Weak non-linear boundary value problems for operator equations in critical cases (Q2906213)

Let \(I\) be an interval, \(B_1\) the Banach space of functions \(z:I\rightarrow\mathbb R^n\), \(B_2\) the Banach space of functions \(f:I\rightarrow \mathbb R^{n_1}\); \(C(\epsilon)\) the space of continuous functions \(c:I_{\epsilon_0}=[0,\epsilon_0]\rightarrow \mathbb R^n\).NEWLINENEWLINEThe following problem is considered NEWLINENEWLINE\[NEWLINELz(\cdot,\epsilon)(t)=f(t)+\epsilon Z(z(t,\epsilon),t,\epsilon),\tag{1}NEWLINE\]NEWLINE NEWLINE\[NEWLINElz(\cdot,\epsilon)=\alpha+J(z(\cdot,\epsilon),\epsilon).\tag{2}NEWLINE\]NEWLINE NEWLINEHere, \(L:B_1\rightarrow B_2\) is the Noether operator \((\mu=\dim(N(L))<\infty,\;\nu=\dim(N(L^*))<\infty\) and \(\nu\neq\mu)\), \(Z:B_1\times I\times I_{\epsilon_0}\rightarrow B_2\times I_{\epsilon_0}\) is a bounded operator, nonlinear with respect to \(z\). It is assumed that \(Z\) is Fréchet-differentiable in some neighborhood of the generating function \(z_0\) (see below) and continuous with respect to \(\epsilon\). Moreover, it is assumed that NEWLINENEWLINE\[NEWLINEZ(0,t,0)=0,\;Z^\prime_z(0,t,0)=0,\;f\in B_2,\;l:B_1\rightarrow \mathbb R^mNEWLINE\]NEWLINENEWLINE is a linear operator, \(J:B_1\times I_{\epsilon_0}\rightarrow\mathbb R^m\) is an operator bounded, nonlinear with respect to \(z\), Fréchet-differentiable in the neighborhood of \(z_0\), continuous with respect to \(\epsilon\) satisfying \(J(0,0)=0\), \(J^\prime_z(0,0)=0\), \(\alpha\in\mathbb R^m\). The above mentioned generating function \(z_0\) is a solution of the linear system NEWLINE\[NEWLINELz_0(t)=f(t)\tag{3}NEWLINE\]NEWLINE NEWLINE\[NEWLINElz_0(\cdot)=\alpha,\;\epsilon=0.\tag{4}NEWLINE\]NEWLINE Critical case means here that the system (3), (4) has more than one solution.NEWLINENEWLINEThe main results are the following ones NEWLINE{\parindent=0.5cmNEWLINE\begin{itemize}\item[-] necessary conditions for the existence of a solution to (1), (2);NEWLINE\item[-] sufficient conditions. NEWLINENEWLINE\end{itemize}}NEWLINETheorem 4 gives sufficient conditions for the existence of a unique solution to (1), (2).