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 [1] [2] ), 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 [2] .

Notice that the principal part of the investigated equation possesses complicated characteristic, not multiple characteristics as in [3] .

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 [2] ).

‖ 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 .

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 [4] 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 [5] ), 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.

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.

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