Mathematics

Yu. I. Lyubich

On Conditions for Uniqueness of the Solution of an Abstract Cauchy Problem

(Presented by Academician S. N. Bernstein on 12 X 1959)

The abstract Cauchy problem, in the terminology of Hille ((^1)), consists in finding a vector function (x(t)) ((t \ge 0)) in a Banach space satisfying the equation

[
\frac{dx(t)}{dt}=Ax(t) \qquad (t>0)
\tag{1}
]

and the initial condition

[
x(0)=x_0.
\tag{2}
]

Here (A) is a given linear operator, (x_0) a given vector.

A vector function (x(t)) ((t \ge 0)) will be called a weak solution of problem (1)—(2) if it: 1) is weakly absolutely continuous and weakly differentiable almost everywhere on the half-axis (t>0); 2) satisfies equation (1) almost everywhere; 3) is weakly continuous at (t=0); 4) satisfies the initial condition (2). Replacing in this definition the word “weak” by the word “strong,” we obtain the definition of a strong solution.

Theorem 1. If on some ray (L) of the positive half-plane the spectrum of the operator (A) is absent, and if the resolvent (R_\lambda) of the operator (A) on the ray (L) is a function of finite degree, i.e.

[
\sigma \equiv \varlimsup_{\lambda\to+\infty}\frac{\ln|R_\lambda|}{\lambda}<\infty,
\tag{3}
]

then the Cauchy problem (1)—(2) cannot have two different weak solutions.

Proof. Let (x(t)) be a weak solution of the problem

[
\frac{dx(t)}{dt}=Ax(t) \qquad (t>0),
]

[
x(0)=0.
]

We shall show that

[
x(t)=0 \qquad (0\le t<\infty).
\tag{4}
]

Take an arbitrary linear functional (f) and form the function

[
\varphi(t,\lambda)=f[R_\lambda x(t)] \qquad (t\ge 0,\ \lambda\in L).
\tag{5}
]

This function on the half-axis (t>0) is an absolutely continuous solution of the equation

[
\frac{\partial \varphi(t,\lambda)}{\partial t}-\lambda\varphi(t,\lambda)=f[x(t)],
]

since

[
R_\lambda A \subset \lambda R_\lambda+E.
]

Moreover, (\varphi(t,\lambda)) is continuous at (t=0) and (\varphi(0,\lambda)=0). Consequently,

[
\varphi(t,\lambda)=\int_{0}^{t} f[x(t-\tau)] e^{\lambda \tau}\,d\tau .
\tag{6}
]

From formula (6) it is clear that, for fixed (t), (\varphi(t,\lambda)) is an entire function of (\lambda) of finite degree. Applying the well-known theorem of Pólya* ((^{2})), it is easy to show that the indicator diagram of the function (\varphi(t,\lambda)) is the smallest interval of the real axis containing all those values (\tau), (0\leqslant \tau\leqslant t), for which (f[x(t-\tau)]\ne 0). However, in view of (5) and (3), the indicator diagram of the function (\varphi(t,\lambda)) lies in the half-plane (\operatorname{Re}\tau\leqslant \sigma). Therefore, for (t>\sigma),

[
f[x(t-\tau)]=0 \qquad (\sigma\leqslant \tau\leqslant t),
]

i.e.

[
f[x(\tau)]=0 \qquad (0\leqslant \tau\leqslant t-\sigma).
]

By virtue of the arbitrariness of (t>\sigma) and of the linear functional (f), identity (4) holds. The theorem is proved.

The theorem proved is sharp in the following sense.

Theorem 2. Let (\rho(\lambda)) ((\lambda>0)) be a positive continuous function increasing as (\lambda\to+\infty) faster than any exponential, i.e. such that

[
\frac{\ln \rho(\lambda)}{\lambda}\longrightarrow +\infty .
]

Then there exists a Hilbert space and, in it, a linear operator (A) such that:

a) the positive half-axis contains no spectrum of the operator (A), and

[
|R_\lambda|\leqslant \rho(\lambda)
\tag{7}
]

for sufficiently large (\lambda>0);

b) the Cauchy problem

[
\frac{dx(t)}{dt}=Ax(t) \qquad (t>0),
\tag{8}
]

[
x(0)=0
\tag{9}
]

has a nontrivial strong solution.

Proof. As the operator (A) we take the differentiation operator (-d/ds) in the space (\mathscr L_{\alpha}^{2}(0,\infty)) of functions square-summable on the half-axis ([0,\infty)) with some positive measurable weight (\alpha(s)). The domain of definition of this operator consists of all absolutely continuous functions (X(s)\in \mathscr L_{\alpha}^{2}(0,\infty)) whose derivative also belongs to (\mathscr L_{\alpha}^{2}(0,\infty)).

In order to obtain a nontrivial strong solution of problem (8)—(9) for the chosen operator (A), it is enough to take any finite continuously differentiable function (X(\tau)) ((-\infty<\tau<\infty)), equal to zero on the negative half-axis, and put (x(t)=X(t-s)).

It remains to choose the weight so as to satisfy condition a). Put

[
\beta(\tau)=\sup_{s\geqslant 0}\frac{\alpha(s)}{\alpha(s+\tau)} \qquad (\tau\geqslant 0)
]

* See, for example, ((^{3})), pp. 113–116.

and require that for any (k>0) the inequality

[
\beta(\tau) \leq C(k)e^{-k\tau}
\tag{10}
]

hold with some constant (C(k)). We shall show that then, for (\lambda>0), the equation

[
AX-\lambda X=Y
\tag{11}
]

is uniquely solvable for any (Y\in \mathcal L_\alpha^2(0,\infty)).

The uniqueness of the solution follows from the fact that (e^{-\lambda s}\notin \mathcal L_\alpha^2(0,\infty)), which in turn follows from the inequality

[
\alpha(s)\geq \frac{\alpha(0)}{\beta(s)} \geq \frac{\alpha(0)}{C(2\lambda)}e^{2\lambda s}.
]

A solution of equation (11) is the function

[
X(s)=e^{-\lambda s}\int_s^\infty Y(\tau)e^{\lambda \tau}\,d\tau
=\int_0^\infty Y(s+\tau)e^{\lambda \tau}\,d\tau.
\tag{12}
]

That the function (12) belongs to (\mathcal L_\alpha^2(0,\infty)) is seen from the estimate

[
|X(s)|^2 \leq \int_0^\infty |Y(s+\tau)|^2 e^{(2\lambda+1)\tau}\,d\tau,
]

by virtue of which

[
\int_0^\infty |X(s)|^2\alpha(s)\,ds
\leq
\int_0^\infty \beta(\tau)e^{(2\lambda+1)\tau}\,d\tau
\int_0^\infty |Y(s)|^2\alpha(s)\,ds.
]

The last inequality also shows that, for the given operator (A),

[
|R_\lambda|\leq
\left[
\int_0^\infty \beta(\tau)e^{(2\lambda+1)\tau}\,d\tau
\right]^{1/2}
\quad(\lambda>0).
\tag{13}
]

Relying on estimate (13), we construct the weight (\alpha(s)) so as to ensure inequality (7). Of course, condition (10) must be observed in doing so. Put

[
M(\lambda)=\frac{(2\lambda+1)\rho^2(\lambda)}{e^{2\lambda+1}-1},
]

and let (\lambda_0>0) be so large that (\min_{\lambda\geq \lambda_0} M(\lambda)>1). Next put (\beta_0=1) and successively define the numbers (\beta_1,\beta_2,\ldots) so that

[
0<\beta_{N+1}<
\min_{\lambda\geq \lambda_0}
\left{
M(\lambda)-\sum_{n=0}^{N}\beta_n e^{(2\lambda+1)n}
\right}
e^{-(2\lambda+1)(N+1)}
\tag{14}
]

and, moreover, so that

[
\beta_{N+1}<\frac{\beta_N^2}{\beta_{N-1}},\qquad
\beta_N<e^{-N^2}\quad (N=1,2,\ldots).
\tag{15}
]

Introduce the step function (B(s)), setting

[
B(s)=\beta_N\qquad (N\leq s<N+1;\; N=0,1,2,\ldots).
]

and for the function (-\ln B(s)) we construct a convex majorant (K(s)) so that (K(0)=0). This is possible by virtue of (15).

Finally, put

[
\alpha(s)=e^{K(s)}.
]

Then

[
\frac{\alpha(s)}{\alpha(s+\tau)}
= e^{K(s)-K(s+\tau)}
\leqslant e^{-K(\tau)}
\leqslant B(\tau).
]

Consequently, for the weight thus chosen we shall have

[
\beta(\tau)\leqslant B(\tau),
]

whence, according to (13), (14), it follows that

[
|R_\lambda|\leqslant
\left[
\sum_{n=0}^{\infty}
\beta_n e^{(2\lambda+1)n}
\cdot
\frac{e^{2\lambda+1}-1}{2\lambda+1}
\right]^{1/2}
\leqslant \rho(\lambda).
]

The theorem is proved.

In conclusion we note that uniqueness of the solution of the Cauchy problem may fail also because of the presence, in the right half-plane, of arbitrarily distant points of the spectrum of the operator (A); an example is provided by the operator

[
-\frac{d}{ds}\quad \text{in } \mathscr{L}^{2}(0,\infty).
]

Kharkov State University
named after A. M. Gorky

Received
11 X 1959

REFERENCES

(^{1}) E. Hille, Proc. Int. Congr. of Math., Amst., 1954, 3, p. 109.
(^{2}) G. Pólya, Math. Zs., 29, 549 (1929).
(^{3}) B. Ya. Levin, Distribution of Zeros of Entire Functions, Moscow, 1956.