Let \(\alpha,\beta\in(1,2]\). For \(( x,s ) \in (0,1]\times[0,1]\), put \(H_{0}(x,s)=G(x,s)\) and
$$ H_{n}(x,s)= \int_{0}^{1}G(x,t)H_{n-1}(t,s)p(t)\,dt, \quad n\geq1. $$
(3.1)
Now, let \(\mathcal{H}:(0,1]\times[0,1]\rightarrow \mathbb{R}\) be defined by
$$ \mathcal{H} ( x,s ) =\sum_{n=0}^{\infty} ( -1 ) ^{n}H_{n}(x,s), $$
(3.2)
provided that the series converges.
Lemma 3.1
Let
\(\alpha,\beta\in(1,2]\)
and
\(m,M>0\)
be as in (2.1). Let
\(p\in\mathcal{J}_{\alpha,\beta}\)
with
\(\kappa_{p}<1\). Then on
\((0,1]\times[0,1]\), we have
-
(i)
\(H_{n}(x,s)\leq\kappa_{p}^{n}G(x,s)\)
for each
\(n\in\mathbb{N}\).
So, \(\mathcal{H} ( x,s ) \)
is well defined in
\((0,1]\times[0,1]\).
-
(ii)
For each
\(n\in\mathbb{N}\),
$$ l_{n}x^{\beta-2}(1-x) ( 1-s ) ^{\alpha-1}\leq H_{n}(x,s)\leq r_{n}x^{\beta-2}(1-x) ( 1-s ) ^{\alpha-1}, $$
(3.3)
where
$$ l_{n}=m^{n+1}\biggl( \int_{0}^{1}t^{\beta-2}(1-t)^{\alpha}p(t)\,dt \biggr)^{n}\quad \textit{and}\quad r_{n}=M^{n+1} \biggl( \int_{0}^{1}t^{\beta -2}(1-t)^{\alpha}p(t)\,dt \biggr)^{n}. $$
-
(iii)
\(H_{n+1}(x,s)=\int_{0}^{1}H_{n}(x,t)G(t,s)p(t)\,dt\)
for each
\(n\in\mathbb{N}\).
-
(iv)
\(\int_{0}^{1}\mathcal{H} ( x,t ) G(t,s)p(t)\,dt=\int_{0}^{1}G ( x,t ) \mathcal{H}(t,s)p(t)\,dt\).
Proof
By simple induction we prove (i), (ii), and (iii).
(iv) By Lemma 3.1(i) we have
$$ 0\leq H_{n}(x,t)G(t,s)p(t)\leq\kappa_{p}^{n}G(x,t)G(t,s)p(t) \quad\mbox{for }n\geq0\mbox{ and all }x,t,s\in(0,1]. $$
Therefore, the series \(\sum_{n\geq0}\int_{0}^{1}H_{n}(x,t)G(t,s)p(t)\,dt\) converges.
So by applying the dominated convergence theorem we deduce that
$$\begin{aligned} \int_{0}^{1}\mathcal{H} ( x,t ) G(t,s)p(t)\,dt =& \sum_{n=0}^{\infty} \int_{0}^{1} ( -1 ) ^{n}H_{n}(x,t)G(t,s)p(t)\,dt \\ =&\sum_{n=0}^{\infty} \int_{0}^{1} ( -1 ) ^{n}G(x,t)H_{n}(t,s)p(t)\,dt \\ =& \int_{0}^{1}G ( x,t ) \mathcal{H}(t,s)p(t)\,dt. \end{aligned}$$
□
Proposition 3.2
Let
\(\alpha,\beta\in(1,2]\)
and
\(p\in \mathcal{J}_{\alpha,\beta}\)
with
\(\kappa_{p}<1\). Then the function
\((x,s)\rightarrow x^{2-\beta}\mathcal{H} ( x,s ) \in C([0,1]\times[0,1])\).
Proof
Clearly, the function \((x,s)\rightarrow x^{2-\beta}H_{0}(x,s)\in C([0,1]\times[0,1])\).
Assume that the function \((x,s)\rightarrow x^{2-\beta }H_{n-1}(x,s)\in C([0,1]\times[0,1])\).
Using Lemma 3.1(i) and (2.1), we have, for all \((x,s,t)\in{}[0,1]\times[0,1]\times(0,1]\),
$$\begin{aligned} x^{2-\beta}G(x,t)H_{n-1}(t,s)p(t) \leq&\kappa _{p}^{n-1}x^{2-\beta }G(x,t)G(t,s)p(t) \\ \leq&M^{2}(1-x) (1-t)^{\alpha-1}t^{\beta-2}(1-t) (1-s)^{\alpha -1}p(t) \\ \leq&M^{2}t^{\beta-2}(1-t)^{\alpha}p(t). \end{aligned}$$
So by (3.1) and the dominated convergence theorem we conclude that the function \((x,s)\rightarrow x^{2-\beta }H_{n}(x,s)\in C([0,1]\times[0,1])\).
From Lemma 3.1(i) and (2.1) we deduce that
$$ x^{2-\beta}H_{n}(x,s)\leq\kappa_{p}^{n}x^{2-\beta}G(x,s) \leq M\kappa_{p}^{n}. $$
(3.4)
Therefore, the series \(\sum_{n\geq0} ( -1 ) ^{n}x^{2-\beta}H_{n}(x,s)\) is uniformly convergent on \([0,1]\times[0,1]\), and so the function \((x,s)\rightarrow x^{2-\beta}\mathcal{H} ( x,s ) \in C([0,1]\times[ 0,1])\). □
Lemma 3.3
Let
\(\alpha,\beta\in(1,2]\)
and
\(p\in \mathcal{J}_{\alpha,\beta}\)
with
\(\kappa_{p}\leq\frac{1}{2}\). Then for
\(( x,s ) \in(0,1]\times[0,1]\), we have
$$ ( 1-\kappa_{p} ) G ( x,s ) \leq \mathcal{H} ( x,s ) \leq G ( x,s ) . $$
(3.5)
Proof
Let \(p\in\mathcal{J}_{\alpha,\beta}\) with \(\kappa_{p}\leq \frac{1}{2}\). By Lemma 3.1(i) we deduce that
$$ \bigl\vert \mathcal{H} ( x,s ) \bigr\vert \leq\sum_{n=0}^{\infty} ( \kappa_{p} ) ^{n}G ( x,s ) =\frac{1}{1-\kappa_{p}}G ( x,s ) . $$
(3.6)
Now, from the expression of \(\mathcal{H}\) we have
$$ \mathcal{H} ( x,s ) =G ( x,s ) -\sum_{n=0}^{\infty} ( -1 ) ^{n}H_{n+1}(x,s). $$
(3.7)
Since the series \(\sum_{n\geq0}\int_{0}^{1}G(x,t)H_{n}(t,s)p(t)\,dt\) converges, we conclude by (3.7) and (3.1) that
$$\begin{aligned} \mathcal{H} ( x,s ) =&G ( x,s ) -\sum_{n=0}^{\infty} ( -1 ) ^{n} \int_{0}^{1}G(x,t)H_{n}(t,s)p(t)\,dt \\ =&G ( x,s ) - \int_{0}^{1}G(x,t) \Bigl(\sum_{n=0}^{\infty} ( -1 ) ^{n}H_{n}(t,s) \Bigr)p(t)\,dt, \end{aligned}$$
namely,
$$ \mathcal{H} ( x,s ) =G ( x,s ) -W \bigl( p\mathcal{H} (\cdot,s ) \bigr) ( x ) . $$
(3.8)
On the other hand, since
$$\begin{aligned} W \bigl( p\mathcal{H} (\cdot,s ) \bigr) ( x ) \leq &\frac{1}{1-\kappa_{p}}W \bigl( pG ( \cdot,s ) \bigr) ( x ) \\ =&\frac{1}{1-\kappa_{p}}H_{1}(x,s)\leq\frac{\kappa_{p}}{1-\kappa _{p}}G ( x,s ) , \end{aligned}$$
(3.9)
we deduce that
$$ \mathcal{H} ( x,s ) \geq G ( x,s ) -\frac{\kappa _{p}}{1-\kappa_{p}}G ( x,s ) = \frac{1-2\kappa_{p}}{1-\kappa_{p}}G ( x,s ) \geq0. $$
Hence, \(\mathcal{H} ( x,s ) \leq G ( x,s ) \), and by (3.8) we have
$$ \mathcal{H} ( x,s ) \geq G ( x,s ) -W \bigl( pG (\cdot,s ) \bigr) ( x ) \geq ( 1- \kappa _{p} ) G ( x,s ) . $$
□
Corollary 3.4
Let
\(\alpha,\beta\in(1,2]\)
and
\(p\in \mathcal{J}_{\alpha,\beta}\)
with
\(\kappa_{p}\leq\frac{1}{2}\).
Let
\(\varphi\in\mathcal{B}^{+} ( (0,1) ) \). Then
$$ W_{p}\varphi\in C_{2-\beta} \bigl( [0,1] \bigr) \quad\textit{if and only if}\quad \int_{0}^{1} ( 1-s ) ^{\alpha-1}\varphi ( s ) \,ds< \infty. $$
Proof
The assertion follows from Proposition 3.2, (3.5), and (2.1). □
Lemma 3.5
Let
\(\alpha,\beta\in(1,2]\)
and
\(p\in \mathcal{J}_{\alpha,\beta}\)
with
\(\kappa_{p}\leq\frac{1}{2}\).
Let
\(h \in\mathcal{B}^{+} ( (0,1) ) \). Then we have, for
\(x\in(0,1]\),
$$ Wh (x)=W_{p}h (x)+W_{p} ( pWh ) (x)=W_{p}h (x)+W ( pW_{p}h ) (x). $$
(3.10)
In particular, if
\(W(ph )<\infty\), then
$$ \bigl(I-W_{p} ( p\cdot ) \bigr) \bigl(I+W ( p\cdot ) \bigr)h =\bigl(I+W ( p\cdot ) \bigr) \bigl(I-W_{p} ( p\cdot ) \bigr)h =h . $$
(3.11)
Proof
Let \((x,s)\in(0,1]\times[0,1]\). Then by (3.8) we have
$$ G ( x,s ) =\mathcal{H} ( x,s ) +W \bigl( p\mathcal{H} ( \cdot,s ) \bigr) ( x ) . $$
Let \(h\in\mathcal{B}^{+} ( (0,1) ) \). Using the Fubini theorem, we obtain
$$\begin{aligned} Wh (x) =& \int_{0}^{1} \bigl( \mathcal{H} ( x,s ) +W \bigl( p\mathcal{H} (\cdot,s ) \bigr) ( x ) \bigr) h (s)\,ds \\ =&W_{p}h (x)+W ( pW_{p}h ) (x). \end{aligned}$$
Using Lemma 3.1(iv) and again the Fubini theorem, we have
$$ \int_{0}^{1} \int_{0}^{1}\mathcal{H} ( x,t ) G(t,s)p(t)h (s)\,dt\,ds= \int_{0}^{1} \int_{0}^{1}G ( x,t ) \mathcal{H}(t,s)p(t)h (s)\,dt\,ds, $$
that is,
$$ W_{p} ( pWh ) (x)=W ( pW_{p}h ) (x). $$
So
$$ Wh (x)=W_{p}h (x)+W ( pW_{p}h ) (x)=W_{p}h (x)+W_{p} ( pWh ) (x). $$
□
Proposition 3.6
Let
\(\alpha,\beta\in(1,2]\)
and
\(p\in \mathcal{J}_{\alpha,\beta}\cap C(\mathcal{(}0,1\mathcal{))}\)
with
\(\kappa_{p}\leq\frac{1}{2}\). Let
\(\varphi\in\mathcal{B}^{+} ( (0,1) ) \)
be such that
\(s\rightarrow(1-s)^{\alpha-1}\varphi(s)\in C((0,1))\cap L^{1}((0,1))\). Then
\(W_{p}\varphi\in C_{2-\beta}([0,1])\), and it is the unique nonnegative solution of the problem
$$ \textstyle\begin{cases} D^{\alpha}(D^{\beta}u)(x)+p(x)u(x)=\varphi(x),\quad 0< x< 1,\\ \lim_{x\rightarrow0^{+}}D^{\beta -1}u(x)=\lim_{x\rightarrow 0^{+}}D^{\alpha-1}(D^{\beta}u)(x)=u(1)=D^{\beta}u(1)=0,\end{cases} $$
(3.12)
satisfying
$$ ( 1-\kappa_{p} ) W\varphi\leq u \leq W\varphi. $$
(3.13)
Proof
By Corollary 3.4 the function \(x\rightarrow p(x)W_{p}\varphi ( x ) \in C((0,1))\).
Using (3.10) and (2.1), we have that there exists \(c\geq 0\) such that
$$ W_{p}\varphi(x)\leq W\varphi(x)\leq M \int_{0}^{1}x^{\beta -2}(1-x) (1-s)^{\alpha-1}\varphi(s)\,ds=cx^{\beta-2}(1-x). $$
(3.14)
Therefore,
$$ \int_{0}^{1}(1-s)^{\alpha-1}p(s)W_{p} \varphi(s)\,ds\leq c \int_{0}^{1}s^{\beta-2}(1-s)^{\alpha}p(s)\,ds< \infty. $$
Hence, by Proposition 2.6 the function \(u=W_{p}\varphi=W\varphi-W ( pW_{p}\varphi ) \) satisfies the equation
$$ \textstyle\begin{cases} D^{\alpha}(D^{\beta}u) ( x ) =\varphi(x)-p(x)u(x),\quad 0< x< 1, \\ \lim_{x\rightarrow0^{+}}D^{\beta -1}u(x)=\lim_{x\rightarrow 0^{+}}D^{\alpha-1}(D^{\beta}u)(x)=u(1)=D^{\beta}u(1)=0.\end{cases} $$
By integration of inequalities (3.5) we obtain (3.13).
Let us prove the uniqueness. Let \(v\in C_{2-\beta }([0,1])\) be another solution of problem (3.12) satisfying \(v\leq W\varphi\).
Put \(\tilde {v}:=v+W(pv)\). Since the function \(s\rightarrow (1-s)^{\alpha-1}p(s)v(s)\in C((0,1))\cap L^{1}((0,1))\), then by Proposition 2.6 it follows that
$$ \textstyle\begin{cases} D^{\alpha}(D^{\beta} \tilde {v}) ( x ) =\varphi(x),\quad 0< x< 1, \\ \lim_{x\rightarrow0^{+}}D^{\beta-1}\tilde {v}(x)=\lim_{x\rightarrow0^{+}}D^{\alpha-1}(D^{\beta} \tilde {v})(x)=\tilde {v}(1)=D^{\beta} \tilde {v}(1)=0.\end{cases} $$
From the uniqueness in Proposition 2.6 we conclude that
$$ \tilde {v}:=v+W(pv)=W\varphi. $$
So
$$ \bigl(I+W(p\cdot)\bigr) \bigl((v-u)^{+}\bigr)=\bigl(I+W(p\cdot)\bigr) \bigl((v-u)^{-}\bigr), $$
where \((v-u)^{+}=\max(v-u,0)\) and \((v-u)^{-}=\max (u-v,0)\).
From (3.13), (3.14), (1.11), and (2.5), there exists a constant \(\tilde {c}>0\), such that
$$ W\bigl(p\vert v-u\vert \bigr)\leq2\tilde {c}W(p\theta)\leq 2 \tilde {c}\kappa_{p}\theta< \infty. $$
Therefore, \(u=v\) by Lemma 3.5. □
Proof of Theorem 1.3
Consider \(\xi\geq0\) and \(\zeta\geq0\) with \(\xi+\zeta >0\). Let \(\alpha,\beta\in(1,2]\) and \(p\in\mathcal{J}_{\alpha,\beta }\cap C((0,1))\) be such that \((A_{2})\) is satisfied.
Let
$$ \mathcal{S}:= \bigl\{ u\in\mathcal{B}^{+} \bigl( (0,1) \bigr) : ( 1- \kappa_{p} ) \theta\leq u\leq\theta \bigr\} , $$
where \(\theta(x):=\xi h_{1}(x)+\zeta h_{2}(x)\), and \(h_{1}\) and \(h_{2} \) are defined respectively by (1.9) and (1.10).
Define the operator \(\mathcal{F}\) on \(\mathcal{S}\) by
$$ \mathcal{F}u=\theta-W_{p} ( p\theta ) +W_{p}\bigl( \bigl( p-f (\cdot,u ) \bigr) u\bigr). $$
By (3.10) and (2.5) we have
$$ W_{p}(p\theta)\leq W ( p\theta ) \leq\kappa_{p}\theta \leq \theta. $$
(3.15)
Using (\(\mathrm{A}_{2}\)), we get
$$ 0\leq f(\cdot,u)\leq p\quad\mbox{for all }u\in\mathcal{S}. $$
(3.16)
Next, we prove that \(\mathcal{FS}\subseteq\mathcal{S}\). Indeed, using (3.16) and (3.15), we have, for \(u\in\mathcal {S}\),
$$ \mathcal{F}u\leq\theta-W_{p} ( p\theta ) +W_{p}(pu)\leq \theta $$
and
$$\begin{aligned} \mathcal{F}u \geq&\theta-W_{p} ( p\theta ) \\ \geq& ( 1-\kappa_{p} ) \theta. \end{aligned}$$
Observe that, by (\(\mathrm{A}_{2}\)), \(\mathcal{F}\) becomes nondecreasing on \(\mathcal{S}\).
Define the sequence \(\{v_{n}\}\) by \(v_{0}= ( 1-\kappa _{p} ) \theta\) and \(v_{n+1}=\mathcal{F}v_{n}\) for \(n\in \mathbb{N}\). Since \(\mathcal{FS}\subseteq\mathcal{S}\), we have \(v_{1}=\mathcal{F}v_{0}\geq\)
\(v_{0}\), and by the monotonicity of \(\mathcal{F}\) we deduce that
$$ ( 1-\kappa_{p} ) \theta=v_{0}\leq v_{1}\leq\cdots \leq v_{n}\leq v_{n+1}\leq\theta. $$
Using (\(\mathrm{A}_{1}\))-(\(\mathrm{A}_{2}\)) and the dominated convergence theorem, we deduce that the sequence \(\{v_{n}\}\) converges to a function \(u\in\mathcal{S}\) satisfying
$$ u= \bigl( I-W_{p} ( p\cdot) \bigr) \theta+W_{p}\bigl( \bigl( p-f ( \cdot,u ) \bigr) u\bigr), $$
that is,
$$ \bigl( I-W_{p} ( p\cdot) \bigr) u= \bigl( I-W_{p} ( p\cdot) \bigr) \theta-W_{p} \bigl( uf (\cdot,u ) \bigr) , $$
and by (3.15) we have \(W ( pu ) \leq W ( p\theta ) \leq\theta<\infty\). Therefore, by Lemma 3.5 we deduce that
$$ u=\theta-W \bigl( uf ( \cdot,u ) \bigr) . $$
(3.17)
We claim that u is a solution.
Indeed, from (3.16) and (1.11), there exists a constant \(c>0\) such that
$$ (1-s)^{\alpha-1}u(s)f \bigl( s,u(s) \bigr) \leq(1-s)^{\alpha -1} \theta(s)p(s)\leq cs^{\beta-2}(1-s)^{\alpha}p(s). $$
(3.18)
So, by Proposition 2.6 the function \(W ( uf ( \cdot,u ) ) \in C_{2-\beta}([0,1])\). This implies by (3.17) that \(u\in C_{2-\beta}([0,1])\).
Now, since the function \(s\rightarrow(1-s)^{\alpha -1}u(s)f ( s,u(s) ) \in C((0,1))\cap L^{1}((0,1))\), we deduce by Proposition 2.6 that u is a solution.
It remains to prove the uniqueness. Let v be another solution in \(C_{2-\beta}([0,1])\) to problem (1.2) satisfying (1.12). Since \(v\leq\theta\), we deduce by (3.18) that
$$ 0\leq v(s)f \bigl( s,v(s) \bigr) \leq\theta(s)p(s)\leq cs^{\beta -2}(1-s)p(s). $$
This implies that \(s\rightarrow(1-s)^{\alpha -1}v(s)f ( s,v(s) ) \in C((0,1))\cap L^{1}((0,1))\). Let \(\tilde {v}:=v+W ( vf (\cdot,v ) ) \). By Proposition 2.6, we have
$$ \textstyle\begin{cases} D^{\alpha}(D^{\beta} \tilde {v})(x)=0,\quad 0< x< 1,\\ \lim_{x\rightarrow0^{+}}D^{\beta-1}\tilde {v}(x)=0,\qquad \lim_{x\rightarrow0^{+}}D^{\alpha-1}(D^{\beta} \tilde {v})(x)=\xi,\\ \tilde {v}(1)=0,\qquad D^{\beta} \tilde {v}(1)=-\zeta.\end{cases} $$
Hence,
$$ v=\theta-W \bigl( vf ( \cdot,v ) \bigr) . $$
(3.19)
Let \(\omega:(0,1)\rightarrow\mathbb{R}\) be defined by
$$ \omega(z)= \textstyle\begin{cases} \frac{v(z)f ( z,v(z) ) -u(z)f ( z,u(z) ) }{v(z)-u(z)} & \mbox{if }v(z)\neq u(z),\\ 0 & \mbox{if }v(z)=u(z).\end{cases} $$
By (\(\mathrm{A}_{3}\)), \(\omega\in\mathcal{B}^{+} ((0,1)) \) and from (3.17) and (3.19) we deduce that
$$ \bigl(I+W(\omega\cdot)\bigr) \bigl((v-u)^{+}\bigr)=\bigl(I+W(\omega\cdot)\bigr) \bigl((v-u)^{-}\bigr), $$
where \((v-u)^{+}=\max(v-u,0)\) and \((v-u)^{-}=\max (u-v,0)\).
From (\(\mathrm{A}_{2}\)) we have \(\omega\leq p\). So by using (1.12) and (2.5) we obtain
$$ W\bigl(\omega \vert v-u\vert \bigr)\leq2W(p\theta)\leq2\kappa _{p} \theta< \infty. $$
Hence, \(u=v\) by (3.11). □
Proof of Corollary 1.4
The statement follows from Theorem 1.3 with \(f ( x,t ) =\lambda q(x)h(t)\), \(\varrho(t)=th(t)\) and \(p(x):=\lambda q(x)\max_{0\leq t\leq\theta ( x ) }\varrho ^{\prime}(t)\). □
Example 3.7
Let \(\sigma\geq0\), \(\nu\geq0\), and \(q\in C^{+}((0,1))\) be such that
$$ \int_{0}^{1}t^{(\beta-2)(1+\sigma+\nu)}(1-t)^{\alpha +\sigma+\nu}q(t)\,dt< \infty. $$
Let \(\varrho(t)=t^{\sigma+1}\ln(1+t^{\nu})\) and \(\tilde {q}(t):=q(t)\max_{0\leq s\leq\theta ( t ) }\varrho^{\prime}(s)\). Since \(\tilde {q}\in\mathcal{J}_{\alpha ,\beta}\), then for \(\xi\geq0\), \(\zeta\geq0\) with \(\xi+\zeta >0\) and \(\lambda\in[0,\frac{1}{2\kappa_{\tilde {q}}})\), the problem
$$ \textstyle\begin{cases} D^{\alpha}(D^{\beta}u)(x)+\lambda q(x)u^{\sigma+1}(x)\ln (1+u^{\nu }(x))=0,\quad 0< x< 1, \\ \lim_{x\rightarrow0^{+}}D^{\beta-1}u(x)=0,\qquad\lim_{x\rightarrow0^{+}}D^{\alpha-1}(D^{\beta}u)(x)=\xi,\\ u(1)=0,\qquad D^{\beta}u(1)=-\zeta,\end{cases} $$
has a unique solution \(u\in C_{2-\beta}([0,1])\) such that
$$ (1-\lambda\alpha_{\tilde {q}})\theta(x)\leq u ( x ) \leq\theta(x)\quad\mbox{for }x \in(0,1]. $$