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.