This section is dedicated to present an application of the measure of noncompactness μ introduced in the previous section.
We will investigate the solvability of a quadratic Hammerstein integral equation of the form

#### Proof.

Let us consider the operator *T* associated with equation (4.1). This means that *T* is defined on the space $R([a,b])$ by the formula

$(Tx)(t)=p(t)+g(t,x(t)){\int}_{a}^{b}k(t,s)f(s,x(s))\mathit{d}s.$(4.2)

For further purposes, let us consider the operators *G*, *F* and *K* defined in the following way:

$(Gx)(t)=g(t,x(t)),(Fx)(t)=f(t,x(t)),(Kx)(t)={\int}_{a}^{b}k(t,s)x(s)\mathit{d}s.$

Then the operator *T* defined by (4.2) can be represented as

$Tx=p+(Gx)(K\cdot F)(x),$(4.3)

where the symbol $K\cdot F$ is understood as the composition of the operators *F* and *K*.

Now, let us fix arbitrarily $\epsilon >0$ and $t\in (a,b]$.
Then, for an arbitrary function $x\in R([a,b])$ and for arbitrary $u,v\in (t-\epsilon ,t)$, on the basis of assumption (ii), we obtain

$|(Gx)(u)-(Gx)(v)|=|g(u,x(u))-g(v,x(v))|$$\le |g(u,x(u))-g(u,x(v))|+|g(u,x(v))-g(v,x(v))|$$\le L|x(u)-x(v)|+{\omega}_{1,\parallel x\parallel}^{-}(g,t;\epsilon ),$(4.4)

where

${\omega}_{1,r}^{-}(g,t;\epsilon )=sup\{|g(u,x)-g(v,x)|:u,v\in (t-\epsilon ,t)\cap [a,b],x\in [-r,r]\}.$

In view of assumption (ii), we infer that ${\omega}_{1,\parallel x\parallel}^{-}(g,t;\epsilon )\to 0$ as $\epsilon \to 0$.
Particularly this implies that the operator *G* transforms the space $R([a,b])$ into itself.

In a similar way, taking into account assumption (iii), we derive

$|(Kx)(u)-(Kx)(v)|\le {\int}_{a}^{b}|k(u,s)-k(v,s)||x(s)|ds\le \parallel x\parallel {\int}_{a}^{b}|k(u,s)-k(v,s)|ds\le \parallel x\parallel {\omega}_{1}^{-}(k,t;\epsilon )(b-a),$(4.5)

where

${\omega}_{1}^{-}(k,t;\epsilon )=sup\{|k(u,s)-k(v,s)|:u,v\in (t-\epsilon ,t)\cap [a,b],s\in [a,b]\}.$

Notice that, in view of assumption (iii), we have that ${\omega}_{1}^{-}(k,t;\epsilon )\to 0$ (similarly, ${\omega}_{1}^{+}(k,t;\epsilon )\to 0$) as $\epsilon \to 0$ for each $t\in [a,b]$.
This implies that the function $t\mapsto {\int}_{a}^{b}k(t,s)\mathit{d}s$ (or the function $t\mapsto {\int}_{a}^{b}|k(t,s)|ds$) is regulated on the interval $[a,b]$. Consequently, it follows that the function $t\mapsto {\int}_{a}^{b}|k(t,s)|ds$ is bounded on the interval $[a,b]$ and simultaneously justifies the fact that $\overline{k}<\mathrm{\infty}$, which was stated before.

Now, from (4.4), (4.5), assumption (iv) and taking into account representation (4.3), we conclude that the operator *T* transforms the space $R([a,b])$ into itself. Obviously, the above reasoning can be repeated for any fixed $t\in [a,b)$ and for $u,v\in (t,t+\epsilon )\cap [a,b]$, if we replace the quantity ${\omega}^{-}$ by ${\omega}^{+}$.

Let us notice that by utilizing our assumptions, for an arbitrarily fixed $x\in R([a,b])$ and $t\in [a,b]$, we have

$|(Tx)(t)|\le |p(t)|+|g(t,x(t))|{\displaystyle {\int}_{a}^{b}}|k(t,s)||f(s,x(s))|ds$$\le |p(t)|+[|g(t,x(t))-g(t,0)|+|g(t,0)|]{\displaystyle {\int}_{a}^{b}}|k(t,s)|[c+d|x(s)|]ds$$\le |p(t)|+[L|x(t)|+|g(t,0)|](c+d\parallel x\parallel ){\displaystyle {\int}_{a}^{b}}|k(t,s)|ds$$\le \overline{p}+(L\parallel x\parallel +\overline{g})(c+d\parallel x\parallel )\overline{k},$(4.6)

where we write $\parallel x\parallel $ in place of ${\parallel x\parallel}_{\mathrm{\infty}}$ and the constants $\overline{p}$, *L*, $\overline{g}$, *a*, *b*, $\overline{k}$ were defined previously or imposed in the assumptions.

The above inequality yields the estimate

$\parallel Tx\parallel \le d\overline{k}L{\parallel x\parallel}^{2}+\overline{k}(cL+d\overline{g})\parallel x\parallel +\overline{p}+c\overline{g}\overline{k}.$

Hence, keeping in mind assumption (v), we deduce that there exists a positive number ${r}_{0}$ such that for all functions $x\in {B}_{{r}_{0}}\subset R([a,b])$, we have that $Tx\in {B}_{{r}_{0}}$, i.e., the operator *T* transforms the ball ${B}_{{r}_{0}}$ into itself.

Keeping in mind the convenience, we will further accept that

${r}_{0}=\frac{1-\overline{k}(cL+d\overline{g})}{2d\overline{k}L}.$(4.7)

To prove the continuity of the operator *T* defined by (4.3) on the ball ${B}_{{r}_{0}}$ let us observe that, in view of assumption (ii), it is sufficient to prove the continuity of the operator $K\cdot F$ on ${B}_{{r}_{0}}$.
Thus, fix arbitrarily $\epsilon >0$ and take $x,y\in {B}_{{r}_{0}}$ such that $\parallel x-y\parallel \le \epsilon $.
Then, for a fixed $t\in [a,b]$, we obtain

$|(K\cdot Fx)(t)-(K\cdot Fy)(t)|\le {\int}_{a}^{b}|k(t,s)||f(s,x(s))-f(s,y(s))|ds\le {\int}_{a}^{b}|k(t,s)|\omega (f,\epsilon )ds\le \overline{k}\omega (f,\epsilon ),$(4.8)

where

$\omega (f,\epsilon )=sup\{|f(t,x)-f(t,y)|:t\in [a,b],x,y\in [-{r}_{0},{r}_{0}],|x-y|\le \epsilon \}.$

Observe that taking into account the uniform continuity of the function *f* on the set $[a,b]\times [-{r}_{0},{r}_{0}]$ (cf. assumption (iv)), we deduce that $\omega (f,\epsilon )\to 0$ as $\epsilon \to 0$.
Combining this fact with (4.8), we infer that the operator $K\cdot F$ is continuous on the ball ${B}_{{r}_{0}}$.

Further on, let us fix an arbitrary nonempty set $X\subset {B}_{{r}_{0}}$ and a number $\epsilon >0$. Then, for $x\in X$ and for $t\in (a,b]$, let us choose arbitrary numbers $u,v\in (t-\epsilon ,t)\cap [a,b]$.
Then we get the estimate

$|(Tx)(u)-(Tx)(v)|\le |p(u)-p(v)|+\left|(Gx)(u)(K\cdot Fx)(u)-(Gx)(v)(K\cdot Fx)(v)\right|$$\le {\omega}^{-}(p,t;\epsilon )+\left|(Gx)(u)(K\cdot Fx)(u)-(Gx)(v)(K\cdot Fx)(u)\right|$$+\left|(Gx)(v)(K\cdot Fx)(u)-(Gx)(v)(K\cdot Fx)(v)\right|$$\le {\omega}^{-}(p,t;\epsilon )+|(K\cdot Fx)(u)||(Gx)(u)-(Gx)(v)|+|(Gx)(v)||(K\cdot Fx)(u)-(K\cdot Fx)(v)|.$

Hence, in view of estimates (4.4) and (4.6), we obtain

$|(Tx)(u)-(Tx)(v)|\le {\omega}^{-}(p,t;\epsilon )+(a+b{r}_{0})\overline{k}\left\{L|x(u)-x(v)|+{\omega}_{1,{r}_{0}}^{-}(g,t;\epsilon )\right\}+(L{r}_{0}+\overline{g})|(K\cdot Fx)(u)-(K\cdot Fx)(v)|.$(4.9)

Further, utilizing estimates (4.5) and (4.6), we derive the following inequality:

$|(K\cdot Fx)(u)-(K\cdot Fx)(v)|\le {\displaystyle {\int}_{a}^{b}}|k(u,s)f(s,x(s))-k(v,s)f(s,x(s))|ds$$\le {\displaystyle {\int}_{a}^{b}}|k(u,s)-k(v,s)||f(s,x(s))|ds$$\le (b-a)(c+d{r}_{0}){\omega}_{1}^{-}(k,t;\epsilon ),$(4.10)

where ${\omega}_{1}^{-}(k,t;\epsilon )$ was introduced previously.
Now, linking estimates (4.9) and (4.10), we obtain

${\omega}^{-}(Tx,t;\epsilon )\le {\omega}^{-}(p,t;\epsilon )+(c+d{r}_{0})\overline{k}\left\{L{\omega}^{-}(x,t;\epsilon )+{\omega}_{1,{r}_{0}}^{-}(g,t;\epsilon )\right\}+(L{r}_{0}+\overline{g})(b-a)(c+d{r}_{0}){\omega}_{1}^{-}(k,t;\epsilon ).$

Next, keeping in mind assumption (i) and the properties of the functions $\epsilon \mapsto {\omega}^{-}(p,t;\epsilon )$, $\epsilon \mapsto {\omega}_{1,{r}_{0}}^{-}(g,t;\epsilon )$ and $\epsilon \mapsto {\omega}_{1}^{-}(k,t;\epsilon )$, and letting $\epsilon \to 0$, we obtain the estimate

${\omega}_{0}^{-}(TX)\le (c+d{r}_{0})\overline{k}L{\omega}_{0}^{-}(X).$(4.11)

In the same way, we can prove that

${\omega}_{0}^{+}(TX)\le (c+d{r}_{0})\overline{k}L{\omega}_{0}^{+}(X).$(4.12)

Combining (4.11) and (4.12) and taking into account formula (3.1), expressing the measure of noncompactness μ , we get

$\mu (TX)\le (c+d{r}_{0})\overline{k}L\mu (X).$(4.13)

Observe that, in view of (4.7) and assumption (v), we obtain

$(c+d{r}_{0})\overline{k}L=\overline{k}(cL-d\overline{g})<\overline{k}(cL+d\overline{g})<1.$

Hence, keeping in mind estimate (4.13) and Theorem 3.3, we conclude that the operator *T* has at least one fixed point *x* in the ball ${B}_{{r}_{0}}$. Obviously, the function $x=x(t)$ is a desired solution of equation (4.1) belonging to the space $R([a,b])$. The proof is complete.
∎

#### Example 4.3.

Let us fix a natural number $n\ge 2$ and consider the function $p=p(t)$ defined on the interval $I=[0,1]$ in the following way:

$p(t)=\sum _{k=1}^{n}\frac{1}{n+k}{\chi}_{k}(t),$(4.14)

where ${\chi}_{k}$ stands for the characteristic function of the interval $[\frac{k-1}{n},\frac{k}{n}]$ for $k=1,2,\mathrm{\dots},n$.
Obviously, the function $p=p(t)$, being the step function, is the regulated function on the interval $[0,1]$.
Moreover, $\overline{p}={\parallel p\parallel}_{\mathrm{\infty}}=\frac{1}{n+1}$.
Particularly, this means that the function *p* satisfies assumption (i) of Theorem 4.2.

Next, consider the function $g(t,x)=g:I\times \mathbb{R}\to \mathbb{R}$ defined by

$g(t,x)=x\sum _{k=1}^{n}\frac{1}{nk+1}[k\mathrm{sin}\pi nt]{\chi}_{k}(t)+\sum _{k=1}^{n}\frac{1}{{n}^{2}+k}{\chi}_{k}(t),$(4.15)

where (similarly as above) the function ${\chi}_{k}$ denotes the characteristic function of the interval $[\frac{k-1}{n},\frac{k}{n}]$ for $k=1,2,\mathrm{\dots},n$ and $[y]$ denotes the integer part of the number *y*.
It is easily seen that the function $g(t,x)$ is Lipschitzian with respect to *x*. Indeed, for an arbitrary $t\in I$ and $x,y\in \mathbb{R}$, we get

$|g(t,x)-g(t,y)|\le |x-y|\sum _{k=1}^{n}\frac{{\chi}_{k}(t)}{kn+1}[k\mathrm{sin}\pi nt]\le |x-y|\frac{n}{{n}^{2}+1}=\frac{n}{{n}^{2}+1}|x-y|\le \frac{1}{n}|x-y|.$

This means that the function $g(t,x)$ satisfies the Lipschitz condition with the constant $L=\frac{1}{n}$.

Now, taking into account the fact that the functions $t\mapsto {\sum}_{k=1}^{n}\frac{1}{kn+1}{\chi}_{k}(t)[k\mathrm{sin}\pi nt]$ and $t\mapsto {\sum}_{k=1}^{n}\frac{1}{{n}^{2}+k}{\chi}_{k}(t)$ are step functions on the interval *I* and keeping in mind that for each fixed $r>0$, the function

${g}_{1}(t,x)=x\sum _{k=1}^{n}\frac{1}{kn+1}{\chi}_{k}(t)[k\mathrm{sin}\pi nt]$

is uniformly regulated for $|x|\le r$ on the interval $I=[0,1]$, in view of Remark 4.1, we conclude that the function $g(t,x)$ satisfies assumption (ii) of Theorem 4.2.

Further, let us take the function $k=k(t,s)$ defined on the set ${I}^{2}$ by

$k(t,s)={s}^{2}+\alpha \sum _{n=1}^{\mathrm{\infty}}\frac{1}{n}{\chi}_{n}^{\prime}(t),$(4.16)

where ${\chi}_{n}^{\prime}$ denotes the characteristic function of the interval $(\frac{1}{n+1},\frac{1}{n}]$ for $n=1,2,\mathrm{\dots}$ and $\alpha >0$ is a constant.

Observe that the function $k=k(t,s)$ satisfies assumption (iii) of Theorem 4.2. Moreover, we have

${\int}_{0}^{1}|k(t,s)|ds\le \frac{1}{3}+\alpha $

for any $t\in I$, so we can accept that $\overline{k}=\frac{1}{3}+\alpha $, where the constant $\overline{k}$ was defined previously.

Further, let us consider the quadratic Hammerstein integral equation

$x(t)=p(t)+g(t,x(t)){\int}_{0}^{1}k(t,s)\frac{s}{{s}^{2}+1}x(s)\mathrm{sin}x(s)\mathit{d}s$(4.17)

for $t\in I=[0,1]$, where the functions $p(t)$, $g(t,x)$ and $k(t,s)$ are defined by formulas (4.14), (4.15) and (4.16), respectively.

Notice that equation (4.17) is a special case of equation (4.1), where

$f(t,x)=\frac{t}{{t}^{2}+1}x\mathrm{sin}x.$

Obviously, the function $f=f(t,x)$ is continuous on the set $I\times \mathbb{R}$ and

$|f(t,x)|\le \frac{t}{{t}^{2}+1}|x|\le \frac{1}{2}|x|$

for $t\in I$ and $x\in \mathbb{R}$.
Thus, the function $f(t,x)$ satisfies assumption (iv) of Theorem 4.2 with $c=0$ and $d=\frac{1}{2}$.

Summing up, we see that assumptions (i)–(iv) of Theorem 4.2 are satisfied. In addition, we have that

$g(t,0)=\sum _{k=1}^{n}\frac{1}{{n}^{2}+k}{\chi}_{k}(t)\le \frac{1}{{n}^{2}+1}.$

Thus, we can take $\overline{g}=\frac{1}{{n}^{2}+1}$.

To verify assumption (v), let us note that the first inequality in (v) has the form

$\frac{1}{2}\left(\frac{1}{3}+\alpha \right)\frac{1}{{n}^{2}+1}<1.$(4.18)

Hence, we see that for each fixed $\alpha >0$, we can choose a number $n\in \mathbb{N}$ such that (4.18) is satisfied.

Further, let us take into account the second inequality in assumption (v).
It has the form

$\frac{1}{n+1}<\frac{1-\frac{\frac{1}{3}+\alpha}{2({n}^{2}+1)}}{2(\frac{1}{3}+\alpha )\cdot \frac{1}{n}}$

or, equivalently,

$\frac{1}{n(n+1)}<\frac{1-\frac{\frac{1}{3}+\alpha}{2({n}^{2}+1)}}{2(\frac{1}{3}+\alpha )}.$(4.19)

It is easily seen that for any $\alpha >0$, we can choose $n\in \mathbb{N}$ so big that both inequalities (4.18) and (4.19) are satisfied.

Thus, on the basis of Theorem 4.2, we infer that equation (4.17) has at least one regulated solution on the interval $[0,1]$, provided that we choose an arbitrary number $\alpha >0$ and a natural number *n* big enough.

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