Let us consider problem (1.1). The following assumptions will be considered:

(H)
\(a,b:\mathbb{T}\times \mathbb{R}\rightarrow \mathbb{R}\) are continuous functions;

(A1)
There exist an open set \(\mathbb{T}_{1}\subseteq \mathbb{T}\), \(a_{1}>0\), and \(p_{1}\in (0,1)\) such that
$$ a_{1}x^{p_{1}}\leq a(t,x),\quad t\in \mathbb{T}_{1}, x\in {\biggl[}0, \frac{N1}{2} {\biggr]}; $$

(A2)
\(0\leq a(t,0)\), \(t\in \mathbb{T}\);

(A3)
There exist \(a_{2}>0\) and \(p_{2}\in (0,1)\) such that
$$ a_{2}x^{p_{2}}\geq a(t,x),\quad t\in \mathbb{T}, x\in { \biggl[}0, \frac{N1}{2} {\biggr]}; $$

(B1)
There exist an open set \(\mathbb{T}_{1}\subseteq \mathbb{T}\), \(b_{1}>0\), and \(q_{1}\in [1,+\infty )\) such that
$$ b_{1}x^{q_{1}}\leq b(t,x),\quad t\in \mathbb{T}_{1}, x\in {\biggl[}0, \frac{N1}{2} {\biggr]}; $$

(B2)
\(0\leq b(t,0)\), \(t\in \mathbb{T}\);

(B3)
There exist \(b_{2}>0\) and \(q_{2}\in (1,+\infty )\) such that
$$ b_{2}x^{q_{2}}\geq b(t,x),\quad t\in \mathbb{T}, x\in { \biggl[}0, \frac{N1}{2} {\biggr]}. $$
Theorem 3.1
Let (H) hold. Then:

(i)
If \(\mu =0\) and (A1)–(A2) are satisfied, then for every \(\lambda >0\) problem (1.1) has at least one positive solution;

(ii)
If \(\lambda =0\) and (B1)–(B2) are satisfied, then there exists \(\mu ^{*}>0\) such that, for every \(\mu >\mu ^{*}\), problem (1.1) has at least one positive solution;

(iii)
If \(\lambda =0\) and (B1)–(B3) are satisfied, then there exists \(\mu ^{*}>0\) such that, for every \(\mu >\mu ^{*}\), problem (1.1) has at least two positive solutions;

(iv)
If (A1)–(A3) and (B1)–(B3) are satisfied, then there exist \(\mu ^{*}>0\) and a function \(\lambda (\cdot ):(\mu ^{*},+\infty )\rightarrow \mathbb{R}\) such that, for every \(\mu >\mu ^{*}\), \(\lambda \in (0,\lambda (\mu ))\), problem (1.1) has at least three positive solutions.
Proof
Step 1. An equivalent problem. Fix \(\lambda \geq 0\) and \(\mu \geq 0\). Assume (H), (A2) (\(\lambda >0\)) and (B2) (\(\mu >0\)). Define the functions \(\bar{a},\bar{b},\bar{f}:\mathbb{T}\times [\frac{N1}{2}, \frac{N1}{2}]\rightarrow \mathbb{R}\) by
$$\begin{aligned}& \bar{a}(t,x)= \textstyle\begin{cases} a(t,0), &\frac{N1}{2}\leq x< 0, \\ a(t,x), &0\leq x\leq \frac{N1}{2}, \end{cases}\displaystyle \\& \bar{b}(t,x)= \textstyle\begin{cases} b(t,0), &\frac{N1}{2}\leq x< 0, \\ b(t,x), &0\leq x\leq \frac{N1}{2}, \end{cases}\displaystyle \end{aligned}$$
and \(\bar{f}=\lambda \bar{a}+\mu \bar{b}\).
Notice that any nontrivial solution u of the Dirichlet problem
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=\lambda \bar{a}(t,u(t))+\mu \bar{b}(t,u(t)), \quad t\in \mathbb{T}, \\ u(1)=u(N)=0 \end{cases} $$
(3.1)
is positive. Indeed, as \(\bar{f}(t,x)\geq 0\) for all \(t\in \mathbb{T}\), \(x\in [\frac{N1}{2},0]\), multiplying the equation in (3.1) by \(\mathbf{u}^{}:=\min \{\mathbf{u},\mathbf{0}\}\) and summing by parts, we have
$$\begin{aligned} 0&\leq \sum_{t=1}^{N} \bar{f}(t,u)u^{}=\sum_{t=1}^{N} \nabla \bigl[\phi (\triangle u)\bigr]u^{}=\sum _{t=1}^{N1} \phi ( \triangle u)\triangle \bigl(u^{}\bigr) \\ &=\sum_{t=1}^{N1} \phi \bigl( \triangle \bigl(u^{}\bigr)\bigr)\triangle \bigl(u^{}\bigr)\leq 0, \end{aligned}$$
and hence \(\triangle (\mathbf{u}^{})=0\). As \(u(1)^{}=u(N)^{}=0\), we conclude that \(\mathbf{u}^{}=0\). Therefore u is a positive solution of (1.1) if and only if it is a nontrivial solution of (3.1).
We set
$$ D=\bigl\{ \mathbf{u}\in Y: \vert \triangle \mathbf{u} \vert _{\infty }< 1\bigr\} $$
and denote \(K:\mathbb{R}^{N}\rightarrow Y\) that sends any vector v onto the unique solution \(\mathbf{w}\in Y\) of
$$ \textstyle\begin{cases} \nabla [\phi (\triangle w(t))]=v(t), \quad t\in \mathbb{T}, \\ w(1)=w(N)=0. \end{cases} $$
The operator \(N_{\lambda ,\mu }:\bar{D}\rightarrow \mathbb{R}^{N}\) is defined by
$$ N_{\lambda ,\mu }(u)=\bar{f}(\cdot ,u), $$
and the operator \(F_{\lambda ,\mu }=K\circ N_{\lambda ,\mu }\) is completely continuous because Y is a finite dimensional space. Obviously, a function u is a positive solution of (3.1) if and only if \(\mathbf{u}\in D\) is a nontrivial fixed point of \(F_{\lambda ,\mu }\).
Step 2. Proof of (i). Take \(\mu =0\) and fix \(\lambda >0\). Assume (H) and (A1)–(A2). We further use the symbols N and F to replace the operators \(N_{\lambda ,0}\) and \(F_{\lambda ,0}\). Set \(\Lambda _{a}=\lambda \bar{a}_{\infty }\) and let \(\delta _{a}\in (0,1)\) be the constant \(\delta _{a}=\delta \) introduced in Lemma 2.3, with \(\Lambda =\Lambda _{a}\).
Let us consider the Dirichlet problem
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=\lambda a_{1}u^{p_{1}}, \quad t\in \mathbb{T}_{1} , \\ u(t_{1})=u(t_{2})=0, \end{cases} $$
(3.2)
where \(t_{1},t_{2}\in \mathbb{R}\), \(\mathbb{T}_{1}\), \(a_{1}\) and \(p_{1}\) are defined in (A1). Suppose that \(\hat{\mathbb{T}}_{1}\subset \mathbb{T}\). By Lemma 2.6 there exists a solution α satisfying \(\boldsymbol {\alpha}_{\infty }\leq \frac{N1}{2}\) and \(\boldsymbol {\alpha}\gg \boldsymbol {0}\). Let us extend α to a function \(\tilde{\boldsymbol {\alpha}}\) satisfying \(\triangle \tilde{\boldsymbol {\alpha}}_{\infty }\leq 1\) and
$$ \frac{N1}{2}< \tilde{\alpha }(t)< 0,\quad t\in \hat{\mathbb{T}}\setminus \hat{\mathbb{T}}_{1}. $$
We define the open bounded subset of Y
$$ \mathrm{U}_{0}=\bigl\{ \mathbf{u}\in Y: \mathbf{u}\gg \tilde{ \boldsymbol {\alpha}}, \vert \triangle \mathbf{u} \vert _{\infty }< 1\delta _{a} \bigr\} , $$
and \(\mathbf{v}_{0}\in \mathbb{R}^{N}\) by setting
$$ v_{0}(t)=\lambda \bar{a}\bigl(t,\tilde{\alpha }(t)\bigr),\quad t\in \mathbb{T}. $$
Note that \(\mathbf{v}_{0}\geq \boldsymbol {0}\) in \(\mathbb{T}\), \(\mathbf{v}_{0}>\boldsymbol {0}\) in \(\mathbb{T}_{1}\) and \(\mathbf{v}_{0}_{\infty }\leq \Lambda _{a}\). Let \(\mathbf{z}_{0}\) be the solution of the problem
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=v_{0}(t), \quad t\in \mathbb{T}, \\ u(1)=u(N)=0. \end{cases} $$
By Lemma 2.4, \(\mathbf{z_{0}}\gg 0\) in \(\hat{\mathbb{T}}\).
Claim
F has no fixed points on \(\partial \mathrm{U}_{0}\) and \(\deg (IF,\mathrm{U}_{0},0)=1\).
We first prove that
$$ \deg (I\mathbf{z}_{0},\mathrm{U}_{0},0)=1. $$
It suffices to verify that \(\mathbf{z}_{0}\in \mathrm{U}_{0}\). The condition \(\triangle \mathbf{z}_{0}_{\infty }<1\delta _{a}\) is satisfied by the definition of \(\delta _{a}\). It remains to prove that \(\mathbf{z}_{0}\geq \tilde{\boldsymbol {\alpha}}\), \(t\in \mathbb{T}\). Since \(z_{0}(t)\geq 0\), \(t\in \mathbb{T}\) and \(\tilde{\alpha }(t)<0\), \(t\in \hat{\mathbb{T}}\setminus \hat{\mathbb{T}}_{1}\), we only need to show that \(z_{0}(t)>\alpha (t)\), \(t\in \hat{\mathbb{T}}_{1}\). Since \(z_{0}(t)\gg 0,t\in \hat{\mathbb{T}}\), we have \(\min _{t\in \hat{\mathbb{T}}_{1}}z_{0}(t)>0\). Moreover, as \(\lambda a_{1}u^{p_{1}}\leq v_{0}\), \(t\in \mathbb{T}\) by (A1), we get, by Lemma 2.1,
$$ \alpha (t)\leq z_{0}(t)\min _{\{t_{1},t_{2} \}}z_{0}(t)< z_{0}(t) $$
for all \(t\in \hat{\mathbb{T}}_{1}\).
Next we consider the homotopy \(H:[0,1]\times \bar{D}\rightarrow Y\) defined by
$$ H(k,u)=K\bigl(kN(u)+(1k)v_{0}\bigr). $$
Obviously, H is completely continuous. Note that
$$ H(0,u)=z_{0},\qquad H(1,u)=F(u) $$
for all \(\mathbf{u}\in \bar{D}\). Fix now \(k\in [0,1]\) and suppose that \(\mathbf{u}\in \bar{\mathrm{U}}_{0}\) is a fixed point of \(H(k,\cdot )\). We will prove that \(\mathbf{u}\in \mathrm{U}_{0}\). Since u is a fixed point of \(H(k,\cdot )\), u is a solution of
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=k\lambda a(t,u(t))+(1k)v_{0}(t), \quad t\in \mathbb{T}, \\ u(1)=u(N)=0. \end{cases} $$
Observe that
$$ k\lambda \bar{a}(t,x)+(1k)v_{0}(t)\geq 0,\quad t\in \mathbb{T}, x\in \biggl[\frac{N1}{2},0 \biggr]. $$
Arguing as in Step 1, we see that \(\mathbf{u}\geq \mathbf{0}\) in \(\mathbb{T}\). Moreover, as
$$ k\lambda \bar{a}\bigl(t,u(t)\bigr)+(1k)v_{0}(t)>0,\quad t\in \mathbb{T}_{1}, $$
by Lemma 2.5 we deduce that \(\min_{t\in \hat{T}_{1}} \mathbf{u}>0\). Let us prove that \(\mathbf{u}\gg \tilde{\boldsymbol {\alpha}}\) in \(\hat{\mathbb{T}}\). As above we observe that, since \(u(t)\geq 0\), \(t\in \mathbb{T}\) and \(\tilde{\alpha }(t)<0\), \(t\in \hat{\mathbb{T}}\setminus \hat{\mathbb{T}}_{1}\), we only need to show that \(u(t)>\alpha (t)\), \(t\in \hat{\mathbb{T}}_{1}\). Note that, using (A1) and \(\mathbf{u}\in \bar{\mathrm{U}}_{0}\), we have
$$ k\lambda \bar{a}\bigl(\cdot ,u(t)\bigr)+(1k)v_{0}(t)\geq k\lambda a_{1}u^{p_{1}}+(1k) \lambda \bar{a}\bigl(\cdot ,\alpha (t) \bigr)\geq \lambda a_{1}\alpha ^{p_{1}},\quad t \in \mathbb{T}_{1}. $$
Applying Lemma 2.1, we get
$$ \alpha (t)\leq u(t)\min_{\{t_{1},t_{2}\}} u< u(t) $$
for all \(t\in \hat{\mathbb{T}}\).
Furthermore, as
$$ \bigl\vert k\lambda \bar{a}\bigl(t,u(t)\bigr)+(1k)v_{0}(t) \bigr\vert _{\infty }\leq \Lambda _{a}, $$
Lemma 2.3 yields
$$ \vert \triangle \mathbf{u} \vert _{\infty }< 1\delta _{a}. $$
In conclusion, \(\mathbf{u}\in \mathrm{U}_{0}\). The homotopy invariance of the degree implies that
$$ \deg (IF,\mathrm{U}_{0},0)=\deg (I\mathbf{z}_{0}, \mathrm{U}_{0},0)=1. $$
This concludes the proof of the claim.
Therefore, for every \(\lambda >0\), there exists a nontrivial fixed point \(\mathbf{u}\in \mathrm{U}_{0}\) of the operator F, i.e., there exists a positive solution u of (3.1) satisfying \(\mathbf{u}\gg \tilde{\boldsymbol {\alpha}}\) in \(\hat{\mathbb{T}}\).
Step 3. Proof of (ii). Take \(\lambda =0\) and \(\mu >0\). Assume (H) and (B1)–(B2). We further use the symbols N and F to replace the operators \(N_{0,\mu }\) and \(F_{0,\mu }\). Set \(\Lambda _{b}=\mu \bar{b}_{\infty }\) and let \(\delta _{b}\in (0,1)\) be the constant \(\delta _{b}=\delta \) introduced in Lemma 2.3, with \(\Lambda =\Lambda _{b}\).
Let us consider the Dirichlet problem
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=\mu b_{1}u^{q_{1}}, \quad t\in \mathbb{T}_{1}, \\ u(t_{1})=u(t_{2})=0, \end{cases} $$
(3.3)
where \(b_{1}\) and \(q_{1}\) are defined in (B1). We still suppose that \(\hat{\mathbb{T}}_{1}\subset \mathbb{T}\). By Lemma 2.6, there exists a constant \(\mu ^{*}>0\) such that, for any \(\mu >\mu ^{*}\), problem (3.3) has at least one solution \(\boldsymbol {\alpha}_{1}\) satisfying \(\boldsymbol {\alpha}_{1}_{\infty }\leq \frac{N1}{2}\) and \(\boldsymbol {\alpha}_{1}\gg \boldsymbol {0}\) in \(\hat{\mathbb{T}}\). As in step 2, we extend \(\boldsymbol {\alpha}_{1}\) to a function \(\tilde{\boldsymbol {\alpha}}_{1}\) satisfying \(\triangle \tilde{\boldsymbol {\alpha}}_{1}_{\infty }<1\) and
$$ \frac{N1}{2}< \tilde{\boldsymbol{\alpha }}_{1}(t)< 0,\quad t\in \hat{\mathbb{T}}\setminus \hat{\mathbb{T}}_{1}. $$
We define the open bounded set
$$ \mathrm{U}_{1}=\bigl\{ \mathbf{u}\in Y: \mathbf{u}\gg \boldsymbol { \alpha}_{1}, \vert \triangle \mathbf{u} \vert _{\infty }< 1\delta _{b}\bigr\} $$
and \(\mathbf{v}_{1}\in \mathbb{R}^{N}\) by setting
$$ v_{1}(t)=\mu \bar{b}\bigl(t,\tilde{\alpha }_{1}(t) \bigr). $$
The proof continues as in step 2, by showing that F has no fixed points on \(\partial \mathrm{U}_{1}\) and
$$ \deg (IF,\mathrm{U}_{1},0)=1. $$
Therefore we conclude that, for all \(\mu >\mu ^{*}\), there exists a nontrivial fixed point \(\mathbf{u}\in \mathrm{U}_{1}\) of F, i.e., there exists a positive solution u of (3.1) satisfying \(\mathbf{u}\gg \tilde{\boldsymbol {\alpha }}_{1}\).
Step 4. Proof of (iii). Take \(\lambda =0\) and \(\mu >0\). Assume (H) and (B1)–(B3). Note that (B1) and (B3) together imply \(q_{1}>1\). Similarly, we use the symbols N and F to denote the operators \(N_{0,\mu }\) and \(F_{0,\mu }\). Let \(\mu ^{*}\) be the constant, whose existence was proved in Step 3, such that problem (3.1) has at least one positive solution for all \(\mu >\mu ^{*}\). Fix \(\mu >\mu ^{*}\) and let \(\mathbf{u}_{1}\in \mathrm{U}_{1}\) be a corresponding solution. Let us prove the existence of a second positive solution.
For each \(r>0\), we set
$$ \mathrm{U}_{2}^{r}=\bigl\{ \mathbf{u}\in Y: \vert \mathbf{u} \vert _{\infty }< r, \vert \triangle \mathbf{u} \vert _{\infty }< 1\delta _{b}\bigr\} $$
with \(\delta _{b}\) defined in Step 3.
Claim
There exists \(\hat{r}>0\) such that, for each \(r\in (0,\hat{r}]\), F has no fixed points on \(\partial \mathrm{U}_{2}^{r}\) and
$$ \deg \bigl(IF,\mathrm{U}_{2}^{r},0\bigr)=1. $$
Consider the homotopy \(H:[0,1]\times \bar{D}\rightarrow Y\) defined by
$$ H(k,u)=K\bigl(kN(u)\bigr). $$
Obviously, H is completely continuous. We have
$$ H(0,u)=0,\qquad H(1,u)=F(u) $$
for all \(\mathbf{u}\in \bar{D}\). Fix \(k\in [0,1]\) and suppose that \(\mathbf{u}\in \bar{\mathrm{U}}_{2}^{r}\) is a fixed point of \(H(k,\cdot )\). We will prove that \(\mathbf{u}\in \mathrm{U}_{2}^{r}\). Since u is a fixed point of \(H(k,\cdot )\), u is a solution of
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=k\mu \bar{b}(t,u(t)),\quad t\in \mathbb{T}, \\ u(1)=u(N)=0, \end{cases} $$
(3.4)
multiplying the equation in (3.4) by \(u(t)\) and summing, we obtain
$$ \begin{aligned} \sum_{t=2}^{N1} \bigl(\triangle u(t)\bigr)^{2}&\leq \sum_{t=2}^{N1} \phi \bigl(\triangle u(t)\bigr)\triangle u(t)=\sum_{t=2}^{N1}k \mu \bar{b}\bigl(t,u(t)\bigr)u(t) \\ &\leq \mu b_{2}\sum_{t=2}^{N1} \vert u \vert ^{q_{2}+1}\leq \mu b_{2}r^{q_{2}1} \sum_{t=2}^{N1}u^{2} \\ &\leq \mu b_{2}r^{q_{2}1} \frac{1}{\lambda _{1}\min_{t\in \mathbb{T}} m(t)}\sum _{t=2}^{N1}\bigl(\triangle u(t) \bigr)^{2}, \end{aligned} $$
where \(\lambda _{1}\) is the smallest eigenvalue of (2.8). Hence there exists a sufficiently small \(\hat{r}>0\) such that, for every \(r\in (0,\hat{r}]\), we have \(\triangle \mathbf{u}=0\), and therefore \(\mathbf{u}=\mathbf{0}\). The homotopy invariance of the degree implies that
$$ \deg \bigl(IF,\mathrm{U}_{2}^{r},0\bigr)=1. $$
This concludes the proof of the claim.
We finally set
$$ \mathrm{U}_{3}=\bigl\{ \mathbf{u}\in Y: \vert \triangle \mathbf{u} \vert _{\infty }< 1 \delta _{b}\bigr\} . $$
Using the definition of \(\delta _{b}\) and arguing as above, we easily see that
$$ \deg (IF,\mathrm{U}_{3},0)=1. $$
Let us fix \(r\in (0,\min \{\tilde{\boldsymbol {\alpha}}_{1}_{\infty },\hat{r}\}]\), with \(\tilde{\boldsymbol {\alpha}}_{1}\) defined in Step 3. Notice that the sets \(\mathrm{U}_{1}\) and \(\mathrm{U}_{2}^{r}\), previously defined, are disjoint and both contained in \(\mathrm{U}_{3}\). Let us define
$$ \mathrm{W}^{r}=\mathrm{U}_{3}\setminus \bigl( \overline{\mathrm{U}_{1}\cup \mathrm{U}_{2}^{r}} \bigr). $$
As F has no fixed point in \(\partial \mathrm{U}_{1}\cup \partial \mathrm{U}_{2}^{r}\cup \partial \mathrm{U}_{3}\), by the excision and additivity properties of the degree, we have
$$ \begin{aligned} \deg (IF,\mathrm{U}_{3},0)&=\deg \bigl(IF,\mathrm{U}_{3}\setminus \bigl( \partial \mathrm{U}_{1}\cup \partial \mathrm{U}_{2}^{r} \bigr),0\bigr) \\ &=\deg (IF,\mathrm{U}_{1},0)+\deg \bigl(IF,\mathrm{U}_{2}^{r},0 \bigr)+\deg \bigl(IF, \mathrm{W}^{r},0\bigr) \end{aligned} $$
and hence
$$ \deg \bigl(IF,\mathrm{W}^{r},0\bigr)=1. $$
In particular, there exists a fixed point \(\mathbf{u}_{2}\) of F such that \(\mathbf{u}_{2}_{\infty }>r\) and for which the condition \(\mathbf{u}_{2}\gg \boldsymbol {\alpha}_{1}\) does not hold. Therefore \(\mathbf{u}_{2}\) is a positive solution of (3.1) which differs from \(\mathbf{u}_{1}\). We conclude that, for all \(\mu >\mu ^{*}\), there exist at least two positive solutions of (3.1).
Step 5. Proof of (iv). Take \(\lambda >0\) and \(\mu >0\). Assume (H), (A1)–(A3) and (B1)–(B3). Again, we can suppose that \(\hat{\mathbb{T}}_{1}\subset \mathbb{T}\). As already noticed in Step 4, we have \(q_{1}>1\). Let \(\mu ^{*}\) be the constant, introduced in Step 3, such that problem (3.1), with \(\lambda =0\), has at least one positive solution for all \(\mu >\mu ^{*}\). Fix \(\mu >\mu ^{*}\), set
$$ \Lambda = \vert \bar{a} \vert _{\infty }+\mu \vert \bar{b} \vert _{\infty }, $$
and let \(\delta \in (0,1)\) be the constant introduced in Lemma 2.3. Let us take \(\hat{\mathbb{T}}_{2}\subset \mathbb{T}_{1}\), and consider, for \(\lambda \in (0,1]\), the Dirichlet problem
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=\lambda a_{1}u^{p_{1}}, \quad t\in \mathbb{T}_{2}, \\ u(t_{3})=u(t_{4})=0, \end{cases} $$
(3.5)
where \(t_{3},t_{4}\in \mathbb{R}\). By Lemma 2.6 there exists a solution \(\boldsymbol {\alpha}_{2}^{\lambda }\) of (3.5) satisfying \(\boldsymbol {\alpha}_{2}^{\lambda }_{\infty }\leq \frac{N1}{2}\) and \(\boldsymbol {\alpha}_{2}^{\lambda }\gg \boldsymbol {0}\).
Denote by \(c_{0}>0\) a constant, dependent on Λ and \(\mathbb{T}_{2}\), such that
$$ \vert \mathbf{u} \vert _{\infty }\leq c_{0} \vert \mathbf{v} \vert _{\infty} $$
holds for all \(\mathbf{v}\in \mathbb{R}^{N}\) satisfying \(\mathbf{v}_{\infty }\leq \Lambda \). Similarly, denote by \(c_{0}'\) a constant, dependent on Λ, \(\mathbb{T}\), such that
$$ \vert \mathbf{u} \vert _{\infty }\leq c_{0}' \vert \mathbf{v} \vert _{\infty} $$
holds for all \(\mathbf{v}\in \mathbb{R}^{N}\) satisfying \(\mathbf{v}_{\infty }\leq \Lambda \). Set \(c_{1}=\Lambda \max \{c_{0},c_{0}'\}\) and \(r_{\lambda }=\lambda (c_{1}+1)\). Observe that, since by (A1), \(a_{1}(\boldsymbol {\alpha}_{2}^{\lambda })^{p_{1}}_{\infty }\leq \Lambda \), we have
$$ \bigl\vert \boldsymbol {\alpha}_{2}^{\lambda } \bigr\vert _{\infty }\leq c_{0}\Lambda \lambda \leq c_{1} \lambda . $$
As in Step 3, let \(\boldsymbol {\alpha}_{1}\) be a solution of the Dirichlet problem
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]= \mu b_{1}u^{q_{1}}, \quad t\in \mathbb{T}_{1}, \\ u(t_{1})=u(t_{2})=0. \end{cases} $$
Since \(\boldsymbol {\alpha}_{1}\gg \mathbf{0},t\in \mathbb{\hat{T}}\), we have \(\min_{t\in \mathbb{\hat{T}}_{2}} \boldsymbol {\alpha }_{1}> \mathbf{0}\). Therefore we can take \(\bar{\lambda }\in (0,1]\) such that, for all \(\lambda \in (0,\bar{\lambda })\), \(r_{\lambda }<\min_{t\in \mathbb{\hat{T}}_{2}} \boldsymbol {\alpha }_{1}\). For all \(\lambda \in (0,\bar{\lambda })\), we extend \(\boldsymbol {\alpha }_{1}\) to a function \(\tilde{\boldsymbol {\alpha }}_{1}^{\lambda }\) and \(\boldsymbol {\alpha }_{2}^{\lambda }\) to a function \(\tilde{\boldsymbol {\alpha }}_{2}^{\lambda }\) such that
$$\begin{aligned}& \bigl\vert \triangle \tilde{\boldsymbol {\alpha }}_{1}^{\lambda } \bigr\vert _{\infty }\leq 1, \qquad \bigl\vert \triangle \tilde{\boldsymbol { \alpha }}_{2}^{\lambda } \bigr\vert _{\infty }\leq 1,\qquad \bigl\vert \triangle \tilde{\boldsymbol {\alpha }}_{2}^{\lambda } \bigr\vert _{\infty }\leq r_{\lambda }, \\& \frac{N1}{2}< \tilde{\alpha }_{2}^{\lambda }(t)< \tilde{\alpha }_{1}^{\lambda }(t)< 0,\quad t\in \mathbb{\hat{T}} \setminus \mathbb{\hat{T}}_{1}, \\& \frac{N1}{2}< \tilde{\alpha }_{2}^{\lambda }(t)< 0,\quad t \in \mathbb{\hat{T}}_{1}\setminus \mathbb{\hat{T}}_{2}. \end{aligned}$$
We define, for every \(\lambda \in (0,\bar{\lambda })\), the open bounded sets
$$ \mathrm{V}_{1}^{\lambda }=\bigl\{ \mathbf{u}\in Y:\mathbf{u} \gg \tilde{\boldsymbol {\alpha }}_{1}^{\lambda }, \vert \triangle \mathbf{u} \vert _{\infty }< 1 \delta \bigr\} $$
and
$$ \mathrm{V}_{2}^{\lambda }=\bigl\{ \mathbf{u}\in Y:\mathbf{u}\gg \tilde{\boldsymbol {\alpha }}_{2}^{\lambda }, \vert \mathbf{u} \vert _{\infty }< r_{\lambda }, \vert \triangle \mathbf{u} \vert _{\infty }< 1\delta \bigr\} . $$
We also set, for \(t\in \mathbb{T}\),
$$ v_{1}^{\lambda }(t)=\mu \bar{b}\bigl(t,\tilde{\alpha }_{1}^{\lambda }(t)\bigr) $$
and
$$ v_{2}^{\lambda }(t)=\lambda \bar{a}\bigl(t,\tilde{\alpha }_{2}^{\lambda }(t)\bigr). $$
For every \(\lambda \in (0,\bar{\lambda })\), let \(\mathbf{z}_{1}^{\lambda }\) be the solution of the Dirichlet problem
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=v_{1}^{\lambda }(t), \quad t\in \mathbb{T}, \\ u(1)=u(N)=0. \end{cases} $$
Arguing as in the claim of Step 2, we easily verify that \(F_{\lambda ,\mu }\) has no fixed points on \(\partial \mathrm{V}_{1}^{\lambda }\) and
$$ \deg \bigl(IF_{\lambda ,\mu },\mathrm{V}_{1}^{\lambda },0 \bigr)=1. $$
For every \(\lambda \in (0,\bar{\lambda })\), let \(\mathbf{z}_{2}^{\lambda }\) be the solution of the Dirichlet problem
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=v_{2}^{\lambda }(t), \quad t\in \mathbb{T}, \\ u(1)=u(N)=0. \end{cases} $$
Claim
There exists \(\lambda (\mu )\in (0,\bar{\lambda }]\) such that, for all \(\lambda \in (0,\lambda (\mu ))\), \(F_{\lambda ,\mu }\) has no fixed points on \(\partial \mathrm{V}_{2}^{\lambda }\) and
$$ \deg \bigl(IF_{\lambda ,\mu },\mathrm{V}_{2}^{\lambda },0 \bigr)=1. $$
We first prove that
$$ \deg \bigl(I\mathbf{z}_{2}^{\lambda },\mathrm{V}_{2}^{\lambda },0 \bigr)=1. $$
It suffices to verify that \(\mathbf{z}_{2}^{\lambda }\in \mathrm{V}_{2}^{\lambda }\). Arguing as in Step 2, we easily see that \(\triangle \mathbf{z}_{2}^{\lambda }_{\infty }<1\delta \) and \(\mathbf{z}_{2}^{\lambda }\gg \boldsymbol {\alpha}_{2}^{\lambda }\) in \(\mathbb{T}\). Furthermore, we have
$$ \bigl\vert \mathbf{z}_{2}^{\lambda } \bigr\vert _{\infty }\leq c_{0}'\Lambda \lambda \leq c_{1} \lambda < r_{\lambda }. $$
Next we consider the homotopy \(H:[0,1]\times \bar{D}\rightarrow Y\) defined by
$$ H(k,u)=K\bigl(kN_{\lambda ,\mu }(u)+(1k)v_{2}^{\lambda } \bigr). $$
Obviously, H is completely continuous. Observe that
$$ H(0,u)=z_{2}^{\lambda }, \qquad H(1,u)=F_{\lambda ,\mu }(u) $$
for all \(\mathbf{u}\in \bar{D}\).
Fix now \(k\in [0,1]\), and suppose that \(\mathbf{u}\in \bar{\mathrm{V}}_{2}^{\lambda }\) is a fixed point of \(H(k,\cdot )\). We will prove that \(\mathbf{u}\in \mathrm{V}_{2}^{\lambda }\). Arguing as in step 2 we easily show that \(\boldsymbol {u}\geq \boldsymbol {0}\) in \(\mathbb{T}\) and \(\min_{\hat{\mathbb{T}}_{2}}\boldsymbol {u}>0\). Let us prove that \(\mathbf{u}\gg \tilde{\boldsymbol {\alpha }}_{2}^{\lambda }\) in \(\mathbb{T}\). Since \(u(t)\geq 0\), \(t\in \mathbb{T}\) and \(\tilde{\alpha }_{2}^{\lambda }(t)<0\), \(t\in \hat{\mathbb{T}}\setminus \hat{\mathbb{T}}_{2}\), we only need to verify that \(u(t)>\alpha _{2}^{\lambda }(t)\), \(t\in \hat{\mathbb{T}}_{2}\). Note that
$$ k\bar{f}(\cdot ,u)+(1k)v_{2}^{\lambda }\geq k\lambda a_{1}u^{p_{1}}+(1k) \lambda \bar{a}\bigl(\cdot ,\alpha _{2}^{\lambda }\bigr)\geq \lambda a_{1}\bigl( \alpha _{2}^{\lambda }\bigr)^{p_{1}}, \quad t\in \mathbb{T}. $$
Applying Lemma 2.1, we get
$$ \alpha _{2}^{\lambda }(t)\leq u(t)\min_{ \{t_{3},t_{4}\}}u< u(t),\quad t \in \hat{ \mathbb{T}}_{2}. $$
Moreover, as
$$ \bigl\vert k\bar{f}(\cdot ,u)+(1k)v_{2}^{\lambda } \bigr\vert _{\infty }\leq \Lambda , $$
Lemma 2.3 yields
$$ \vert \triangle \mathbf{u} \vert _{\infty }< 1\delta . $$
Finally, we verify that \(\mathbf{u}_{\infty }< r_{\lambda }\) if λ is sufficiently small. Since both \(\mathbf{u}_{\infty }\leq r_{\lambda }\) and \(\boldsymbol {\alpha}_{2}^{\lambda }_{\infty }\leq r_{\lambda }\) hold, we have
$$ \begin{aligned} k\bar{f}(\cdot ,u)+(1k)v_{2}^{\lambda }& \leq k\bigl(\lambda a_{2} \vert u \vert _{\infty }^{p_{2}}+ \mu b_{2} \vert u \vert _{\infty }^{q_{2}} \bigr)+(1k)\lambda a_{2} \bigl\vert \tilde{\alpha }_{2}^{\lambda } \bigr\vert _{\infty }^{p_{2}} \\ &\leq k(a_{2}(c_{1}+1)^{p_{2}}\lambda ^{p_{2}+1}+\mu b_{2}(c_{1}+1)^{q_{2}} \lambda ^{q_{2}}+(1k)a_{2}(c_{1}+1)^{p_{2}} \lambda ^{p_{2}+1} \\ &\leq a_{2}(c_{1}+1)^{p_{2}}\lambda ^{p_{2}+1}+\mu b_{2}(c_{1}+1)^{q_{2}} \lambda ^{q_{2}}\leq c_{2}\lambda ^{1+\epsilon }, \quad t\in \mathbb{T}, \end{aligned} $$
where \(c_{2}>0\) is a constant independent of λ and \(\epsilon =\min \{p_{2},q_{2}1\}\). Applying Lemma 2.3, we obtain
$$ \vert \mathbf{u} \vert _{\infty }\leq c_{3}\lambda ^{1+\epsilon }, $$
where \(c_{3}>0\) is a constant independent of λ. Let \(\lambda (\mu )\in (0,\bar{\lambda })\) be such that
$$ \lambda (\mu )\leq \biggl(\frac{c_{1}+1}{c_{3}}\biggr)^{\frac{1}{\epsilon }}. $$
Then, for each \(\lambda \in (0,\lambda (\mu ))\), the inequality \(\mathbf{u}_{\infty }< r_{\lambda }\) holds and hence \(\mathbf{u}\in \mathrm{V}_{2}^{\lambda }\). The homotopy invariance of the degree implies then that
$$ \deg \bigl(IF_{\lambda ,\mu },\mathrm{V}_{2}^{\lambda },0 \bigr)=1. $$
This concludes the proof the claim.
Note that \(\mathrm{V}_{1}^{\lambda }\) and \(\mathrm{V}_{2}^{\lambda }\) are disjoint because of the choice of λ. Therefore problem (3.1) has at least two positive solutions \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) such that \(\mathbf{u}_{1}\gg \boldsymbol {\alpha}_{1}^{\lambda }\) and \(\mathbf{u}_{2}_{\infty }< r_{\lambda }\). Finally we define, for all \(\lambda \in (0,\lambda (\mu ))\),
$$ \mathrm{V}_{3}^{\lambda }=\bigl\{ \mathbf{u}\in Y: \mathbf{u}_{1}\gg \tilde{\boldsymbol {\alpha }}_{2}^{\lambda }, \vert \triangle \mathbf{u} \vert _{\infty }< 1 \delta \bigr\} . $$
We also set \(\mathrm{W}^{\lambda }=\mathrm{V}_{3}^{\lambda }\setminus ( \overline{\mathrm{V}_{1}^{\lambda }\cup V_{2}^{\lambda }})\). Fix \(\lambda \in (0,\lambda (\mu ))\). Arguing as in the first part of the previous claim, we easily verify that
$$ \deg \bigl(IF_{\lambda ,\mu },\mathrm{V}_{3}^{\lambda },0 \bigr)=1. $$
By the excision and the additivity properties of the degree, we obtain
$$ \deg \bigl(IF_{\lambda ,\mu },\mathrm{W}^{\lambda },0\bigr)=1. $$
Hence, there exists a fixed point \(\mathbf{u}_{3}\) of \(F_{\lambda ,\mu }\) such that \(\mathbf{u}_{3}_{\infty }>r_{\lambda }\) and for which the condition \(\mathbf{u}_{3}\gg \tilde{\boldsymbol {\alpha }}_{1}^{\lambda }\) in \(\hat{\mathbb{T}}\) does not hold. Therefore \(\mathbf{u}_{3}\) is a positive solution of (3.1) which differs both from \(\mathbf{u}_{1}\) and from \(\mathbf{u}_{2}\). We conclude that, for every \(\mu >\mu ^{*}\), \(\lambda \in (0,\lambda (\mu ))\), problem (3.1) has at least three positive solutions. □
Finally, we give an example to illustrate the conclusion of Theorem 3.1.
Consider the Dirichlet problem
$$ \textstyle\begin{cases} \nabla [\phi (\triangle u(t))]=\frac{3}{2}\lambda u^{\frac{1}{2}}+2 \mu u^{2}, \quad t\in \mathbb{T}, \\ u(1)=u(N)=0. \end{cases} $$
(3.6)
In this case, \(a(t,u)=\frac{3}{2}u^{\frac{1}{2}}\), \(b(t,u)=2u^{2}\).
Take \(a_{1}=1\), \(p_{1}=\frac{1}{2}\). It is easy to check that the assumptions of (i) of Theorem 3.1 are satisfied. Hence, when \(\mu =0\), for every \(\lambda >0\), problem (3.6) has at least one positive solution.
Take \(b_{1}=1\), \(p_{1}=2\). It is easy to check that the assumptions of (ii) of Theorem 3.1 are satisfied. Hence, when \(\lambda =0\), there exists \(\mu ^{*}>0\) such that, for every \(\mu >\mu ^{*}\), problem (3.6) has at least one positive solution.
Take \(b_{2}=3\), \(p_{2}=2\). It is easy to check that the assumptions of (iii) of Theorem 3.1 are satisfied. Hence, when \(\lambda =0\), then there exists \(\mu ^{*}>0\) such that, for every \(\mu >\mu ^{*}\), problem (3.6) has at least two positive solutions.
Take \(a_{2}=2\), \(p_{2}=\frac{1}{2}\). It is easy to check that all the assumptions of (iv) of Theorem 3.1 are satisfied. Then there exist \(\mu ^{*}>0\) and a function \(\lambda (\cdot ):(\mu ^{*},+\infty )\rightarrow \mathbb{R}\) such that, for every \(\mu >\mu ^{*}\), \(\lambda \in (0,\lambda (\mu ))\), problem (3.6) has at least three positive solutions.