UDC 513.88 : 517.941

MATHEMATICS

Yu. I. LYUBICH

INVESTIGATION OF THE DEFECT OF AN ABSTRACT CAUCHY PROBLEM

(Presented by Academician S. N. Bernstein on 21 V 1965)

Consider the abstract Cauchy problem

[
dx(t)/dt=Ax(t)\quad (0<t<\infty),\qquad x(0)=x_0,
\tag{1}
]

where (A) is a closed linear operator in a Banach space (\mathfrak B). The set of those vectors (x_0) for which problem (1) has a solution* (strongly continuously differentiable for (t>0) and strongly continuous for (t=0)) is called the initial manifold and is denoted by (I_A). In [1] some criteria are given for the density of the manifold (I_A) in the whole space under the condition that the domain (D_A) of the operator (A) is dense. If (D_A) is not dense, then the manifold (I_A) is all the more not dense, as is shown by the following simple

Lemma 1. The inclusion always holds* *

[
\overline{I}_A\subseteq \overline{D}_A .
\tag{2}
]

Indeed, if (x_0\in I_A), then (x_0=\lim\limits_{t\to 0}x(t)), where (x(t)) is the corresponding solution. But (x(t)\in D_A) ((t>0)).

Relation (2) gives reason to introduce the following general

Definition. The dimension of the quotient space (\overline{D}_A/\overline{I}_A) is called the defect of the Cauchy problem (1) and is denoted by (d_A).

Now the subject of the present article is clear from its title.

Recall that, for any subspace (\mathcal L), the dimension of the quotient space (\mathfrak B/\mathcal L) is called the defect of the subspace (\mathcal L) and is denoted by (\operatorname{def}\mathcal L).

Lemma 2. If the resolvent set of the operator (A) is nonempty, then the inequality holds

[
\dim\left(\overline{D}{A^k}/\overline{D}_A}}\right)\leq \operatorname{def}\overline{D
\quad (k=1,2,\ldots).
\tag{3}
]

Proof. If ({f_i}1^p) is a linearly independent system of linear functionals on (\overline{D}}), orthogonal to (\overline{D{A^{k+1}}), then ({(R\mu^k)^*f_i}1^p) is a linearly independent system of linear functionals on all of (\mathfrak B), orthogonal to (\overline{D}_A). Here (R\mu) denotes, as usual, the resolvent of the operator (A), i.e. (R_\mu=(A-\mu E)^{-1}); (\mu) is some point of the resolvent set.

We can now establish the following central result:

Theorem 1. Let some half-plane (\operatorname{Re}\lambda\geq a) belong to the resolvent set of the operator (A), and in this half-plane

[
|R_\lambda|=O(1+|\lambda|^m)
\tag{4}
]

for some (m\geq 0). Then**

[
d_A\leq [m+1]\operatorname{def}\overline{D}_A .
\tag{5}
]

* Not necessarily unique.
* The bar denotes closure.
** ([\,]) denotes the integer part.

This theorem is, in essence, a generalization of Theorem 2 ((^1)) and is proved in a similar way.*

Proof. Let (\operatorname{Re}\mu>\alpha). Take an arbitrary vector (x) and put

[
u(t)=-\frac{1}{2\pi i}\int_{\alpha-i\infty}^{\alpha+i\infty}
\frac{e^{\lambda t}R_\lambda x}{(\lambda-\mu)^{[m+2]}}\,d\lambda
\qquad (t\geqslant 0).
]

The vector-function (u(t)) is strongly continuous. Put

[
u_h(t)=\frac{1}{h}\int_t^{t+h}u(\tau)\,d\tau
=\frac{1}{2\pi i}\int_{\alpha-i\infty}^{\alpha+i\infty}
\frac{e^{\lambda h}-1}{\lambda}\,
\frac{e^{\lambda t}R_\lambda x}{(\lambda-\mu)^{[m+2]}}\,d\lambda
\qquad (h>0).
]

The vector-function (u_h(t)), for every (h>0), is a solution of problem (1) with (x_0=u_h(0)). Therefore (u_h(0)\in I_A). Consequently, (u(0)\in \overline{I}A). But
(u(0)=R\mu^{[m+2]}x). In view of the arbitrariness of the vector (x),

[
\overline{D}_{A^{[m+2]}}\subseteq \overline{I}_A .
]

Hence, by Lemma 2,

[
d_A\leqslant \dim(\overline{D}A/\overline{D})}
=\sum_{k=1}^{[m+1]}\dim(\overline{D}{A^k}/\overline{D})}
\leqslant [m+1]\operatorname{def}\overline{D}_A,
]

as was required.

Theorem 1 is sharp first of all in the sense that the following is true.

Theorem 2. For any integers (m\geqslant 0), (p>0), there exists an operator (A) satisfying the conditions of Theorem 1 such that

[
\operatorname{def}\overline{D}_A=p,\qquad d_A=(m+1)p.
\tag{6}
]

Moreover, the following is true.

Theorem 3. Let (\rho(r)>1\;(r\geqslant 0)) be an arbitrary function growing faster than any power in the sense that (r^{-m}\rho(r)\to\infty) ((r\to\infty)), whatever (m>0). Then there exists an operator (A), having no spectrum in the half-plane (\operatorname{Re}\lambda\geqslant 0), such that

[
|R_\lambda|\leqslant \rho(|\lambda|)
\qquad (\operatorname{Re}\lambda\geqslant 0)
\tag{7}
]

and at the same time
[
\operatorname{def}\overline{D}_A=1,\qquad d_A=\infty.
]

The proof of Theorem 2 is reduced to the case (p=1) with the aid of the following simple proposition.

Lemma 3. Let closed linear operators (A_1,A_2) act in Banach spaces (\mathfrak{B}_1,\mathfrak{B}_2), respectively. Form the direct sums (\mathfrak{B}=\mathfrak{B}_1+\mathfrak{B}_2), (A=A_1+A_2). Then

[
\operatorname{def}\overline{D}A
=\operatorname{def}\overline{D}
+\operatorname{def}\overline{D}{A_2},
\qquad
d_A=d.}+d_{A_2
\tag{8}
]

Proof of Theorem 2 for the case (p=1). Let
(\mathfrak{B}=C^{m+1}[0,1])—the space of functions (\varphi(s)) ((0\leqslant s\leqslant 1)) having a continuous ((m+1))-st derivative;

[
|\varphi|=\max_{0\leqslant n\leqslant m+1}\left(\max_{0\leqslant s\leqslant 1}|\varphi^{(n)}(s)|\right).
]

Let (A) be the differentiation operator (-d/ds) on functions having a continuous ((m+2))-nd derivative and satisfying the boundary condition (\varphi(0)=0). Obviously,
(\operatorname{def}\overline{D}_A=1).

The operator (A) has no spectrum: for any (\psi\in\mathfrak{B}) and any (\lambda),

[
R_\lambda\psi(s)=-e^{-\lambda s}\int_0^s \psi(\tau)e^{\lambda\tau}\,d\tau
\qquad (0\leqslant s\leqslant 1).
]

* Theorem 2 ((^1)) corresponds to the case (\operatorname{def}\overline{D}_A=0). Then (d_A=0).

Hence, by integration by parts, one obtains the estimate

[
\max_{0\leq s\leq 1}|R_\lambda\psi(s)|
\leq
\frac{1}{|\lambda|}
\left(2\max_{0\leq s\leq 1}|\psi(s)|+\max_{0\leq s\leq 1}|\psi'(s)|\right)
\quad(\operatorname{Re}\lambda>0).
\tag{9}
]

But, since

[
(R_\lambda\psi(s))^{n}
=
-\lambda(R_\lambda\psi(s))^{(n-1)}-\psi^{(n-1)}(s)
]

[
(n=1,2,\ldots,m+1;\ 0\leq s\leq 1),
]

it follows, in view of (9), that

[
\max_{0\leq s\leq 1}|(R_\lambda\psi(s))^{(n)}|
=
O\left(|\psi|(1+|\lambda|^{n-1})\right)
]

[
(n=0,1,\ldots,m+1;\ \operatorname{Re}\lambda>0),
]

i.e. condition (4) is satisfied in the half-plane (\operatorname{Re}\lambda>0).

At the same time, the solution of problem (1) with initial function (x_0=\xi(s)) has the form
[
x(t)=\xi(s-t)\max(0,\operatorname{sign}(s-t)).
]
In order that (\xi(s)\in L_A), it is necessary and sufficient that
[
\xi^{(n)}(0)=0\quad(n=0,1,\ldots,m+2).
]
Therefore
[
\Gamma_A={\xi\mid \xi^{(n)}(0)=0\ (n=0,1,\ldots,m+1)},
]
whence it is clear that (d_A=m+1).

We now prove Theorem 3. Let ({M_n}_0^\infty) be a sequence of positive numbers ((M_0=1,\ M_1\geq 1)) satisfying the conditions

[
\sum_{k=0}^{\infty}\frac{r^k}{M_k}\leq \rho(r)\quad(r\geq 0);\qquad
M_{p+q}\geq M_pM_q\quad(p,q=0,1,2,\ldots).
\tag{10}
]

It is easy to construct it inductively. Consider the Banach space (C{M_n}) of infinitely differentiable functions (\varphi(s)) ((0\leq s\leq 1)) with norm

[
|\varphi|=\sup_n \frac{1}{M_n}\max_{0\leq s\leq 1}|\varphi^{(n)}(s)|.
]

In this space, take the operator (\widetilde A=-d/ds) on those functions (\varphi(s)) for which (\varphi'(s)\in C{M_n}), (\varphi(0)=0).

Let (\psi(s)\in C{M_n}). Then the equation

[
-\varphi'(s)-\lambda\varphi(s)=\psi(s)\quad(0\leq s\leq 1)
]

under the condition (\varphi(0)=0) has the unique solution

[
\varphi(s)=-e^{-\lambda s}\int_{0}^{s}\psi(\tau)e^{\lambda\tau}\,d\tau.
\tag{11}
]

This solution belongs to (C{M_n}) in the half-plane (\operatorname{Re}\lambda\geq 0), since, recursively,

[
\max_{0\leq s\leq 1}|\varphi^{(n)}(s)|
\leq
|\lambda|^n\max_{0\leq s\leq 1}|\psi(s)|
+
\sum_{k=1}^{n}|\lambda|^{\,n-k}\max_{0\leq s\leq 1}|\psi^{(k-1)}(s)|
\quad(n=0,1,2,\ldots),
]

whence, by (10),

[
|\varphi|
\leq
|\psi|\sum_{k=0}^{n}\frac{|\lambda|^k}{M_k}
\leq
\rho(|\lambda|)|\psi|.
]

We see that the operator (\widetilde A) has no spectrum in the half-plane (\operatorname{Re}\lambda\geq 0), and its resolvent satisfies inequality (7). When is (\operatorname{def}\overline{D}_{\widetilde A}=1)? This question appears difficult (see, for example, (2), pp. 240–243). We shall avoid it by considering, instead of the operator (\widetilde A), a certain restriction (\bar A). Let (\Pi) be the linear manifold of all polynomials. Obviously, (\Pi\subset C{M_n}). Put
[
\Pi_0=\Pi\cap{\varphi\mid \varphi(0)=0}.
]
The operator (\widetilde A) everywhere on (\Pi_0)

is defined and maps (\Pi_0) into (\Pi). Restrict it to (\Pi_0) and close it. This gives a certain operator (A \subset \widetilde A), acting in the Banach space (\mathfrak B = \overline{\Pi} \subset C{M_n}). Obviously, (\overline{D}_A = \overline{\Pi}_0) and, therefore, (\operatorname{def}\overline{D}_A = 1). Further, if (\xi(s) \subset I_A), then (\xi^{(n)}(0)=0) ((n=0,1,2,\ldots)) and, consequently, (d_A=\infty). It remains to verify that, when the operator (\widetilde A) is restricted to the operator (A), no residual spectrum appears. For this it is enough to prove that if in (11) the function (\psi) is a polynomial, then (\varphi \subset \Pi). This assertion, in turn, reduces to the inclusions (s^k e^{-\lambda s} \subset \overline{\Pi}) ((k=0,1,2,\ldots;\ \operatorname{Re}\lambda \geqslant 0)). But the point is that the expansion of (e^{-\lambda s}) in a Taylor series, as is not difficult to verify, converges in the space (C{M_n}), and the expansions of the remaining functions (s^k e^{-\lambda s}) are obtained by repeated differentiation with respect to (\lambda). Thus Theorem 3 is proved.

The author expresses gratitude to V. A. Tkachenko for a useful discussion.

Kharkov State University
named after A. M. Gorky

Received
19 V 1965

References

1. Yu. I. Lyubich, DAN, 155, No. 2 (1964).
2. S. Mandelbrojt, Adherent Series, IL, 1955.