Suppose that H is a separable Hilbert space with a scalar product , , A is a self-adjoint non negative definite operator on H ( , E is the identity operator). By ( ), we denote the scale of Hilbert spaces generated by the operator A, i.e. , , . For we consider that , , .
By ( ), we denote the space of measurable functions (see  ) with values in H and the norm
is the space of all functions with values in H such that ,
with the norm
For more details about the space , see [  , Ch.1]. We assume that
For the whole article, all the derivatives are understood in the sense of distributions.
Consider the following problem on the semiaxis :
where , , .
Definition 1. If the vector function satisfies Equation (1) nearly everywhere in , then it is called a regular solution of Equation (1).
Definition 2. For any there exists a regular solution of Equation (1) satisfying the boundary Conditions (2) in the sense of relation
and the following inequality holds:
Then we say that say that problem (1), (2) is regularly solvable.
In this paper, the regular solvability of the initial boundary value problem (1), (2) is established.
In spite of the fact that, since 50th years of the last century, there are sufficiently many papers and books dedicated to the solvability of parabolic operator-differential equations in Banach spaces, in particular, Hilbert spaces, interest in such equations doesn’t abate until recently (see, for example,    ). However, in the mathematical literature studies on the parabolic operator-differential equations with multiple characteristics are almost not mentioned, although they have a wide application in some problems in mechanics and mathematical physics. Note that Equation (1) is an abstract parabolic equation in Hilbert space that has multiple characteristics.
And denote by the operator acting from the space to the
space as follows:
Then the following theorem holds
Theorem 1. The operator is an isomorphism between the spaces and .
Proof. Obviously, the equation , has only the trivial solution
. Really, the equation has a solution in the space
of the form , where the vectors . Taking into account conditions (??), we have and . Therefore, the equation
has only the trivial solution in the space .
Now we shall show that the equation for any
has a solution in the space . We extend the function to
zero for , then, applying the direct and inverse Fourier transform, it becomes clear that
Satisfies the equation almost everywhere in R. We show that . From Parseval’s equality we obtain:
where and are the Fourier transforms of the functions and , respectively. denotes the spectrum of the operator A. From the spectral decomposition of the operator A for , we have:
Taking into account (4) and (5) into (3) we obtain:
Therefore, . Now we denote the restriction of the function to by . Obviously, . By the theorem on traces [  , Ch.1] . Then we will find a solution to the equation of the form
where the vectors . For a clear determination of from conditions (2) we obtain that ,
It is clear that the operator is bounded and acts from the space to the space . Indeed, taking into account the theorem on intermediate derivatives [  , Ch.1], we have:
Thus, the operator is bijective and bounded. Hence, using the Banach inverse operator theorem, the operator is bounded. Therefore, the operator is an isomorphism between the spaces and .
The theorem is proved. □
Corollary 2. It follows from Theorem 1 that initial-boundary value problem (1), (2) is regularly solvable.
Corollary 3. It follows from Theorem 1 that the norms and are equivalent on the space .