In this section we consider the existence of uncountably many positive solutions for Eq. (1) which are bounded by two positive functions. We use the notation m=max\{\tau ,\sigma \}.

**Theorem 2.1** *Suppose that there exist bounded from below and from above by the functions* *u* *and* v\in {C}^{1}([{t}_{0},\mathrm{\infty}),(0,\mathrm{\infty})) *constants* c>0, {K}_{2}>{K}_{1}\ge 0 *and* {t}_{1}\ge {t}_{0}+m *such that*

u(t)\le v(t),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},

(2)

v(t)-v({t}_{1})-u(t)+u({t}_{1})\ge 0,\phantom{\rule{1em}{0ex}}{t}_{0}\le t\le {t}_{1},

(3)

\begin{array}{r}\frac{1}{u(t-\tau )}(u(t)-{K}_{1}+{\int}_{t}^{\mathrm{\infty}}p(s)f(v(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds)\\ \phantom{\rule{1em}{0ex}}\le a(t)\le \frac{1}{v(t-\tau )}(v(t)-{K}_{2}+{\int}_{t}^{\mathrm{\infty}}p(s)f(u(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds)\le c<1,\phantom{\rule{1em}{0ex}}t\ge {t}_{1}.\end{array}

(4)

*Then Eq*. (1) *has uncountably many positive solutions which are bounded by the functions* *u*, *v*.

*Proof* Let C([{t}_{0},\mathrm{\infty}),R) be the set of all continuous bounded functions with the norm \parallel x\parallel ={sup}_{t\ge {t}_{0}}|x(t)|. Then C([{t}_{0},\mathrm{\infty}),R) is a Banach space. We define a closed, bounded and convex subset Ω of C([{t}_{0},\mathrm{\infty}),R) as follows:

\mathrm{\Omega}=\{x=x(t)\in C([{t}_{0},\mathrm{\infty}),R):u(t)\le x(t)\le v(t),t\ge {t}_{0}\}.

For K\in [{K}_{1},{K}_{2}] we define two maps {S}_{1} and {S}_{2}:\mathrm{\Omega}\to C([{t}_{0},\mathrm{\infty}),R) as follows:

({S}_{1}x)(t)=\{\begin{array}{ll}K+a(t)x(t-\tau ),& t\ge {t}_{1},\\ ({S}_{1}x)({t}_{1}),& {t}_{0}\le t\le {t}_{1},\end{array}

(5)

({S}_{2}x)(t)=\{\begin{array}{ll}-{\int}_{t}^{\mathrm{\infty}}p(s)f(x(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds,& t\ge {t}_{1},\\ ({S}_{2}x)({t}_{1})+v(t)-v({t}_{1}),& {t}_{0}\le t\le {t}_{1}.\end{array}

(6)

We will show that for any x,y\in \mathrm{\Omega}, we have {S}_{1}x+{S}_{2}y\in \mathrm{\Omega}. For every x,y\in \mathrm{\Omega} and t\ge {t}_{1} with regard to (4), we obtain

\begin{array}{rcl}({S}_{1}x)(t)+({S}_{2}y)(t)& =& K+a(t)x(t-\tau )-{\int}_{t}^{\mathrm{\infty}}p(s)f(y(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds\\ \le & K+a(t)v(t-\tau )-{\int}_{t}^{\mathrm{\infty}}p(s)f(u(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds\\ \le & K+v(t)-{K}_{2}\le v(t).\end{array}

For t\in [{t}_{0},{t}_{1}] we have

\begin{array}{rcl}({S}_{1}x)(t)+({S}_{2}y)(t)& =& ({S}_{1}x)({t}_{1})+({S}_{2}y)({t}_{1})+v(t)-v({t}_{1})\\ \le & v({t}_{1})+v(t)-v({t}_{1})=v(t).\end{array}

Furthermore, for t\ge {t}_{1} we get

\begin{array}{rcl}({S}_{1}x)(t)+({S}_{2}y)(t)& \ge & K+a(t)u(t-\tau )-{\int}_{t}^{\mathrm{\infty}}p(s)f(v(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds\\ \ge & K+u(t)-{K}_{1}\ge u(t).\end{array}

(7)

Let t\in [{t}_{0},{t}_{1}]. With regard to (3), we get

v(t)-v({t}_{1})+u({t}_{1})\ge u(t),\phantom{\rule{1em}{0ex}}{t}_{0}\le t\le {t}_{1}.

Then, for t\in [{t}_{0},{t}_{1}] and any x,y\in \mathrm{\Omega}, we obtain

\begin{array}{rcl}({S}_{1}x)(t)+({S}_{2}y)(t)& =& ({S}_{1}x)({t}_{1})+({S}_{2}y)({t}_{1})+v(t)-v({t}_{1})\\ \ge & u({t}_{1})+v(t)-v({t}_{1})\ge u(t).\end{array}

Thus we have proved that {S}_{1}x+{S}_{2}y\in \mathrm{\Omega} for any x,y\in \mathrm{\Omega}.

We will show that {S}_{1} is a contraction mapping on Ω. For x,y\in \mathrm{\Omega} and t\ge {t}_{1}, we have

|({S}_{1}x)(t)-({S}_{1}y)(t)|=|a(t)||x(t-\tau )-y(t-\tau )|\le c\parallel x-y\parallel .

This implies that

\parallel {S}_{1}x-{S}_{1}y\parallel \le c\parallel x-y\parallel .

Also, for t\in [{t}_{0},{t}_{1}] the inequality above is valid. We conclude that {S}_{1} is a contraction mapping on Ω.

We now show that {S}_{2} is completely continuous. First, we show that {S}_{2} is continuous. Let {x}_{k}={x}_{k}(t)\in \mathrm{\Omega} be such that {x}_{k}(t)\to x(t) as k\to \mathrm{\infty}. Because Ω is closed, x=x(t)\in \mathrm{\Omega}. For t\ge {t}_{1} we have

\begin{array}{rcl}|({S}_{2}{x}_{k})(t)-({S}_{2}x)(t)|& \le & |{\int}_{t}^{\mathrm{\infty}}p(s)[f({x}_{k}(s-\sigma ))-f(x(s-\sigma ))]\phantom{\rule{0.2em}{0ex}}ds|\\ \le & {\int}_{{t}_{1}}^{\mathrm{\infty}}p(s)|f({x}_{k}(s-\sigma ))-f(x(s-\sigma ))|\phantom{\rule{0.2em}{0ex}}ds.\end{array}

According to (7), we get

{\int}_{{t}_{1}}^{\mathrm{\infty}}p(s)f(v(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds<\mathrm{\infty}.

(8)

Since |f({x}_{k}(s-\sigma ))-f(x(s-\sigma ))|\to 0 as k\to \mathrm{\infty}, by applying the Lebesgue dominated convergence theorem, we obtain that

\underset{k\to \mathrm{\infty}}{lim}\parallel ({S}_{2}{x}_{k})(t)-({S}_{2}x)(t)\parallel =0.

This means that {S}_{2} is continuous.

We now show that {S}_{2}\mathrm{\Omega} is relatively compact. It is sufficient to show by the Arzela-Ascoli theorem that the family of functions \{{S}_{2}x:x\in \mathrm{\Omega}\} is uniformly bounded and equicontinuous on [{t}_{0},\mathrm{\infty}). The uniform boundedness follows from the definition of Ω. For the equicontinuity, we only need to show, according to the Levitan result [7], that for any given \epsilon >0, the interval [{t}_{0},\mathrm{\infty}) can be decomposed into finite subintervals in such a way that on each subinterval all functions of the family have a change of amplitude less than *ε*. Then, with regard to condition (8), for x\in \mathrm{\Omega} and any \epsilon >0, we take {t}^{\ast}\ge {t}_{1} large enough so that

{\int}_{{t}^{\ast}}^{\mathrm{\infty}}p(s)f(x(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds<\frac{\epsilon}{2}.

Then, for x\in \mathrm{\Omega}, {T}_{2}>{T}_{1}\ge {t}^{\ast}, we have

\begin{array}{rl}|({S}_{2}x)({T}_{2})-({S}_{2}x)({T}_{1})|\le & {\int}_{{T}_{2}}^{\mathrm{\infty}}p(s)f(x(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds\\ +{\int}_{{T}_{1}}^{\mathrm{\infty}}p(s)f(x(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon .\end{array}

For x\in \mathrm{\Omega} and {t}_{1}\le {T}_{1}<{T}_{2}\le {t}^{\ast}, we get

\begin{array}{rl}|({S}_{2}x)({T}_{2})-({S}_{2}x)({T}_{1})|& \le {\int}_{{T}_{1}}^{{T}_{2}}p(s)f(x(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds\\ \le \underset{{t}_{1}\le s\le {t}^{\ast}}{max}\{p(s)f(x(s-\sigma ))\}({T}_{2}-{T}_{1}).\end{array}

Thus there exists {\delta}_{1}=\frac{\epsilon}{M}, where M={max}_{{t}_{1}\le s\le {t}^{\ast}}\{p(s)f(x(s-\sigma ))\}, such that

|({S}_{2}x)({T}_{2})-({S}_{2}x)({T}_{1})|<\epsilon \phantom{\rule{1em}{0ex}}\text{if}0{T}_{2}-{T}_{1}{\delta}_{1}.

Finally, for any x\in \mathrm{\Omega}, {t}_{0}\le {T}_{1}<{T}_{2}\le {t}_{1}, there exists a {\delta}_{2}>0 such that

\begin{array}{rcl}|({S}_{2}x)({T}_{2})-({S}_{2}x)({T}_{1})|& =& |v({T}_{1})-v({T}_{2})|=|{\int}_{{T}_{1}}^{{T}_{2}}{v}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}ds|\\ \le & \underset{{t}_{0}\le s\le {t}_{1}}{max}\left\{|{v}^{\prime}(s)|\right\}({T}_{2}-{T}_{1})<\epsilon \phantom{\rule{1em}{0ex}}\text{if}0{T}_{2}-{T}_{1}{\delta}_{2}.\end{array}

Then \{{S}_{2}x:x\in \mathrm{\Omega}\} is uniformly bounded and equicontinuous on [{t}_{0},\mathrm{\infty}), and hence {S}_{2}\mathrm{\Omega} is a relatively compact subset of C([{t}_{0},\mathrm{\infty}),R). By Lemma 1.1 there is an {x}_{0}\in \mathrm{\Omega} such that {S}_{1}{x}_{0}+{S}_{2}{x}_{0}={x}_{0}. We conclude that {x}_{0}(t) is a positive solution of (1).

Next we show that Eq. (1) has uncountably many bounded positive solutions in Ω. Let the constant \overline{K}\in [{K}_{1},{K}_{2}] be such that \overline{K}\ne K. We infer similarly that there exist mappings {\overline{S}}_{1}, {\overline{S}}_{2} satisfying (5), (6), where *K*, {S}_{1}, {S}_{2} are replaced by \overline{K}, {\overline{S}}_{1}, {\overline{S}}_{2}, respectively. We assume that x,y\in \mathrm{\Omega}, {S}_{1}x+{S}_{2}x=x, {\overline{S}}_{1}y+{\overline{S}}_{2}y=y, which are the bounded positive solutions of Eq. (1), that is,

\begin{array}{c}x(t)=K+a(t)x(t-\tau )-{\int}_{t}^{\mathrm{\infty}}p(s)f(x(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\ge {t}_{1},\hfill \\ y(t)=\overline{K}+a(t)y(t-\tau )-{\int}_{t}^{\mathrm{\infty}}p(s)f(y(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\ge {t}_{1}.\hfill \end{array}

From condition (8) it follows that there exists a {t}_{2}>{t}_{1} satisfying

{\int}_{{t}_{2}}^{\mathrm{\infty}}p(s)[f(x(s-\sigma ))+f(y(s-\sigma ))]\phantom{\rule{0.2em}{0ex}}ds<|K-\overline{K}|.

(9)

In order to prove that the set of bounded positive solutions of Eq. (1) is uncountable, it is sufficient to verify that x\ne y. For t\ge {t}_{2} we get

\begin{array}{r}|x(t)-y(t)|\\ \phantom{\rule{1em}{0ex}}=|K+a(t)x(t-\tau )-{\int}_{t}^{\mathrm{\infty}}p(s)f(x(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds\\ \phantom{\rule{2em}{0ex}}-\overline{K}-a(t)y(t-\tau )+{\int}_{t}^{\mathrm{\infty}}p(s)f(y(s-\sigma ))\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}\ge |K-\overline{K}+a(t)[x(t-\tau )-y(t-\tau )]\\ \phantom{\rule{2em}{0ex}}-{\int}_{t}^{\mathrm{\infty}}p(s)[f(x(s-\sigma ))-f(y(s-\sigma ))]\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}\ge |K-\overline{K}|-a(t)\parallel x-y\parallel -|{\int}_{t}^{\mathrm{\infty}}p(s)[f(x(s-\sigma ))-f(y(s-\sigma ))]\phantom{\rule{0.2em}{0ex}}ds|\\ \phantom{\rule{1em}{0ex}}\ge |K-\overline{K}|-c\parallel x-y\parallel -{\int}_{t}^{\mathrm{\infty}}p(s)[f(x(s-\sigma ))+f(y(s-\sigma ))]\phantom{\rule{0.2em}{0ex}}ds.\end{array}

Then we have

(1+c)\parallel x-y\parallel \ge |K-\overline{K}|-{\int}_{t}^{\mathrm{\infty}}p(s)[f(x(s-\sigma ))+f(y(s-\sigma ))]\phantom{\rule{0.2em}{0ex}}ds,\phantom{\rule{1em}{0ex}}t\ge {t}_{2}.

According to (9) we get that x\ne y. Since the interval [{K}_{1},{K}_{2}] contains uncountably many constants, then Eq. (1) has uncountably many positive solutions which are bounded by the functions u(t), v(t). This completes the proof. □

**Corollary 2.1** *Suppose that there exist bounded from below and from above by the functions* *u* *and* v\in {C}^{1}([{t}_{0},\mathrm{\infty}),(0,\mathrm{\infty})) *constants* c>0, {K}_{2}>{K}_{1}\ge 0 *and* {t}_{1}\ge {t}_{0}+m *such that* (2), (4) *hold and*

{v}^{\prime}(t)-{u}^{\prime}(t)\le 0,\phantom{\rule{1em}{0ex}}{t}_{0}\le t\le {t}_{1}.

(10)

*Then Eq*. (1) *has uncountably many positive solutions which are bounded by the functions* *u*, *v*.

*Proof* We only need to prove that condition (10) implies (3). Let t\in [{t}_{0},{t}_{1}] and set

H(t)=v(t)-v({t}_{1})-u(t)+u({t}_{1}).

Then, with regard to (10), it follows that

{H}^{\prime}(t)={v}^{\prime}(t)-{u}^{\prime}(t)\le 0,\phantom{\rule{1em}{0ex}}{t}_{0}\le t\le {t}_{1}.

Since H({t}_{1})=0 and {H}^{\prime}(t)\le 0 for t\in [{t}_{0},{t}_{1}], this implies that

H(t)=v(t)-v({t}_{1})-u(t)+u({t}_{1})\ge 0,\phantom{\rule{1em}{0ex}}{t}_{0}\le t\le {t}_{1}.

Thus all the conditions of Theorem 2.1 are satisfied. □

**Example 2.1** Consider the nonlinear neutral differential equation

{[x(t)-a(t)x(t-2)]}^{\prime}=p(t){x}^{3}(t-1),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},

(11)

where p(t)={e}^{-t}. We will show that the conditions of Corollary 2.1 are satisfied. The functions u(t)=0.5, v(t)=2 satisfy (2) and also condition (10) for t\in [{t}_{0},{t}_{1}]=[0,4]. For the constants {K}_{1}=0.5, {K}_{2}=1, condition (4) has the form

16{e}^{-t}\le a(t)\le \frac{1}{2}+\frac{1}{16}{e}^{-t},\phantom{\rule{1em}{0ex}}t\ge {t}_{1}=4.

(12)

If the function a(t) satisfies (12), then Eq. (11) has uncountably many positive solutions which are bounded by the functions *u*, *v*.

**Example 2.2** Consider the nonlinear neutral differential equation

{[x(t)-a(t)x(t-\tau )]}^{\prime}=p(t){x}^{2}(t-\sigma ),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},

(13)

where \tau ,\sigma \in (0,\mathrm{\infty}), p(t)={e}^{-3t}. We will show that the conditions of Corollary 2.1 are satisfied. The functions u(t)={e}^{-2t}, v(t)={e}^{\tau}+{e}^{-t}, t\ge 1, satisfy (2) and since

{v}^{\prime}(t)-{u}^{\prime}(t)={e}^{-t}(2{e}^{-t}-1)<0\phantom{\rule{1em}{0ex}}\text{for}t\in [1,2],

condition (10) is also satisfied. For the constants {K}_{1}=0, {K}_{2}={e}^{\tau}-1, condition (4) has the form

{e}^{-2\tau}+\frac{1}{3}{e}^{-t}+\frac{1}{2}{e}^{-2t+\sigma -\tau}+\frac{1}{5}{e}^{-3t+2(\sigma -\tau )}\le a(t)\le {e}^{-\tau}+\frac{{e}^{-7t+4\sigma -\tau}}{7(1+{e}^{-t})},\phantom{\rule{1em}{0ex}}t\ge 2.

For \tau =\sigma =1 we get

{e}^{-2}+\frac{1}{3}{e}^{-t}+\frac{1}{2}{e}^{-2t}+\frac{1}{5}{e}^{-3t}\le a(t)\le {e}^{-1}+\frac{{e}^{-7t+3}}{7(1+{e}^{-t})},\phantom{\rule{1em}{0ex}}t\ge {t}_{1}=2.

(14)

If the function a(t) satisfies (14), then Eq. (13) has uncountably many solutions which are bounded by the functions *u*, *v*.

**Example 2.3** Consider the nonlinear neutral differential equation

{[x(t)-a(t)x(t-\tau )]}^{\prime}=p(t){x}^{3}(t-\sigma ),\phantom{\rule{1em}{0ex}}t\ge {t}_{0},

(15)

where \tau ,\sigma \in (0,\mathrm{\infty}), p(t)={e}^{-t}. We will show that the conditions of Corollary 2.1 are satisfied. The functions u(t)={e}^{-t}, v(t)={e}^{\tau}+2{e}^{-t}, t\ge 1 satisfy (2) and also (10)

{v}^{\prime}(t)-{u}^{\prime}(t)=-{e}^{-t}<0\phantom{\rule{1em}{0ex}}\text{for}t\in [1,3.2].

For the constants {K}_{1}=1, {K}_{2}={e}^{\tau}-1, where \tau >ln2, t\ge 3.2, condition (4) has the form

{e}^{-\tau}(1-{e}^{t}+{e}^{3\tau}+3{e}^{2\tau +\sigma -t}+4{e}^{\tau +2\sigma -2t}+2{e}^{3\sigma -3t})\le a(t)\le {e}^{-\tau}+\frac{{e}^{3\sigma -\tau -4t}}{4(1+2{e}^{-t})}.

For \tau =\sigma =1 and t\ge 3.2, we have

{e}^{-1}(1-{e}^{t}+{e}^{3}+3{e}^{3-t}+4{e}^{3-2t}+2{e}^{3(1-t)})<0.

Then for a(t), which satisfies the inequalities

0<a(t)\le {e}^{-1}+\frac{{e}^{2(1-2t)}}{4(1+2{e}^{-t})},\phantom{\rule{1em}{0ex}}t\ge {t}_{1}\ge 3.2,

(16)

Eq. (11) has uncountably many solutions which are bounded by the functions *u*, *v*.