**4. Examples**

Through several examples we will illustrate the application of the obtained results.

**Example 1.** *Consider the BVPs for the equation*

$$\mathbf{x}^{\prime\prime\prime}(t) = \exp(\mathbf{x}^{\prime\prime} - \mathbf{3}) + \dots \mathbf{5} \mathbf{x}^{\prime\prime}(\mathbf{x}^{\prime 2} + 1) - t \sin \mathbf{x}, t \in (0, 1),$$

*with boundary conditions* (2) *or* (3)*.*

For *F*- 2 = −| *C*| − 2, *F*- 1 = −| *C*| − 1, *L*- 1 = max{| *<sup>C</sup>*|, 3} + 1, *L*- 2 = max{| *<sup>C</sup>*|, 3} + 2 and *σ* = 0.1, for example, each of these problems has a solution in *C*<sup>3</sup>[0, 1] by Theorem 1.

**Example 2.** *Consider the BVP*

$$\mathbf{x}^{\prime\prime\prime}(t) = \wp(t, \mathbf{x}, \mathbf{x}^{\prime}) \Big( \lg\big( (\mathbf{x}^{\prime\prime} + 50)(60 - \mathbf{x}^{\prime\prime}) \big) - 3 \Big), t \in (0, 1),$$

$$\mathbf{x}(0) = \mathbf{5}, \mathbf{x}^{\prime}(0) = 10, \mathbf{x}^{\prime}(1) = 40,$$

*where ϕ* : [0, 1] × R<sup>2</sup> → R *is continuous and does not change its sign.*

If *ϕ*(*<sup>t</sup>*, *x*, *p*) ≥ 0 on [0, 1] × R2, the assumptions of Theorem 3 are satisfied for *F*2 = −36, *F*1 = −35, *F*- 2 = −46, *F*- 1 = −45, *L*- 1 = 40, *L*- 2 = 41, *L*1 = 55, *L*2 = 56 and *σ* = 0.01, for example, and if *ϕ*(*<sup>t</sup>*, *x*, *p*) ≤ 0 on [0, 1] × R2, they are satisfied for *F*- 2 = −36, *F*- 1 = −35, *F*2 = −46, *F*1 = −45, *L*1 = 40, *L*2 = 41, *L*- 1 = 55, *L*- 2 = 56 and *σ* = 0.01, for example; it is clear, *K* = 30. Thus, the considered problem has at least one solution in *C*<sup>3</sup>[0, 1]. Let us note, here *Dq* = ( −50, <sup>60</sup>).

**Example 3.** *Consider the BVP*

$$\mathbf{x}^{\prime\prime\prime}(t) = \frac{t(\mathbf{x}^{\prime\prime} + 8)(\mathbf{x}^{\prime\prime} + 3)\sqrt{625 - \mathbf{x}^{\prime2}}}{\sqrt{900 - \mathbf{x}^2}\sqrt{100 - \mathbf{x}^{\prime\prime2}}}, t \in (0, 1),$$

$$\mathbf{x}(0) = 9, \mathbf{x}(1) = 1, \mathbf{x}^{\prime\prime}(1) = -4.$$

For *F*- 2 = −6, *F*- 1 = −5, *L*- 1 = −3, *L*- 2 = −2 and *σ* = 0.1, for example, this problem has a positive, decreasing, concave solution in *C*<sup>3</sup>[0, 1] by Theorem A1; notice, here *Dx*, *Dp* and *Dq* are bounded.

**Author Contributions:** All authors contributed equally. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Conflicts of Interest:** The authors declare no conflict of interest.
