In this section, we extend some results in the previous section to the following equation:
\{\begin{array}{l}{x}^{\u2033}(t)=g(x(t))f(t,x(t)),\phantom{\rule{1em}{0ex}}t\in J,\\ {U}_{1}x=0,\phantom{\rule{2em}{0ex}}{U}_{2}x=0,\end{array}
(4.1)
where {U}_{1} and {U}_{2} are linear operators defined as
{U}_{1}x=\sum _{i=1}^{{n}_{1}}{a}_{i}x({\lambda}_{i})+\sum _{j=1}^{{n}_{2}}{b}_{j}{x}^{\prime}({\mu}_{j}),\phantom{\rule{2em}{0ex}}{U}_{2}x=\sum _{s=1}^{{m}_{1}}{c}_{s}x({\nu}_{s})+\sum _{l=1}^{{m}_{2}}{d}_{l}{x}^{\prime}({\kappa}_{l}),
where {a}_{i},{b}_{j},{c}_{s},{d}_{l}\in \mathbb{R} and 0<{\lambda}_{i},{\mu}_{j},{\nu}_{s},{\kappa}_{l}\le 1 for 1\le i\le {n}_{1}, 1\le j\le {n}_{2}, 1\le s\le {m}_{1}, 1\le l\le {m}_{2}.
Consider the boundary value problem for the linear differential equation:
\{\begin{array}{l}{x}^{\u2033}(t)+px(t)=h(t),\phantom{\rule{1em}{0ex}}t\in J,\\ {U}_{1}x=0,\phantom{\rule{2em}{0ex}}{U}_{2}x=0,\end{array}
(4.2)
where p>0 is sufficiently large, h:J\to \mathbb{R}. We introduce the following assumptions.
(P_{1}) The condition h\in C(J,\mathbb{R}) implies that boundary value problem (4.2) has a unique solution {x}_{h}\in {C}^{2}(J,\mathbb{R}).
(P_{2}) The condition h\equiv C\in \mathbb{R} implies that {x}_{h}\equiv C/p on J.
(P_{3}) The condition h\in C(J,\mathbb{R}):h\ge 0 (t\in J) implies that {x}_{h}\ge 0 on J.
(P_{4}) The condition h\in C(J,\mathbb{R}):h(t)\le C (C>0) implies that {x}_{h}\le C/p on J.
We say that the boundary condition {U}_{1}x={U}_{2}x=0 satisfies P_{123} if (4.2) satisfies conditions (P_{1}), (P_{2}), (P_{3}), and P_{134} if (4.2) satisfies conditions (P_{1}), (P_{3}), (P_{4}).
Theorem 4.1

(1)
Assume that the boundary condition {U}_{1}x={U}_{2}x=0 satisfies P_{123} and (H_{1}) holds. Then (4.1) has at least one solution x with m\le x\le M.

(2)
Assume that the boundary condition {U}_{1}x={U}_{2}x=0 satisfies P_{134} and (H_{2}) holds. Then (4.1) has at least one solution x with 0\le x\le M.
The proof of Theorem 4.1 is similar to that of Theorem 3.1 and we omit it.
Remark 4.1 The solution obtained in Corollary 3.2 or (2) of Theorem 4.1 may be trivial. Further suppose that
g(0)f(t,0)\not\equiv 0,\phantom{\rule{1em}{0ex}}t\in J.
Then the solution obtained is nonnegative and nontrivial.
Remark 4.2 The boundary condition {U}_{1}x={U}_{2}x=0 satisfies P_{134} if it satisfies P_{123}.
Remark 4.3 Consider the twopoint boundary conditions:
{x}^{\prime}(0)={x}^{\prime}(1)=0,
(4.3)
x(0)=x(1),\phantom{\rule{2em}{0ex}}{x}^{\prime}(0)={x}^{\prime}(1),
(4.4)
{x}^{\prime}(0)=x(1)=0,
(4.6)
x(0)={x}^{\prime}(1)=0.
(4.7)
One can easily check that boundary conditions (4.3), (4.4) satisfy P_{123}, and conditions (4.5), (4.6), (4.7) satisfy P_{134}.
Next, we consider the boundary conditions
\begin{array}{r}{U}_{1}x:=\alpha x(0)\beta {x}^{\prime}(0)\sum _{i=1}^{k}{b}_{i}x({\eta}_{i})=0,\\ {U}_{2}x:=\lambda x(1)+\mu {x}^{\prime}(1)\sum _{j=1}^{n}{c}_{j}x({\xi}_{j})=0,\end{array}
(4.8)
where α, β, λ, μ, {b}_{i} (1\le i\le k), {c}_{j} (1\le j\le n) are constants and \alpha ,\lambda \in (0,+\mathrm{\infty}), \beta ,\mu ,{b}_{i},{c}_{j}\in [0,+\mathrm{\infty}), {\eta}_{i},{\xi}_{j}\in (0,1).
Theorem 4.2 Set b={\sum}_{i=1}^{k}{b}_{i}, c={\sum}_{j=1}^{n}{c}_{j}.

(1)
If b=\alpha, c=\lambda, then the boundary condition {U}_{1}x={U}_{2}x=0 satisfies P_{123}.

(2)
If b\in [0,\alpha ], c\in [0,\lambda ) or b\in [0,\alpha ), c\in [0,\lambda ], then the boundary condition {U}_{1}x={U}_{2}x=0 satisfies P_{134}.
Proof Without loss of generality, we assume that m=n=1. For any sufficiently large p>0 and h\in C(J,\mathbb{R}), the linear differential equation
\{\begin{array}{l}{x}^{\u2033}(t)+px(t)=h(t),\phantom{\rule{1em}{0ex}}t\in J,\\ {U}_{1}x=0,\phantom{\rule{2em}{0ex}}{U}_{2}x=0\end{array}
(4.9)
has a unique solution {x}_{h}\in {C}^{2}(J,\mathbb{R}).
Suppose that h\ge 0 on J; if {x}_{h}\ge 0 is not true, by the maximum principle, we get that {x}_{h}(0)={min}_{t\in J}{x}_{h}(t)<0 or {x}_{h}(1)={min}_{t\in J}x(t)<0. If {x}_{h}(0)={min}_{t\in J}{x}_{h}(t)<0, then {x}_{h}^{\prime}(0)\ge 0. From the boundary conditions, we have
0>{x}_{h}(0)=\frac{\beta}{\alpha}{x}_{h}^{\prime}(0)+\frac{{b}_{1}}{\alpha}{x}_{h}({\eta}_{1})\ge \frac{{b}_{1}}{\alpha}{x}_{h}({\eta}_{1})\ge {x}_{h}({\eta}_{1}).
By the maximum principle, we obtain that {x}_{h}({\eta}_{1})\ge 0, which is a contradiction. If {x}_{h}(1)={min}_{t\in J}{x}_{h}(t)<0, then {x}_{h}^{\prime}(1)\le 0. From the boundary conditions, we have
0>{x}_{h}(1)=\frac{{c}_{1}}{\lambda}{x}_{h}({\xi}_{1})\frac{\mu}{\lambda}{x}_{h}^{\prime}(1)\ge \frac{{c}_{1}}{\lambda}{x}_{h}({\xi}_{1})\ge {x}_{h}({\xi}_{1}).
By the maximum principle, we obtain that {x}_{h}({\xi}_{1})\ge 0, a contradiction.
If h\equiv C and b=\alpha, c=\lambda, then {x}_{h}\equiv C/p.
Now suppose that 0\le h\le C on J and h\not\equiv 0.
If there is \theta \in (0,1) such that {x}_{h}(\theta )={max}_{t\in J}x(t)>0, noting that {x}_{h}^{\u2033}(\theta )\le 0, we have
0\le p{x}_{h}(\theta )\le {x}_{h}^{\u2033}(\theta )+p{x}_{h}(\theta )=h(\theta )\le C.
If {x}_{h}(0)={max}_{t\in J}x(t)>0, from the boundary conditions, we obtain that b=\alpha, \beta {x}_{h}^{\prime}(0)=0, which implies that {x}_{h}({\eta}_{1})={x}_{h}(0)={max}_{t\in J}x(t). The case has been discussed. If {x}_{h}(1)={max}_{t\in J}x(t)>0, from the boundary conditions, we obtain that c=\lambda, \mu {x}_{h}^{\prime}(1)=0, which implies that {x}_{h}({\xi}_{1})={x}_{h}(1)={max}_{t\in J}x(t). The case has also been discussed. The proof is complete. □
Example 4.1 Consider the differential equation
\{\begin{array}{l}{x}^{\u2033}(t)={sin}^{1}x(t){x}^{\lambda}(t)+tx(t),\phantom{\rule{1em}{0ex}}t\in J,\\ x(0)=x(\eta ),\phantom{\rule{2em}{0ex}}x(1)=x(\xi ),\end{array}
(4.10)
where \lambda >0, 0<\eta ,\xi <1 are constants.
Let g(u)={sin}^{1}u and f(t,u)={u}^{\lambda}tu. Set {m}_{n}=2n\pi +{(4n\pi )}^{\lambda 1}, {M}_{n}=2n\pi +0.5\pi. If n is a sufficiently large, positive integer, then for any t\in J, u\in [{m}_{n},{M}_{n}],
1=g({M}_{n})\le f(t,u)\le g({m}_{n})\approx {(4n\pi )}^{\lambda +1}.
By Theorems 4.1 and 4.2, (4.10) has a solution {m}_{n}\le x\le {M}_{n}. Hence, (4.10) has infinitely many solutions.
Example 4.2 Consider the differential equation
\{\begin{array}{l}{x}^{\u2033}(t)+{x}^{\mu}(t)cosx(t)={t}^{2},\phantom{\rule{1em}{0ex}}t\in J,\\ x(0){x}^{\prime}(0)=x({\xi}_{1}),\phantom{\rule{2em}{0ex}}x(1)=\frac{1}{2}x({\xi}_{2})+\lambda x({\xi}_{3}),\end{array}
(4.11)
where \mu >0, 0<{\xi}_{1},{\xi}_{2},{\xi}_{3}<1, 0\le \lambda \le 0.5 are constants.
In fact, g(u)={u}^{\mu}cosu and f(t)={t}^{2}. The boundary conditions in (4.11) satisfy P_{134} for \lambda \in [0,0.5] and P_{123} for \lambda =0.5.

(1)
Equation (4.11) has a solution 0\le \tilde{x}\le 1 and \tilde{x}(t)>0, t\in (0,1] for all 0\le \lambda \le 0.5. Set M=1, then g(M)=1\le f(t)\le g(0) for all t\in J. By Theorems 4.1 and 4.2, (4.11) has a solution 0\le \tilde{x}\le 1. Now we show that \tilde{x}(t)>0 for t\in (0,1]. Assume that there exists r\in (0,1) with \tilde{x}(r)=0. Since \tilde{x}(t)\ge 0, \tilde{x}(r) is minimum value and {\tilde{x}}^{\prime}(r)=0, {\tilde{x}}^{\u2033}(r)\ge 0. On the other hand, {\tilde{x}}^{\u2033}(r)={\tilde{x}}^{\mu}(r)cos\tilde{x}(r){r}^{2}={r}^{2}<0, a contradiction. If \tilde{x}(1)=0, then \tilde{x}({\xi}_{2})=0. This is impossible.

(2)
Equation (4.11) has infinitely many solutions for \lambda =0.5. Set {M}_{n}=2n\pi +2\pi, {m}_{n}=2n\pi +1.5\pi, where n>0 is an integer. Since g({M}_{n})={M}_{n}^{\mu}\le f(t)\le g({m}_{n})={m}_{n}^{\mu} for all t\in J, (4.11) has a solution {m}_{n}\le x\le {M}_{n}. Hence, (4.11) has infinitely many solutions.