On the solvability of a Neumann boundary value problem for the differential equation $f(t,x,x^{\mathrm{\prime }},x^{\mathrm{\prime }\mathrm{\prime }})=0$

Abstract

Using barrier strip arguments, we investigate the existence of $C^2[0,1]$-solutions to the Neumann boundary value problem $f(t,x,x^{\mathrm{\prime }},x^{\mathrm{\prime }\mathrm{\prime }})=0$, $x^{\mathrm{\prime }}(0)=a$, $x^{\mathrm{\prime }}(1)=b$.

MSC:34B15.

1 Introduction

The purpose of this paper is to establish the existence of $C^2[0,1]$-solutions to the scalar Neumann boundary value problem (BVP)

$$\{\begin{array}{c}f(t,x,x^{\prime },x^{″})=0,t\in [0,1],\hfill \\ x^{\prime }(0)=a,x^{\prime }(1)=b,a\ne b,\hfill \end{array}$$

(N)

where the function $f(t,x,p,q)$ and its first derivatives are continuous only on suitable subsets of the set $[0,1]\times R^3$.

The literature devoted to the solvability of singular and nonsingular Neumann BVPs for second order ordinary differential equations whose main nonlinearities do not depend on the second derivative is vast. We quote here only [1–5] for results and references.

The solvability of the homogeneous Neumann problem for the equation ${(p(t)x^{\mathrm{\prime }})}^{\mathrm{\prime }}+f(t,x,x^{\mathrm{\prime }},x^{\mathrm{\prime }\mathrm{\prime }})=y(t)$, under appropriate conditions on f, has been studied in [6–8]. Results, concerning the existence of solutions to the homogeneous and nonhomogeneous Neumann problem for the equation $x^{\mathrm{\prime }\mathrm{\prime }}=f(t,x,x^{\mathrm{\prime }},x^{\mathrm{\prime }\mathrm{\prime }})-y(t)$ can be found in [9] and [10] respectively. BVPs for the same equation with various linear boundary conditions have been studied in [9, 11–13]. The results of [14] guarantee the solvability of BVPs for the equation $x^{\mathrm{\prime }\mathrm{\prime }}=f(t,x,x^{\mathrm{\prime }},x^{\mathrm{\prime }\mathrm{\prime }})$ with fully linear boundary conditions. BVPs for the equation $f(t,x,x^{\mathrm{\prime }},x^{\mathrm{\prime }\mathrm{\prime }})=0$ with fully nonlinear boundary conditions have been studied in [15]. For results, which guarantee the solvability of the Dirichlet BVP for the same equation, in the scalar and in the vector cases, see [12] and [16] respectively.

Concerning the kind of the nonlinearity of the function $f(t,x,p,q)$, we note that it is assumed sublinear in [6], semilinear in [11] and linear with respect to x, p and q in [8, 12]. Finally, in [9] and [17]f is a linear function with respect to q, while with respect to p, it is a quadratic function or satisfies Nagumo type growth conditions respectively.

As in [10, 15, 18, 19], we use sign conditions to establish a priori bounds for x, $x^{\mathrm{\prime }}$ and $x^{\mathrm{\prime }\mathrm{\prime }}$, where $x(t)\in C^2[0,1]$ is a solution to a suitable family of BVPs similar to that in [10, 19]. Using these a priori bounds and applying the topological transversality theorem from [20], we prove our main existence result.

2 Basic hypotheses

To formulate our hypotheses, we use the sets

$$\begin{array}{c}{J}_x=[min\{0,\frac{a+b}{2},\frac{a^2}{2(a-b)}\},max\{0,\frac{a+b}{2},\frac{a^2}{2(a-b)}\}]\text{and}\hfill \\ J_p=[min\{a,b\},max\{a,b\}].\hfill \end{array}$$

So, we assume that there are positive constants K, M and a sufficiently small $\epsilon >0$ such that: H1.

$$\begin{array}{c}f(t,x,p,q)\text{ is continuous with respect to }x\in R\text{ for each }(t,p,q)\in [0,1]\times R^2,\hfill \\ f(t,x,p,q)\text{ is continuous with respect to }q\in R\text{ for each }(t,x,p)\in [0,1]\times R^2,\hfill \end{array}$$

there are constants $K_x$ and $K_q$ such that

$$\begin{array}{c}{f}_x(t,x,p,q)\ge K_x>0\text{for }(t,x,p,q)\in [0,1]\times R^3,\hfill \\ f_q(t,x,p,q)\le -K_q<0\text{for }(t,x,p,q)\in [0,1]\times [-M_0-\epsilon ,M_0+\epsilon ]\times R^2,\hfill \end{array}$$

where

$$M_0=max\{\frac{e}{e^2-1}(|a-be|+|ae-b|),\frac{L}{min\{K,K_x,K_q\}}+max\{\frac{|a+b|}{2},\frac{a^2}{2|a-b|}\}\},$$

$f(t,x,p,b-a-(1-\lambda )x)$ is bounded for $(\lambda ,t,x,p)\in {[0,1]}^2\times J_x\times J_p$ and $L=max\{sup|f(t,x,p,b-a-(1-\lambda )x)|,maxK|b-a-(1-\lambda )x|\}$ for $(\lambda ,t,x,p)\in {[0,1]}^2\times J_x\times J_p$.H2.

$$f(t,x,p,q)+Kq\ge 0\text{for }(t,x,p,q)\in [0,1]\times [-M_0-\epsilon ,M_0+\epsilon ]\times R\times (-\mathrm{\infty },-M)$$

and

$$f(t,x,p,q)+Kq\le 0\text{for }(t,x,p,q)\in [0,1]\times [-M_0-\epsilon ,M_0+\epsilon ]\times R\times (M,\mathrm{\infty }),$$

where $M_0$ is as in H1.

H3. The functions $f(t,x,p,q)$ and $f_q(t,x,p,q)$ are continuous for $(t,x,p,q)\in [0,1]\times [-M_0-\epsilon ,M_0+\epsilon ]\times [-M_1-\epsilon ,M_1+\epsilon ]\times [-M_2-\epsilon ,M_2+\epsilon ]$, where $M_1=min\{|a|,|b|\}+M_0+M$, $M_2=M_0+M$ and $M_0$ is as in H1.

3 Auxiliary lemmas

In order to obtain our main existence results, we use the constant K from the hypotheses to construct the family of BVPs

where $\lambda \in [0,1]$ and prove the following three auxiliary results.

Lemma 3.1 Let H 1 hold and $x(t)\in C^2[0,1]$ be a solution to (3.1) _λ , $\lambda \in [0,1]$. Then

$$|x(t)|\le M_0,t\in [0,1].$$

Proof For $\lambda =0$, problem (3.1)₀ is of the form

$$x^{\mathrm{\prime }\mathrm{\prime }}-x=0,x^{\mathrm{\prime }}(0)=a,x^{\mathrm{\prime }}(1)=b.$$

The unique solution to this BVP satisfies the bound

$$|x(t)|\le \frac{e}{e^2-1}(|a-be|+|ae-b|),t\in [0,1].$$

Let now $\lambda \in (0,1]$. Then the function

$$y(t)=x(t)-s(t),t\in [0,1],\text{ where }s(t)=\frac{1}{2}(b-a)t^2+at,t\in [0,1],$$

is a solution to the homogeneous boundary value problem

$$\begin{array}{c}K(y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s))\hfill \\ =\lambda (K(y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s))+f(t,y+s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s))),\hfill \\ y^{\mathrm{\prime }}(0)=0,y^{\mathrm{\prime }}(1)=0.\hfill \end{array}$$

The equation is equivalent to the following one

$$\begin{array}{c}(1-\lambda )K{y}^{\mathrm{\prime }\mathrm{\prime }}\hfill \\ ={(1-\lambda )}^{2}Ky-(1-\lambda )K(b-a-(1-\lambda )s)\hfill \\ +\lambda f(t,y+s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s))\hfill \\ -\lambda f(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s))\hfill \\ +\lambda f(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s)).\hfill \end{array}$$

Hence, by the intermediate value theorem, we obtain consecutively

$$\begin{array}{c}(1-\lambda )K{y}^{\mathrm{\prime }\mathrm{\prime }}\hfill \\ ={(1-\lambda )}^{2}Ky-(1-\lambda )K(b-a-(1-\lambda )s)\hfill \\ +\lambda f_x(t,s+{\theta }_{1}y,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s))y\hfill \\ +\lambda f(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s))\hfill \\ -\lambda f(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )s)\hfill \\ +\lambda f(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )s),\hfill \end{array}$$

for any ${\theta }_1\in (0,1)$ depending on $\lambda \in [0,1]$, $t\in [0,1]$ and $y(t)$,

$$\begin{array}{c}(1-\lambda )K{y}^{\mathrm{\prime }\mathrm{\prime }}\hfill \\ ={(1-\lambda )}^{2}Ky-(1-\lambda )K(b-a-(1-\lambda )s)\hfill \\ +\lambda f_x(t,s+{\theta }_{1}y,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s))y\hfill \\ +\lambda f_q(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )s-{\theta }_2(1-\lambda )y)((\lambda -1)y)\hfill \\ +\lambda f(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )s)-\lambda f(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},b-a-(1-\lambda )s)\hfill \\ +\lambda f(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},b-a-(1-\lambda )s)\hfill \end{array}$$

and

$$\begin{array}{c}(1-\lambda )K{y}^{\mathrm{\prime }\mathrm{\prime }}\hfill \\ ={(1-\lambda )}^{2}Ky-(1-\lambda )K(b-a-(1-\lambda )s)\hfill \\ +\lambda f_x(t,s+{\theta }_{1}y,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y+s))y\hfill \\ +\lambda f_q(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},y^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )s-{\theta }_2(1-\lambda )y)(-(1-\lambda )y)\hfill \\ +\lambda f_q(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},b-a-(1-\lambda )s+{\theta }_3{y}^{\mathrm{\prime }\mathrm{\prime }})y^{\mathrm{\prime }\mathrm{\prime }}+\lambda f(t,s,y^{\mathrm{\prime }}+s^{\mathrm{\prime }},b-a-(1-\lambda )s),\hfill \end{array}$$

for any ${\theta }_2,{\theta }_3\in (0,1)$ depending on $\lambda \in [0,1]$, $t\in [0,1]$ and $y(t)$,

$$\{\begin{array}{c}((1-\lambda )K-\lambda f_q(t,s,y^{\prime }+s^{\prime },b-a-(1-\lambda )s+{\theta }_3{y}^{″}))y^{″}\hfill \\ =\{{(1-\lambda )}^{2}K+\lambda f_x(t,s+{\theta }_{1}y,y^{\prime }+s^{\prime },y^{″}+b-a-(1-\lambda )(y+s))\hfill \\ -\lambda (1-\lambda )f_q(t,s,y^{\prime }+s^{\prime },y^{″}+b-a-(1-\lambda )s-{\theta }_2(1-\lambda )y)\}y\hfill \\ +\lambda f(t,s,y^{\prime }+s^{\prime },b-a-(1-\lambda )s)-(1-\lambda )K(b-a-(1-\lambda )s).\hfill \end{array}$$

(3.2)

Next, suppose that $|y(t)|$ achieves its maximum at $t_0\in (0,1)$. Then the function $z=y^2(t)$ has also a maximum at $t_0$. Consequently, we have

$$0\ge z^{\mathrm{\prime }\mathrm{\prime }}(t_0)=2y(t_0)y^{\mathrm{\prime }\mathrm{\prime }}(t_0).$$

(3.3)

Using the fact that $y^{\mathrm{\prime }}(t_0)=0$, from (3.2) we obtain

$$\{\begin{array}{c}\{(1-\lambda )K-\lambda f_q(t_0,s_0,s_0^{\prime },b-a-(1-\lambda )s_0+{\theta }_{3,0}{y}_0^{″})\}{y}_0^{″}\hfill \\ =\{(1-\lambda )\{(1-\lambda )K-\lambda f_q(t_0,s_0,s_0^{\prime },y_0^{″}+b-a-(1-\lambda )s_0-{\theta }_{2,0}(1-\lambda )y_0)\}\hfill \\ +\lambda f_x(t_0,s_0+{\theta }_{1,0}{y}_0,s_0^{\prime },y_0^{″}+b-a-(1-\lambda )(y_0+s_0))\}{y}_0\hfill \\ +\lambda f(t_0,s_0,s_0^{\prime },b-a-(1-\lambda )s_0)-(1-\lambda )K(b-a-(1-\lambda )s_0),\hfill \end{array}$$

(3.4)

where ${\theta }_{1,0}={\theta }_1(t_0,s_0^{\mathrm{\prime }},y_0^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y_0+s_0))$, ${\theta }_{2,0}={\theta }_2(t_0,s_0,s_0^{\mathrm{\prime }})$, ${\theta }_{3,0}={\theta }_3(t_0,s_0,s_0^{\mathrm{\prime }})$, and $s_0=s(t_0)$, $s_0^{\mathrm{\prime }}=s^{\mathrm{\prime }}(t_0)$, $y_0=y(t_0)$, $y_0^{\mathrm{\prime }\mathrm{\prime }}=y^{\mathrm{\prime }\mathrm{\prime }}(t_0)$.

In view of H1, from (3.4) we have

$$\{\begin{array}{c}(1-\lambda )\{(1-\lambda )K-\lambda {\overline{f}}_q\}+\lambda {\overline{f}}_x\ge min\{(1-\lambda )K-\lambda {\overline{f}}_q,{\overline{f}}_x\}\hfill \\ \phantom{(1-\lambda )((1-\lambda )K-\lambda {\overline{f}}_q)+\lambda {\overline{f}}_x}\ge min\{K,-{\overline{f}}_q,{\overline{f}}_x\}\ge min\{K,K_x,K_q\},\hfill \end{array}$$

(3.5)

where

$$\begin{array}{c}{\overline{f}}_q=f_q(t_0,s_0,s_0^{\mathrm{\prime }},y_0^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )s_0-{\theta }_{2,0}(1-\lambda )y_0),\hfill \\ {\overline{f}}_x=f_x(t_0,s_0+{\theta }_{1,0}{y}_0,s_0^{\mathrm{\prime }},y_0^{\mathrm{\prime }\mathrm{\prime }}+b-a-(1-\lambda )(y_0+s_0)).\hfill \end{array}$$

Suppose now that $|y(t_0)|>L{(min\{K,K_x,K_q\})}^{-1}$. Then, from (3.4) and (3.5) it follows that

$$\{\begin{array}{c}\{(1-\lambda )K-\lambda f_q(t_0,s_0,s_0^{\prime },b-a-(1-\lambda )s_0+{\theta }_{3,0}{y}_0^{″})\}{y}_0^{″}\hfill \\ \ge min\{K,K_x,K_q\}y(t_0)\hfill \\ +\lambda f(t_0,s_0,s_0^{\prime },b-a-(1-\lambda )s_0)-(1-\lambda )K(b-a-(1-\lambda )s_0)\hfill \end{array}$$

(3.6)

if $y(t_0)>L{(min\{K,K_x,K_q\})}^{-1}$ or

$$\{\begin{array}{c}\{(1-\lambda )K-\lambda f_q(t_0,s_0,s_0^{\prime },b-a-(1-\lambda )s_0+{\theta }_{3,0}{y}_0^{″})\}{y}_0^{″}\hfill \\ \le min\{K,K_x,K_q\}y(t_0)\hfill \\ +\lambda f(t_0,s_0,s_0^{\prime },b-a-(1-\lambda )s_0)-(1-\lambda )K(b-a-(1-\lambda )s_0)\hfill \end{array}$$

(3.7)

if $y(t_0)<-L{(min\{K,K_x,K_q\})}^{-1}$. Multiplying (3.6) and (3.7) by $y(t_0)$, we obtain

$$\begin{array}{c}\{(1-\lambda )K-\lambda f_q(t_0,s_0,s_0^{\mathrm{\prime }},b-a-(1-\lambda )s_0+{\theta }_{3,0}{y}_0^{\mathrm{\prime }\mathrm{\prime }})\}{y}_0^{\mathrm{\prime }\mathrm{\prime }}{y}_0\hfill \\ \ge y_0(min\{K,K_x,K_q\}{y}_0-L)>0,\hfill \\ \{(1-\lambda )K-\lambda f_q(t_0,s_0,s_0^{\mathrm{\prime }},b-a-(1-\lambda )s_0+{\theta }_{3,0}{y}_0^{\mathrm{\prime }\mathrm{\prime }})\}{y}_0^{\mathrm{\prime }\mathrm{\prime }}{y}_0\hfill \\ \ge |y_0|(min\{K,K_x,K_q\}|y_0|-L)>0,\hfill \end{array}$$

respectively. Finally, since $(t_0,s_0,s_0^{\mathrm{\prime }},b-a-(1-\lambda )s_0+{\theta }_{3,0}{y}_0^{\mathrm{\prime }\mathrm{\prime }})\in [0,1]\times [-M_0-\epsilon ,M_0+\epsilon ]\times R^2$ we have $f_q(t_0,s_0,s_0^{\mathrm{\prime }},b-a-(1-\lambda )s_0+{\theta }_{3,0}{y}_0^{\mathrm{\prime }\mathrm{\prime }})<0$. So

$$y_0^{\mathrm{\prime }\mathrm{\prime }}{y}_0>0,$$

which contradicts (3.3). Thus, we infer that if $|y(t)|$ achieves its maximum on $(0,1)$, then

$$|y(t)|\le \frac{L}{min\{K,K_x,K_q\}}\text{for }t\in [0,1]\text{ and }\lambda \in (0,1].$$

Let $|y(1)|$ be the maximum of $|y(t)|$ and suppose that $|y(1)|>L{(min\{K,K_x,K_q\})}^{-1}$. Following the above reasoning and using the fact that $y^{\mathrm{\prime }}(1)=0$, we obtain

$$y(1)y^{\mathrm{\prime }\mathrm{\prime }}(1)>0.$$

If $y(1)>0$, then $y^{\mathrm{\prime }\mathrm{\prime }}(1)>0$ and so $y^{\mathrm{\prime }}(t)$ is a strictly increasing function for $t\in U_1$, where $U_1\subset [0,1]$ is a sufficiently small neighbourhood of $t=1$. So, we see that

$$y^{\mathrm{\prime }}(t)<y^{\mathrm{\prime }}(1)=0\text{for }t\in U_1\setminus \{1\},$$

i.e., $y(t)$ is a strictly decreasing function for $t\in U_1$. Therefore, $y(1)=|y(1)|$ can not be the maximum of $|y(t)|$ on $[0,1]$, which is a contradiction. Assume next that $y(1)<0$. Then similar to the above arguments lead again to a contradiction. Thus, we see that

$$|y(1)|\le \frac{L}{min\{K,K_x,K_q\}}.$$

The inequality

$$|y(0)|\le \frac{L}{min\{K,K_x,K_q\}}$$

can be obtained in the same manner. Consequently, the eventual solutions of (3.1) _λ , $\lambda \in (0,1]$ satisfy the bound

$$|x(t)|\le |y(t)|+|s(t)|\le \frac{L}{min\{K,K_x,K_q\}}+max\{\frac{a^2}{2|a-b|},\frac{|a+b|}{2}\},t\in [0,1],$$

and the proof of the lemma is completed. □

Lemma 3.2 Let H 1 and H 2 hold and $x(t)\in C^2[0,1]$ be a solution to (3.1) _λ , $\lambda \in [0,1]$. Then:

• (a) $|x^{\mathrm{\prime }\mathrm{\prime }}(t)-(1-\lambda )x(t)|\le M$, $|x^{\mathrm{\prime }\mathrm{\prime }}(t)|\le M_0+M$, $t\in [0,1]$.

• (b) $|x^{\mathrm{\prime }}(t)|\le min\{|a|,|b|\}+M_0+M$, $t\in [0,1]$.

Proof (a) Suppose there exists a $({\lambda }_0,t_0)\in {[0,1]}^2$ or a $({\lambda }_1,t_1)\in {[0,1]}^2$ such that

$$x^{\mathrm{\prime }\mathrm{\prime }}(t_0)-(1-{\lambda }_0)x(t_0)<-M\text{or}{x}^{\mathrm{\prime }\mathrm{\prime }}(t_1)-(1-{\lambda }_1)x(t_1)>M.$$

By Lemma 3.1, we have

$$|x(t)|\le M_0\text{for }t\in [0,1].$$

(3.8)

In particular, (3.8) holds for $t_0$ and $t_1$. Thus, in view of H2, we have

$$\begin{array}{rcl}0& >& K(x^{\mathrm{\prime }\mathrm{\prime }}(t_0)-(1-{\lambda }_0)x(t_0))\\ =& {\lambda }_0\{K(x^{\mathrm{\prime }\mathrm{\prime }}(t_0)-(1-{\lambda }_0)x(t_0))+f(t_0,x(t_0),x^{\mathrm{\prime }}(t_0),(x^{\mathrm{\prime }\mathrm{\prime }}(t_0)-(1-{\lambda }_0)x(t_0)))\}\ge 0\end{array}$$

or

$$\begin{array}{rcl}0& <& K(x^{\mathrm{\prime }\mathrm{\prime }}(t_1)-(1-{\lambda }_1)x(t_1))\\ =& {\lambda }_1\{K(x^{\mathrm{\prime }\mathrm{\prime }}(t_1)-(1-{\lambda }_1)x(t_1))+f(t_1,x(t_1),x^{\mathrm{\prime }}(t_1),(x^{\mathrm{\prime }\mathrm{\prime }}(t_1)-(1-{\lambda }_1)x(t_1)))\}\le 0,\end{array}$$

respectively. The obtained contradictions show that

$$-M\le x^{\mathrm{\prime }\mathrm{\prime }}(t)-(1-\lambda )x(t)\le M\text{for }t\in [0,1]\text{ and }\lambda \in [0,1],$$

and therefore

$$-(M_0+M)\le x^{\mathrm{\prime }\mathrm{\prime }}(t)\le M_0+M\text{for }t\in [0,1],$$

which proves (a).

• (b) By the mean value theorem, for each $t\in (0,1]$ there is a $\xi \in (0,t)$ such that

  $$x^{\mathrm{\prime }}(t)-x^{\mathrm{\prime }}(0)=x^{\mathrm{\prime }\mathrm{\prime }}(\xi )t.$$

Since, in view of (a), we have $|x^{\mathrm{\prime }\mathrm{\prime }}(\xi )|\le M_0+M$, from the last formula we find that

$$|x^{\mathrm{\prime }}(t)|\le |x^{\mathrm{\prime }}(0)|+|x^{\mathrm{\prime }\mathrm{\prime }}(\xi )|\le min\{|a|,|b|\}+M_0+M,t\in [0,1],$$

which proves (b) and completes the proof of the lemma. □

Lemma 3.3 Let H 1, H 2 and H 3 hold. Then there exists a function $G(\lambda ,t,x,p)$ continuous for $(\lambda ,t,x,p)\in {[0,1]}^2\times [-M_0-\epsilon ,M_0+\epsilon ]\times [-M_1-\epsilon ,M_1+\epsilon ]$ and such that

• (a) the BVP

  $$\begin{array}{c}{x}^{\mathrm{\prime }\mathrm{\prime }}-(1-\lambda )x=G(\lambda ,t,x,x^{\mathrm{\prime }}),t\in [0,1],\hfill \\ x^{\mathrm{\prime }}(0)=a,x^{\mathrm{\prime }}(1)=b,\hfill \end{array}$$

is equivalent to BVP (3.1) _λ .

• (b) $G(0,t,x,p)=0$ for $(t,x,p)\in {\mathrm{\Pi }}_q\equiv [0,1]\times [-M_0-\epsilon ,M_0+\epsilon ]\times [-M_1-\epsilon ,M_1+\epsilon ]$.

Proof (a) We write the differential equation from (3.1) _λ as

$$\lambda f(t,x,x^{\mathrm{\prime }},(x^{\mathrm{\prime }\mathrm{\prime }}-(1-\lambda )x))-(1-\lambda )K(x^{\mathrm{\prime }\mathrm{\prime }}-(1-\lambda )x)=0$$

(3.9)

and consider the function

$$F(\lambda ,t,x,p,q)=\lambda f(t,x,p,q)-(1-\lambda )Kq\text{for }(\lambda ,t,x,p,q)\in [0,1]\times \mathrm{\Pi },$$

where $\mathrm{\Pi }=[0,1]\times [-M_0-\epsilon ,M_0+\epsilon ]\times [-M_1-\epsilon ,M_1+\epsilon ]\times [-M_2-\epsilon ,M_2+\epsilon ]$. Since $-M_2-\epsilon <-M$ and $M_2+\epsilon >M$, we can use H2 to conclude that

$$F(\lambda ,t,x,p,-M_2-\epsilon )F(\lambda ,t,x,p,M_2+\epsilon )<0\text{for }(\lambda ,t,x,p)\in [0,1]\times {\mathrm{\Pi }}_q.$$

(3.10)

On the other hand, for $(\lambda ,t,x,p,q)\in [0,1]\times \mathrm{\Pi }$ we have

$$F_q(\lambda ,t,x,p,q)=\lambda f_q(t,x,p,q)-(1-\lambda )K\le max\{-K,-K_q\}<0.$$

(3.11)

Finally, from H3 we have that

$$F(\lambda ,t,x,p,q)\text{ and }{F}_q(\lambda ,t,x,p,q)\text{ are continuous for }(\lambda ,t,x,p,q)\in [0,1]\times \mathrm{\Pi }.$$

(3.12)

So, (3.10), (3.11) and (3.12) allow us to apply a well-known theorem to conclude that there is a unique function $G(\lambda ,t,x,p)$ which is continuous for $(\lambda ,t,x,p)\in [0,1]\times {\mathrm{\Pi }}_q$ and such that the equations

$$q=G(\lambda ,t,x,p),(\lambda ,t,x,p)\in [0,1]\times {\mathrm{\Pi }}_q$$

and

$$F(\lambda ,t,x,p,q)=0,(\lambda ,t,x,p,q)\in [0,1]\times \mathrm{\Pi }$$

are equivalent. Now from Lemma 3.1 we have

$$-M_0-\epsilon \le x(t)\le M_0+\epsilon \text{for }t\in [0,1],$$

and Lemma 3.2 yields

$$-M_1-\epsilon \le x^{\mathrm{\prime }}(t)\le M_1+\epsilon \text{and}-M_2-\epsilon <-M\le x^{\mathrm{\prime }\mathrm{\prime }}(t)-(1-\lambda )x(t)\le M<M_2+\epsilon$$

for $t\in [0,1]$ and $\lambda \in [0,1]$. Consequently, equation (3.9) is equivalent to the equation

$$x^{\mathrm{\prime }\mathrm{\prime }}-(1-\lambda )x=G(\lambda ,t,x,x^{\mathrm{\prime }}),t\in [0,1],$$

which yields the first assertion.

• (b) It follows immediately from $F(0,t,x,p,0)=0$ for $(t,x,p)\in {\mathrm{\Pi }}_q$. □

4 The main result

Our main result is the following existence theorem, the proof of which is based on the lemmas of the previous sections and the Topological transversality theorem [20].

Theorem 4.1 Let H 1, H 2 and H 3 hold. Then problem (N) has at least one solution in $C^2[0,1]$.

Proof First, we observe that according to Lemma 3.3, the family of boundary value problems

is equivalent to the family (3.1) _λ for $\lambda \in [0,1]$. Next define the set

$$U=\{x\in C_B^2[0,1]:|x|<M_0+\epsilon ,\left|x^{\mathrm{\prime }}\right|<M_1+\epsilon ,\left|x^{\mathrm{\prime }\mathrm{\prime }}\right|<M_2+\epsilon \},$$

where $C_B^2[0,1]=\{x(t)\in C^2[0,1]:x^{\mathrm{\prime }}(0)=a,x^{\mathrm{\prime }}(1)=b\}$, and the maps

$$\begin{array}{c}j:C_B^2[0,1]\to C^1[0,1]\text{by }jx=x,\hfill \\ G_{\lambda }:C^1[0,1]\to C[0,1]\text{by }(G_{\lambda }x)(t)=G(\lambda ,t,x(t),x^{\mathrm{\prime }}(t))-x(t),\hfill \end{array}$$

where $t\in [0,1]$, $\lambda \in [0,1]$, $x(t)\in j(\overline{U})$ and

$$L_{\lambda }:C_B^2[0,1]\to C[0,1]\text{by }{L}_{\lambda }x=x^{\mathrm{\prime }\mathrm{\prime }}-(2-\lambda )x,\lambda \in [0,1].$$

Since $L_{\lambda }$, $\lambda \in [0,1]$, is a continuous, linear, one-to-one map of $C_B^2[0,1]$ onto $C[0,1]$, the map $L_{\lambda }^{-1}$, $\lambda \in [0,1]$ exists and is continuous. In addition, $G_{\lambda }$, $\lambda \in [0,1]$, is a continuous and j is a completely continuous embedding. Since $j(\overline{U})$ is a compact subset of $C^1[0,1]$, and $G_{\lambda }$, $\lambda \in [0,1]$, and $L_{\lambda }^{-1}$, $\lambda \in [0,1]$, are continuous on $j(\overline{U})$ and $G_{\lambda }(j(\overline{U}))$ respectively, the homotopy

$$H:\overline{U}\times [0,1]\to C^2[0,1]\text{defined by }H(x,\lambda )\equiv H_{\lambda }(x)\equiv L_{\lambda }^{-1}{G}_{\lambda }j(x)$$

is compact. Besides, the equation

$$L_{\lambda }^{-1}{G}_{\lambda }j(x)=x\text{for }x\in \overline{U}\text{yields}{L}_{\lambda }x=G_{\lambda }j(x),$$

which coincides with BVP (3.13) _λ . Thus, the fixed points of $H_{\lambda }(x)$ are solutions to (3.13) _λ . But, from Lemma 3.1 and Lemma 3.2 it follows that the solutions to (3.13) _λ are elements of U. Consequently, $H_{\lambda }(x)$, $\lambda \in [0,1]$, is a fixed point free on ∂U, i.e., $H_{\lambda }(x)$ is an admissible map for all $\lambda \in [0,1]$. Finally, we see that the map $H_0$ is a constant map, i.e., $H_0(x)\equiv l$, where l is the unique solution to the BVP

$$x^{\mathrm{\prime }\mathrm{\prime }}-2x=-x,x^{\mathrm{\prime }}(0)=a,x^{\mathrm{\prime }}(1)=b.$$

From the fact that $l\in U$, it follows that $H_0$ is an essential map (see, [20]). By the Topological transversality theorem (see, [20]), $H_1=L_1^{-1}{G}_{1}j$ is also essential, i.e., problem (3.13)₁ has a $C^2[0,1]$-solution. It is also a solution to (3.1)₁, by Lemma 3.3. To complete the proof, remark that problem (3.1)₁ coincides with the problem (N). □

We conclude with the following example, which illustrates our main result.

Example 4.2 Consider the boundary value problem

$$\begin{array}{c}1-(1.5+t)x^{\mathrm{\prime }\mathrm{\prime }}-t{x^{\mathrm{\prime }\mathrm{\prime }}}^5-cos{x}^{\mathrm{\prime }}+x=0,\hfill \\ x^{\mathrm{\prime }}(0)=0,x^{\mathrm{\prime }}(1)={10}^{-4}.\hfill \end{array}$$

It is clear that for $(t,x,p,q)\in [0,1]\times R^3$ the function

$$f(t,x,p,q)=1-(1.5+t)q-t{q}^5-cosp+x$$

is continuous and $f_x(t,x,p,q)=1$ and $f_q(t,x,p,q)=-1.5-t-5t{q}^4$. Thus H1 holds for $K_x=1$ and $K_q=1.5$.

To verify H2 we choose, for example, $K=0.5$, $M=5$ and $\epsilon =3\cdot {10}^{-5}$. Next we need the constants L and $M_0$. Having in mind that $J_x=[0,5\cdot {10}^{-5}]$ and $J_p=[0,{10}^{-4}]$, from

$$5\cdot {10}^{-5}\le {10}^{-4}-(1-\lambda )x\le {10}^{-4}\text{for }(\lambda ,x)\in [0,1]\times J_x$$

it follows that $maxK|b-a-(1-\lambda )x|=0.5max({10}^{-4}-(1-\lambda )x)=5\cdot {10}^{-5}$. On the other hand, from

$$-2,5\cdot {10}^{-4}-{10}^{-20}\le -(1,5+t)({10}^{-4}-(1-\lambda )x)-t{({10}^{-4}-(1-\lambda )x)}^5\le -7,5\cdot {10}^{-5}$$

for $(\lambda ,t,x)\in {[0,1]}^2\times J_x$ and

$$0\le 1-cosp\le 5\cdot {10}^{-9}\text{for }p\in J_p$$

we have

$$\begin{array}{rcl}-26\cdot {10}^{-5}& <& 1-(1,5+t)({10}^{-4}-(1-\lambda )x)-t{({10}^{-4}-(1-\lambda )x)}^5-cosp+x\\ \le & -2,5\cdot {10}^{-5}+5\cdot {10}^{-9}\end{array}$$

for $(\lambda ,t,x,p)\in {[0,1]}^2\times J_x\times J_p$, which means that for $(\lambda ,t,x,p)\in {[0,1]}^2\times J_x\times J_p$

$$\begin{array}{c}max\left|f(t,x,p,b-a-(1-\lambda )x)\right|=\hfill \\ =max|1-(1,5+t)({10}^{-4}-(1-\lambda )x)-t{({10}^{-4}-(1-\lambda )x)}^5-cosp+x|\le 26\cdot {10}^{-5}.\hfill \end{array}$$

So, $L=max\{26\cdot {10}^{-5},5\cdot {10}^{-5}\}=26\cdot {10}^{-5}$. Then

$$M_0=max\{\frac{e}{e^2-1}({10}^{-4}e+{10}^{-4}),\frac{26\cdot {10}^{-5}}{min\{0.5,1,1.5\}}+5\cdot {10}^{-5}\}=57\cdot {10}^{-5}$$

and we see that for $(t,x,p,q)\in [0,1]\times [-M_0-\epsilon ,M_0+\epsilon ]\times R\times (-\mathrm{\infty },-M)$

$$f(t,x,p,q)+Kq=-(1+t)q-t{q}^5+1-cosp+x>0$$

and

$$f(t,x,p,q)+Kq<0\text{for }(t,x,p,q)\in [0,1]\times [-M_0-\epsilon ,M_0+\epsilon ]\times R\times (M,\mathrm{\infty }).$$

Thus, H2 also holds.

Finally, H3 holds since $f(t,x,p,q)$ and $f_q(t,x,p,q)$ are continuous for $(t,x,p,q)\in [0,1]\times R^3$.

Thus, we can apply Theorem 4.1 to conclude that the considered problem has a solution in $C^2[0,1]$.