Abstract
This paper investigates the wellposedness of a boundary value problem on the semiaxis for a class of thirdorder operatordifferential equations whose principal part has multiple real characteristics. We obtain sufficient conditions for the existence and uniqueness of the solution of a boundary value problem in the Sobolevtype space . These conditions are expressed in terms of the operator coefficients of the investigated equation. We find relations between the estimates of the norms of intermediate derivatives operators in the subspace and the solvability conditions. Furthermore, we calculate the exact values of these norms. The results are illustrated with an example of the initialboundary value problems for partial differential equations.
MSC: 34G10, 47A50, 47D03, 47N20.
Keywords:
wellposed and unique solvability; operatordifferential equation; multiple characteristic; selfadjoint operator; the Sobolevtype space; intermediate derivatives operators; factorization of pencils1 Introduction
The paper is dedicated to the formulation and study of the wellposedness of a boundary value problem for a class of thirdorder operatordifferential equations with a real and real multiple characteristic. Note that the differential equations whose characteristic equations have real different or real multiple roots find a wide application in modeling problems of mechanics and engineering, such as problems of heat mass transfer and filtration [1], dynamics of arches and rings [2], etc.
Suppose that H is a separable Hilbert space with a scalar product , , A is a selfadjoint positive definite operator on H (, , E is the identity operator), and () is the scale of Hilbert spaces generated by the operator A, i.e., , , . For we consider that , , . Here the operator is determined from the spectral decomposition of the operator A, i.e.,
where is the resolution of the identity for A.
By we denote the Hilbert space of all vectorfunctions defined on with values in H and the norm
Similarly, we define the space , where ,
Define the following spaces:
(for more details about these spaces, see [[3], Ch.1]). Here and further, the derivatives are understood in the sense of distributions (see [3]).
The spaces and become Hilbert spaces with respect to the norms
and
respectively.
Consider the following subspaces of :
By the theorem on intermediate derivatives, both of these spaces are complete [3].
Now let us state the boundary value problem under study.
In the Hilbert space H, we consider the following thirdorder operatordifferential equation whose principal part has multiple characteristic:
where , A is the selfadjoint positive definite operator defined above, and , are linear, in general, unbounded operators on H. Assuming , we attach to equation (1.1) boundary conditions at zero of the form
Definition 1.1 If the vector function satisfies equation (1.1) almost everywhere in , then it is called a regular solution of equation (1.1).
Definition 1.2 If for any there exists a regular solution of equation (1.1) satisfying the boundary conditions (1.2) in the sense of relation , and the following inequality holds:
then we say that problem (1.1), (1.2) is regularly solvable.
The solvability of boundary value problems for operatordifferential equations has been studied by many authors. Among such works, we should especially mention the papers by Gasymov, Kostyuchenko, Gorbachuk, Dubinskii, Shkalikov, Mirzoev, Jakubov, Aliev and their followers (see, e.g., [411]) that are close to our paper. Allowing to treat both ordinary and partial differential operators from the same point of view, these equations are also interesting from the aspect that the wellposedness of boundary value problems for them is closely related to the spectral theory of polynomial operator pencils [4] (for comprehensive survey, see Shkalikov [12]). And, of course, wellposed solvability of the Cauchy problem and nonlocal boundary value problems for operatordifferential equations as well as related spectral problems (see, e.g., Shkalikov [13], Gorbachuk and Gorbachuk [14], Agarwal et al.[15]) are also of great interest.
In this paper, we obtain conditions for the regular solvability of boundary value problem (1.1) (1.2), which are expressed only in terms of the operator coefficients of equation (1.1). We also show the relationship between these conditions and the exact estimates for the norms of intermediate derivatives operators in the subspaces and with respect to the norm of the operator generated by the principal part of equation (1.1). Mirzoev [16] was the first who paid detailed attention to such relation (for more details about the calculation of the norms of intermediate derivatives operators, see [17]). To estimate these norms, he used the method of factorization of polynomial operator pencils which depend on a real parameter. Further these results have been developed in [18,19].
It should be noted that all the abovementioned works, unlike equation (1.1), consider the operatordifferential equations with a simple characteristic. Although similar matters of solvability and related problems have already been studied for fourthorder operatordifferential equations whose principal parts have multiple characteristic (see, for example, [20,21], also [22] and some references therein), but they have not been studied for odd order operatordifferential equations with multiple characteristic, including those of thirdorder. One of the goals of the present paper is to fill this gap.
2 Equivalent norms and conditional theorem on solvability of boundary value problem (1.1), (1.2)
We show that the norm of the operator generated by the principal part of equation (1.1) is equivalent to the initial norm on the space .
Let denote the operator acting from the space to the space as follows:
Then the following theorem holds.
Theorem 2.1The operatoris an isomorphism between the spacesand.
Proof First, we note that if , then and if , then (see, e.g., [23]), where is the strongly continuous semigroup of bounded operators generated by the operator −A.
Obviously, the homogeneous equation has only the trivial solution in . But the equation for any has a solution of the form
where
In fact, such a solution satisfies the equation
and the conditions at zero , therefore, it belongs to (see, e.g., [4,5]).
Let us now show the boundedness of the operator . We have
Since ,
then (2.1) takes the form
Further, by theorems on intermediate derivatives and traces [[3], Ch.1], we obtain
Thus, the operator is bounded and bijective from the space to the space . Therefore, due to the Banach inverse operator theorem, is an isomorphism between these spaces. The theorem is proved. □
Corollary 2.2It follows from Theorem 2.1 that the normon the spaceis equivalent to the initial norm.
Before we state the conditional theorem on solvability of boundary value problem (1.1) (1.2), we prove the following lemma.
Lemma 2.3Let the operators, be bounded onHand let the operatoract from the spaceto the spaceas follows:
Proof Since for , by virtue of the intermediate derivatives theorem [[3], Ch.1], we obtain
The lemma is proved. □
Theorem 2.4Let the operators, be bounded onHand the following inequality hold:
where
Then boundary value problem (1.1), (1.2) is regularly solvable.
Proof We represent boundary value problem (1.1), (1.2) in the form of the operator equation , where , . Since by Theorem 2.1 the operator has the bounded inverse which acts from to , then, after the replacement , we obtain the equation in the space .
Under the conditions of Theorem 2.4, we obtain
Therefore, if the inequality holds, then the operator is invertible and we can define by the formula . Moreover,
The theorem is proved. □
Naturally, there arises a problem of finding exact values or estimates for the numbers , , and it is very important for extending the class of operatordifferential equations of the form (1.1) for which our boundary value problem is solvable. We will make the calculations for , in Section 4.
3 On spectral properties of some polynomial operator pencils and basic equalities for the functions in the space
Consider the following polynomial operator pencils depending on the real parameter β:
We need to clarify considering naturally arising pencils (3.1). Obviously, for , we obtain
then, as a result, we obtain
If we use the Fourier transform in (3.2), then
where is the Fourier transform of the function . Therefore, if , then
That is why to estimate , , it is necessary to study some properties of pencils (3.1).
The following theorem on factorization of pencils (3.1) holds.
Theorem 3.1Let. Then polynomial operator pencils (3.1) are invertible on the imaginary axis and the following representations are true:
where
here, , the numbers, are positive and satisfy the following systems of equations:
Proof It is clear that
are the characteristic polynomial operator pencils (3.1), where ( denotes the spectrum of the operator A). Let , . Then characteristic polynomials (3.4) satisfy the following relations:
Since
we obtain
for . Inequalities (3.5) imply that polynomials (3.4) have no roots on the imaginary axis for . Besides, it can be seen from (3.4) that each of the characteristic polynomial has exactly three roots in the left halfplane for . Since polynomials (3.4) are homogeneous with respect to the arguments λ and σ, then the following factorization is true for them:
where
, , and the numbers , are positive according to Vieta’s formulas and satisfy systems of equations (3.3) derived from (3.6) during the comparison of same degree coefficients. Further, using the spectral decomposition of the operator A, from equalities (3.6) we obtain the assertions of the theorem. The theorem is proved. □
Now, we state a theorem playing a significant role in the subsequent study. Let us introduce another notation, which will be used in the proof of that theorem: will denote the linear set of infinitely differentiable functions with values in and compact support in . As is well known, the space is everywhere dense in (see [[3], Ch.1]).
Theorem 3.2Let. Then, for any, the following relation holds:
where
Proof It suffices to prove the theorem for functions . We have
Integrating by parts, we obtain
Making similar calculations for , we obtain
From (3.8), taking into account (3.9) and applying Theorem 3.1, we get the validity of (3.7). The theorem is proved. □
whereis obtained fromby discarding the first two rows and columns, here.
4 On the values of the numbers ,
It is easy to check that the norms and are equivalent on . Then it follows from the theorem on intermediate derivatives [[3], Ch.1] that the following numbers are finite:
Proof Passing to the limit as in (3.10), we see that, for any function , the following inequality holds:
i.e., . We need to show that here we have the equality. To do this, it suffices to show that for any , there exists a function such that
Note that the procedure of constructing such functions is thoroughly described in [17] (in addition, the one for fourthorder equations with multiple characteristic is available in [20]). This method is applicable to our case, too. Therefore, we omit the respective part of the proof. So lemma is proved. □
Remark 4.2 Since , then , . Therefore, there arises the question: When do we have , ?
Denote by the root of the equation in the interval if such exists.
Theorem 4.3The following relation holds:
Proof If , then it follows from equation (3.11) that for any function and for all , the following inequality holds:
Now let us consider the Cauchy problem
Since by Theorem 3.1, for , the polynomial operator pencil is of the form , where , , then Cauchy problem (4.2)(4.4) has a unique solution , which can be expressed as:
where are uniquely determined by the conditions at zero (4.3), (4.4). Therefore, if we rewrite the inequality (4.1) for function , then for we will have
This means that , and therefore, for all . Now let for all . Then it follows from (3.11) that for any function and for all ,
Passing here to the limit as , we obtain that , and hence, .
Continuing the proof of the theorem, we suppose that . Then . Note that for , we have
Therefore, from (3.11) we find that for , the following inequality holds:
Applying this inequality again to the solution of Cauchy problem (4.2)(4.4), we obtain for all . On the other hand, it follows from the definition of that, for any , there exists a function such that
Taking into account this inequality in (3.11), we obtain
where . Therefore, there exists a vector such that
Thus, for . And since is a continuous function of the argument β in the interval , then . It follows from these arguments that the equation has a root in the interval . Now let have a root in the interval . This means that the inequality cannot be satisfied for any . Therefore, according to our earlier reasonings in the proof of this theorem, we have . Obviously, for the root of the equation , we have that , because the proof of the theorem for implies that . And since , we obtain . The theorem is proved. □
Remark 4.4 From Theorem 4.3, it becomes clear that to find the numbers , , we must solve the equations , , together with systems (3.3) respectively. In this case, it is necessary to take into account the properties of the numbers , , .
The following theorem holds.
Proof In view of Remark 4.4, in the case , we have due to the negativity of , despite . In the case , we have or . Then or , respectively. Therefore, . The theorem is proved. □
5 Solvability of boundary value problem (1.1), (1.2). Example
The results obtained above allow us to establish exact conditions for regular solvability of boundary value problem (1.1), (1.2). These conditions are expressed in terms of the operator coefficients of equation (1.1).
Theorem 5.1Let the operators, be bounded onHand the following inequality hold:
Then boundary value problem (1.1), (1.2) is regularly solvable.
Note that the above conditions for regular solvability of boundary value problem (1.1), (1.2) are easily verified in applications because they are expressed in terms of the operator coefficients of equation (1.1).
Let us illustrate our solvability results with an example of an initialboundary value problem for a partial differential equation.
Example 5.2 On the halfstrip , consider the problem
where , are bounded functions on and . Note that problem (5.1)(5.3) is a special case of boundary value problem (1.1), (1.2). In fact, here we have , and . The operator A is defined on by the relation and the conditions .
Applying Theorem 5.1, we obtain that under the condition
problem (5.1)(5.3) has a unique solution in the space .
Remark 5.3 Using the same procedure, we can obtain similar results for equation (1.1) on the semiaxis with boundary conditions or .
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors read and approved the final manuscript.
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