Defining the operators:

$\begin{array}{c}\begin{array}{r}F:C[0,T]\to C[0,T],\\ [Fu](t)=f(t,u(t)),\phantom{\rule{1em}{0ex}}u\in C[0,T],t\in [0,T],\end{array}\hfill \\ \begin{array}{r}K:C[0,T]\to C[0,T],\\ [K\sigma ](t)={\int}_{0}^{T}k(t,s)\sigma (s)\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}\sigma \in C[0,T],t\in [0,T],\end{array}\hfill \\ \begin{array}{r}L:C[0,T]\to C[0,T],\\ [Lu](t)=\frac{t}{T}u(T),\phantom{\rule{1em}{0ex}}u\in C[0,T],t\in [0,T],\end{array}\hfill \end{array}$

the nonlinear problem is equivalent to

where $N=K\circ F+L$.

We note that (6) can be written as

$u(t)-\frac{t}{T}u(T)={\int}_{0}^{T}k(t,s)\sigma (s)\phantom{\rule{0.2em}{0ex}}ds$

and the nonlinear problem (1) as

$u(t)-\frac{t}{T}u(T)={\int}_{0}^{T}k(t,s)f(s,u(s))\phantom{\rule{0.2em}{0ex}}ds.$

This suggests to introduce the new function $v(t)=u(t)-\frac{t}{T}u(T)$. To find a solution *u*, we have to find *v* and $u(T)$.

For every constant

$c\in \mathbb{R}$, we solve

$v(t)={\int}_{0}^{T}k(t,s)f(s,v(s)+\frac{s}{T}c)\phantom{\rule{0.2em}{0ex}}ds$

(7)

and let

$\phi (c)$ 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

${v}_{c}\in \phi (c)$, we consider

${u}_{c}(t)={v}_{c}(t)+\frac{t}{T}c,$

and hence

${u}_{c}(t)={\int}_{0}^{T}k(t,s)f(s,{u}_{c}(s))\phantom{\rule{0.2em}{0ex}}ds+\frac{t}{T}c.$

If

$c={u}_{c}(T)$, then

${u}_{c}$ is a solution of the nonlinear problem (1). We then look for fixed points of the map

$c\in \mathbb{R}\u27f6{u}_{c}(T)\in \mathbb{R}.$

For $c\in \mathbb{R}$ fixed, we try to solve the integral equation (7).

Assume that there exist

$a,b\in C[0,T]$ and

$\alpha \in [0,1)$ such that

$|f(t,u)|\le a(t)+b(t){|u|}^{\alpha}$

(8)

for every $t\in [0,T]$, $u\in \mathbb{R}$.

For

$v\in C[0,T]$, define

${F}_{c}v\in C[0,T]$ as

$[{F}_{c}v](t)=f(t,v(t)+\frac{t}{T}c).$

Thus, a solution of (7) is precisely a fixed point of $K\circ {F}_{c}={K}_{c}$. Note that ${K}_{c}$ is a compact operator. For $v\in C[0,T]$, let $\parallel v\parallel ={sup}_{t\in [0,T]}|v(t)|$.

For

$\lambda \in (0,1)$, if

$v=\lambda {K}_{c}(v)$ we have

$v(t)=\lambda {\int}_{0}^{T}k(t,s)f(s,v(s)+\frac{s}{T}c)\phantom{\rule{0.2em}{0ex}}ds,$

and

$|v(t)|\le \parallel k\parallel {\int}_{0}^{T}f(s,v(s)+\frac{s}{T}c)\phantom{\rule{0.2em}{0ex}}ds\le \parallel k\parallel \cdot T[\parallel a\parallel +\parallel b\parallel {(\parallel v\parallel +c)}^{\alpha}].$

Hence there exist constants

${a}_{0}$,

${b}_{0}$ such that

$\parallel v\parallel \le {a}_{0}+{b}_{0}{(\parallel v\parallel +c)}^{\alpha}$

(9)

for any $v\in C[0,T]$ and $\lambda \in (0,1)$ solution of $v=\lambda {K}_{c}(v)$. This implies that *v* is bounded independently of $\lambda \in (0,1)$, and hence by Schaefer’s fixed point theorem (Theorem 4.3.2 of [22]), ${K}_{c}$ has at least a fixed point, *i.e.*, for given *c*, equation (7) is solvable.

Now suppose *f* is Lipschitz continuous.

Then there exists

$l>0$ such that

$|f(t,x)-f(t,y)|\le l|x-y|$

(10)

for every $t\in [0,T]$ and $x,y\in \mathbb{R}$.

Then, for

$v,w\in C[0,T]$, we have

$|[{K}_{c}v](t)-[{K}_{c}w](t)|\le {\int}_{0}^{T}k(t,s)l|v(s)-w(s)|\phantom{\rule{0.2em}{0ex}}ds$

and

$\parallel {K}_{c}v-{K}_{c}w\parallel \le \parallel k\parallel \cdot l\cdot T\parallel v-w\parallel .$

Thus, for $l>0$ small, equation (7) has a unique solution in view of the classical Banach contraction fixed point theorem.

Now, under conditions (8) and (10), set

$c\in \mathbb{R}\u27f6{v}_{c}\in C[0,T],$

where ${v}_{c}$ is the unique solution of (7), and as a consequence of the contraction principle, this map is continuous.

Define the map

$\begin{array}{c}\phi :\mathbb{R}\u27f6\mathbb{R},\hfill \\ \phi (c)={v}_{c}(T).\hfill \end{array}$

If there exists

$c\in \mathbb{R}$ such that

$\phi (c)=0$, then for that

*c* we have

${v}_{c}(T)$, and the function

${u}_{c}(t)={v}_{c}(t)+\frac{t}{T}c$

is such that ${u}_{c}(T)=c$, and therefore ${u}_{c}$ is a solution of the original nonlinear problem (1).

Now, assume that

$\underset{u\to \pm \mathrm{\infty}}{lim}f(t,u)=\pm \mathrm{\infty}$

(11)

uniformly on $t\in [0,T]$.

Then the growth of

$\parallel v\parallel $ is sublinear in view of estimate (9). However,

*c* growths linearly. Hence the norm of the function

${v}_{c}(s)+\frac{s}{T}c$

growths asymptotically as *c*.

This implies that ${lim}_{c\to \pm \mathrm{\infty}}\phi (c)=\pm \mathrm{\infty}$, and there exists $c\in \mathbb{R}$ with $\phi (c)=0$.

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 $f(t,u)=\sigma (t)$ and, in view of (5), the problem (1) may have no solution.