examples of growth of perturbations in chemical organizations

examples of growth of perturbations in chemical organizations

We will examine several simple examples of chemical systems where we start with one species
of molecules (or a

closed subset

of species) then intoruduce a

small perturbation

and evolve
the system using mass action dynamics. We want to know whether this perturbation will grow
and, if so, at what rate. Ultimately, we would like to link the behavor to some feature of
the reaction system, perhaps related to Rosen’s theory of M-R systems.

To get started, consider a

trivial

case,

A

+

B

→

B

+

B

. The system of equations which
describes this system is:

d

⁢

x

d

⁢

t

=

-

k

⁢

x

⁢

y

d

⁢

y

d

⁢

t

=

k

⁢

x

⁢

y

It is easy enough to solve this system. We begin by noting that

d

d

⁢

t

⁢

(

x

+

y

)

=

0

, hence

x

+

y

=

x

0

+

y

0

.
Substituting this back in to the second equation, we conclude
that

d

⁢

y

d

⁢

t

=

k

⁢

(

x

0

+

y

0

-

y

)

⁢

(

y

)

.

This equation can readily be solved to yield the implicit solution

k

⁢

t

=

1

x

0

+

y

0

⁢

log

⁡

y

x

0

+

y

0

-

y

⋅

x

0

y

0

which can be solved to produce the explicit solution

y

=

x

0

+

y

0

-

x

0

+

y

0

1

+

y

0

x

0

⁢

exp

⁡

(

k

⁢

(

x

0

+

y

0

)

⁢

t

)

.

Looking at the solution, we see that it starts out at

y

=

y

0

and grows
towards

y

=

x

0

+

y

0

as

t

→

∞

. This is as we expect — as
time goes on, whatever A’s there are left react with B’s to turn into B’s
until we are left with nothing but B’s.

If we suppose that, at the initial time

t

=

0

, there is only a tiny
proportion of B’s, i.e.

y

0

≪

x

0

, then we may make an

expansion

of the fraction and conclude that

y

grows exponentially for small
values of

t

:

1

1

+

y

0

x

0

⁢

exp

⁡

(

k

⁢

(

x

0

+

y

0

)

⁢

t

)

≈

1

-

y

0

x

0

⁢

exp

⁡

(

k

⁢

(

x

0

+

y

0

)

⁢

t

)

y

≈

y

0

x

0

⁢

exp

⁡

(

k

⁢

(

x

0

+

y

0

)

⁢

t

)

We can also come to this

conclusion

by bounding

y

without solving the
equation first. For a simple equation like this which is readily solved,
this is hardly needed but, for larger, more complicated equations, it
becomes important and this simple example can serve as a illustration of
the general technique.

Theorem 1

.

Let

C

be a

real number

such that

0

<

C

<

1

and let

x

0

and

y

0

be
strictly positive real numbers such that

0

<

y

0

<

C

⁢

(

x

0

+

y

0

)

.
Set

t

1

=

1

k

⁢

(

x

0

+

y

0

)

⁢

log

⁡

C

⁢

x

0

+

y

0

y

0

. Then
there exists a

function

f

:

[

0

,

t

1

)

→

[

0

,

C

⁢

(

x

0

+

y

0

)

)

such that

f

satisfies

the

differential equation

d

⁢

f

⁢

(

t

)

d

⁢

t

=

k

⁢

(

x

0

+

y

0

-

f

⁢

(

t

)

)

⁢

f

⁢

(

t

)

.

and, for all

t

∈

[

0

,

t

1

)

,

y

0

⁢

exp

⁡

(

(

1

-

C

)

⁢

k

⁢

(

x

0

+

y

0

)

⁢

t

)

≤

f

⁢

(

t

)

≤

y

0

⁢

exp

⁡

(

k

⁢

(

x

0

+

y

0

)

⁢

t

)

.

Proof.

By the

existence theorem

, there exists a positive real number

t

0

and a function

f

:

[

0

,

t

0

)

→

ℝ

such that

f

⁢

(

0

)

=

y

0

and

f

satifies the differential equation.
Since

f

⁢

(

0

)

=

y

0

<

C

⁢

(

x

0

+

y

0

)

, by continuity there exists
a positive real number

t

2

and a function

f

:

[

0

,

t

0

)

→

[

0

,

C

⁢

(

x

0

+

y

0

)

)

which satisfies the same differential equation with the same

initial condition

. Furthermore, we assume that

t

2

is maximal.

Starting with this condition

f

⁢

(

t

)

<

C

⁢

(

x

0

+

y

0

)

and
doing some algebra, we conclude that

(

1

-

C

)

⁢

k

⁢

(

x

0

+

y

0

)

≤

1

y

⁢

d

⁢

y

d

⁢

t

≤

k

⁢

(

x

0

+

y

0

)

Now,

1

y

⁢

d

⁢

y

d

⁢

t

=

d

d

⁢

t

⁢

(

log

⁡

y

)

so, by the mean
value theorem, we conclude that

(

1

-

C

)

⁢

k

⁢

(

x

0

+

y

0

)

⁢

t

≤

log

⁡

y

y

0

≤

k

⁢

(

x

0

+

y

0

)

⁢

t

.

Exponentiating,

y

0

⁢

exp

⁡

(

(

1

-

C

)

⁢

k

⁢

(

x

0

+

y

0

)

⁢

t

)

≤

y

≤

y

0

⁢

exp

⁡

(

k

⁢

(

x

0

+

y

0

)

⁢

t

)

.

We can ensure that the bound on

y

is satisfied if the condition

C

⁢

(

x

0

+

y

0

)

≥

y

0

⁢

exp

⁡

(

k

⁢

(

x

0

+

y

0

)

⁢

t

)

is met, which amounts to
demanding that

0

≤

t

≤

t

1

where

t

1

=

1

k

⁢

(

x

0

+

y

0

)

⁢

log

⁡

C

⁢

x

0

+

y

0

y

0

.

∎

Title

examples of growth of perturbations in chemical organizations

Canonical name

ExamplesOfGrowthOfPerturbationsInChemicalOrganizations

Date of creation

2014-05-31 15:55:20

Last modified on

2014-05-31 15:55:20

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

33

Author

rspuzio (6075)

Entry type

Example

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