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UDC 517.934
EIGENVALUE PROBLEMS FOR DIFFERENTIAL EQUATIONS OF THE SECOND ORDER IN BANACH AND HILBERT SPACES
G. I. LAPTEV
The present article is devoted to the consideration of the following problem of finding eigenfunctions and eigenvalues:
\[ \frac{d^2u}{dt^2}=Au-\lambda B(t)u \quad (0\leq t\leq T), \tag{0.1} \]
\[ \begin{aligned} L_1(u)&\equiv \alpha_{11}u_0+\alpha_{12}u'_0+\beta_{11}u_T+\beta_{12}u'_T=0,\\ L_2(u)&\equiv \alpha_{21}u_0+\alpha_{22}u'_0+\beta_{21}u_T+\beta_{22}u'_T=0. \end{aligned} \tag{0.2} \]
Here \(u(t)\) is the unknown function with values in a complex Banach space \(E\); \(A\) is a closed linear operator with everywhere dense domain of definition \(D(A)\) in the space \(E\), possessing a completely continuous inverse; \(B(t)\) is a certain family of operators, the conditions on which will be formulated below; \(\alpha_{ij}, \beta_{ij}\) \((i,j=1,2)\) are complex numbers.
In [1, 2] similar boundary-value problems without the parameter \(\lambda\) were studied. We shall make essential use of the results of these works and, in particular, shall apply the operator Green function obtained there. Let us recall that in the cited works it was assumed that the operator \(A\) has the property: for all \(\lambda\geq 0\) there exists, defined on the whole space \(E\), the operator \((A+\lambda I)^{-1}\), and
\[ \left\|(A+\lambda I)^{-1}\right\|\leq \frac{C}{1+\lambda}. \tag{0.3} \]
This assumption is retained here as well. The Green function for the equation
\[ \frac{d^2u}{dt^2}=Au-f(t) \tag{0.4} \]
and the regular boundary conditions (0.2) was obtained in [2] in the form of the determinant
\[ G(t,\tau)=D^{-1} \begin{vmatrix} U_1(t) & U_2(t) & U_0(t,\tau)\\ L_1(U_1) & L_1(U_2) & L_1(U_0)_t\\ L_2(U_1) & L_2(U_2) & L_2(U_0)_t \end{vmatrix}, \tag{0.5} \]
where \(U(t)\) is the semigroup constructed from the operator \(-A^{1/2}\),
\[ U_1(t)=U(t)A^{-1/2},\qquad U_2(t)=U(T-t)A^{-1/2}, \]
\[ U_0(t,\tau)=\frac{1}{2}U(|t-\tau|)A^{-1/2}, \tag{0.6} \]
and \(D^{-1}\) is an operator whose boundedness was established under certain conditions (see [2], Theorems 1 and 4).
§ 1. PROBLEMS IN A BANACH SPACE
Let us introduce for consideration the Banach space \(C([0,T];E)\) of functions \(f(t)\), continuous on the interval \([0,T]\), with values in \(E\), with the usual norm
\[ \|f\|_C=\max_{0\le t\le T}\|f(t)\|_E . \]
Lemma 1. Suppose that, for problem (0.1), (0.2), the following assumptions are satisfied:
1) the operator \(A^{-1}\) is completely continuous;
2) the boundary conditions (0.2) are regular;
3) for some \(0\le \alpha<1/2\), the function \(A^{-\alpha}B(t)\) is uniformly continuous on the interval \([0,T]\);
4) the number \(\lambda=0\) is not an eigenvalue of problem (0.1), (0.2), and let \(G(t,\tau)\) denote the Green’s function of the boundary-value problem under consideration.
Then the integral operator
\[ F(t)=\int_0^T G(t,\tau)B(\tau)f(\tau)\,d\tau \]
is completely continuous in the space \(C([0,T];E)\).
Proof. If in equation (0.1) we set \(\lambda=0\), then we obtain a homogeneous boundary-value problem which, by the hypothesis of the lemma, can have only the zero solution. Hence the Green’s function \(G(t,\tau)\) exists and is representable in the form of the determinant (0.5). In expanded form one obtains
\[ G(t,\tau)=Q_1(\tau)U_1(t)+Q_2(\tau)U_2(t)+U_0(t,\tau), \]
where \(Q_i(\tau)\) are certain operators. Since the boundary conditions (0.2) are regular, it follows from Theorem 1 in [2] that the operators \(Q_i(\tau)\) are bounded. Using formulas (0.6), we then have
\[ G(t,\tau)= \]
\[ =\left[Q_1U(t)+Q_2U(T-t)+\frac{1}{2}U(|t-\tau|)\right]A^{-1/2} \equiv H(t,\tau)A^{-1/2}. \tag{1.1} \]
Hence
\[ F(t)=\int_0^T H(t,\tau)A^{-1/2}B(\tau)f(\tau)\,d\tau . \tag{1.2} \]
The number \(\alpha\) in the conditions of the lemma is strictly less than \(1/2\), i.e. \(-\frac{1}{2}+\alpha<0\), and the operator \(A^{-1/2+\alpha}\) turns out to be completely continuous. We take it outside the integral sign in (1.2):
\[ F(t)=A^{-\frac12+\alpha}\left\{\int_0^T H(t,\tau)\,[A^{-\alpha}B(\tau)]f(\tau)\,d\tau\right\}. \tag{1.3} \]
The function \(A^{-\alpha}B(t)\) is uniformly continuous by assumption and, consequently, uniformly bounded: \(\|A^{-\alpha}B(t)\|\leq K\). If \(f(t)\) is an element of the bounded set in \(C([0,T];E)\)
\[ \|f\|_C\leq M, \tag{1.4} \]
then the function in braces in (1.3) will take its values in some bounded set \(E_0\) in the space \(E\) (recall that the function \(H(t,\tau)\) is uniformly bounded in the square \(0\leq t,\tau\leq T\)). The completely continuous operator \(A^{-\frac12+\alpha}\) maps \(E_0\) into a compact set of the space \(E\).
Let us also show that, under condition (1.4), the functions \(F(t)\) are equicontinuous. Then, to prove the theorem, it will suffice to use Arzelà’s theorem. For this purpose we shall have to use the inequalities following from the fact that the semigroup \(U(t)\) is analytic and is generated by the operator \(-A^{1/2}\) [3]:
\[ \|A^\alpha U(t)\|\leq \frac{C_1}{t^{2\alpha}},\qquad \|[U(t)-U(s)]A^{-\frac12+\alpha}\|\leq C_2|t-s|^{1-2\alpha}\quad (0<t,s<T). \]
Estimate the difference
\[ \begin{aligned} \|F(t)-F(s)\|_E &\leq \int_0^T \|[H(t,\tau)-H(s,\tau)]A^{-\frac12+\alpha}\|\cdot \|A^{-\alpha}B(\tau)\|\cdot \|f(\tau)\|\,d\tau \\ &\leq KM\int_0^T \|[H(t,\tau)-H(s,\tau)]A^{-\frac12+\alpha}\|\,d\tau . \end{aligned} \]
By formula (1.1),
\[ H(t,\tau)=Q_1U(t)+Q_2U(T-t)+\frac12\,U(|t-\tau|), \]
\[ H(s,\tau)=Q_1U(s)+Q_2U(T-s)+\frac12\,U(|s-\tau|). \]
Now one can subtract the second equality from the first term by term and estimate in norm each of the three differences obtained on the right-hand side (with the multiplier \(A^{-\frac12+\alpha}\)).
For example, for the first difference one obtains
\[ \|Q_1[U(t)-U(s)]A^{-\frac12+\alpha}\| \leq C_2\|Q_1\|\cdot |t-s|^{1-2\alpha}. \]
The second of the differences is estimated similarly. Finally, for the third we have
\[ \|[U(|t-\tau|)-U(|s-\tau|)]A^{-\frac12+\alpha}\| \leq \]
\[ \leq C_2\bigl||t-\tau|-|s-\tau|\bigr|^{1-2\alpha} \leq C_2|t-s|^{1-2\alpha}. \]
Finally,
\[ \|F(t)-F(s)\|\leq KMC_2\int_0^T\left(\|Q_1\|+\|Q_2\|+\frac{1}{2}\right)\times \]
\[ {}\times |t-s|^{1-2\alpha}\,d\tau=C|t-s|^{1-2\alpha}, \]
where the constant \(C\) does not depend on \(t\) and \(s\). Since \(\alpha<1/2\), it follows that \(1-2\alpha=\beta>0\), and the established inequality means the uniform continuity of the functions \(F(t)\).
The lemma is proved.
Theorem 1. Under the assumptions of Lemma 1, problem (0.1), (0.2) reduces to the problem of eigenfunctions of a completely continuous operator in the Banach space \(C([0,T];E)\).
Proof. In equation (0.1) denote \(\lambda B(t)u(t)=f(t)\). The solution of the problem is written with the aid of the Green function \(G(t,\tau)\) as
\[ u(t)=\int_0^T G(t,\tau)f(\tau)\,d\tau \]
or, substituting its value for \(f(t)\),
\[ u(t)=\lambda\int_0^T G(t,\tau)B(\tau)u(\tau)\,d\tau. \]
Application of the lemma completes the proof.
§ 2. HILBERT SPACE
In this section we shall assume that the original space is a Hilbert space. It will be denoted by \(H\). Let a function \(f(t)\) with values in \(H\) be defined on the interval \([0,T]\) and be Bochner integrable on this interval. Suppose that it also satisfies the condition
\[ \int_0^T \|f(t)\|^2\,dt<\infty. \]
The collection of such functions, which will be denoted by \(L_2([0,T];H)\), forms a Banach space ([4], Ch. III). It is not difficult to verify that if in this space one introduces the scalar product by the formula
\[ (f,g)_{L_2}=\int_0^T (f(t),g(t))_H\,dt, \]
then \(L_2([0,T];H)\) becomes a Hilbert space.
Lemma 2. Let \(K(t,\tau)\), for almost all \((t,\tau)\) from the square \([0,T;0,T]\), be a linear completely continuous operator in the space \(H\). If the function \(K(t,\tau)\) is Bochner integrable in the square \(0\leq t,\tau\leq T\) and
\[ \int_0^T\int_0^T \|K(t,\tau)\|^2\,dt\,d\tau<\infty, \]
then the linear operator
\[ \mathbf{K}u=\int_0^T K(t,\tau)u(\tau)\,d\tau \]
is completely continuous in the space \(L_2([0,T];H)\).
Proof. We may assume that the operator \(K(t,\tau)\) is completely continuous in \(H\) for all \(0\leq t,\tau\leq T\), and not merely almost everywhere, since sets of measure zero are of no interest to us.
Next, the function \(K(t,\tau)\) is Bochner integrable and therefore almost separably valued ([4], Ch. III). Without loss of generality it may be regarded as precisely separably valued. This means that there exists a sequence of completely continuous operators \(K(t_n,\tau_n)\) that is dense in the range of values of the function \(K(t,\tau)\) \((0\leq t,\tau\leq T)\).
Choosing an arbitrary \(\varepsilon>0\), set
\[ e_n=\{(t,\tau):\|K(t,\tau)-K(t_n,\tau_n)\|<\varepsilon\}. \tag{2.1} \]
It is clear that the sets \(e_n\) cover the square \(0\leq t,\tau\leq T\). With their aid we construct disjoint sets \(E_n=e_n-\sum_{k<n}e_k\), which, obviously, also cover the square \([0,T;0,T]\), and define the function
\[ K_\varepsilon(t,\tau)=K(t_n,\tau_n)\quad \text{for }(t,\tau)\in E_n . \]
It follows from (2.1) that throughout the square
\[ \|K(t,\tau)-K_\varepsilon(t,\tau)\|<\varepsilon . \tag{2.2} \]
It is convenient to represent the function \(K_\varepsilon(t,\tau)\) as the sum of a countable number of two-valued functions:
\[ K_\varepsilon(t,\tau)=\sum_{n=1}^{\infty}K^{(n)}(t,\tau),\quad \text{where }K^{(n)}(t,\tau)= \begin{cases} K(t_n,\tau_n), & (t,\tau)\in E_n,\\ 0, & (t,\tau)\in \overline{E}_n . \end{cases} \tag{2.3} \]
Denote by \(\mathbf{B}_2(H)\) the collection of operator-valued functions \(K(t,\tau)\) satisfying the condition
\[ \int_0^T\int_0^T \|K(t,\tau)\|^2\,dt\,d\tau<\infty . \]
Inequality (2.2) means that if \(K(t,\tau)\in\mathbf{B}_2(H)\), then \(K_\varepsilon(t,\tau)\in\mathbf{B}_2(H)\) and \(\|K-K_\varepsilon\|_{\mathbf{B}_2(H)}<\varepsilon T^2\). Hence, in turn, it follows that the series in (2.3) converges to the function \(K_\varepsilon(t,\tau)\) in the norm of \(\mathbf{B}_2(H)\). Indeed, the sets \(E_n\) are disjoint and together form the square \([0,T;0,T]\). By the complete additivity of measure,
\[ \sum_{n=1}^{\infty}\operatorname{mes}E_n=T^2, \]
whence, as \(N\to\infty\),
\[ \sum_{n=N}^{\infty}\operatorname{mes}E_n\to 0. \]
Then
\[ \left\|K_\varepsilon-\sum_{n=1}^{N-1}K^{(n)}\right\|_{\mathbf{B}_2}^{2} = \int_0^T\int_0^T \left\|K_\varepsilon(t,\tau)-\sum_{n=1}^{N-1}K^{(n)}(t,\tau)\right\|^2 \,dt\,d\tau = \]
\[ = \iint_{\sum_{n=N}^{\infty} E_n} \|K_{\varepsilon}(t,\tau)\|^2\,dt\,d\tau \to 0 \quad (N\to\infty). \]
The latter follows from the fact that the function \(K_{\varepsilon}(t,\tau)\) belongs to the class \(\mathbf{B}_2(H)\). Thus, for a given \(\varepsilon>0\) one can find a number \(N\) such that
\[ \left\|K_{\varepsilon}-\sum_{n=1}^{N} K^{(n)}\right\|_{\mathbf{B}_2(H)}<\varepsilon, \]
and, combining this with inequality (2.2),
\[ \left\|K-\sum_{n=1}^{N} K^{(n)}\right\|_{\mathbf{B}_2(H)}<2\varepsilon . \tag{2.4} \]
We shall now show that the integral operator with kernel \(K^{(n)}(t,\tau)\) (for fixed \(n=1,2,\ldots\)) is completely continuous in the space \(L_2([0,T];H)\). Consider the function
\[ L(t,\tau)= \begin{cases} L, & \text{for } \alpha<t<\beta,\quad \gamma<\tau<\delta,\\ 0, & \text{for all other } (t,\tau), \end{cases} \tag{2.5} \]
where \(\alpha,\beta,\gamma,\delta\in[0,T]\), and \(L\) is a completely continuous operator in \(H\). For the integral operator \(\mathbf{L}\) generated by the kernel (2.5), we have
\[ \mathbf{L}v=\int_0^T L(t,\tau)v(\tau)\,d\tau =\int_{\gamma}^{\delta} Lv(\tau)\,d\tau = L\left[\int_{\gamma}^{\delta} v(\tau)\,d\tau\right]. \]
Let the set of functions \(\{v(t)\}\) be bounded in \(L_2([0,T];H)\). Then the elements
\[ \int_{\gamma}^{\delta} v(\tau)\,d\tau \]
are uniformly bounded in \(H\):
\[ \left\|\int_{\gamma}^{\delta} v(\tau)\,d\tau\right\|_H \le \int_0^T \|v(\tau)\|_H\,d\tau \le T^{1/2}\|v\|_{L_2}, \]
therefore the completely continuous operator \(L\) maps them into a compact set of the space \(H\). If the functions
\[ x=L\left[\int_{\gamma}^{\delta} v(\tau)\,d\tau\right] \]
are regarded as elements of the space \(L_2([0,T];H)\), then they also form a compact set there, since from \(\|x_n-x_m\|_H\to0\) as \(n,m\to\infty\) it follows that
\[ \|x_n-x_m\|_{L_2}^2 =\int_0^T \|x_n-x_m\|_H^2\,dt =T\|x_n-x_m\|_H^2\to0. \]
Thus, the operator \(\mathbf{L}\) with kernel (2.5) is completely continuous in \(L_2([0,T];H)\). By Vitali’s theorem (see, for example, [5]) the fixed set \(E_n\) can be approximated arbitrarily accurately in measure by a finite number of nonintersecting squares, which corresponds to approximating the two-variable function \(K^{(n)}(t,\tau)\) by a sum
\[ \left\|K^{(n)}-\sum_{k=1}^{m}L_k^{(n)}\right\|_{\mathbf{B}_2}<\varepsilon, \tag{2.6} \]
where each \(L_k^{(n)}(t,\tau)\) has the form (2.5).
To complete the proof it remains to verify the following: suppose that some sequence of operator functions \(K_n(t,\tau)\) converges in the norm \(\mathbf{B}_2(H)\) to the function \(K(t,\tau)\), and that the integral operators \(K_n\) with kernels \(K_n(t,\tau)\) are completely continuous in \(L_2([0,T];H)\); then the operator \(K\) with kernel \(K(t,\tau)\) is also completely continuous.
Indeed,
\[ \|(K-K_n)v\|_{L_2}^{2} = \int_{0}^{T} \left\{ \left\| \int_{0}^{T} [K(t,\tau)-K_n(t,\tau)]v(\tau)\,d\tau \right\|_{H} \right\}^{2}dt. \]
By Hölder’s inequality,
\[ \left\| \int_{0}^{T}[K(t,\tau)-K_n(t,\tau)]v(\tau)\,d\tau \right\| \le \int_{0}^{T}\|K(t,\tau)-K_n(t,\tau)\|_{H}\cdot \|v(\tau)\|_{H}\,d\tau \le \]
\[ \le \left\{ \int_{0}^{T}\|K(t,\tau)-K_n(t,\tau)\|_{H}^{2}\,d\tau \right\}^{1/2} \left\{ \int_{0}^{T}\|v(\tau)\|_{H}^{2}\,d\tau \right\}^{1/2}. \]
Finally,
\[ \|(K-K_n)v\|_{L_2}^{2} \le \int_{0}^{T} \left\{ \left[ \int_{0}^{T}\|K(t,\tau)-K_n(t,\tau)\|^{2}\,d\tau \right] \left[ \int_{0}^{T}\|v(\tau)\|^{2}\,d\tau \right] \right\}dt = \]
\[ = \|K(t,\tau)-K_n(t,\tau)\|_{\mathbf{B}_2}^{2}\cdot \|v\|_{L_2}^{2} \to 0 \quad (n\to\infty). \]
Thus, from inequality (2.6) it follows that the operators with kernels \(K^{(n)}(t,\tau)\) are completely continuous, and from inequality (2.4) the assertion of the lemma follows.
Theorem 2. Suppose that the assumptions of Lemma 1 are satisfied, but condition 3 of this lemma is replaced by the following: the function \(A^{-\alpha}B(t)\) is uniformly continuous on \([0,T]\) for some \(0\le \alpha<1\). Then problem (0.1), (0.2) is reduced to an eigenfunction problem for a completely continuous operator in the Hilbert space \(L_2([0,T];H)\).
Proof. By the arguments used in the proof of Theorem 1, we reduce problem (0.1), (0.2) to the integral equation
\[ u(t)=\lambda\int_{0}^{T}G(t,\tau)B(\tau)u(\tau)\,d\tau. \]
We decompose the kernel of this integral operator into the product
\[ G(t,\tau)B(\tau) = H(t,\tau)A^{-1/2}B(\tau) = \left[H(t,\tau)A^{-1/2+\alpha}\right]\cdot [A^{-\alpha}B(\tau)]. \]
The second factor is bounded, by assumption, on the whole interval \([0,T]\). To estimate the first, recall the expression for \(H(t,\tau)\):
\[ H(t,\tau) = Q_1(\tau)U(t)+Q_2(\tau)U(T-t)+\frac{1}{2}U(|t-\tau|), \]
where \(Q_i(\tau)=R_i^{(1)}U(\tau)+R_i^{(2)}U(T-\tau)\) and the operators \(R_i\) are bounded.
Using the inequality
\[ \|A^\beta U(t)\|\leq \frac{C}{t^{2\beta}}, \]
we note that
\[ \left\|H(t,\tau)A^{-\frac12+\alpha}\right\|\leq \frac{C_1}{(t+\tau)^{2\alpha-1}} +\frac{C_2}{(T+t-\tau)^{2\alpha-1}} + \frac{C_3}{(T-t+\tau)^{2\alpha-1}} +\frac{C_4}{(2T-t-\tau)^{2\alpha-1}} +\frac{C_5}{|t-\tau|^{2\alpha-1}} . \]
By assumption \(\alpha<1\); therefore all singularities here are quadratically summable, i.e.
\[ \int_0^T\int_0^T \|G(t,\tau)B(\tau)\|^2\,d\tau\,dt<\infty . \]
Further, the operator
\[
A^\beta U(t)=[A^\beta U(t-\varepsilon)]U(\varepsilon)
\]
is completely continuous as the product of the bounded operator \(A^\beta U(t-\varepsilon)\) (for \(\varepsilon<t\)) and the completely continuous \(U(\varepsilon)\). Thus the kernel \(G(t,\tau)B(\tau)\) will be a completely continuous operator for all \(0\leq t,\tau\leq T\), except for the diagonal \(t=\tau\) of this square and its vertices.
Application of the lemma completes the proof of the theorem.
§ 3. SELF-ADJOINT BOUNDARY VALUE PROBLEMS
If the operator \(A\) is self-adjoint, then the differential expression \(Lu\equiv u''-Au\) will also be self-adjoint: \(L'v\equiv v''-Av\). Green’s formula can be written in the form
\[ \int_0^T [(v,Lu)-(L'v,u)]\,dt=Q(v,u), \tag{3.1} \]
where the bilinear form \(Q(v,u)\) is a combination of scalar products:
\[ Q(v,u)=(v_T,u_T')-(v_T',u_T)-(v_0,u_0')+(v_0',u_0). \]
The boundary conditions
\[ \begin{aligned} L_1(u)&\equiv \alpha_{11}u_0+\alpha_{12}u_0'+\beta_{11}u_T+\beta_{12}u_T'=0,\\ L_2(u)&\equiv \alpha_{21}u_0+\alpha_{22}u_0'+\beta_{21}u_T+\beta_{22}u_T'=0 \end{aligned} \tag{3.2} \]
will be called self-adjoint if they coincide with their adjoints. From the definition of adjoint boundary conditions (see [1]) it is not difficult to note that they do not depend on the operator \(A\), but are determined entirely by the matrix of the original boundary conditions
\[ \begin{pmatrix} \alpha_{11} & \alpha_{12} & \beta_{11} & \beta_{12}\\ \alpha_{21} & \alpha_{22} & \beta_{21} & \beta_{22} \end{pmatrix}. \]
Theorem 3. Suppose that, for the differential equation
\[ \frac{d^2u}{dt^2}=Au-\lambda B(t)u \qquad (0<t<T), \tag{3.3} \]
in the Hilbert space \(H\), under the boundary conditions (3.2), the following constraints are satisfied:
1) the operator \(A\) is self-adjoint, positive definite, and has a completely continuous resolvent;
2) the boundary conditions are self-adjoint;
3) the function \(B(t)\) is uniformly continuous on the interval \([0,T]\) and for each \(0\leq t\leq T\) is a positive definite bounded operator having a bounded inverse;
4) the number \(\lambda=0\) is not an eigenvalue of the problem.
Then the boundary-value problem (0.3), (0.2) has a countable number of real eigenvalues \(\lambda_n\), tending to infinity, and a complete system of eigenfunctions \(\{u_n(t)\}\) orthonormal with respect to the scalar product
\[ (Bu,v)=\int_0^T \bigl(B(t)u(t),v(t)\bigr)_H\,dt . \]
Proof. For differential equations of the second order, self-adjoint boundary conditions are a special case of regular ones; therefore, by Theorem 2, the problem under consideration is equivalent to the integral equation
\[ u(t)=\lambda\int_0^T G(t,\tau)B(\tau)u(\tau)\,d\tau \tag{3.4} \]
with a completely continuous operator.
We symmetrize the kernel of this operator by the substitution \(B^{1/2}(t)u(t)=y(t)\). Equation (3.4) is transformed into the following one (after multiplying it on the left by \(B^{1/2}(t)\)):
\[ y(t)=\lambda\int_0^T B^{1/2}(t)G(t,\tau)B^{1/2}(\tau)y(\tau)\,d\tau . \tag{3.5} \]
Let us verify that the resulting integral operator \(\mathbf G\) in (3.5) is self-adjoint in the space \(L_2([0,T];H)\). For this purpose consider the scalar product
\[ (\mathbf G f,g)_{L_2} = \int_0^T \left( \int_0^T B^{1/2}(t)G(t,\tau)B^{1/2}(\tau)f(\tau)\,d\tau, g(t) \right)_H dt = \]
\[ = \int_0^T\int_0^T \bigl( G(t,\tau)B^{1/2}(\tau)f(\tau), B^{1/2}(t)g(t) \bigr)_H \,d\tau\,dt . \tag{3.6} \]
Let the functions \(u(t)\) and \(v(t)\) satisfy the boundary conditions (3.2). Denote \(Lu=\varphi\), \(Lv=\psi\); then
\[ u(t)=\int_0^T G(t,\tau)\varphi(\tau)\,d\tau,\qquad v(t)=\int_0^T G(t,\tau)\psi(\tau)\,d\tau . \tag{3.7} \]
These expressions for \(u\) and \(v\) can be substituted into Green’s formula (3.1), which leads to the relation (recall that \(Q(v,u)=0\))
\[ \int_0^T \int_0^T (G(t,\tau)\psi(\tau), \varphi(t))_H\,d\tau\,dt = \int_0^T \int_0^T (\psi(\tau), G(\tau,t)\varphi(t))_H\,dt\,d\tau . \]
This equality makes it possible in (3.6) to transfer the operator \(G(t,\tau)\) to the second factor of the scalar product
\[ (\mathbf{G}f,g)_{L_2} = \int_0^T \int_0^T \bigl(B^{1/2}(\tau)f(\tau),\,G(\tau,t)B^{1/2}(t)g(t)\bigr)_H\,dt\,d\tau . \]
Finally transferring the operator \(B^{1/2}(\tau)\), we see that
\((\mathbf{G}f,g)_{L_2}=(f,\mathbf{G}g)_{L_2}\), i.e. the operator \(\mathbf{G}\) is self-adjoint. By Theorem 2 it is also completely continuous and, as an invertible operator, has a countable number of positive eigenvalues \(\lambda_n\) and a complete orthonormal system of eigenfunctions \(y_i(t)\)
\[ (y_i,y_j)_{L_2}=\delta_{ij}, \tag{3.8} \]
where \(\delta_{ij}\) is the Kronecker symbol. Relation (3.8) for the functions \(u_i(t)\) is written in the form
\[ (y_i,y_j)_{L_2} = \int_0^T \bigl(B^{1/2}(t)u_i(t),\,B^{1/2}(t)u_j(t)\bigr)_H\,dt = \]
\[ = \int_0^T \bigl(B(t)u_i(t),\,u_j(t)\bigr)_H\,dt = (\mathbf{B}u_i,u_j)_{L_2} = \delta_{ij}, \]
which proves the theorem.
Up to now it has been assumed throughout that the point zero belongs to the resolvent set of the operator \(A\). However, the following is true.
Theorem 4. If, in the conditions of Theorem 3, one assumes that the operator \(A\) has the point zero as its eigenvalue of finite multiplicity \(n\), then the assertions of the theorem remain valid.
In conclusion, the author expresses his deep gratitude to S. G. Krein, under whose supervision the work was carried out.
References
- Krein S. G., Laptev G. I. Differential Equations, 2, No. 3, 382—390, 1966.
- Krein S. G., Laptev G. I. Differential Equations, 2, No. 7, 919—926, 1966.
- Solomyak M. Z. On differential equations in Banach spaces. Izv. vuzov, ser. matem., No. 1, 1960.
- Hille E., Phillips R. Functional Analysis and Semigroups. IL, Moscow, 1962.
- Natanson I. P. Theory of Functions of a Real Variable. Moscow, 1957.
- Naimark M. A. Linear Differential Operators. Moscow, 1954.
Received by the editors
October 5, 1965
Computing Center
Latvian State University