Defining the operators:
the nonlinear problem is equivalent to
We note that (6) can be written as
and the nonlinear problem (1) as
This suggests to introduce the new function . To find a solution u, we have to find v and .
For every constant
, we solve
be the set of solutions of (7). This set may be empty (no solution), a singleton (unique solution) or with more than one element (multiple solutions). For every
, we consider
is a solution of the nonlinear problem (1). We then look for fixed points of the map
For fixed, we try to solve the integral equation (7).
Assume that there exist
for every , .
Thus, a solution of (7) is precisely a fixed point of . Note that is a compact operator. For , let .
Hence there exist constants
for any and solution of . This implies that v is bounded independently of , and hence by Schaefer’s fixed point theorem (Theorem 4.3.2 of ), has at least a fixed point, i.e., for given c, equation (7) is solvable.
Now suppose f is Lipschitz continuous.
Then there exists
for every and .
, we have
Thus, for small, equation (7) has a unique solution in view of the classical Banach contraction fixed point theorem.
Now, under conditions (8) and (10), set
where is the unique solution of (7), and as a consequence of the contraction principle, this map is continuous.
Define the map
If there exists
, then for that c
, and the function
is such that , and therefore is a solution of the original nonlinear problem (1).
Now, assume that
uniformly on .
Then the growth of
is sublinear in view of estimate (9). However, c
growths linearly. Hence the norm of the function
growths asymptotically as c.
This implies that , and there exists with .
We have the following result.
Theorem 3.1 Suppose that f satisfies the growth conditions (8) and (10). If (11) holds, then (1) is solvable for l sufficiently small.
Note that condition (11) is crucial since for and, in view of (5), the problem (1) may have no solution.