**Appendix A**

In this part we show how the barrier strips can be used for studying the existence of positive or non-negative, monotone, convex or concave *C*<sup>3</sup>[0, 1] - solutions. Here, we demonstrate this on problem (1), (4) but it can be done for the rest of the BVPs considered in this paper. Similar results for various other two-point boundary conditions can be found in R. Agarwal and P. Kelevedjiev [16] and P. Kelevedjiev and T. Todorov [15].

**Lemma A1.** *Let A*, *B* ≥ 0, *C* ≤ 0. *Suppose* **(H1)** *holds for K* = *C with L*-1 ≤ 0. *Then each solution x* ∈ *C*<sup>3</sup>[0, 1] *to* (1)*<sup>λ</sup>,* (4) *satisfies the bounds*

$$
\min\{A, B\} \le \mathfrak{x}(t) \le A + |B - A| + |F\_1'| \; t \in [0, 1].
$$

$$
B - A + F\_1' \le \mathfrak{x}'(t) \le B - A - F\_1' \; t \in [0, 1].
$$

**Proof.** From Lemma 2 we know that *F*-1 ≤ *x*--(*t*) ≤ *L*-1 for *t* ∈ [0, 1]. Besides, for some *μ* ∈ (0, 1) we have *x*-(*μ*) = *B* − *A*. Then,

$$\int\_{t}^{\mu} F\_{1}' ds \le \int\_{t}^{\mu} \mathfrak{x}''(s) ds \le \int\_{t}^{\mu} L\_{1}' ds, t \in [0, \mu).$$

gives

$$B - A \le \mathfrak{x}'(t) \le B - A - F\_{1'}'t \in [0, \mu]\_{\prime \prime}$$

and

$$\int\_{\mu}^{t} F\_1' ds \le \int\_{\mu}^{t} \mathbf{x}^{\prime\prime}(s) ds \le \int\_{\mu}^{t} L\_1^{\prime} ds, t \in (\mu, 1]\_{\star}$$

implies

$$B - A + F\_1' \le \mathfrak{x}'(t) \le B - A, t \in [\mathfrak{u}, 1].$$

As a result,

$$B - A + F\_1' \le \mathfrak{x}'(t) \le B - A - F\_{1'}'t \in [0, 1].$$

Using Lemma 5, conclude

$$|\mathbf{x}(t)| \le A + |B - A| + |F\_1'| \text{ for } t \in [0, 1].$$

From *x*--(*t*) ≤ *L*-1 ≤ 0 for *t* ∈ [0, 1] it follows that *x*(*t*) is concave on [0, 1] and so, in view of *A*, *B* ≥ 0, *x*(*t*) ≥ min{*<sup>A</sup>*, *B*} on [0, 1], which completes the proof.

**Theorem A1.** *Let A* ≥ *B* ≥ 0 *and C* ≤ 0 *(A* ≥ *B* > 0 *and C* < <sup>0</sup>). *Suppose* **(H1)** *holds for K* = *C with B* − *A* ≤ *F*-1 (*B* − *A* < *F*-1) *and L*-1 ≤ 0, *and* **(H3)** *holds for*

$$m\_0 = B,\\ M\_0 = 2A - B + |F\_1'| \, \prime$$

$$m\_1 = B - A + F\_1',\\ M\_1 = B - A - F\_1',\\ m\_2 = F\_1',\\ M\_2 = L\_1'$$

*Then BVP* (1)*,* (4) *has at least one non-negative, non-increasing (positive, decreasing), concave solution in C*<sup>3</sup>[0, 1].

1.

**Proof.** By Lemma 5, for every solution *x* ∈ *C*<sup>3</sup>[0, 1] to (1)*<sup>λ</sup>*, (4) we have *F*-1 ≤ *x*--(*t*) ≤ *L*-1 on [0, 1], and Lemma A1 yields

$$B - A + F\_1' \le \mathbf{x}'(t) \le B - A - F\_{1'}' \ t \in [0, 1]$$

min{*<sup>A</sup>*, *B*} ≤ *x*(*t*) ≤ *A* + |*B* − *A*| + |*F*-1|, *t* ∈ [0, 1].

Because of *A* ≥ *B*, the last inequality gets the form

$$B \le \mathfrak{x}(t) \le 2A - B + |F\_1'|\_\prime \ t \in [0, 1].$$

So, *x*(*t*) satusfies the bounds

$$m\_0 \le \mathfrak{x}^{(i)}(t) \le M\_{0\prime}t \in [0,1], i = 0,1,2.$$

Essentially the same reasoning as in the proof of Theorem 1 establishes that (1), (4) has a solution in *C*<sup>3</sup>[0, 1]. Since *m*0 = *B* ≥ <sup>0</sup>(*<sup>m</sup>*0 > <sup>0</sup>), *M*1 = *B* − *A* − *F*-1 ≤ <sup>0</sup>(*<sup>M</sup>*1 < 0) and *M*2 = *L*-1 ≤ 0, this solution has the desired properties.
