OALibJOpen Access Library Journal2333-9705Scientific Research Publishing10.4236/oalib.1104631OALibJ-85547ArticlesBiomedical&Life Sciences Business&Economics Chemistry&Materials Science Computer Science&Communications Earth&Environmental Sciences Engineering Medicine&Healthcare Physics&Mathematics Social Sciences&Humanities Solvability of a Class of Operator-Differential Equations of Third Order with Complicated Characteristic on the Whole Real Axis AbdelBaset I. Ahmed1MohamedA. Labeeb1Egyptian Russian University, Badr City, Egypt050620180506152, May 201823, June 2018 26, June 2018© Copyright 2014 by authors and Scientific Research Publishing Inc. 2014This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/

On the whole real axis, we demonstrate sufficient conditions of regular solvability of third order operator-differential equations with complicated characteristics. These conditions were formulated only by the oper ator coefficients of the equation. In addition, by the principal part of the equation, the norms of the operators of intermediate derivative were estimated.

Operator-Differential Equation Hilbert Space Self-Adjoint Operator Intermediate Derivative Operator
1. Introduction

In a separable Hilbert space H, we have the following equation:

P u ( x ) ≡ p 0 u ( x ) + p 1 u ( x ) = f ( x ) , x ∈ R , (1)

where

p 0 u ( x ) = ( d d t − A ) ( d d t + A ) 2 u ( x ) . p 1 u ( x ) = ∑ s = 1 2   A s d 3 − s u ( x ) d x 3 − s ,

A is a self-adjoint positive-definite operator, and A s , s = 1 , 2 are generally linear unbounded operators. All derivatives are understood in the sense of distributions theory.

We consider f ( x ) ∈ L 2 ( R ; H ) , where

L 2 ( R ; H ) = { f ( x ) : ‖ f ( x ) ‖ L 2 ( R ; H ) 2 ( ∫ − ∞ + ∞ ‖ f ( x ) ‖ H 2 d t ) 1 2 < + ∞ }

(see   ), and u ( x ) ∈ W 2 3 ( R ; H ) , which are determined as follows:

W 2 3 ( R ; H ) = { u ( x ) : d 3 u ( x ) d x 3 ∈ L 2 ( R + ; H ) , A 3 u ( x ) ∈ L 2 ( R ; H ) } .

With the norm

‖ u ‖ W 2 3 ( R + ; H ) = ( ‖ d 3 u d x 3 ‖ L 2 ( R ; H ) 2 + ‖ A 3 u ‖ L 2 ( R ; H ) 2 ) 1 2 ,

See  .

Notice that the principal part of the investigated equation possesses complicated characteristic, not multiple characteristics as in  .

Definition 1. If for any f ( x ) ∈ L 2 ( R ; H ) there exists a vector function u ( x ) ∈ W 2 2 ( R ; H ) that satisfies Equation (1) almost everywhere in R, then it is called a regular solution of Equation (1)

Definition 2. If for any f ( x ) ∈ L 2 ( R ; H ) there exists a regular solution of Equation (1), and satisfies the inequality

‖ u ‖ W 2 3 ( R ; H ) ≤ c o n s t ‖ f ‖ L 2 ( R ; H ) , (2)

then Equation (1) is called regularly solvable.

It is known that if p 1 : W 2 3 ( R ; H ) → L 2 ( R ; H ) , then A 3 − s d s u ( x ) d x s ∈ L 2 ( R ; H ) , s = 1 , 2 .

And the following inequalities are valid (see  ).

‖ A 3 − s d s u ( x ) d x s ‖ L 2 ( R ; H ) ≤ c s ‖ u ‖ W 2 2 ( R ; H ) ,     s = 1 , 2. (3)

Definition 3. Parseval’s equality

∫ − ∞ + ∞ | f ( x ) | 2 d x = 1 2π ∫ − ∞ + ∞ | f ˜ ( ζ ) | 2 d ζ

where,

f ˜ ( ζ ) = ∫ − ∞ + ∞ f ( x ) e − i x ζ d x .

2. Main Results

Theorem 1. The operator P 0 is an isomorphism from the space W 2 3 ( R ; H ) to the space L 2 ( R ; H ) .

Proof. From (2), it is easy to prove that the operator P 0 acts from W 2 3 ( R ; H ) to L 2 ( R ; H ) be bounded. Using Fourier transforms for the equation P 0 u ( x ) = f ( x ) , we obtain

( − i ξ E − A ) ( − i ξ E + A ) 2 u ˜ ( ξ ) = f ˜ ( ξ ) . (4)

(E is the unit operator), where u ˜ ( ξ ) , f ˜ ( ξ ) are Fourier transform for the functions u ( x ) , f ( x ) , respectively. The operator pencil ( − i ξ E − A ) ( − i ξ E + A ) 2 is invertible and moreover

u ˜ ( ξ ) = ( − i ξ E − A ) − 1 ( − i ξ E + A ) − 2 f ˜ ( ξ ) , (5)

Hence,

u ( x ) = 1 2π ∫ − ∞ + ∞ ( − i ξ E − A ) − 1 ( − i ξ E + A ) − 2 f ˜ ( ξ ) e i ζ x d ζ . (6)

We show that u ( x ) ∈ W 2 3 ( R ; H ) . By using the Parseval equality and (3), we obtain:

‖ u ‖ W 2 3 ( R ; H ) 2 = ‖ d 3 u d t 3 ‖ L 2 ( R ; H ) 2 + ‖ A 3 u ‖ L 2 ( R ; H ) 2 = ‖ − i ζ 3 u ˜ ( ξ ) ‖ L 2 ( R ; H ) 2 + ‖ A 3 u ˜ ( ξ ) ‖ L 2 ( R ; H ) 2 = ‖ − i ζ 3 ( − i ξ E − A ) − 1 ( − i ξ E + A ) − 2 f ˜ ( ξ ) ‖ L 2 ( R ; H ) 2     + ‖ A 3 ( − i ξ E − A ) − 1 ( − i ξ E + A ) − 2 f ˜ ( ξ ) ‖ L 2 ( R ; H ) 2 ≤ sup ζ ∈ R ‖ − i ζ 3 ( − i ζ E − A ) − 1 ( − i ζ E + A ) − 2 ‖ H → H 2 ‖ f ˜ ( ζ ) ‖ L 2 ( R ; H ) 2     + sup ζ ∈ R ‖ A 3 ( − i ζ E − A ) − 1 ( − i ζ E + A ) − 2 ‖ H → H 2 ‖ f ˜ ( ζ ) ‖ L 2 ( R ; H ) 2 (7)

If σ ( A ) is a spectrum of the operator A, then we consider

sup ξ ∈ R ‖ − i ζ 3 ( − i ζ E − A ) − 1 ( − i ζ E + A ) − 2 ‖ H → H ≤ sup ξ ∈ R sup σ ∈ σ ( A ) | − i ζ 3 ( − i ζ − σ ) − 1 ( − i ζ + σ ) − 2 | = sup ξ ∈ R | ζ | 3 ( ζ 2 + σ 2 ) 3 2 ≤ 1 , (8)

sup ξ ∈ R ‖ A 3 ( − i ζ E − A ) − 1 ( i ζ E + A ) − 2 ‖ H → H ≤ sup ξ ∈ R sup σ ∈ σ ( A ) | σ 3 ( − i ζ − σ ) − 1 ( − i ζ + σ ) − 2 | = sup σ ∈ σ ( A ) σ 3 ( ζ 2 + σ 2 ) 3 2 ≤ 1 (9)

Taking into account (5) and (6) into (4) we obtain:

‖ u ‖ W 2 3 ( R ; H ) 2 ≤ 2 ‖ f ¯ ( ζ ) ‖ L 2 ( R ; H ) 2 = 2 ‖ f ( x ) ‖ L 2 ( R ; H ) 2 . (10)

Consequently, u ( x ) ∈ W 2 3 ( R ; H ) .

Applying Banach theorem on the inverse operator, we get that the operator P 0 is an isomorphism from W 2 3 ( R ; H ) to L 2 ( R ; H ) .

Now, we estimate the norms of intermediate derivative operators participating in the main part of the Equation (1) for finding exact conditions on regular solvability of the given equation, expressed only by its operator coefficients.

From theorem 1, we have that the norms ‖ p 0 u ‖ L 2 ( R ; H ) and ‖ u ‖ W 2 3 ( R ; H ) are equivalent in the space W 2 3 ( R ; H ) . Therefore by the norm ‖ p 0 u ‖ L 2 ( R ; H ) the theorem on intermediate derivatives is valid as well.

Theorem 2. Let u ( x ) ∈ W 2 3 ( R ; H ) . Then there hold the following inequalities:

‖ A 3 − s d s u ( x ) d x s ‖ L 2 ( R ; H ) ≤ a s ‖ p 0 u ‖ L 2 ( R ; H ) ,     s = 1,2. (11)

where a 1 = a 2 = 2 3 3 .

Proof. To establish the validity of inequality (11) we make change p 0 u ( x ) = f ( x ) and apply the Fourier transformation. We get

‖ A 3 − s ( − i ζ ) s ( − i ζ E − A ) − 1 ( − i ζ E + A ) − 2 f ¯ ( ζ ) ‖ L 2 ( R ; H ) ≤ sup ξ ∈ R ‖ A 3 − s ( − i ζ ) s ( − i ζ E − A ) − 1 ( − i ζ E + A ) − 2 ‖ H → H ‖ f ¯ ( ζ ) ‖ L 2 ( R ; H ) ,     s = 1,2. (12)

For ζ ∈ R we estimate the following norms:

‖ A 3 − s ( − i ζ ) s ( − i ζ E − A ) − 1 ( − i ζ E + A ) − 2 ‖ H → H ≤ sup σ ∈ σ ( A ) | σ 3 − s ( − i ζ ) s ( − i ζ − σ ) − 1 ( i ζ + σ ) − 2 | = sup σ ∈ σ ( A ) | σ − s ( − i ζ ) s ( − i ζ σ − 1 ) − 1 ( − i ζ σ + 1 ) − 2 | ≤ sup μ = ξ 2 σ 2 ≥ 0 μ s / 2 ( μ + 1 ) 3 / 2 = 1 3 3 s s / 2 ( 3 − s ) ( 3 − s ) / 2 ,     s = 1 , 2. (13)

Finally, from (12), we have

‖ A 3 − s ( − i ζ ) s ( − i ζ E − A ) − 1 ( − i ζ E + A ) − 2 f ˜ ( ζ ) ‖ L 2 ( R ; H ) ≤ a s ‖ f ˜ ( ξ ) ‖ L 2 ( R ; H ) ,     s = 1 , 2. (14)

Lemma. The operator P 1 continuously acts from W 2 3 ( R ; H ) to L 2 ( R ; H ) provided that the operators A s A − s , s = 1 , 2 are bounded in H.

Taking into account the results found up  to now we get possibility to establish regular solvability conditions of Equation (1).

Theorem 3. Let the operators A s A − s , s = 1 , 2 be bounded in H and it holds the inequality ∑ s = 1 2   a s ‖ A 3 − s A − ( 3 − s ) ‖ H → H ≺ 1 , where the numbers a s , s = 1 , 2 are determined in theorem 2. Then the Equation (1) is regularly solvable.

Proof. By theorem 1, provided that the operator P 0 has a bounded inverse operator P 0 − 1 acting from L 2 ( R ; H ) to W 2 3 ( R ; H ) , then after replacing p 0 u ( x ) = v ( x ) in Equation (1) can be written as ( E + p 1 p 0 − 1 ) v ( x ) = f ( x ) .

Now we prove under the theorem conditions (see  ), that the norm

‖ p 1 p 0 − 1 ‖ L 2 ( R ; H ) → L 2 ( R ; H ) < 1.

By theorem (2), we have:

‖ p 1 p 0 − 1 v ‖ L 2 ( R ; H ) = ‖ p 1 u ‖ L 2 ( R ; H ) ≤ ∑ s = 1 2 ‖ A s d 3 − s u d x 3 − s ‖ L 2 ( R ; H ) ≤ ∑ s = 1 2 ‖ A s A − s ‖ H → H ‖ A s d 3 − s u d x 3 − s ‖ L 2 ( R ; H ) ≤ ∑ s = 1 2   a 3 − s ‖ A s A − s ‖ H → H ‖ p 0 u ‖ L 2 ( R ; H ) = ∑ s = 1 2   a 3 − s ‖ A s A − s ‖ H → H ‖ v ‖ L 2 ( R ; H )

Consequently,

‖ p 1 p 0 − 1 ‖ L 2 ( R ; H ) → L 2 ( R ; H ) ≤ ∑ s = 1 2   a 3 − s ‖ A s A − s ‖ H → H ≺ 1.

Thus, the operator E + p 1 p 0 − 1 is invertible in L 2 ( R ; H ) and hence u ( x ) can be determined by u ( x ) = p 0 − 1 ( E + p 1 p 0 − 1 ) − 1 f ( x ) , moreover

‖ u ‖ W 2 3 ( R ; H ) ≤ ‖ p 0 − 1 ‖ L 2 ( R ; H ) → W 2 3 ( R ; H ) ‖ ( ( E + p 1 p 0 − 1 ) ) − 1 ‖ L 2 ( R ; H ) → L 2 ( R ; H ) ‖ f ‖ L 2 ( R ; H ) ≤ c o n s t ‖ f ‖ L 2 ( R ; H ) . (15)

The theorem is proved.

3. Conclusion

We formulated exact conditions on regular solvability of Equation (1), expressed only by its operator coefficients. We estimated the norms of intermediate derivative operators participating in the principle part of the given equation. In the case when in the perturbed part of Equation (1), the participant variable operator coefficients, i.e. A s ( x ) , s = 1 , 2 are linear operators, which determined for all x ∈ R , are investigated in a similar way.

Cite this paper

Ahmed, A.B.I. and Labeeb, M.A. (2018) Solvability of a Class of Operator-Differential Equations of Third Order with Complicated Characteristic on the Whole Real Axis. Open Access Library Journal, 5: e4631. https://doi.org/10.4236/oalib.1104631

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