In this section, it is provided an existence and location theorem for the problem (1)-(2). More precisely, sufficient conditions are given for, not only the existence of a solution *u*, but also to have information about the location of *u*, and all its derivatives up to the $(n-1)$ order.

The arguments of the proof require the following lemma, given on [29].

**Lemma 4** *For*$v,w\in C(I)$*such that*$v(x)\le w(x)$,

*for every*$x\in I$,

*define*$q(x,u)=max\{v,min\{u,w\}\}.$

*Then*,

*for each*$u\in {C}^{1}(I)$*the next two properties hold*:

- (a)
$\frac{d}{dx}q(x,u(x))$ *exists for a*.*e*. $x\in I$.

- (b)
*If*
$u,{u}_{m}\in {C}^{1}(I)$
*and*
${u}_{m}\to u$
*in*
${C}^{1}(I)$
*then*
$\frac{d}{dx}q(x,{u}_{m}(x))\to \frac{d}{dx}q(x,u(x))\phantom{\rule{1em}{0ex}}\mathit{\text{for a.e.}}\phantom{\rule{1em}{0ex}}x\in I.$

Now, we are in a position to prove the main result of this paper.

**Theorem 5** *Assume that there exists a pair of lower and upper solutions*$(\alpha ,\beta )$*of problem* (1)-(2).

*If*$f:I\times {(C(I))}^{n-2}\times {\mathbb{R}}^{2}\to \mathbb{R}$*is a*${L}^{1}$-

*Carathéodory function*,

*satisfying a Nagumo*-

*type condition in*${E}_{\ast}=\{\begin{array}{c}(x,{y}_{0},\dots ,{y}_{n-1})\in I\times {\mathbb{R}}^{n-1}:{\alpha}_{i}(x)\le {y}_{i}\le {\beta}_{i}(x),\phantom{\rule{1em}{0ex}}i=0,\dots ,n-3,\hfill \\ {\alpha}^{(n-2)}(x)\le {y}_{n-2}\le {\beta}^{(n-2)}(x),\hfill \end{array}$

*then problem* (1)-(2)

*has at least one solution* *u* *such that**for every*$x\in I$,

*and*$|{u}^{(n-1)}(x)|\le K$,

$\mathrm{\forall}x\in I$,

*where*$K=max\{R,|{\alpha}^{(n-1)}(x)|,|{\beta}^{(n-1)}(x)\left|\right\}$

(10)

*and*$R>0$*is given by* (7).

*Proof* Define the continuous functions, for

$i=0,\dots ,n-3$,

$\begin{array}{r}{\delta}_{i}(x,{y}_{i})=max\{{\alpha}_{i}(x),min\{{y}_{i},{\beta}_{i}(x)\}\},\\ {\delta}_{n-2}(x,{y}_{n-2})=max\{{\alpha}^{(n-2)}(x),min\{{y}_{n-2},{\beta}^{(n-2)}(x)\}\},\end{array}$

(11)

and the truncation, not necessarily continuous,

$\xi (z)=max\{-K,min\{z,K\}\},$

with *K* given by (10).

Consider the modified problem composed by the equation

${u}^{(n)}(x)=f\left(\begin{array}{c}x,{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-3}(\cdot ,{u}^{(n-3)}),\\ {\delta}_{n-2}(x,{u}^{(n-2)}(x)),\xi (\frac{d}{dx}({\delta}_{n-2}(x,{u}^{(n-2)}(x))))\end{array}\right)$

(12)

and the boundary conditions, for

$i=0,\dots ,n-2$,

$\begin{array}{r}{u}^{(i)}(a)={\delta}_{i}(a,{u}^{(i)}(a)+{L}_{i}\left(\begin{array}{c}{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-2}(\cdot ,{u}^{(n-2)}),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(\cdot ,{u}^{(n-2)}))),{u}^{(i)}(a)\end{array}\right)),\\ {u}^{(n-2)}(b)={\delta}_{n-2}(b,{u}^{(n-2)}(b)+{L}_{n-1}\left(\begin{array}{c}{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-2}(\cdot ,{u}^{(n-2)}),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(\cdot ,{u}^{(n-2)}))),{u}^{(n-2)}(b)\end{array}\right)).\end{array}$

(13)

The proof will follow the next steps:

Step 1. Every solution

*u* of problem (12)-(13), satisfies

and $|{u}^{(n-1)}(x)|<K$, for every $x\in I$, with $K>0$ given in (10).

Let

*u* be a solution of the modified problem (12)-(13). Assume, by contradiction, that there exists

$x\in I$ such that

${\alpha}^{(n-2)}(x)>{u}^{(n-2)}(x)$ and let

${x}_{0}\in I$ be such that

$\underset{x\in I}{min}{(u-\alpha )}^{(n-2)}(x):={(u-\alpha )}^{(n-2)}({x}_{0})<0.$

As, by (13),

${u}^{(n-2)}(a)\ge {\alpha}^{(n-2)}(a)$ and

${u}^{(n-2)}(b)\ge {\alpha}^{(n-2)}(b)$, then

${x}_{0}\in (a,b)$. So, there is

$({x}_{1},{x}_{2})\subset (a,b)$ such that

$\begin{array}{r}{u}^{(n-2)}(x)<{\alpha}^{(n-2)}(x),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in ({x}_{1},{x}_{2}),\\ {(u-\alpha )}^{(n-2)}({x}_{1})={(u-\alpha )}^{(n-2)}({x}_{2})=0.\end{array}$

(14)

Therefore,

${\delta}_{n-2}(x,{u}^{(n-2)}(x))={\alpha}^{(n-2)}(x),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in ({x}_{1},{x}_{2}),$

and

$\frac{d}{dx}{\delta}_{n-2}(x,{u}^{(n-2)}(x))={\alpha}^{(n-1)}(x),\phantom{\rule{1em}{0ex}}\text{a.e.}x\in ({x}_{1},{x}_{2}).$

Now, since for all

$u\in {C}^{n-2}(I)$ it is satisfied that

$({\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-3}(\cdot ,{u}^{\mathrm{\prime}}))\in A$, we deduce that

$\begin{array}{rcl}{u}^{(n)}(x)& =& f\left(\begin{array}{c}x,{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-3}(\cdot ,{u}^{(n-3)}),{\delta}_{n-2}(x,{u}^{(n-2)}(x)),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(x,{u}^{(n-2)}(x))))\end{array}\right)\\ =& f(x,{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-3}(\cdot ,{u}^{(n-3)}),{\alpha}^{(n-2)}(x),{\alpha}^{(n-1)}(x))\\ \le & {\alpha}^{(n)}(x)\phantom{\rule{1em}{0ex}}\text{for a.e.}x\in ({x}_{1},{x}_{2}).\end{array}$

As ${(u-\alpha )}^{(n-1)}({x}_{0})=0$ and ${(u-\alpha )}^{(n-1)}$ is nonincreasing in $({x}_{1},{x}_{2})$, this contradicts the definitions of ${x}_{0}$ and ${x}_{2}$.

The inequality

${u}^{(n-2)}(x)\le {\beta}^{(n-2)}(x)$, in

*I*, can be proved in same way and so,

${\alpha}^{(n-2)}(x)\le {u}^{(n-2)}(x)\le {\beta}^{(n-2)}(x),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in I.$

(15)

By (13) and (3), the following inequalities hold for every

$x\in I$:

$\begin{array}{rcl}{u}^{(n-3)}(x)& =& {u}^{(n-3)}(a)+{\int}_{a}^{x}{u}^{(n-2)}(s)\phantom{\rule{0.2em}{0ex}}ds\ge {\alpha}_{n-3}(a)+{\int}_{a}^{x}{\alpha}^{(n-2)}(s)\phantom{\rule{0.2em}{0ex}}ds\\ \ge & min\{{\alpha}^{(n-3)}(a),{\beta}^{(n-3)}(a)\}+{\int}_{a}^{x}{\alpha}^{(n-2)}(s)\phantom{\rule{0.2em}{0ex}}ds={\alpha}_{n-3}(x).\end{array}$

Analogously, it can be obtained ${u}^{(n-3)}(x)\le {\beta}_{n-3}(x)$, for $x\in I$.

The remaining inequalities are obtained by the same integration process.

Applying previous bounds in Lemma 2, and remarking that

${\int}_{r}^{K}\frac{s}{\phi (s)}\phantom{\rule{0.2em}{0ex}}ds\ge {\int}_{r}^{R}\frac{s}{\phi (s)}\phantom{\rule{0.2em}{0ex}}ds,$

for *K* given by (10), it is obtained, by Lemma 2, the *a priori* bound $|{u}^{(n-1)}(x)|<K$, for $x\in I$. For details, see [[33], Lemma 2].

Step 2. Problem (12)-(13) has at least one solution.

For

$\lambda \in [0,1]$ let us consider the homotopic problem given by

${u}^{(n)}(x)=\lambda f\left(\begin{array}{c}x,{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-3}(\cdot ,{u}^{(n-3)}),\\ {\delta}_{n-2}(x,{u}^{(n-2)}(x)),\xi (\frac{d}{dx}({\delta}_{n-2}(x,{u}^{(n-2)}(x))))\end{array}\right)$

(16)

and the boundary conditions, for

$i=0,\dots ,n-2$,

$\begin{array}{r}{u}^{(i)}(a)=\lambda {\delta}_{i}(a,{u}^{(i)}(a)+{L}_{i}\left(\begin{array}{c}{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-2}(\cdot ,{u}^{(n-2)}),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(\cdot ,{u}^{(n-2)}))),{u}^{(i)}(a)\end{array}\right)):=\lambda {L}_{{A}_{i}},\\ {u}^{(n-2)}(b)=\lambda {\delta}_{n-2}(b,{u}^{(n-2)}(b)+{L}_{n-1}\left(\begin{array}{c}{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-2}(\cdot ,{u}^{(n-2)}),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(\cdot ,{u}^{(n-2)}))),{u}^{(n-2)}(b)\end{array}\right))\\ \phantom{{u}^{(n-2)}(b)}:=\lambda {L}_{B}.\end{array}$

(17)

Let us consider the norms in

${C}^{n-1}(I)$ and in

${L}^{1}(I)\times {\mathbb{R}}^{n}$, respectively,

${\parallel v\parallel}_{{C}^{n-1}}=max\{{\parallel v\parallel}_{\mathrm{\infty}},\dots ,{\parallel {v}^{(n-1)}\parallel}_{\mathrm{\infty}}\}$

and

$|(h,{h}_{1},\dots ,{h}_{n})|=max\{{\parallel h\parallel}_{{L}^{1}},max\{|{h}_{1}|,\dots ,|{h}_{n}|\}\}.$

Define the operators

$\mathcal{L}:{W}^{n,1}(I)\subset {C}^{n-1}(I)\to {L}^{1}(I)\times {\mathbb{R}}^{n}$ by

$\mathcal{L}u=({u}^{(n)},u(a),\dots ,{u}^{(n-2)}(a),{u}^{(n-2)}(b))$ and, for

$\lambda \in [0,1]$,

$i=0,\dots ,n-2$,

${\mathcal{N}}_{\lambda}:{C}^{n-1}(I)\to {L}^{1}(I)\times {\mathbb{R}}^{n}$ by

${\mathcal{N}}_{\lambda}u=\left(\begin{array}{c}\lambda f\left(\begin{array}{c}x,{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-3}(\cdot ,{u}^{(n-3)}),{\delta}_{n-2}(x,{u}^{(n-2)}(x)),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(x,{u}^{(n-2)}(x))))\end{array}\right),\\ \lambda {L}_{{A}_{1}},\dots ,\lambda {L}_{{A}_{n-2}},\lambda {L}_{B}\end{array}\right).$

Since ${L}_{0},\dots ,{L}_{n-1}$ are continuous and *f* is a ${L}^{1}$-Carathéodory function, then, from Lemma 4, ${\mathcal{N}}_{\lambda}$ is continuous. Moreover, as ${\mathcal{L}}^{-1}$ is compact, it can be defined the completely continuous operator ${\mathcal{T}}_{\lambda}:{C}^{n-1}(I)\to {C}^{n-1}(I)$ by ${\mathcal{T}}_{\lambda}u={\mathcal{L}}^{-1}{\mathcal{N}}_{\lambda}(u)$.

It is obvious that the fixed points of operator ${\mathcal{T}}_{\lambda}$ coincide with the solutions of problem (16)-(17).

As

${\mathcal{N}}_{\lambda}u$ is bounded in

${L}^{1}(I)\times {\mathbb{R}}^{n}$ and uniformly bounded in

${C}^{n-1}(I)$, we have that any solution of the problem (16)-(17), verifies the following

*a priori* bound

${\parallel u\parallel}_{{C}^{n-1}}\le {\parallel {\mathcal{L}}^{-1}\parallel}_{{C}^{n-1}}|{\mathcal{N}}_{\lambda}(u)|\le \overline{K},$

for some $\overline{K}>0$ independent of *λ*.

In the set $\mathrm{\Omega}=\{u\in {C}^{n-1}(I):{\parallel u\parallel}_{{C}^{n-1}}<\overline{K}+1\}$, the degree $d(\mathcal{I}-{\mathcal{T}}_{\lambda},\mathrm{\Omega},0)$ is well defined for every $\lambda \in [0,1]$ and, by the invariance under homotopy, $d(\mathcal{I}-{\mathcal{T}}_{0},\mathrm{\Omega},0)=d(\mathcal{I}-{\mathcal{T}}_{1},\mathrm{\Omega},0)$.

As the equation

$x={\mathcal{T}}_{0}(x)$ is equivalent to the problem

$\{\begin{array}{c}{u}^{(n)}(x)=0,\hfill \\ {u}^{(i)}(a)={u}^{(n-2)}(b)=0,\phantom{\rule{1em}{0ex}}i=0,\dots ,n-2,\hfill \end{array}$

which has only the trivial solution, then $d(\mathcal{I}-{\mathcal{T}}_{0},\mathrm{\Omega},0)=\pm 1$. So, by degree theory, the equation $x={\mathcal{T}}_{1}(x)$ has at least one solution, that is, the problem (12)-(13) has at least a solution in Ω.

Step 3. Every solution *u* of problem (12)-(13) is a solution of (1)-(2).

Let

*u* be a solution of the modified problem (12)-(13). By previous steps, function

*u* fulfills equation (

1). So, it will be enough to prove the following inequalities, for

$i=0,\dots ,n-3$:

and

$\begin{array}{rcl}{\alpha}^{(n-2)}(b)& \le & {u}^{(n-2)}(b)+{L}_{n-1}\left(\begin{array}{c}{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-2}(x,{u}^{(n-2)}(x)),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(x,{u}^{(n-2)}(x)))),{u}^{(n-2)}(b)\end{array}\right)\\ \le & {\beta}^{(n-2)}(b).\end{array}$

Assume that

$u(a)+{L}_{0}\left(\begin{array}{c}{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-2}(x,{u}^{(n-2)}(x)),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(x,{u}^{(n-2)}(x)))),u(a)\end{array}\right)>{\beta}_{0}(a).$

(18)

Then, by (13),

$u(a)={\beta}_{0}(a)$. By previous steps, it is obtained the following contradiction with (18):

Applying similar arguments, it can be proved that

${\alpha}_{0}(a)\le u(a)+{L}_{0}\left(\begin{array}{c}{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-2}(x,{u}^{(n-2)}(x)),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(x,{u}^{(n-2)}(x)))),u(a)\end{array}\right)$

and analogously, for

$j=1,\dots ,n-3$,

${\alpha}_{j}(a)\le {u}^{(j)}(a)+{L}_{j}\left(\begin{array}{c}{\delta}_{0}(\cdot ,u),\dots ,{\delta}_{n-2}(x\cdot ,{u}^{(n-2)}(x)),\\ \xi (\frac{d}{dx}({\delta}_{n-2}(x,{u}^{(n-2)}(x)))),{u}^{(j)}(a)\end{array}\right)\le {\beta}_{j}(a).$

Also, using the same arguments and the same techniques, it can be proved that

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