1.1 Modification of the LyapunovSchmidt method
In the Hilbert space {H}_{T} defined above, we consider the boundaryvalue problem
\frac{d\phi (t)}{dt}=i{H}_{0}\phi (t)+\epsilon Z(\phi (t),t,\epsilon )+f(t),
(10)
\phi (0,\epsilon )\phi (w,\epsilon )=\alpha .
(11)
We seek a generalized solution \phi (t,\epsilon ) of the boundaryvalue problem (10), (11) that becomes one of the solutions of the generating equation (1), (2) {\phi}_{0}(t,\overline{c}) in the form (6) for \epsilon =0.
To find a necessary condition for the operator function Z(\phi ,t,\epsilon ), we impose the joint constraints
Z(\cdot ,\cdot ,\cdot )\in C([0;w],{H}_{T})\times C[0,{\epsilon}_{0}]\times C[\parallel \phi {\phi}_{0}\parallel \le q],
where q is a positive constant.
The main idea of the next results was used in [15] for the investigation of bounded solutions.
Let us show that this problem can be solved with the use of the following operator equation for generating amplitudes:
F(\overline{c})={U}_{0}(w){\int}_{0}^{w}{U}^{1}(\tau )Z({\phi}_{0}(\tau ,\overline{c}),\tau ,0)\phantom{\rule{0.2em}{0ex}}d\tau =0.
(12)
Theorem 2 (Necessary condition)
Suppose that the nonlinear boundaryvalue problem (10), (11) has a generalized solution \phi (\cdot ,\epsilon ) that becomes one of the solutions {\phi}_{0}(t,\overline{c}) of the generating equation (1), (2) with constant \overline{c}={c}^{0} and \phi (t,0)={\phi}_{0}(t,{c}^{0}) for \epsilon =0. Then this constant must satisfy the equation for generating amplitudes (12).
Proof If the boundaryvalue problem (10), (11) has classical generalized solutions, then, by Lemma 1, the following solvability condition must be satisfied:
{U}_{0}(w)(\alpha +{\int}_{0}^{w}{U}^{1}(\tau )\{f(\tau )+\epsilon Z(\phi (\tau ,\epsilon ),\tau ,\epsilon )\}\phantom{\rule{0.2em}{0ex}}d\tau )=0.
(13)
By using condition (5), we establish that condition (13) is equivalent to the following:
{U}_{0}(w){\int}_{0}^{w}{U}^{1}(\tau )Z(\phi (\tau ,\epsilon ),\tau ,\epsilon )\phantom{\rule{0.2em}{0ex}}d\tau =0.
Since \phi (t,\epsilon )\to {\phi}_{0}(t,{c}^{0}) as \epsilon \to 0, we finally obtain [by using the continuity of the operator function Z(\phi ,t,\epsilon )] the required assertion.
To find a sufficient condition for the existence of solutions of the boundaryvalue problem (10), (11), we additionally assume that the operator function Z(\phi ,t,\epsilon ) is strongly differentiable in a neighborhood of the generating solution (Z(\cdot ,t,\epsilon )\in {C}^{1}[\parallel \phi {\phi}_{0}\parallel \le q]).
This problem can be solved with the use of the operator
{B}_{0}=\frac{dF(\overline{c})}{d\overline{c}}{}_{\overline{c}={c}_{0}}={U}_{0}(w){\int}_{0}^{w}{U}^{1}(t){A}_{1}(t)\phantom{\rule{0.2em}{0ex}}dt:H\to H,
where {A}_{1}(t)={Z}^{1}(v,t,\epsilon ){}_{v={\phi}_{0},\epsilon =0} (Fréchet derivative). □
Theorem 3 (Sufficient condition)
Suppose that the operator {B}_{0} satisfies the following conditions:

(1)
The operator {B}_{0} is MoorePenrose pseudoinvertible;

(2)
{\mathcal{P}}_{N({B}_{0}^{\ast})}{U}_{0}(w)=0.
Then, for an arbitrary element c={c}^{0}\in {H}_{T} satisfying the equation for generating amplitudes (12), there exists at least one solution of (10), (11).
This solution can be found by using the following iterative process:
\begin{array}{c}{\overline{v}}_{k+1}(t,\epsilon )=\epsilon G[Z({\phi}_{0}(\tau ,{c}^{0})+{v}_{k},\tau ,\epsilon ),\alpha ](t),\hfill \\ {c}_{k}={B}_{0}^{+}{U}_{0}(w){\int}_{0}^{w}{U}^{1}(\tau )\{{A}_{1}(\tau ){\overline{v}}_{k}(\tau ,\epsilon )+\mathcal{R}({v}_{k}(\tau ,\epsilon ),\tau ,\epsilon )\}\phantom{\rule{0.2em}{0ex}}d\tau ,\hfill \\ \mathcal{R}({v}_{k}(t,\epsilon ),t,\epsilon )=Z({\phi}_{0}(t,{c}^{0})+{v}_{k}(t,\epsilon ),t,\epsilon )Z({\phi}_{0}(t,{c}^{0}),t,0){A}_{1}(t){v}_{k}(t,\epsilon ),\hfill \\ \mathcal{R}(0,t,0)=0,\phantom{\rule{2em}{0ex}}{\mathcal{R}}_{x}^{1}(0,t,0)=0,\hfill \\ {v}_{k+1}(t,\epsilon )=U(t){U}_{0}(w){c}_{k}+{\overline{v}}_{k+1}(t,\epsilon ),\hfill \\ {\phi}_{k}(t,\epsilon )={\phi}_{0}(t,{c}^{0})+{v}_{k}(t,\epsilon ),\phantom{\rule{1em}{0ex}}k=0,1,2,\dots ,\phantom{\rule{2em}{0ex}}{v}_{0}(t,\epsilon )=0,\phi (t,\epsilon )=\underset{k\to \mathrm{\infty}}{lim}{\phi}_{k}(t,\epsilon ).\hfill \end{array}
1.2 Relationship between necessary and sufficient conditions
First, we formulate the following assertion:
Corollary Suppose that a functional F(\overline{c}) has the Fréchet derivative {F}^{(1)}(\overline{c}) for each element {c}^{0} of the Hilbert space H satisfying the equation for generating constants (12). If {F}^{1}(\overline{c}) has a bounded inverse, then the boundaryvalue problem (10), (11) has a unique solution for each {c}^{0}.
Remark 2 If the assumptions of the corollary are satisfied, then it follows from its proof that the operators {B}_{0} and {F}^{(1)}({c}^{0}) are equal. Since the operator {F}^{(1)}(\overline{c}) is invertible, it follows that assumptions 1 and 2 of Theorem 3 are necessarily satisfied for the operator {B}_{0}. In this case, the boundaryvalue problem (10), (11) has a unique bounded solution for each {c}^{0}\in {H}_{T} satisfying (12). Therefore, the invertibility condition for the operator {F}^{1}(\overline{c}) expresses the relationship between the necessary and sufficient conditions. In the finitedimensional case, the condition of invertibility of the operator {F}^{(1)}(\overline{c}) is equivalent to the condition of simplicity of the root {c}^{0} of the equation for generating amplitudes [5].
In this way, we modify the wellknown LyapunovSchmidt method. It should be emphasized that Theorems 2 and 3 give us a condition for the chaotic behavior of (10) and (11) [16].
1.3 Example
We now illustrate the obtained assertion. Consider the following differential equation in a separable Hilbert space H:
\ddot{y}(t)+Ty(t)=\epsilon (1{\parallel y(t)\parallel}^{2})\dot{y}(t),
(14)
y(0)=y(w),\phantom{\rule{2em}{0ex}}\dot{y}(0)=\dot{y}(w),
(15)
where T is an unbounded operator with compact {T}^{1}. Then there exists an orthonormal basis {e}_{i}\in H such that y(t)={\sum}_{i=1}^{\mathrm{\infty}}{c}_{i}(t){e}_{i} and Ty(t)={\sum}_{i=1}^{\mathrm{\infty}}{\lambda}_{i}{c}_{i}(t){e}_{i}, {\lambda}_{i}\to \mathrm{\infty}. In this case, the operator system (10), (11) for the boundaryvalue problem (14), (15) is equivalent to the following countable system of ordinary differential equations ({c}_{k}(t)={x}_{k}(t)):
\begin{array}{c}{\dot{x}}_{k}(t)=\sqrt{{\lambda}_{k}}{y}_{k}(t),\phantom{\rule{1em}{0ex}}k=1,2,\dots ,\hfill \\ {\dot{y}}_{k}(t)=\sqrt{{\lambda}_{k}}{x}_{k}(t)+\epsilon \sqrt{{\lambda}_{k}}(1\sum _{j=1}^{\mathrm{\infty}}{x}_{j}^{2}(t)){y}_{k}(t),\hfill \end{array}
(16)
{x}_{k}(0)={x}_{k}(w),\phantom{\rule{2em}{0ex}}{y}_{k}(0)={y}_{k}(w).
(17)
We find the solutions of these equations in the space {W}_{2}^{1}([0;w]) that, for \epsilon =0, turn into one of the solutions of the generating equation. Consider the critical case {\lambda}_{i}=4{\pi}^{2}{i}^{2}/{w}^{2}, i\in N. Let w=2\pi. In this case, the set of all periodic solutions of (16), (17) has the form
\begin{array}{c}{x}_{k}(t)=cos(kt){c}_{1}^{k}+sin(kt){c}_{2}^{k},\hfill \\ {y}_{k}(t)=sin(kt){c}_{1}^{k}+cos(kt){c}_{2}^{k}\hfill \end{array}
for all pairs of constants {c}_{1}^{k},{c}_{2}^{k}\in R, k\in N. The equation for generating amplitudes (12) is equivalent in this case to the following countable systems of algebraic nonlinear equations:
\begin{array}{c}{\left({c}_{1}^{k}\right)}^{3}+2\sum _{j=1,j\ne k}({c}_{1}^{k}{\left({c}_{1}^{j}\right)}^{2}+{c}_{1}^{k}{\left({c}_{2}^{j}\right)}^{2})+{c}_{1}^{k}{\left({c}_{2}^{k}\right)}^{2}4{c}_{1}^{k}=0,\hfill \\ {\left({c}_{2}^{k}\right)}^{3}+2\sum _{j=1,j\ne k}({c}_{2}^{k}{\left({c}_{1}^{j}\right)}^{2}+{c}_{2}^{k}{\left({c}_{2}^{j}\right)}^{2})+{\left({c}_{1}^{k}\right)}^{2}{c}_{2}^{k}4{c}_{2}^{k}=0,\phantom{\rule{1em}{0ex}}k\in N.\hfill \end{array}
Then we can obtain the next result.
Theorem 4 (Necessary condition for the van der Pol equation)
Suppose that the boundaryvalue problem (16), (17) has a bounded solution \phi (\cdot ,\epsilon ) that becomes one of the solutions of the generating equations with pairs of constants ({c}_{1}^{k},{c}_{2}^{k}), k\in N. Then only a finite number of these pairs are not equal to zero. Moreover, if ({c}_{1}^{{k}_{i}},{c}_{2}^{{k}_{i}})\ne (0,0), i=\overline{1,N}, then these constants lie on an Ndimensional torus in the infinitedimensional space of constants:
{\left({c}_{1}^{{k}_{i}}\right)}^{2}+{\left({c}_{2}^{{k}_{i}}\right)}^{2}={\left(\frac{2}{\sqrt{2N1}}\right)}^{2},\phantom{\rule{1em}{0ex}}i=\overline{1,N}.
Remark Similarly, we can study the Schrödinger equation with a variable operator and more general boundary conditions (as noted in the introduction).
Consider the differential Schrödinger equation
\frac{d\phi (t)}{dt}=iH(t)\phi (t)+f(t),\phantom{\rule{1em}{0ex}}t\in J
(18)
in a Hilbert space H with the boundary condition
Q\phi (\cdot )=\alpha ,
(19)
where, for each t\in J\subset R, the unbounded operator H(t) has the form H(t)={H}_{0}+V(t), {H}_{0}={H}_{0}^{\ast} is an unbounded selfadjoint operator with domain D=D({H}_{0})\subset H, and the mapping t\to V(t) is strongly continuous. The operator Q is linear and bounded and acts from the Hilbert space H to {H}_{1}. As in [12], we define the operatorvalued function
\tilde{V}(t)={e}^{it{H}_{0}}V(t){e}^{it{H}_{0}}.
In this case, \tilde{V}(t) admits the Dyson representation [[12], p.311]; denote its propagator by \tilde{U}(t,s). If U(t,s)={e}^{it{H}_{0}}\tilde{U}(t,s){e}^{is{H}_{0}}, then {\psi}_{s}(t)=U(t,s)\psi is a weak solution of (14) with the condition {\phi}_{s}(s)=\psi in the sense that, for any \eta \in D({H}_{0}), the function (\eta ,{\psi}_{s}(t)) is differentiable and
\frac{d}{dt}(\eta ,{\psi}_{s}(t))=i({H}_{0}\eta ,{\psi}_{s}(t))i(V(t)\eta ,{\psi}_{s}(t))+(f(t),{\psi}_{s}(t)),\phantom{\rule{1em}{0ex}}t\in J.
A detailed study of the boundaryvalue problem (18), (19) will be given in a separate paper.