On a functional equation

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Authors

D.V. Ionescu
Institutul de Calcul

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D.V. Ionescu, Sur une équation fonctionnelle. (French) Mathematica (Cluj) 1 (24) 1959 11–26.

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Journal

Mathematica Cluj

Publisher Name

Published by the Romanian Academy Publishing House

DOI

Print ISSN

1222-9016

Online ISSN

2601-744X

MR0144095

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1959-DVIonescu-On-an-equation

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TIBERIU POPOVICIU

ON A FUNCTIONAL EQUATION

by

DV IONESCU
in Cluj

In analyzing a paper by H. Löwner [1] on matrix functions, T. Popovici drew attention to the functional equation

\begin{matrix} (1) & Δ_{n, h} | f (x) | = | \begin{array}{lllll} f (x) & f (x + h) & \cdot & f (x + n h) \\ f (x + h) & f (x + 2 h) & \cdot & \cdot \\ \cdot & \cdot & \cdot & \cdot & \cdot \\ \cdot & ⋱ & \cdot & \cdot & \cdot \\ f (x + n h) & f (x + (n + 1) h) & \cdot & \cdot & \cdot \end{array} (x + 2 n h) \end{matrix}

for which solutions are being sought

f (x)

real and continuous regardless of

x

, and such that the functional equation is verified regardless of

x

And

h

, real.

Integrating the functional equation (1) presents rather significant difficulties for

n

any. We were able to overcome these difficulties in the case

n = 2

and in this work we provide the solution to this particular case.

§ 1. The case $n = 1$

It is obvious that the identically zero function is a solution of the functional equation (1). Setting aside this commonplace solution, we will consider equation (1), corresponding to $n = 1$

\begin{matrix} (2) & Δ_{1, h} [f (x)] = | \begin{array}{ll} f (x) & f (x + h) \\ f (x + h) & f (x + 2 h) \end{array} | = 0 \end{matrix}

Note that if the solution

f (x)

of the functional equation (2) has a zero

x_{0}

It boils down to the banal solution.

Indeed, the functional equation (2) gives for

x = x_{0}

f (x_{0} + h) = 0

regardless of

h

, and consequently

f (x) ≐ 0

It
follows that a solution of the functional equation (2) that is not trivial does not vanish and retains its sign regardless of

x

.

By asking

x = p h, f (p h) = t_{p}

Or

h

is a number, and

p

for any integer, the functional equation becomes
or
(3')

\begin{matrix} (3) & | \begin{array}{ll} f_{p} & f_{p + 1} \\ f_{p + 1} & f_{p + 2} \end{array} | = 0 \\ f_{p} f_{p + 2} = f_{p + 1}^{2} \end{matrix}

It therefore follows that

g_{p} = \log | f_{p} |

checks the recurrence equation
whose solution is

g_{p + 2} - 2 g_{p + 1} + g_{p} = 0

g_{p} = g_{0} + (g_{1} - g_{0}) p

Equation (3) therefore has the solution
(4)

\log | f_{p} | = \log | f_{0} | + p \log | \frac{f_{1}}{f_{0}} |

Let us determine this solution such that
(5)

f (0) = f_{0}, f (h) = f_{1}

Or

f_{0}

And

f_{1}

are given numbers.
By setting
(6)
we have

C_{1} = f_{0}

C_{1} e^{{has}_{1} h} = f_{1}

and equation (4) gives us

α_{1} h = \log \frac{f_{1}}{f_{0}}

(7)

f (p h) = C_{1} e^{{has}_{1} p h}

This formula is valid regardless of the integer.

p

It is also true when we replace

p

by a rational number

r

any.

Let us return to the considerations that led us to formula (7), starting from

h_{1} = \frac{h}{s}

, Or

s

is any natural number. We will have

\begin{matrix} (8) & f (p h_{1}) = C_{1}^{'} e^{{has}_{1}^{'} p h_{1}} \end{matrix}

whatever the integer

p

We can determine the constants

C_{1}^{'}

And

α_{1}^{'}

by the conditions

f (0) = f_{h}, f (s h_{1}) = f (h) = f_{1}

We will have the equations

\begin{matrix} C_{1}^{'} = f_{0} \\ C_{1}^{'} e^{a_{1}^{'} h} = f_{1} \end{matrix}

which coincide with equations (6), from which it follows that

C_{1}^{'} = C_{1}, α_{1}^{'} = α_{1} .

and consequently formula (8) becomes
or

\begin{matrix} f (\frac{p}{s} h) = C_{1} e^{a_{1} \frac{p}{s} h} \\ (9) & f (r h) = C_{1} e^{a_{r} r h} \end{matrix}

r

being any rational number.
The function

f (x)

being continuous regardless of

x

From equation (9) we deduce that the solution of the non-identically zero functional equation (2) is

\begin{matrix} (10) & f (x) = C_{1} e^{α_{1} x} \end{matrix}

Or

C_{1}

And

α_{1}

are constants that can be determined by conditions (5).

§ 2. The case $n = 2$ Considerations on the zeros of a solution $f (x)$ .

By discarding the simple solution, we note that any solution to the functional equation $Δ_{1, h} [f (x)] = 0$ , is also a solution of the functional equation $Δ_{2, h} [f (x)] = 0$ It is enough to check that

Δ_{2, h} [C_{1} e^{a_{1} x}] = 0

By also eliminating these solutions, we will look for solutions to the functional equation

\begin{matrix} (11) & Δ_{2, h} [f (x)] = | \begin{array}{lll} f (x) & f (x + h) & f (x + 2 h) \\ f (x + h) & f (x + 2 h) & f (x + 3 h) \\ f (x + 2 h) & f (x + 3 h) & f (x + 4 h) \end{array} | = 0 \end{matrix}

which are not solutions to the functional equation

Δ_{1, h} [f (x)] = 0

.
It is essential for the integration of the functional equation (11) to perform an analysis of the zeros of a continuous solution of this equation which is not identically zero.

Either

x_{0}

a point where

f (x_{0}) \neq 0

We can then circle

x_{0}

of an interval

(α, β)

Or

f (x) \neq 0

Several scenarios are possible:

1^{\circ}

The interval

(α, β)

is the interval

(- \infty, + \infty)

So we have

f (x) \neq 0

, regardless of

x

.

2^{\circ}

The interval

(α, β)

is the interval

(α, + \infty)

And

f (α) = 0

In this case, the solution

f (x)

cannot have any other zeros

α_{0} < α

.

Let's suppose the opposite, that is to say

f (α_{0}) = 0

, with

α_{0} < α

By choosing in the functional equation (11),

x = α_{0}

And

h = α - α_{0}

We have

| \begin{array}{lll} 0 & 0 & f (2 α - α_{0}) \\ 0 & f (2 α - α_{0}) & f (3 α - 2 α_{0}) \\ f (2 α - α_{0}) & f (3 α - 2 α_{0}) & f (4 α - 3 α_{0}) \end{array} | = 0

from which it follows that

f (α - α_{0}) = 0; f (x)

therefore has a zero

2 α - α_{0} > α

, which is impossible since in the interval (

α, + \infty

) We have

f (x) \neq 0

.

The solution

f (x)

Therefore, in this case, there is only one zero.

α

and we have

f (x) \neq 0

, For

x \neq α

.

3^{\circ}

The interval

(α, β)

is finished

α

And

β

are two zeros of the solution

f (x)

which we will call consecutive zeros. In this case, the solution

f (x)

has an infinite number of zeros

x_{p} = α + p h

., Or

h = β - α

, And

p

is any integer.

Indeed, let's do

x = α + p h

in the functional equation (11), where

h = β - α

and let's ask

We will have

f (α + p h) = f_{p}

By doing

p = 0

, We have

f_{0} = 0, f_{1} = 0

, and consequently

f_{2} = 0

then by doing

p = 1

we will have

f_{3} = 0, \dots

and so on.

Similarly, by doing

p = - 3

, We have

f_{0} = 0, f_{1} = 0

and consequently

j_{- 1} = 0

then by doing

p = - 4

, We have

f_{- 2} = 0, \dots

and so on.

It therefore follows that the solution

f (x)

has an infinite number of zeros

x_{p} = α + p h

These are the only zeros of

f (x)

.

For the integration of the functional equation (11) we need some auxiliary theorems.
3. Theorem A. If f(x) is a solution of the functional equation (11) that is not a solution of the functional equation (2), there exists a number h such that

f (r h)

And

F (\frac{h}{s})

be different from zero for any rational number

r

and for any natural number

s

, Or

\begin{matrix} (13) & F (h) = | \begin{array}{ll} f (0) & f (h) \\ f (h) & f (2 h) \end{array} | \end{matrix}

We will demonstrate this theorem by successively examining the cases where the solution

f (x)

has an infinite number of zeros, a single zero, or has no zeros.

1^{\circ}

The solution

f (x)

has an infinite number of zeros. The zeros being

x_{0} + p h

, Or

p

is any integer, let's place the origin at the point

x_{0}

.

The set

M = {r h}

Or

r

is any rational number, is countable. The complement of the set

M

contains no zeros in the solution

f (x)

. If

h^{'}

is an element of the complementary set and

r^{'}

is any non-zero rational number,

r^{'} h^{'}

is also an element of the complementary set, because otherwise we would have

h^{'} r^{'} = r h

, Or

h^{'} = r^{''} h

which is impossible,

h^{'}

not belonging to the set

M

.

The number

h^{'}

Having been thus chosen, we have

f (r^{'} h^{'}) \neq 0

, regardless of the rational number

r^{'}

.

We can add that

F (r^{'} h^{'}) = | \begin{array}{ll} f (0) & f (r^{'} h^{'}) \\ f (r^{'} h^{'}) & f {(2 r^{'} h)}^{'} \end{array} | = - f^{2} (r^{'} h^{'}) \neq 0

and consequently

F (\frac{h^{'}}{s}) \neq 0

whatever the natural number

s

Theorem A is therefore proven in this case .

2^{\circ}

The solution

f (x)

has a single zero

x_{0}

By placing the origin at the point

x_{0}

, And

h^{'}

given any non-zero number, we have

f (r^{'} h^{'}) \neq 0

regardless of the rational number

r^{'}

.

We will have, as above

F (r^{'} h^{'}) \neq 0

, Or

F (\frac{h^{'}}{s}) \neq 0

whatever the natural number

s

.

Theorem A is therefore proven in this case.

3^{\circ}

The solution

f (x) n^{'}

has no zeros. We have

f (x) \neq 0

regardless of

x

The number still needs to be chosen.

h^{'}

such as

F (r^{'} h^{'}) \neq 0

for any rational number

r^{'}

.

Let's do this in the functional equation (11)

x = p h

and let's ask

f_{p} = f (p h)

.

We will have

\begin{matrix} (14) & | \begin{array}{lll} f_{p} & f_{p + 1} & f_{p + 2} \\ f_{p + 1} & f_{p + 2} & f_{p + 3} \\ f_{p + 2} & f_{p + 3} & f_{p + 4} \end{array} | = 0 \end{matrix}

The equations can easily be deduced from this.

\begin{matrix} (15) & | \begin{array}{ll} f_{p} & f_{p + 1} \\ f_{p + 1} & f_{p + 2} \end{array} | | \begin{array}{ll} f_{p} & f_{p + 1} \\ f_{p + 2} & f_{p + 3} \end{array} | = 0 \end{matrix}

And

\begin{matrix} (16) & | \begin{array}{ll} f_{p + 2} & f_{p + 3} \\ f_{p + 3} & f_{p + 4} \end{array} | | \begin{array}{ll} f_{p + 1} & f_{p + 2} \\ f_{p + 3} & f_{p + 4} \end{array} | = 0 \end{matrix}

whatever the integer

p

Let us now
demonstrate that if

h_{1}

is a non-zero zero of

F (h)

, the function

F (h)

has an infinite number of zeros

h = p h_{1}

, Or

p

is any integer.

Let's ask

f (p h_{1}) = {\bar{f}}_{p}

Having

\begin{matrix} (17) & | \begin{array}{ll} \overset{―}{f_{0}} & \overset{―}{f_{1}} \\ \overset{―}{f_{1}} & \overset{―}{f_{2}} \end{array} | = 0, \end{matrix}

equations (15) and (16) give for

p = 0

And

p = - 2

\begin{matrix} (18) & | \begin{array}{ll} {\bar{f}}_{0} & {\bar{f}}_{1} \\ {\bar{f}}_{2} & {\bar{f}}_{3} \end{array} | = 0, | \begin{array}{ll} {\bar{f}}_{- 1} & {\bar{f}}_{0} \\ {\bar{f}}_{1} & {\bar{f}}_{2} \end{array} | = 0 \end{matrix}

We can determine the number

λ \neq 0

, by the equation

{\bar{f}}_{1} = λ {\bar{f}}_{0}

And then equations (17) and (18) give us:

\begin{aligned} \overset{―}{f_{2}} = λ {\bar{f}}_{1} \\ {\bar{f}}_{3} = λ {\bar{f}}_{2} \\ {\bar{f}}_{0} = λ {\bar{f}}_{- 1} \end{aligned}

From these relationships, we deduce that

| \begin{array}{ll} {\bar{f}}_{1} & {\bar{f}}_{2} \\ {\bar{f}}_{2} & {\bar{f}}_{3} \end{array} | = 0, | \begin{array}{ll} {\bar{f}}_{- 1} & {\bar{f}}_{0} \\ {\bar{f}}_{0} & {\bar{f}}_{1} \end{array} | = 0 .

Similarly, it is demonstrated that in general one has

{\bar{f}}_{p} = λ {\bar{f}}_{p - 1}

whatever the integer

p

The
resulting identities

\begin{aligned} (19) & | \begin{array}{ll} {\bar{f}}_{0} & {\bar{f}}_{p + 1} \\ {\bar{f}}_{p + 1} & {\bar{f}}_{2 p + 2} \end{array} | & = λ^{2} | \begin{array}{ll} {\bar{f}}_{0} & {\bar{f}}_{p} \\ {\bar{f}}_{p} & {\bar{f}}_{2 p} \end{array} | \\ (20) & | \begin{array}{ll} {\bar{f}}_{0} & {\bar{f}}_{- p - 1} \\ {\bar{f}}_{- p - 1} & {\bar{f}}_{- 2 p - 2} \end{array} | & = \frac{1}{λ^{2}} | \begin{array}{ll} {\bar{f}}_{0} & {\bar{f}}_{- p} \\ {\bar{f}}_{- p} & {\bar{f}}_{- 2 p} \end{array} | \end{aligned}

For

p \neq 0

.

Pous

p = 1

, the second member of formula (19) is zero,

h_{1}

being a zero of

F (h)

It therefore follows that

2 h_{1}

is also a zero of

F (h), \dots

and, so on, it is demonstrated that the

p h_{1}

are zeros of

F (h), p

being any natural number.

We also have identity

| \begin{array}{ll} {\bar{f}}_{0} & {\bar{f}}_{- 1} \\ {\bar{f}}_{- 1} & {\bar{f}}_{- 2} \end{array} | = \frac{1}{λ^{4}} | \begin{array}{cc} {\bar{f}}_{0} & {\bar{f}}_{1} \\ {\bar{f}}_{1} & {\bar{f}}_{2} \end{array} |

which shows that -

h_{1}

is a zero of

F (h)

and if we focus on identity (20)

p = 1, 2, \dots

we deduce that the

- p h_{1}

are zeros of

F (h)

,

p

being any natural number.
zeros.
Let's return to the function

F (h)

and note that

h = 0

is one of his.
It can happen that for

h > 0

, we have

F (h) \neq 0

In this case we will take any number

h^{'} > 0

and we will have

F (h^{'}) \neq 0

and also

F (\frac{h^{'}}{s}) \neq 0

whatever the natural number

s

Theorem A is proven in this case.

It can also happen that

F (h)

be different from zero in the interval

(0, h_{1})

Or

h_{1}

is a finite number and

F (h_{1}) = 0

We will then take a number

h^{'}

any interval (

0, h_{1}

) and we will have

F (h^{'}) \neq 0

, as well as

F (\frac{h^{'}}{s}) \neq\neq 0

, regardless of the rational number

s

Theorem A is also proven in this case.

However, it is possible that the origin is a point of accumulation of zeros of the continuous function.

F (h)

We will demonstrate that in this case, the function

F (h)

is zero for all h and that

f (x)

is a solution of the functional equation (2).

Indeed, either

h

any point and

ε

any positive number. The function

F (h)

being continuous at the point

h

we can determine the positive number

η

, such as
(21)

| F (\bar{h}) - F (h) | < ε

For
(

21^{'}

)

| \bar{h} - h | < η .

The number η being thus determined, either

h_{1}, 0 < h_{1} < η

, a zero of

F (h)

But then all the numbers

p h_{1}

are zeros of

F (h), p

being any integer. It can happen that

h

either in the form

h = p h_{1}

and in this case we have

F (h) = 0

If not, the number

h

will belong to an interval

(p h_{1}, (p + 1) h_{1})

length

h_{1} < η

By taking

\bar{h} = p h_{1}

From inequality (21) we deduce that

| F (h) | < ε

which shows that the function

F (h)

is identically zero.

Taking into account the equation

| \begin{array}{ll} f_{0} & f_{1} \\ f_{1} & f_{2} \end{array} | = 0

regardless of

h

Equations (15) and (16) give us

| \begin{array}{ll} f_{0} & f_{1} \\ f_{2} & f_{3} \end{array} | = 0, | \begin{array}{ll} f_{- 1} & f_{0} \\ f_{1} & f_{2} \end{array} | = 0

and by proceeding as above, we demonstrate that

| \begin{array}{ll} f_{1} & f_{2} \\ f_{2} & f_{3} \end{array} | = 0, | \begin{array}{ll} f_{- 1} & f_{0} \\ f_{0} & f_{1} \end{array} | = 0

and in general

\begin{matrix} (21) & | \begin{array}{ll} f_{p} & f_{p + 1} \\ f_{p + 1} & f_{p + 2} \end{array} | = 0 \end{matrix}

whatever the integer

p

But
then, this equation is identical to equation (3) and consequently it follows that

\begin{matrix} (22) & f (p h) = C_{1} e^{α_{1} p h} \end{matrix}

Equation (21) being valid regardless of

h

, we can substitute in formula (22)

h

by

\frac{h}{s}

, from which it follows that we have

f (r h) = C_{1} e^{a_{1} r h}

r

being any rational number.
The function

f (x)

being continuous, at every point

x

From this, we deduce that

f (x) = C_{1} e^{ε_{1} x}

which shows that the solution

f (x)

is also a solution of the functional equation
(2). This case was excluded from the beginning (nr. 2) and consequently Theorem A is completely proven.
4. Theorem B. Assuming that the number ha was chosen according to the requirements of Theorem A, all the determinants

\begin{matrix} (23) & | \begin{array}{ll} f_{p} & f_{p + 1} \\ f_{p + 1} & f_{p + 2} \end{array} | \end{matrix}

Or

f_{p} = f (p h)

, are different from zero regardless of the integer

p

To
demonstrate this, let us revisit equations (15) and (16), which we write in the form

\begin{matrix} (') & | \begin{array}{ll} | \begin{array}{ll} f_{p - 2} & f_{p - 1} \\ f_{p - 1} & f_{p} \end{array} | & | \begin{array}{ll} f_{p - 2} & f_{p - 1} \\ f_{p} & f_{p + 1} \end{array} | \\ | \begin{array}{ll} f_{p - 2} & f_{p - 1} \\ f_{p} & f_{p + 1} \end{array} | & | \begin{array}{ll} f_{p - 2} & f_{p} \\ f_{p} & f_{p + 2} \end{array} | = 0 \end{array} | = 0 \end{matrix}

And

| \begin{array}{ll} | \begin{array}{ll} f_{p} & f_{p + 1} \\ f_{p + 1} & f_{p + 2} \end{array} | & | \begin{array}{ll} f_{p - 1} & f_{p} \\ f_{p + 1} & f_{p + 2} \end{array} | \\ | \begin{array}{ll} f_{p - 1} & f_{p} \\ f_{p + 1} & f_{p + 2} \end{array} | & | \begin{array}{ll} f_{p - 2} & f_{p} \\ f_{p} & f_{p + 2} \end{array} | = 0 \end{array} | = 0

and suppose that the determinant

\begin{matrix} (24) & | \begin{array}{ll} f_{p - 1} & f_{p} \\ f_{p} & f_{p + 1} \end{array} | \end{matrix}

be different from zero, which is the case for

p = 1

Suppose that the determinant (23) is zero. Then
equation (16') shows us that

| \begin{array}{ll} f_{p - 1} & f_{p} \\ f_{p + 1} & f_{p + 2} \end{array} | = 0

and, by expanding the determinant,

| \begin{array}{lll} f_{p - 2} & f_{p - 1} & f_{p} \\ f_{p - 1} & f_{p} & f_{p + 1} \\ f_{p} & f_{p + 1} & f_{p + 2} \end{array} | = 0

based on the elements in the first column, we will have

f_{p} | \begin{array}{ll} f_{p - 1} & f_{p} \\ f_{p} & f_{p + 1} \end{array} | = 0

which is impossible since both factors are non-zero. The determinants (23) are therefore non-zero for

p = 1, 2, 3, \dots

Assuming that the determinant (24) is different from zero, which is the case for

p = 1

, let's demonstrate that the determinant

\begin{matrix} (25) & | \begin{array}{ll} f_{p - 2} & f_{p - 1} \\ f_{p - 1} & f_{p} \end{array} | \end{matrix}

is also different from zero.
Suppose, on the contrary, that the determinant (25) is zero. Equation (15') then shows us that we also have

| \begin{array}{ll} f_{p - 2} & f_{p - 1} \\ f_{p} & f_{p + 1} \end{array} | = 0

By developing the determinant

| \begin{array}{lll} f_{p - 2} & f_{p - 1} & f_{p} \\ f_{p - 1} & f_{p} & f_{p + 1} \\ f_{p} & f_{p + 1} & f_{p + 2} \end{array} | = 0

based on the elements of the last column we obtain

f_{p} | \begin{array}{ll} f_{p - 1} & f_{p} \\ f_{p} & f_{p + 1} \end{array} | = 0

which is impossible. Therefore, the determinants (23) are different from zero, for

p = 0, - 1, - 2, - 3, \dots

and consequently theorem B is completely proven.

§ 3. Integration of the functional equation (11).

Let's take the functional equation (11) and substitute $x$ by ph, where $h$ is a number chosen according to the requirements of Theorem A. We will have

\begin{matrix} (26) & | \begin{array}{lll} f_{p} & f_{p + 1} & f_{p + 2} \\ f_{p + 1} & f_{p + 2} & f_{p + 3} \\ f_{p + 2} & f_{p + 3} & f_{p + 4} \end{array} | = 0 \end{matrix}

whatever the integer

p

and according to Theorem B, the determinant

\begin{matrix} (27) & Δ_{p} = | \begin{array}{ll} f_{p} & f_{p + 1} \\ f_{p + 1} & f_{p + 2} \end{array} | \end{matrix}

which is different from zero, regardless of the integer

p

.

We determine the numbers

λ_{0}

And

λ_{1}

through equations

\begin{aligned} (28) & λ_{0} f_{0} + λ_{1} f_{1} = f_{2} \\ λ_{0} f_{1} + λ_{1} f_{2} = f_{3} \end{aligned}

which is possible since the determinant of the system is

Δ_{0} \neq 0

.

By doing

p = 0

In equation (26) and taking into account equations (28), we also deduce

\begin{matrix} (') & λ_{0} f_{2} + λ_{1} f_{3} = f_{4} \end{matrix}

Then, by doing this in equation (26)

p = 1

, and taking into account equations (28) and (28') and the fact that

Δ_{1} \neq 0

From this, we deduce
and... so on.

λ_{0} f_{3} + λ_{1} f_{4} = f_{5}

In general, we have the recurrence equation

\begin{matrix} (29) & λ_{0} f_{n} + λ_{1} f_{n + 1} = f_{n + 2} \end{matrix}

valid for

n = 0, 1, 2, \dots

If we now consider the equation

| \begin{array}{ccc} f_{- 1} & f_{0} & f_{1} \\ f_{0} & f_{1} & f_{2} \\ f_{1} & f_{2} & f_{3} \end{array} | = 0

and if we take into account equations (28), we deduce

λ_{0} f_{- 1} + λ_{1} f_{0} = f_{1}

which proves that equation (29) is also valid for

n = - 1

It is demonstrated step by step that it is also valid for

n = - 2, - 3, \dots

that is to say, it is valid regardless of the integer

n

.

The characteristic equation of the recurrence equation (29) is

\begin{matrix} (30) & ρ^{2} - λ_{1} ρ - λ_{0} = 0 \end{matrix}

By eliminating

λ_{0}

And

λ_{1}

Between equations (28) and (30) we find that the characteristic equation is

\begin{matrix} (') & | \begin{array}{ccc} 1 & ρ & ρ^{2} \\ f_{0} & f_{1} & f_{2} \\ f_{1} & f_{2} & f_{3} \end{array} | = 0 \end{matrix}

This equation is also encountered in the determination of numbers.

C_{1}, C_{2}, α_{1}, α_{2}

through equations

\begin{aligned} C_{1} + C_{2} = f_{0} \\ (31) & C_{1} e^{a_{1} h} + C_{2} e^{a_{2} h} = f_{1} \\ C_{1} e^{2 a_{1} h} + C_{2} e^{2 a_{2} h} = f_{2} \\ C_{1} e^{3 a_{1} h} + C_{2} e^{3 a_{2} h} = f_{3} \end{aligned}

Let's assume for a moment that the numbers

α_{1}

And

α_{2}

let's be different and eliminate

C_{1}

And

C_{2}

between these equations. We will have the equations

| \begin{array}{lll} 1 & 1 & f_{0} \\ e^{a_{1} h} & e^{a_{2} h} & f_{1} \\ e^{2 a_{1} h} & e^{2 a_{3} h} f_{2} \end{array} | = 0, | \begin{array}{ccc} 1 & 1 & f_{1} \\ e^{a_{1} h} & e^{a_{2} h} & f_{2} \\ e^{2 a_{1} h} & e^{2 a_{3} h} f_{3} \end{array} | = 0

Or

\begin{aligned} f_{0} e^{(α_{1} + α_{2}) h} - f_{1} (e^{a_{5} h} + e^{a_{3} h}) + f_{2} = 0 \\ (32) & f_{1} e^{(a_{1} + a_{5}) h} - f_{2} (e^{a_{1} h} + e^{a_{3} h}) + f_{3} = 0 \end{aligned}

By setting
(33)

μ_{1} = e^{a_{1} h}, μ_{2} = e^{a_{2} h}

we see that

μ_{1}

And

μ_{2}

are the roots of the equation

\begin{matrix} (34) & e^{(a_{1} + a_{2}) h} - (e^{a_{1} h} + e^{a_{2} h}) μ + μ^{2} = 0 \end{matrix}

By eliminating

e^{(a_{1} + a_{2}) h}, e^{a_{1} h} + e^{a_{2} h}

Between equations (32) and (34) we obtain the equation in

μ

\begin{matrix} (35) & | \begin{array}{ccc} 1 & μ & μ^{2} \\ f_{0} & f_{1} & f_{2} \\ f_{1} & f_{2} & f_{3} \end{array} | = 0 \end{matrix}

which is identical to the characteristic equation (

30^{'}

).
It follows that the roots of the characteristic equation

(30^{'})

are

ρ_{1} = e^{a_{1} h}, ρ_{2} = e^{a_{2} h}

assuming it has distinct roots.
The solution to the recurrence equation (29) is then
or

t_{n} = A_{1} ρ_{1}^{n} + A_{2} ρ_{2}^{n}

f_{n} = A_{1} e^{a_{1} n h} + A_{2} e^{a_{2} n h}

where the constants

A_{1}

And

A_{2}

are given by the equations

\begin{aligned} A_{1} + A_{2} = f_{0} \\ A_{1} e^{a_{1} h} + A_{2} e^{a_{2} h} = f_{1} \end{aligned}

By comparing this system with the first two equations (31), we deduce that
so that

A_{1} = C_{1}, A_{2} = C_{2}

\begin{matrix} (36) & f (n h) = C_{1} e^{a_{1} n h} + C_{2} e^{a_{2} n h} \end{matrix}

whatever the integer

n

.

If the characteristic equation (30') has coincident roots, equal to

ρ_{1}

, the solution of the recurrence equation (29) is

\begin{matrix} (37) & f_{n} = ρ_{1}^{n} (A + B n) \end{matrix}

Or

A

And

B

are constants.
The characteristic equation is encountered with the roots coincident. Also in the determination of numbers

C_{1}, C_{2}, α_{1}

through equations

\begin{matrix} (38) & \begin{array}{cc} C_{1} & = f_{0} \\ e^{α_{1} h} (C_{1} + C_{2} h) & = f_{1} \\ e^{2 α_{1} h} (C_{1} + 2 C_{2} h) & = f_{2} \\ e^{3 α_{1} h} (C_{1} + 3 C_{2} h) & = f_{3} \end{array} \end{matrix}

analogous to equations (31).
Indeed, by eliminating

C_{1}

Between these equations, we find the equations

\begin{aligned} e^{a_{1} h} C_{2} h = f_{1} - f_{0} e^{a_{1} h} \\ e^{2 a_{1} h} C_{2} h = f_{2} - f_{1} e^{a_{1} h} \\ e^{3 a_{1} h} C_{2} h = f_{3} - f_{2} e^{a_{1} h} \end{aligned}

By then eliminating

C_{2}

, we find the equations
or

\begin{matrix} f_{2} - f_{1} e^{a_{1} h} - e^{a_{1} h} (f_{1} - f_{0} e^{a_{1} h}) = 0 \\ f_{3} - f_{2} e^{a_{1} h} - e^{a_{1} h} (f_{2} - f_{1} e^{a_{1} h}) = 0 \\ f_{2} - 2 e^{a_{1} h} f_{1} + e^{2 a_{1} h} f_{0} = 0 \\ f_{3} - 2 e^{2 a_{1} h} f_{2} + e^{2 a_{1} h} f_{1} = 0 \end{matrix}

These two equations show that

e^{a_{1} h}

is a double root of the characteristic equation (

30^{'}

because the two equations

| \begin{array}{lll} 1 & ρ & ρ^{2} \\ f_{0} & f_{1} & f_{2} \\ f_{1} & f_{2} & f_{3} \end{array} | = 0, | \begin{array}{ccc} 0 & 1 & 2 ρ \\ f_{0} & f_{1} & f_{2} \\ f_{1} & f_{2} & f_{3} \end{array} | = 0

are verified by

ρ = e^{α_{1} h}

We can therefore
substitute into equation (37)

ρ_{1}

by

e^{a_{2} h}

and we will have

f_{n} = e^{n a_{1} h} (A + B n) .

The constants

A

And

B

are determined by the equations

\begin{matrix} A = f_{0} \\ e^{α_{3} h} (A + B) = f_{1} \end{matrix}

By comparing these equations with the first two equations (38), we deduce

A = C_{1}, B = C_{2} h .

and consequently the solution of the recurrence equation (29) is

\begin{matrix} (39) & f (n h) = e^{a_{1} n h} (C_{1} + C_{2} n h) \end{matrix}

Or

α_{1}, C_{1}

And

C_{2}

are given by equations (38).
6. Let us show that formulas (36) and (39) are also valid when we replace

n

by a rational number

r

any.

Indeed, if we replace

h

by

h_{1} = \frac{h}{s}

, Or

s

is any natural number, all previous calculations are valid. We will have

\begin{matrix} (40) & f (p h_{1}) = C_{1}^{'} e^{α_{1}^{'} p h_{1}} + C_{2}^{'} e^{α_{2}^{'} p h_{1}} \end{matrix}

Or

\begin{matrix} (41) & f {(p h_{1})}^{'} = e^{α_{1}^{'} p h_{1}} (C_{1}^{'} + C_{2}^{'} p h_{1}) \end{matrix}

depending on whether the corresponding characteristic equation has distinct or coincident roots, and where.

p

is any integer.

To determine the constants

C_{1}^{'}, C_{2}^{'}, α_{1}^{'} α_{2}^{'}

of formula (40), or

C_{1}^{'}, C_{2}^{'}, α_{1}^{'}

From formula (41) we can use systems of equations (31) or (38) where we substitute

h

by

h_{1}

But since this is the solution

f (x)

of the functional equation (11) determined by the conditions

f (0) = f_{0}, f (h) = f_{1}, f (2 h) = f_{2}, f (3 h) = f_{3}

and since equations (40) and (41) are valid for any integer

p

, we can form four successive equations to determine the constants, by giving to

p

the values

0, s, 2 s, 3 s

.

Since

s h_{1} = h

, these equations are
and

\begin{matrix} C_{1}^{'} + C_{2}^{'} = f (0) = f_{0} \\ C_{1}^{'} e^{α_{1}^{'} h} + C_{2}^{'} e^{α_{2}^{'} h} = f (h) = f_{1} \\ C_{1}^{'} e^{2 α_{1}^{'} h} + C_{2}^{'} e^{2 α_{2}^{'} h} = f (2 h) = f_{2} \\ C_{1}^{'} e^{3 α_{1}^{'} h} + C_{2}^{'} e^{3 α_{2}^{'} h} = f (3 h) = f_{3} \\ C_{1}^{'} = f (0) = f_{0} \\ e_{1}^{α_{1}^{'} h} (C_{1}^{'} + C_{2}^{'} h) = f (h) = f_{1} \\ e^{2 α_{1}^{'} h} (C_{1}^{'} + 2 C_{2}^{'} h) = f (2 h) = f_{2} \\ e^{3 α_{1}^{'} h} (C_{1}^{'} + 3 C_{2}^{'} h) = f (3 h) = f_{3} \end{matrix}

These systems of equations are identical to equations (31) and (38); we can then deduce for the first system

C_{1}^{'} = C_{1}, C_{2}^{'} = C_{2}, α_{1}^{'} = α_{1}, α_{2}^{'} = α_{2}

and for the second system

C_{1}^{'} = C_{1}, C_{2}^{'} = C_{2}, α_{1}^{'} = α_{1}

It follows that in formulas (40) and (41) we can replace

C_{1}^{'}, C_{2}^{'}

,

α_{1}^{'}, α_{2}^{'} par C_{1}, C_{2}, α_{1}, α_{2}

And

C_{1}^{'}, C_{2}^{'}, α_{1}^{'} par C_{1}, C_{2}, α

By also replacing

h_{1}

by

\frac{h}{s}

, and then

\frac{p}{s}

by

r

, we can write equations (40) and (41) in the form

\begin{matrix} (42) & f (r h) = C_{1} e^{a_{1} r h} + C_{2} e^{a_{2} r h} \end{matrix}

Or

\begin{matrix} (43) & f (r h) == e^{a_{1} r h} (C_{1} + C_{2} r h) . \end{matrix}

The functions

f (x), e^{a_{1} x}, e^{a_{2} x}, x e^{a_{1} x}

being continuous regardless of

x

, we deduce that, for

x

whatever, we have

\begin{matrix} (44) & f (x) = C_{1} e^{a_{1} x} + C_{2} e^{a_{2} x} \end{matrix}

Or

\begin{matrix} (45) & f (x) = e^{a_{1} x} (C_{1} + C_{2} x) \end{matrix}

In equation (44),

e^{α_{1} h}

And

e^{α_{2} h}

are the roots of the characteristic equation (

30^{'}

). These roots can be real or complex conjugates. In the latter case, the first two equations (31) give for

C_{1}

And

C_{2}

conjugate complex values.

By asking

\begin{array}{ll} α_{1} = β_{1} + i β_{2}, & α_{2} = β_{1} - i β_{2} \\ C_{1} = \frac{K_{1} - i K_{2}}{2}, & C_{2} = \frac{K_{1} + i K_{2}}{2} \end{array}

formula (44) becomes
or

f (x) = e^{β_{1} x} [\frac{K_{1} - i K_{2}}{2} (\cos β_{2} x + i \sin β_{2} x) + \frac{K_{1} + i K_{2}}{2} (\cos β_{2} x - i \sin β_{2} x)]]

\begin{matrix} (') & f (x) = e^{β_{1} x} [K_{1} \cos β_{2} x + K_{2} \sin β_{2} x] \end{matrix}

In summary, the real solutions, continuous regardless of

x

, of the functional equation (11) are of the form

\begin{aligned} f (x) = C_{1} e^{a_{1} x} + C_{2} e^{a_{2} x} \\ (46) & f (x) = e^{β_{1} x} (K_{1} \cos β_{2} x + K_{2} \sin β_{2} x) \\ f (x) = e^{a_{1} x} (C_{1} + C_{2} x) \end{aligned}

The results of this work were communicated to the Academy of the Romanian People's Republic. [2].

H. Löwner, Über monotone Matrixfunctionen Mat. Zeit. 38. 1934. p. 177-216 2. DV Ionescu, Integrarea ecuației funcționale

| \begin{array}{lll} f (x) & f (x + h) & f (x + 2 h) \\ f (x + h) & f (x + 2 h) & f (x + 3 h) \\ f (x + 2 h) & f (x + 3 h) & f (x + 4 h) \end{array} | = 0

Sesiunea științifică a Academiei RPR Filiala Cluj
Received March 15, 1958.

ON A LIMITATION PROBLEM THAT ARISES IN A STEAM BOILER PROJECT

by

C. KALIK

in Cluj

In his work [1], L. NÉMETI addresses a problem that arises when designing a tubular steam boiler with a forced passage. The author proposes a method for calculating the thermal stress in the tube walls. However, calculating the thermal stress requires knowing the temperature, which necessitates solving a boundary value problem. The aim of this work is to provide the solution to this boundary value problem.

First, we introduce the symbols used below and formulate the boundary value problem. Let

r, φ, z

cylindrical coordinates in three-dimensional space. The boiler tube is determined by the following inequalities:

r_{0} ⩽ r ⩽ r_{0} + s; 0 ⩽ φ ⩽ 2 π; - L ⩽ x ⩽ + L

Or

r_{0}

is the inner radius,

s

the thickness and

2 L

the length of the tube. The thermal field can be considered to be the same in each section of the tube with the plane

φ =

constant. The phenomenon becomes static at a certain point, therefore the function

u

which gives us the temperature values must satisfy the following partial differential equation:

\begin{matrix} (1.1) & A (u) = \frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial u}{\partial r}) + \frac{\partial^{2} u}{\partial z^{2}} = 0 \end{matrix}

at each point on the tube wall.
The boundary conditions are determined by the following data: outside the tube, a constant regime is maintained such that the heat flux is constant

Q

Let's assume that heat does not pass through the extremities

z = \pm L

of the tube, which means that here the

1959

On a functional equation

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About this paper

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ON A FUNCTIONAL EQUATION

DV IONESCU
in Cluj

§ 1. The case $n = 1$

§ 2. The case $n = 2$ Considerations on the zeros of a solution $f (x)$ .

§ 3. Integration of the functional equation (11).

ON A LIMITATION PROBLEM THAT ARISES IN A STEAM BOILER PROJECT

Related Posts

On a functional equation

Abstract

Authors

Keywords

Paper coordinates

PDF

About this paper

Journal

Publisher Name

DOI

Print ISSN

Online ISSN

References

Paper (preprint) in HTML form

ON A FUNCTIONAL EQUATION

DV IONESCU in Cluj

§ 1. The case n = 1 n = 1 n=1n=1

§ 2. The case n = 2 n = 2 n=2n=2Considerations on the zeros of a solution f ( x ) f ( x ) f(x)f(x).

§ 3. Integration of the functional equation (11).

ON A LIMITATION PROBLEM THAT ARISES IN A STEAM BOILER PROJECT

Related Posts

DV IONESCU
in Cluj

§ 1. The case $n = 1$

§ 2. The case $n = 2$ Considerations on the zeros of a solution $f (x)$ .