3. The Upper Bound for a Number of Periodic Solutions
Theorem 1. If the right-hand side of Equation (1) for each fixed t is a function increasing in and there exists a time point such that
strictly increases, then Equation (1) can have no more than one periodic solution.
Proof. Suppose Equation (1) has two periodic solutions and contrary to the statement of the theorem. Let us show these solutions have no common points. □
Since
, then there exists a point
such that
. Without loss of generality, it can be assumed that
and
Assume the contrary, namely, that the solutions and intersect. This means that there exists , such that . Let denote an exact lower bound of set . The continuity of functions and and condition (2) implies that .
From inequality (2) and the point
selection it follows that for
.
. By the hypothesis of the theorem, the function
increases with respect
for each fixed
, and, consequently, for
the following the inequality is realized
The subtraction
can be represented as:
From (3) it follows that
but due to (2) the following is obtained
which contradicts the selection of the point
.
Thus, for the solutions and do not intersect. These solutions do not have common points for , which follows from their -periodicity.
So, if Equation (1) has two
-periodic solutions and inequality (2) is realised, then for all
and, consequently, inequality (3) is realized for all
. And then
Moreover, since
is continuous and
is strictly increasing in
, in some neighborhood of the point
a strict inequality is realised:
therefore, we can assert that
Taking into account the last inequality, the following contradictory chain of relations is obtained:
This contradiction approves the assumption that Equation (1) can have two different -periodic solutions.
Example 1. After approving a single periodic solution for Equation (1), consider the case: The first derivative of the right side of this equation is positive and therefore, this equation can have at most one periodic solution. In our case this solution:
Theorem 2. If the right-hand side of (1) is a convex function for each fixed , and is strictly convex with respect to
for some , then Equation (1) can have no more than two different -periodic solutions.
Proof. First of all, let us proove that the Equation (1) has the property of existence and uniqueness for the solution under the conditions of the theorem. Assume for that purpose that if is convex and continuous, then it is a Lipschitz function. □
Indeed, let
be an arbitrary point of the plane, and let
and
be some positive numbers. Let us show that for the domain
there exists
, such that for all
and for all
In the proof of Theorem 14 [
31] it is noted that for each fixed
function
has a bounded right derivative, which is an increasing function with respect to
. Let us show that there exists
, such that for all
Since for each fixed function increases monotonically, it can take its extreme values only at the ends of the interval. Therefore, in order to prove the required inequality, it suffices to show that the functions and as are bounded.
Let us prove that
is bounded from above. Consider some
. Then, taking into account the fact that for any fixed
function
increases with respect to
, for all
The function
is continuous, and consequently,
and
are bounded on
. Thus, it can be concluded that there exists some constant
, such that for all
and all
But then for all
The fact that is bounded from below can be proved in a similar way.
Combining these two results and taking into account that for each function is increasing with respect to , the statement on the boundedness of the function on is valid becomes clear.
Consider two arbitrary points
and
in
and an arbitrary moment
in
. The function
is continuous on
and has a finite right derivative. But due to Theorem 12 [
33] it implies existence of points
, such that
Now, taking into account the boundedness of the function
on
and arbitrariness of
,
and
, it follows that for any
and arbitrary
which means that
is a Lipschitzian function.
Let the conditions of the theorem be satisfied. Let us show that in this case the Equation (1) can not have more than two
-periodic solutions. Assume, by contradiction, that the Equation (1) has three
-periodic solutions
,
and
. As shown above, the function
satisfies the Lipschitz condition, and consequently, the Equation (1) has the property of existence and uniqueness for solutions, and therefore, if for
then for any
Consider two obvious identities, which in view of (4) are valid for all
:
and
Subtracting the second equality from the first one, the following is obtained:
Let us represent the solution of Equation (1)
in the form:
By (4) it is obvious that for all .
Substituting the representation (6) into the right-hand side of (5), the following is obtained:
The function
is continuous, since the functions
and
are continuous and since it follows from inequality (4) and the definition of
that there exists a number
, such that for all
Moreover, taking into account the convexity of the function
, i.e., that for all
it can be concluded that for all
But, by the assertion of the theorem,
is strictly convex for
, which implies that as at the point
itself, as in some neighborhood of it, by continuity the following strict inequality is realized:
Thus, the function is continuous, non-negative and there is an interval which the function takes only positive values in.
Considering this, let the identity (5) be integrated in the range from
to
. Since the functions
,
and
are
-periodic, the following is obtained:
This contradiction shows that if the conditions of the theorem are satisfied, the Equation (1) cannot have three -periodic solutions.
Example 2. For the following statement, consider the case:
In this example, the second derivative is positive, and therefore this equation can have no more than two periodic solutions. These decisions: and
Theorem 3. If the right-hand side of the Equation (1) has a derivative continuous in the set of arguments, which is a function convex with respect to for each and there exists a moment such that is strictly convex, then the Equation (1) cannot have more than three -periodic solutions.
Proof. Let denote a solution of the Equation (1), that starts at the point at . Then is a point which this solution at gets in. Let an auxiliary function be introduced, considering that where prime mark means . □
Since the function
is defined and continuous, then the function
is also defined and continuous, and from the equation in variations it follows that for any
and
Thus, the function is specified correctly.
If it is additionally known that is strictly convex, then the theorem is very simple to prove.
Indeed, let the Equation (1) have four
-periodic solutions
beginning at points
,
. Let, for definiteness,
Let a function
be constructed. It is continuous at each interval
takes the same values at the ends of the interval, since
,
due to
-periodicity, and finally, it has a derivative at each point
. Thus, for the function
all the conditions of Rolle’s theorem are satisfied, and, consequently, there exist points
such that
This implies that
from which it is finally obtained that
So, if the Equation (1) has four -periodic solutions, then the function takes the same values at three different points, which contradicts the assumption of its convexity. Therefore, in order to prove the theorem, it is sufficient to show that the function is strictly convex on .
From the equation in variations it follows that
By stated conditions, the function
is convex with respect to
and, therefore, has a finite right derivative at any point
at each fixed
due to by Theorem 14 [
33]. The function
, by virtue of the existence and uniqueness of the solutions of Equation (1), strictly increases with respect to each fixed. It follows that the right derivative of the function, which we denote by, can be written in the form.
The function
strictly increases with respect to
at each fixed
by virtue of the existence and uniqueness of solutions of the Equation (1). This implies that the right derivative of the function
denoted here by
can be represents in the following form
where
denotes the right derivative of the function
with respect to the variable
at the point
. Then
On the set
the function
is bounded, since it increases with respect to
by virtue of the existence and uniqueness of the solutions of Equation (1), and, consequently,
where
and
are bounded on
as the continuous
-periodic solutions of the Equation (1). Then the function
is also bounded on
as a function continuous in the set of arguments, and this entails the boundedness of the function
on
.
Further, the proof of Theorem 2 implies that the function is bounded on , since it is a right derivative of a function convex in and continuous in the set of arguments. Thus, all the functions entering the right-hand side of (7) are bounded, and therefore, the function is bounded on .
Thus,
is a function continuous on
that has a bounded right derivative on this interval. Then, according to Theorem 14 [
33], in order to prove the strict convexity of the function
on
, it is sufficient to show that
strictly increases on
.
Let
be an arbitrary right-hand derived number of the function
at the point
,
be a sequence which this number is realized on, and
as
,
. Let us prove that
Indeed, considering that function
increases with respect to
,
and using the equation in variations, the following is obtained:
Suppose here that
and consider that
The limiting transition under the integral sign is admissible, since the right derivative of the function exists and it is finite on .
Repeating these arguments, for an arbitrary left derived number of the function
at a point
, the following estimation is obtained:
Since the function
is bounded on
then the function
is bounded below on
.
Thus, all the right derivatives of the function
are nonnegative on
, and all its left derived numbers are uniformly bounded below. Therefore, according to Theorem 9 [
32], the function
increases on
.
Let us show that
strictly increases on
. First of all, note that from the estimation obtained above
it follows that if the point
is not a root of the function
, then
and if there are no roots of the function
in a neighborhood of the point
, then the function
strictly increases in this neighborhood.
Suppose that
increases, but not strictly. Then there exists a segment
which the function
is constant on. But, as follows from the remark made above, this segment can be only zero, since if there is a neighborhood with
, then
increases strictly in it. Then,
Then for the right derivative of the function
at an arbitrary point
As mentioned above, the second term is , therefore, the first term denoted by I(x) is also equal to zero.
Estimating
from below, the following is realized:
where
Let us show that .
By virtue of the theorem on integral continuity for an arbitrarily chosen
consider
, such that for
for all
. It can assumed that the chosen
and
are also suitable for the solution
. The solutions of the Equation (1)
and
are bounded on
as periodic. Considering that
increases with respect to
for each fixed
, it follows that
is bounded on
. Therefore, for
considered above
is limited on
Further, the function
is continuous in the set of arguments by condition and, therefore, it is also bounded on
. But then there exists
, such that for any
,
and
Considering that
it follows that
The positiveness of
and fairness of equality
imply that
Consider a function
setting that
From the equality obtained above it follows that
It is easy to see that the left derivatives of the function are nonnegative. Besides, repeating for the left derivative of the function the arguments similar with ones for the right derivative of this function, it is obvious that the left derivative of the function is also bounded above. Thus, on the basis of Theorem 1 we conclude that the function is continuous on , and then it follows that is constant on .
due to uniqueness. Hence,
since the function
is convex by condition, then
increases by Theorem 14 [
33]. Moreover, by the condition of the theorem there exists a point
at which the function
is strictly convex, and therefore at this point
This inequality must also hold in some neighborhood of the point .
Indeed, if this is not so, then there exists a sequence
converging to
, such that for any
But then, for any
and for all
the following equality is realized:
Hence, in view of the continuity of the functions
and
in the set of arguments, it follows that
but this contradicts the strict convexity of the function
at the point
.
Considering the arguments given above, it follows that
This contradiction shows that there is no interval in which is constant, and consequently, strictly increases in . As noted above, strict increase of the function implies a strict convexity of the function on , which is incompatible with the assumption of the theorem that there exist four -periodic solutions of Equation (1). of the proof of the theorem
Let
be a function continuous in the set of arguments and
-periodic with respect to
. Consider
Then all the obtained results can be represented in a unified form.
Example 3. For the following statement, consider the case:
In this example, the third derivative is positive, and therefore this equation can have no more than three periodic solutions. These solutions to his statements consider the case: , and
Theorem 4. If for some function is continuous in the set of arguments and convex in for each fixed , and there exists a time point , such that is strictly convex, then Equation (1) can have no more than -periodic solutions.
4. Lower Bound of -Periodic Solutions Number
In [
3] V.A. Pliss constructed an example demonstrating that this series of theorems cannot be continued. Here we present another approach, which let obtain the information on a number of
-periodic solutions of the Equation (1).
Theorem 5. Let the right-hand side of the Equation (1) be such that equationhas solutions , with the following property: For any,
in the domain
function
is of constant sign, and the sign of the function changes when transferring to the neighboring domain. Then, if
then the Equation (1) has no less than
-periodic solutions. Here it is assumed that
,
.
Proof. Consider an arbitrary and, for definiteness, assume that in the domain and in . □
Therefore, for selected
in
and in
But then, as shown by H. W. Knobloch в [
30] и J. Mawhin в [
31], in the interval
there exists at least one
-periodic solution of the Equation (1). Since there are
such intervals, then Equation (1) has no less than
-periodic solutions.
Combining Theorem 5 with the statements in [
32,
33], a number of statements can be proved giving more exact information on the number of
-periodic solutions of the Equation (1). As and example, let us prove one such statement.
Example 4. Choose of arbitrary functions satisfying all the requirements of Theorem 5. Then the equationwill have at least periodic solutions. Theorem 6. Let the right-hand side of the Equation (1) satisfy all the requirements of Theorem 5, and for anyis a function increasing or decreasing inin the interval Then the Equation (1) has -periodic solutions.
Proof. Consider an arbitrary
and, for definiteness, assume that in the interval □
increases in for each fixed . Let us show that in the selected interval there is only one -periodic solution of the Equation (1).
Let, on the contrary, in the interval
there are two
-periodic solutions of the Equation (1)
and
. As shown in the proof of Theorem 1, these solutions do not intersect anywhere. Therefore, without loss of generality, it can be assumed that for all
which implies the fulfillment of inequality
by the assumption of the monotonicity of the function
Taking this fact into account, it can be concluded that inequality
To be periodic, the solution
from the domain above the curve
must necessarily transfer into the domain located under the curve
. This means that there exists a time point
such that
But by theorem condition
and a function
graph is the boundary of the domains of sign change for the function
. Therefore, in time point
the following inequality must be realized:
From this inequality and continuity of the function it follows that in (8) there is a sign of strict inequality, and this contradicts the assumption that the solutions and are -periodic.
Thus, the obtained contradiction shows that in the interval
there exists no more than one
-periodic solution of Equation (1). But, as follows from the Theorem 5, in this interval there necessarily exists at least one
-periodic solution of the Equation (1). Hence it can be finally concluded that in each interval
there is only one
-periodic solution of Equation (1). Since there are
such intervals, then the Equation (1) has exactly
-periodic solutions.
In particular, it follows from Theorem 5 that if the right-hand side of the Equation (1) is representable in the form
where
for
,
is of constant sign for
,
, then the Equation (1) has no less than
-periodic solutions, if
,
satisfy the requirements of Theorem 5.