x (t) = a ` ¸y (t) ˙ y (t) y (t) = `b + ˛x (t) (27)

Since solutions are periodic of period, say, T, if we integrate over 0; T ] we have ( 0 = lnx (T )_{x (0)} =R₀Ta ` ¸y (s )ds = aT ` ¸R₀T y (s )ds 0 = lny (T )_{y (0)} =R₀T `b + ˛yxs)ds = bT ` ˛RT 0 x (s )ds and we obtain 1 T RT 0 x (s )ds = b ˛ , 1 T RT 0 y (s )ds = a ¸

Predator-Prey Model with constant harvesting

We must now study what happens when a constant harvesting is introduced.

Let h the harvesting rate of prey and k the harvesting rate of predators 8 > > > < > > > : ˙ x (t) = ax (t) ` ¸y (t)x (t) ` hx (t) ˙ y (t) = `by (t) + ˛x (t)y (t) ` ky (t) x (t0) = x0 y (t0) = y0 (28)

Assuming that a ` h > 0 and b + k > 0 the problem can be treated as we have done before and we find the new steady solutions and the new average values.

¯

x = b + k

˛ y =¯

a ` h ¸

Therefore, if the fishing rate decreases, the equilibrium point moves upward and to the left so as occurs in the data that we have considered 1914 1916 1918 19201922 1915 1917 1919 1921 1923 60 70 80 90 10 20 30 40 (¯x1, ¯y1) = (b+kβ1, a−h1 α ) (¯x2, ¯y2) = (b+k_β2,a−h_α2) h1> h2 k1> k2

Two species x and y share the resources of a single territory Each species would have a growth rate of logistics if he lived in an isolated environment, ( ˙ x (t) = (a ` Ax (t))x (t) ˙ y (t) = (b ` By (t))y (t) ( x (t0) = x0 y (t0) = y0

Because of the mutual effects the growth rate of the population x is

a ` Ax ` ¸y (it decreases proportionally wheny increases) Similarly, the growth rate of the population y è

b ` By ` ˛x (it decreases proportionally when x increases)

Each species grows subtracting resources to the other one

The system that describes the development of two populations is ( ˙ x (t) = (a ` Ax (t) ` ¸y (t))x (t) ˙ y (t) = (b ` By (t) ` ˛x (t))y (t) (29) where a; b; A; B; ¸; ˛ > 0, with initial data

(

x (t0) = x0 y (t0) = y0

Dividing the first equation by x and integrating we obtain ˙ x (t) x (t) = (a ` Ax (t) ` ¸y (t)) Z x (t) x0 ds s = Z t t0 (a ` Ax (s ) ` ¸y (s ))ds lnx (t) x0 = Z t t0 (a ` Ax (s ) ` ¸y (s ))ds and x (t) = x0e Rt t0(a`Ax (s )`¸y (s ))ds Similarly y (t) = y0e Rt t0(b`By (s )`˛x (s ))ds

So we can assert that

x (t); y (t) > 0 If we define k = maxfa; bg we have

(x + y )0 = (a ` Ax (t) ` ¸y (t))x (t)+

+ (b ` By (t) ` ˛x (t))y (t) » k (x + y ) therefore x and y are bounded functions because

0 » (x + y ) » ekt » ekT

and using a classic argument on prolongability we can assert that they are defined on [0; T ] for every positive T .

Solving the algebraic system (

0 = (a ` Ax ` ¸y )x 0 = (b ` By ` ˛x )y we can find the steady solutions

Therefore, the stable solutions are identified as the intersection between axes an the straight lines

R1 : (a ` Ax ` ¸y ) = 0

R2 : (b ` By ` ˛x ) = 0

R1 intersects the axes in ( ˆ‰; 0) = (a

A; 0) ; (0; ”

#_{) = (0;} a ¸) while R2 intersects the axes in

(‰#; 0) = (b ˛; 0) ; (0; ˆ”) = (0; b B) ‰# = b ˛ ” =ˆ b B

R1 and R2 intersects in

(˜x ; ˜y ) = (aB ` ¸b

AB ` ¸˛;

Ab ` a˛

AB ` ¸˛)

Therefore the steady solutions are

(0; 0) ; (0; ˆ”) ; ( ˆ‰; 0) ; (˜x ; ˜y ) The last one can only be considered if

1 _{R1 is not parallel to R2 (AB ` ¸˛ 6= 0)}

2 _{it lies in the first quadrant (˜x > 0 , ˜y > 0).}

The following pictures show the phase portrait and the field directions associated to the differential system

The stability of each constant solution depends upon mutual position of R1 and R2

Two species x and y share the resources of a single territory; each of them decreases exponentially in absence of the other one, but the growth rate of one population increases proportionally to the number of individuals of the other population.

In other words

˙

x (t) = `ax (t) ˙

y (t) = `by (t) when the populations live isolated, while

( ˙

x (t) = (`a + ˛y (t))x (t) ˙

y (t) = (`b + ¸x (t))y (t)

The system ( ˙ x (t) = (`a + ˛y (t))x (t) ˙ y (t) = (`b + ¸x (t))y (t) has constant solutions:

(0; 0) ; (a

˛; b ¸) The picture show the phase portrait and the field

directions associated to the differential system

The non trivial solution is instable.

Two species x and y share the resources of a single territory; each of them has a logistic growth in absence of the other one, but the growth rate of one population increases proportionally to the number of individuals of the other population.

In other words

˙

x (t) = (a ` b(x (t))x (t) ˙

y (t) = (c ` dy (t))y (t) when the populations live isolated, while

˙

x (t) = (a ` b(x (t) + ‚y (t))x (t) ˙

y (t) = (c ` dy (t) + ‹x (t))y (t) when the populations live in the same environment.

The system ( ˙ x (t) = (a ` b(x (t) + ‚y (t))x (t) ˙ y (t) = (c ` dy (t) + ‹x (t))y (t) has the following constant solutions (0; 0) ; (a b; 0) ; (0; c d) (c ‚ + ad bd ` ‚‹; cb + a‹ bd ` ‚‹)

Stability

The examples of differential systems that we have previously studied show that the constant solutions of the system are in fact equilibrium solutions.

If we use the initial data other than the points of equilibrium the solutions can approach asymptotically to them or may depart from them

This fact allows to distinguish the equilibrium points between those for which the solutions are approaching and those from which the solutions turn away.

We will call the first stable solutions and the second unstable solutions.

We begin to examine the stability of 2 ˆ 2 linear homogeneous differential systems with constant coefficients

( ˙

x (t) = ax (t) + by (t) ˙

y (t) = cx (t) + dy (t) a; b; c ; d 2 R (30)

i.e., using vector notation ˙ u(t) = Au(t) (31) where u(t) = „x (t) y (t) « ed A =„a b c d «

If u(t) = (x (t); y (t)) is a solution of the differential system, we define orbit, or (trajectory), the line

(

x = x (t) y = y (t)

The orbits are represented in the plane (x ; y ) (Phase Plan diagram)

The differential system is autonomous and we have already seen that for every point on the phase plane passes one and only one orbit.

Given a 2 ˆ 2 matrix A it is possible to find a non-singular matrix P such that

C = P`1AP

is a matrix, called Jordan Canonical Form, which falls into one of the following types.

Let –1; –2 be the eigenvalues of A. The Jordan Canonical form

reduces to

1

„–1 0

0 –2

«

if –1, –2 are real and distinct.

2 _–

1= ¸ + i ˛, –2 = ¸ ` i ˛,

„ ¸ ˛

`˛ ¸

«

if –1= ¸ + i ˛ are complex and conjugate.

3 „– 0 0 – « oppure„– 0 ‚ – «

Using the Jordan Canonical Form The differential system can be transformed in a simpler system having the coefficient matrix in Jordan Form

To this end we use the following transormation

u = Pv o equivalentemente v = P`1u Indeed P ˙v = ˙u = Au = APv whence ˙ v = P`1APv = Cv

If we are able to study the orbits of the system ˙v = Cv , we also get information about the original one.

The transformation u = Pv is linear and non-singular; therefore it cause only a deformation of the orbits.

So it is enough to study the orbits of the differential systems having coefficient matrix in Jordan Canonical Form.

The following pictures show how the transformation u = Pv acts on axes and on unit circle.

( ‰ = ax + by ” = cx + dy ovvero „ ‰ ” « =„a b c d « „x y «

1)- Real and distinct eigenvalues.

The system in canonical form is ( ˙_{‰ = –1‰}

˙

” = –2” We can solve the system and we get

(

‰(t) = c1e–1t

”(t) = c2e–2t

If at least one eigenvalue is not 0 Trajectories are given by y = x

–2 –1

j–2j > j–1j > 0

j–2j < j–1j ; j–1j > 0

j–2j > j–1j > 0

When one of the eigenvalues is 0, the system becomes ( ˙_{‰ = –1‰} ˙ ” = 0 or ( ˙_{‰ = 0} ˙ ” = –2”

The trajectories are straight lines; the direction of travel depends on the sign of the nonzero eigenvalue

2)- Real and coincident eigenvalues

The system can be reduced to the following ones ( ˙_{‰ = –‰} ˙ ” = –” or ( ˙_{‰ = –‰} ˙ ” = ‚‰ + –” When – 6= 0 the solutions of the first system are

( ‰(t) = c1e–t ”(t) = c2e–t ; ‰ = c2 c1 ” while the solution of the second system are

( ‰(t) = c1e–t ”(t) = (‚c1t + c2)e–t ; 8 < : t = 1_–ln“‰ c1 ” ” = ”(t) = (‚1_–ln“‰ c1 ” + c2 c1)‰

When – = 0 the solution of the first system are (

‰(t) = c1 ”(t) = c2 while the solution of the second system are

(

‰(t) = c1

A =„0 0 0 0 « A =„0 0 1 0 «

Complex conjugate eigenvalues

When eigenvalues are complex conjugate the system can be reduced to:

( ˙_{‰ = ¸‰ ` ˛”} ˙

” = ˛‰ + ¸”

and can solve the system using a change of variables. We set ( ‰ =  cos „ ” =  sin „ and we get 2(t) = ‰2(t) + ”2(t) tan „(t) = ”(t) ‰(t)

Deriving we have

2(t) ˙(t) = 2‰(t) ˙‰(t) + 2”(t) ˙”(t) and

(1 + tan2„(t)) ˙„(t) = ‰(t) ˙”(t) ` ”(t) ˙‰(t) ‰2_(t)

and taking account of the system it follows that ( ˙  = ¸ ˙ „ = ˛ whence ( (t) = C1e¸t „(t) = ˛t + C2

The trajectories are spirals or, in case ¸ = 0, circles.

The sign of beta determines whether the rotation is clockwise or counter-clockwise while the sign of ¸ indicates whether the radius tends to 0 or to +1

Stability Criteria; The linear case

Let A(t) be a n ˆ n matrix valued function, which is continuous over [t0; +1) and let us consider the linear differential system

˙

x (t) = A(t)x (t)

Let G be a fundamental matrix for the system. Then x (t) = 0 is a stable solution if and only if

kG (t)k » K ; 8t – t0

Moreover x (t) = 0 is asymptotically stable if and only if lim

Indeed the solution x (t) of the Cauchy Problem: ( ˙ x (t) = A(t)x (t)) x (t0) = x0 can be expressed as x (t) = G (t)G`1(t0)x0 (32) and kx (t)k » kG (t)k kG`1(t0)k kx0k:

Conversely if x (t) = 0 is s stable solution then exists ‹ > 0 such that for kx0k < ‹ we have kx (t)k < 1

Let D = f‹ei_{; i = 1::ng; then D is a basis for R}n _{abd every solution} x‹ei satisfying x0 = ‹ei, is norm-bounded;

But using x‹ei as columns of a matrix we obtain a fundamental

matrix which is bounded.

Similarly, if x0= 0 is asymptotically stable we have lim t!+1 kx‹ei(t)k = 0: whence lim t!+1 kG (t)k = 0

If the differential system has constant coefficients, then every solution can be written as a linear combination of functions of the type:

t‚e¸tcos(˛t) ; t‚e¸tsin(˛t)

which correspond to the eigenvalues ¸ ˚ {˛, of the matrix of the coefficientsà and ‚ is their multiplicity.

We have that:

1 _{If all eigenvalues of A have negative real part, then x (t) = 0}

is asymptotically stable

2 _{If at least one eigenvalue of A has positive real part , then}

x (t) = 0 is not asymptotically stable neither stable

3 _{If all eigenvalues of A have non-positive real part and the}

multiplicity of eigenvalues with zero real part is 1 , then x (t) = 0 is stable but it is not asymptotically stable

4 _{If at lest one eigenvalues of A have zero real part and it has}

multiplicity greater than 1 then x (t) = 0 is not asymptotically stable neither stable

The stability of a nonlinear system is a much more complicated matter

Many results can be achieved through the use of Lyapunov functions;

However we can find interesting results by means of the study of the linearized system, i.e. studying the linear differential system associated to the Jacobian matrix of the function on the right. T

Stability by linearization Le us consider ˙ x (t) = Ax (t) + f (x (t)) (33) where Ais a n ˆ n matrix, S = fy 2 Rn _{: ky k < ag}

f : S ! R, is a continuous function such that f (0) = 0 and lim

x !0 f (x )

kx k = 0

Then if all eigenvalues of A have negative real part then x (t) = 0 is asymptotically stable.

On the contrary, if A has at least one eigenvalue with positive real part then x (t) = 0 is unstable.

Proof.

Let G the principal fundamental matrix of the linearized system ˙

x = Ax

and let x the solution of 33; we assume that x is defined over a maximal interval [t0; b).

Since all eigenvalues of A have negative real part, we can find ¸ > 0 and K > ` such that

kG (t)k » Ke`¸t

For suitable ff > 0 , when kx k < ff we have

kf (x )k » ¸

2Kkx k; We can also assume that ¸ < ff.

Let z (t) = G (t ` t0)x0+ Z t t0 G (t ` s )f (x (s ))ds we have z0(t) = G0(t ` t0)x0+ Z t t0 G0(t ` s )f (x (s ))ds + G (0)f (x (t)) = = A G (t ` t0)x0+ Z t t0 G (t ` s )f (x (s ))ds ! + f (x (t)) = = Az (t) + f (x (t)) and z (t0) = x0: Taking in account that

˙

x (t) = Ax (t) + f (x (t)) ; x (t0) = x0 we have

Therefore x (t) = G (t ` t0)x0+ Z t t0 G (t ` s )f (x (s ))ds So we have kx (t)k » Ke`¸(t`t0)_{kx0k +} Z t t0 ¸ 2e `¸(t`s)_{kx (s)kds} whence e¸(t`t0)<sub>kx (t)k » K kx0k +</sub> Z t t0 ¸ 2e ¸(s `t0)_{kx (s)kds}

and, by Gronwall’s Lemma

e¸(t`t0)<sub>kx (t)k » K kx0ke</sub>¸(t`t0)=2

and

kx (t)k » K kx0ke`¸(t`t0) _{» K kx}

0k ; 8t – t0 (34)

Therefore if 0 < " < ff and kx0k < "=K , we have kx (t)k < " ; 8t 2 [t0; b) It follows that x (t) is prolongable and b = +1.

Taking the limit for t ! +1 in 34 we can assert that x (t)) = 0 is asymptotically stable.

So the first assertion is proved. The second assertion is a bit more tricky and we omit the proof.

About the sign of eigenvalues of a 2 ˆ 2 matrix.

Let us consider a2 ˆ 2 matrix

A =„a b

c d

«

We can find its eigenvalues solving the equation

det„a ` – b

c d ` –

«

= –2` (a + d)– + (ad ` bc ) (35)

Let T = a + d be the trace of the matrix and D = ad ` bc be its determinant; 35 can be rewritten as

–2` T – + D and we can calculate

– = T

2 ˚

s T2

Therefore When D > 0 T2 4 > T2 4 ` D If T2 4 ` D – 0 I.E. ˛ ˛ ˛ ˛ T 2 ˛ ˛ ˛ ˛ > s T2 4 ` D

then –1; –2 have the same sign as T If

T2

4 ` D < 0

eigenvalues are complex conjugate and the real part has the same sign as T

When D = 0 –1 = 0 ; –2= T When D < 0 We have T2 4 ` D > T2 4

Stability for Intraspecific Competition Population Growth

Two species x and y share the resources of a single territory The growth is described by the following system of differential equations: ( ˙ x (t) = (a ` Ax (t) ` ¸y (t))x (t) ˙ y (t) = (b ` By (t) ` ˛x (t))y (t) (37) with ( x (t0) = x0 y (t0) = y0 where a; b; A; B; ¸; ˛ are positive

We have already seen that x ed y are bounded functions and therefore they are defined over [0; T ], for every T > 0

Let

R1 : (a ` Ax ` ¸y ) = 0

R2 : (b ` By ` ˛x ) = 0

The constant solutions can be found by intersecting R1 and R2

with the axes and with each other.

The straight line R1 intersects the axes in „ a A; 0 « = ( ˆ‰; 0) ; „ 0; a ¸ « = (0; ”#) The straight line R2 intersects the axes in

„ b ˛; 0 « = (‰#; 0) ; „ 0; b B « = (0; ˆ”)

The straight line R1 intersects The straight line R2 in „ aB ` ¸b AB ` ¸˛; Ab ` a˛ AB ` ¸˛ « = (˜x ; ˜y ) therefore the constant solutions are

(0; 0) ; (0; ˆ”) ; ( ˆ‰; 0) ; (˜x ; ˜y ) Obviously (˜x ; ˜y ) can only be considered when

1 _{R1 is not parallel to R2 , i.e. AB ` ¸˛ 6= 0} 2 _{it lies in the first quadrant, i.e. ˜x > 0 , ˜y > 0.}

In order to study the stability of the constant solutions we can linearize the differential system using the taylor expansion centered in (¯x ; ¯y ).

The coefficient matrix of the linearized system ( in (¯x ; ¯y ) is

M =„a ` 2A¯<sub>`˛ ¯</sub>x ` ¸ ¯y `¸¯x y b ` 2B ¯y ` ˛ ¯x « When (¯x ; ¯y ) = (0; 0) we have M = „a 0 0 b «

When (¯x ; ¯y ) = ( ˆ‰; 0) M =„a ` 2A a A `¸ a A 0 b ` ˛a A « = `a `¸a A 0 ˛ “ b ˛ ` a A ” ! = =„`a `¸ a A 0 ˛( ˆ‰ ` ‰#₎ «

The eigenvalues of M are `a and ˛( ˆ‰ ` ‰#_{). The solution is}

stable when ‰# _{< ˆ‰.} When (¯x ; ¯y ) = (0; ˆ”) M =„a ` ¸ b B 0 `˛_Bb b ` 2B<sub>B</sub>b « =„¸ `a ¸ ` b B ´ 0 `˛_Bb `b « = =„¸( ˆ” ` ” #_{) 0} `˛b B `b «

The eigenvalues of M are `b and ¸( ˆ” ` ”#). The solution is

When (¯x ; ¯y ) = (˜x ; ˜y )

M = „`A˜x `¸˜x

`˛ ˜y `B ˜y «

Let D the determinant of M and let T the trace of M; we have

D = ˜x ˜y (AB ` ¸˛) 6= 0 T = `(A˜x + B ˜y )(< 0)

When D > 0 we have T2

4 ` D = (˜x A ` ˜y B)

2_{+ 4˜x ˜y ¸˛ > 0}

since ˜x > 0, ˜y > 0, M has real distinct eigenvalues and they have the same sign as T . Therefore the solution is stable

When D < 0 ,M the eigenvalues of M are real and they have opposite sign. Therefore the solution is unstable.

Compulsory Cooperation

The differential system that describes the model is (

˙

x (t) = (`a + ˛y (t))x (t) ˙

y (t) = (`b + ¸x (t))y (t) It has the following constant solutions

(0; 0) ; „ b

¸; a ˛

«

We can study the stability of the solutions using linearization and the coefficient matrix of the linearized system is

r„ (`a + ˛y )x (`b + ¸x )y « =„`a + ˛y ˛x ¸y `b + ˛x «

When x = 0, y = 0 the coefficient matrix becomes

„`a 0

0 `b

«

Both its eigenvalues are negative so that the solution is stable When x = _˛a, y = _¸b the coefficient matrix becomes

„ 0 b˛_¸

b¸_˛ 0 «

The eigenvalues are p

ab ; `pab

Voluntary Cooperation

The differential system that describes the model is (

˙

x (t) = (a ` b(x (t) + ‚y (t))x (t) ˙

y (t) = (c ` dy (t) + ‹x (t))y (t) The system ha s the following constant solutions

(0; 0) ; (a b; 0) ; (0; c d) (c ‚ + ad bd ` ‚‹; cb + a‹ bd ` ‚‹)

The coefficient matrix of the linearized system is M = „a ` 2b ¯x + ‚ ¯y ‚ ¯x ‹ ¯y c ` 2d ¯y + ‹ ¯x « When (¯x ; ¯y ) = E1 = (0; 0) M =„a 0 0 c «

When (¯x ; ¯y ) = E2 = (_ba; 0) M =„`a ‚ a b 0 c + ‹<sub>b</sub>a « ` a < 0 c + ‹a b > 0 The eigenvalues of M are `a < 0, c + ‹<sub>b</sub>a > 0 so that E2= E2= (a<sub>b</sub>; 0) is a saddle point . When (¯x ; ¯y ) = E3 = (0;c<sub>d</sub>) M = „a + ‚ c d 0 ‚c d `c «

The eigenvalues are a + ‚c_d > 0, `c < 0 so that E3= (0;c_d) is a saddle point.

When (¯x ; ¯y ) = E4 = (_DA;B D)

E4= (_DA;B_D) lies in the first quadrant iff D > 0 and

M = „`b A D `‚ A D ‹B D `d B D «

det M = (bd ` ‚‹)AB = DAB > 0

Calculus of Variations

The classical problem of the calculus of variations consists in minimizing an integral functional depending on a real function and its derivative.

More precisely let

f : R2! R

be a smooth function and let C1 _{the space of all differentiable}

functions

x : [a; b] ! R

whose derivative is continuous and let us consider the problem to find min x 2C1_{;x (a)=¸;x (b)=˛} (Z b a f (x (t); ˙x (t))dt )

The curve of shortest length connecting two fixed points.

Let

P0 = (a; y0) ; P1 = (b; y1)

two fixed points in the plane

We want to determine the curve of shortest length starting in P0

and ending in P1.

The length of the curve defined by y is

Z b

a q

1 + (y0_{(x ))}2_dx

and we want to find

min y 2C1_{;y (a)=y} 0;y (b)=y1 (Z b a q 1 + (y0_{(x ))}2_dx )

The Brachistochrone

Let us consider two fixed points

A = (a; h) ; B = (b; 0) with h > 0

in a vertical plane and a material point P sliding without friction along a curve, starting in A and ending in B, under the sole influence of gravity

We look for the curve such that the material point reaches B in the shortest time.

We choose a coordinate system directed upward so that, by the Conservation of Energy Principle, we can calculate the velocity of the material point as

1

2mv

2_{= mg(h ` y ) e v =</sub>q<sub>2g(h ` y (x ))}

while an element of distance traversed can be calculated using arc length as

ds = q

Therefore the element of time dt to transverse ds is dt = p 1 + (y0_{(x ))}2 p 2g(h ` y (x )) dx

Integrating over [a; b] we get

t = Z b a p 1 + (y0_{(x ))}2 p 2g(h ` y (x )) dx (38)

So that we have to find

min y 2C1_{;y (a)=h;y (b)=0} Z b a p 1 + (y0_{(x ))}2 p 2g(h ` y (x )) dx

Euler’s Equations

Let

f : R2! R

continuous with its first derivative and let C1 _{be the space of all} differentiable functions

x : [a; b] ! R whose derivative is continuous.

Let us consider the problem to find.

min x 2C1_{;x (a)=¸;x (b)=˛}

Z b

a

Let us assume that x02 C1 is such that x0(a) = ¸ ; x0(b) = ˛ and Z b a f (x0(t); ˙x0(t))dt = min x 2C1_{;x (a)=¸;x (b)=˛} Z b a f (x (t); ˙x (t))dt

Let h 2 C1 _{such that h(a) = h(b) = 0;} Then ffi(–) = Z b a f (x0(t) + –h(t); ˙x0(t) + – ˙h(t))dt reaches its minimum for – = 0

so that

Deriving under integral sign and integrating by parts we get ffi0(0) = = Z b a `fx(x0(t); ˙x0(t))h(t) + fv(x0(t); ˙x0(t)) ˙h(t)´ dt = = Z b a fx(x0(t); ˙x0(t))h(t)dt + fv(x0(t); ˙x0(t))h(t) ˛ ˛ ˛ b a ` ` Z b a „ d dtfv(x0(t); ˙x0(t)) « h(t)dt = = Z b a „ fx(x0(t); ˙x0(t)) ` d dtfv(x0(t); ˙x0(t)) « h(t))dt = 0

and we can assert that

fx(x0(t); ˙x0(t)) ` d

dtfv(x0(t); ˙x0(t)) = 0 (39)

Equation 39 is known as Euler’ Equation We also have fx(x (t); ˙x (t)) ` d dtfv(x (t); ˙x (t)) = fx(x (t); ˙x (t)) ` fv ;x(x (t); ˙x (t)) ˙x (t)` ` fv ;v(x (t); ˙x (t))¨x (t)

multiplying by ˙x (t) fx(x (t); ˙x (t)) ˙x (t) ` fv ;x(x (t); ˙x (t)) ˙x2(t)` ` fv ;v(x (t); ˙x (t))¨x (t) ˙x (t) = = fx(x (t); ˙x (t)) ˙x (t) + fv(x (t); ˙x (t))¨x (t) ` fv(x (t); ˙x (t))¨x (t)` ` fv ;x(x (t); ˙x (t)) ˙x2(t) ` fv ;v(x (t); ˙x (t))¨x (t) ˙x (t) = = d dt “ f (x (t); ˙x (t)) ` ˙x (t)fv(x (t); ˙x (t))”= 0 and f (x (t); ˙x (t)) ` ˙x (t)fv(x (t); ˙x (t)) = constant (40)

The curve of shortest length. We have f (y ; v ) =p1 + v2 so that fy(y ; v ) = 0 fv(y ; v ) = v p 1 + v2 The Euler’s equation 39 becomes

d dx y0_{(x )} p 1 + (y0_{(x ))}2 ! = 0

therefore y00(x )p1 + (y0_{(x ))}2_{` y}0_{(x )p}y0(x )y00(x ) 1+(y0_(t))2 1 + (y0_{(x ))}2 = = y00(x ) 1 (1 + (y0_{(x ))}2₎p_{1 + (y}0_{(x ))}2 = 0 whence y00(x ) = 0 and y (x ) = ax + b

The Brachistochrone We have f (y ; v ) = p 1 + v2 p (h ` y ) By 40 p 1 + (y0<sub>(x ))</sub>2 p (h ` y (x )) ` y0(x ) y 0<sub>(x )</sub> p (h ` y (x ))p1 + (y0_{(x ))}2 = constant We get 1 p (h ` y (x ))p1 + (y0<sub>(x ))</sub>2 = constant and h ` y (x ) = k (1 + (y0_{(x ))}2₎

We got an equation in the form:

y (x ) = f (y0(x )) (41)

with f smooth.

Deriving both sides of the equation we get y0(x ) = f0(y0(x ))y00(x ) i.e.

1 = f

0_(y0_{(x ))y}00_{(x )} y0_{(x )}

integrating x = c + Z x x0 f0_(y0_(t))y00_(t) y0_(t) dt = c + Z y0(x ) y0_(x 0) f0_{(s )} s ds = F (y 0 (x ))

Assuming that F is invertible we get y (x ) = f (F`1(x )) and if we set F`1(x ) = p, we have

( x = F (p) = c +R_pp 0 f0_{(s )} s ds y = f (p)

Setting F`1(x ) = ’(t), we have a more general parametric representation in the form

( x = F (’(t)) = c +R_’(t’(t) 0) f0(s ) s ds y = f (’(t))

and integrating by substitution ( x = c +R_tt 0 f0_{(’(s ))} ’(s ) ’(s )ds˙ y = f (’(t))

Therefore if we set z (x ) = h ` y (x ) we have z0(x ) = k (1 + (z0<sub>(x ))</sub>2<sub>)</sub> so that f (p) = k 1 + p2 and ( x (t) = `kR_tt 0 2 (1+ffi2_{(s ))}2ffi 0_{(s )ds + c0} z (t) = _1+ffik2_(t)

If we set

ffi(t) = cot t = 1= tan t we have x (t) = k Z t t0 2 cos2sds = x0+ k Z t t0 (1 ` cos 2s )ds + c0 = = k (t `1 2sin 2t) + c y (t) = h ` k cos2t = h ` k1 ` cos(2t) 2 so that ( x (t) = k (t ` 1₂sin 2t) + c y (t) = h ` k1`cos(2t)₂

We can satisfy the boundary conditions y (a) = h and y (b) = 0, with suitable choices for constants k and c .

Indeed if t = 0 we have

x (0) = a ; y (0) = h

so that , for c = a,

(x (0); y (0)) = (a; h)

moreover we can choose ¯t 2 [0; ı] in such a way that

’(¯t) = ¯ t ` 1<sub>2</sub>sin(2¯t) 1 2(1 ` cos(2¯t)) = 2¯t ` sin(2¯t) (1 ` cos(2¯t)) = b ` a h and if we set k = h 1 2(1 ` cos(2¯t)) we get 8 < : x (¯t) = 1 h 2(1`cos(2¯t)) (¯t ` 1 2sin 2¯t) + a = b ` a + a = b y (¯t) = h ` 1 h 1`cos(2¯t) = h ` h = 0

Therefore 8 < : x (t) = 1 h 2(1`cos(2¯t)) (t ` 1₂sin 2t) + a y (t) = h ` 1 h 2(1`cos(2¯t)) 1`cos(2t) 2 ; t 2 [0; ¯t] ; ¯t 2 [0; ı]

satisfies both the differential equation and the boundary conditions. It represents an arc of cycloid which is the curve traced by a point fixed on a circumference which rolls along a straight line.

The existence and the uniqueness of ¯t 2 [0; ı] such that ’(¯t) = ¯ t ` 1<sub>2</sub>sin(2¯t) 1 2(1 ` cos(2¯t)) = b ` a h can be proved as follows. We have

lim

Moreover ’0(t) = = 2(1 ` cos(2t)) 2<sub>` sin(2t)(2t ` sin(2t))</sub> (1 ` cos(2t))2 Let

N(t) = (1 ` cos(2t))2<sub>` sin(2t)(2t ` sin(2t)) =</sub> = 2(1 ` cos(2t) ` 2t sin(2t))

we have N(0) = N(ı) = 0 and

We have N0(t) – 0 if and only if ( tan(2t) – 2t ; t 2 [0; ı=2] tan(2t) » 2t ; t 2 [ı=2; ı]

Therefore there is a unique ˆt 2 [ı=2; (3=4)ı] such that N0(t) – 0 in [0; ˆt] and N0(t) » 0 in [ˆt; ı]

As a consequence N(t) > 0 for all t 2 (0; ı) and ’ is strictly increasing.

Moon Landing

Let us consider a spacecraft at an altitude h0 from the moon

surface which is falling down with initial velocity v0.

The spacecraft moves downward under the sole effect of lunar gravity g

Its mass is M and it carries a quantity F of fuel, so that its initial mass is m0= M + F

the thrust vector u is less that ¸ and fuel consumption is ku

The spacecraft altitude at time t is h(t)

The spacecraft at time t is directed downward and is v (t) The mass of spacecraft at time t is m(t)

We want to reach the moon surface with zero speed so that the fuel used is minimized.

the lunar module will move according to the following equations 8 > < > : ˙ h(t) = v (t) ˙ v (t) = `g + <sub>m(t)</sub>u(t) ˙ m(t) = `ku(t) 8 > < > : h(0) = h0 v (0) = v0 m(0) = m0 = M + F

We want to find T ,(h; v ; m) and u in such a way that they

Minimize

Z T

0

u(t)dt

under the constraints (

h(T ) = 0

v (T ) = 0 0 » u(t) » ¸

To solve this problem we need the Pontryagin’s maximum principle.

The control Problem Let f0 : Rnˆ Rk ! R F : Rnˆ Rk ! Rn be smooth functions Let x : [0; T ] ! Rn

piecewise continuous with its first derivative and u : [0; T ] ! Rk

piecewise continuous.

We want to determine T 2 R, x piecewise continuous with its first derivative and u piecewise continuous such that the minimum of the functional

Z T

0

f0(x (t); u(t))dt

is attained and the following constraints are satisfied ˙

x (t) = F (x (t); u(t))

x (0) = x0 ; x (T ) = x1

x is usually called state function u is usually called control function U is the control set.

Pontryagin’s maximum principle

Let (x˜_{; u}˜_{) be a solution of the control problem then there are} –0 2 R , – : R ! Rn such that, if we set

H(x ; u) = –0f0(x ; u) + h–; F (x ; u)i = –0f0(x ; u) + n X i =1 –iFi(x ; u) it must be ˙ x˜(t) = F (x˜(t); u˜(t)) x˜(0) = x0 x˜(T ) = x1 ˙ –(t) = `rxH(x˜(t); u˜(t)) H(x˜(t); u˜(t)) = max juj»¸H(x ˜ (t); u) –0 = `1 H(x˜(t); u˜(t)) = 0

Minimum time problem

The minimum time problem is a very simple control problem which is useful to understand how Pontryagin principle can be used.

Let us consider a material point P of mass m which moves along a straight line and let us call x (t) its position with respect to a fixed point O chosen on the line.

We can apply a force to the material point so that the motion equation is

¨

x (t) = u(t) ju(t)j » 1 (42)

where u(t) can be controlled .

We want to find u(t) in such a way that the material point reaches O in the minimum time, starting from a given initial position with initial given speed.

The corresponding control problem can be defined as follows.

Minimize

Z T

0

dt (43)

Over all piecewise differentiable functions x ; u such that ( ˙ x (t) = y (t) ˙ y (t) = u(t) ( x (0) = x0 y (0) = v0 ( x (T ) = 0 y (T ) = 0

With reference to the enunciation of the maximum principle f0((x ; y ); u) = 1 F ((x ; y ); u) =„y u « so that H(x ; u) = –0+ –y + —u (44)

By the maximum principle we have „_˙ – ˙ — « = `„ 0 – « =„Hx Hy « whence ( –(t) = h —(t) = `ht + k

Since H is linear with respect to u, it reaches its maximum value in u = ˚1; more precisely in u = 1 when — > 0 in u = `1 when — < 0 As —(t) = `ht + k

vanishes one and only one time, say in ¯t, even u can change sign

only one time. Therefore

u(t) = (

˚1 ; t 2 [0; ¯t]

When u = 1 we have ( ˙ x (t) = y (t) ˙ y (t) = 1 and ( x (t) = t2_{=2 + at + b} y (t) = t + a ( x (t) = (t+a) 2 2 + b ` a2 2 y (t) = t + a

The figure shows the trajectories of the material point when u = 1 y = v x x =y₂2+ (b −a2 2) u = 1

When u = `1 we have ( ˙ x (t) = y (t) ˙ y (t) = `1 and ( x (t) = `t2<sub>=2 + at + b</sub> y (t) = `t + a = `(t ` a) ( x (t) = `(t`a) 2 2 + b + a2 2 y (t) = `t + a = `(t ` a) The figure shows the trajectories of the material point when u = `1 y = v x x = −y₂2+ (b +a2 2) u = −1

y = v

x

The figure shows the trajectories of the material point when u = ˚1; x (t) represents the position and y (t) represents the speed of the material point. We have

x = ˚y2+ (b ˇ a

2 2)

y = v x  x(t) = t2_{/2 + at + b =}(t+a)2 2 + b − a2 2 y(t) = t + a  x(t) = −t2_{/2 + at + b =}−(t−a)2 2 + b + a2 2 y(t) = −t + a = −(t − a)

The origin can be achieved only by moving along one of the two trajectories that are traveled toward the origin.

We get trajectory ‚ when u = 1

We have y = t + a so that y and t have the same sign. We get trajectory ‹ when u = `1

We have y = `t + a so that y and t have opposite sign. We can reach the origin only traveling along one of these arcs.

y = v

x

δ

When (x0; v0) = (x0; y0) lies on ‚ or ‹ we choose

u(t) = ˚1 to reach the origin; Otherwise we travel along one of the two trajectories which contains (x0; y0) until it intersects ‚ or ‹ and from there we use ‚ or ‹ to reach the origin

The figure shows the optimal trajectories.

y = v

x δ

Moon landing problem

Let us recall that we consider a spacecraft at an altitude h0 from

the moon surface which is falling down with initial velocity v0. The spacecraft altitude at time t is h(t)

The spacecraft at time t is directed downward and is v (t) The mass of spacecraft at time t is m(t)

The spacecraft moves downward under the sole effect of lunar gravity g Its mass is M and it carries a quantity F of fuel, so that its initial

mass is m0 = M + F

the thrust vector u is less that ¸ and fuel

consumption is ku We want to reach 0 altitude with 0 speed with minimum fuel consumption. Therefore h(0) = h0 v (0) = v0 m(0) = m0 = M + F m(t)¨h(t) = `m(t)g + u(t) ˙ m(t) = `ku(t) 0 » u(t) » ¸

The fuel consumption in the time interval [0; t] is

k Z t

We can set a control problem as follows: we have to find T ,(h; v ; m) and u such that

Minimize

Z T

0

u(t)dt (45)

under the constraints 8 > < > : ˙ h(t) = v (t) ˙ v (t) = `g + <sub>m(t)</sub>u(t) ˙ m(t) = `ku(t) 8 > < > : h(0) = h0 v (0) = v0 m(0) = m0 = M + F (46) ( h(T ) = 0 v (T ) = 0 0 » u(t) » ¸

With reference to the maximum principle we have f0((h; v ; m); u) = u F ((h; v ; m); u) = 0 @ v `g + u m `ku 1 A so that H((h; v ; m); u) = –0u + –v + ”(`g + u m) + —(`ku) = = (–0+ ” m ` k —)u + –v ` ”g (47) and Hh = 0 Hv = – Hm = ` ”u m2

Using maximum principle we get 8 > < > : ˙ –(t) = 0 ˙ ”(t) = `–(t) ˙ —(t) = ”(t)u(t)_m2_(t)

Moreover u must be chosen in such a way that

max u2[0;¸] H((h; v ; m); u) = max u2[0;¸] = (–0+ ” m ` k —)u + –v ` ”g

Since H is linear in u, it reaches its maximum when u = 0 or when u = ¸; according to the sign of

–0+ ”

m` k —

More precisely we have:

u = 0 when –0+_m” ` k — < 0

When u = 0 we have free fall. If h(0) = ¯h ; v (0) = ¯v ; m(0 = ¯m we have 8 > < > : ˙ h(t) = v (t) ˙ v (t) = `g ˙ m(t) = 0 da cui 8 > < > : h(t) = `1₂gt2_{+ ¯v t + ¯h} v (t) = ¯v ` gt m(t) = ¯m

Substituting t = v ` v¯ g in h = `1 2gt 2_{+ ¯v t + ¯h} we get h = ` 1 2gv 2₊ „ ¯ h + 1 2gv¯ 2 « h v u = 0

When u = ¸ with h(0) = ¯h ; v (0) = ¯v ; m(0) = ¯m we have 8 > < > : ˙ h(t) = v (t) ˙ v (t) = `g + <sub>m(t)</sub>¸ ˙ m(t) = `k ¸ and m(t) = ¯m ` k ¸t m t M + F F kα m(t) = M + F − kαt m(t) = ¯m − kαt M Since M + F ` k ¸t – ¯m ` k ¸t – M it must be 0 » t » F