2.1. Auxiliary Functions
Let assumption (1.7) hold, and let us introduce auxiliary functions , and as
where ,
Here, the positive constants and are identical with those used in boundary conditions (1.6b). Note that the function is used in the analysis of positive and pseudo dead core solutions of problem (1.6a)(1.6b), while the function for its dead core solutions.
Properties of are described in the following lemma.
Lemma 2.1.
Let assumption (1.7) hold and let . Then , and is increasing on .
Proof.
Let be arbitrary, . Then , and is increasing on by [4, Lemma (where is replaced by )]. Since is arbitrary, the result immediately follows.
In the following lemma, we introduce functions and and discuss their properties.
Lemma 2.2.
Let assumption (1.7) hold. Then the following statements follow.
(i)The function is continuous on , and for .
(ii)For each , there exists a unique such that
and , for , .
(iii)The function
is continuous on .
Proof.

(i)
Let us define on by
Then . Let and define . Then, by (1.7), . Hence
and consequently , which means that is continuous at . Let . We now show that is continuous at the point . Let us choose an arbitrary in the interval . Then for and , where
Since by [4, Lemma (where 1 was replaced by )], it follows that is continuous at . The continuity of at now follows from the fact that is continuous at this point. Hence is continuous on , and from we conclude . Since
we have for .

(ii)
Consider the equation , that is,
The function is increasing on , , and, for , . Hence, for each , there exists a unique such that and . Clearly, for . In order to prove that , suppose the contrary, that is, suppose that is discontinuous at a point , . Then there exist sequences in such that , and the sequences are convergent, , , . Let in and in . This means , , and by the definition of the function , which contradicts .

(iii)
By (ii),
and for . Hence, the function is continuous on and positive on . From
we now deduce that . Since
where , and on , , we conclude . Hence is continuous at , and consequently .
Let be the function from Lemma 2.2(ii) defined on the interval . From now on, denotes the value of at , that is,
In the following lemma, we prove a property of which is crucial for discussing multiple positive solutions of problem (1.6a)(1.6b).
Lemma 2.3.
Let assumption (1.8) hold and let the function be given by (2.5). Then there exists such that
Proof.
Note that d. We deduce from [4, Lemma (with 1 replaced by )] that there exists an such that
If for some , then (2.16) yields
Consequently, inequality (2.15) holds for such an . If the statement of the lemma were false, then some would exist such that and
From the following equalities, compare (2.4),
and from , we conclude that
Finally, from
we have , which contradicts (2.18).
In order to discuss dead core solutions of problem (1.6a)(1.6b) and their dead cores, we need to introduce two additional functions and related to and study their properties.
Lemma 2.4.
Assume that (1.7) holds and let be given by (2.3). Then for each , there exists a unique such that
The function is continuous and decreasing on , and the function
is continuous and increasing on . Moreover, .
Proof.
It follows from (1.7) that . Also, is increasing w.r.t. both variables, for any , and , for any . Hence, for each , there exists a unique such that . In order to prove that is decreasing on , assume on the contrary that for some . Then which contradicts for . Hence, is decreasing on . If was discontinuous at a point , then there would exist sequences and in such that and , are convergent, , and with . Taking the limits in and , we obtain , . Consequently, by the definition of the function , which is not possible.
By (2.22),
and therefore,
It follows from the properties of that the functions , are continuous, positive, and increasing on . Hence (2.25) implies that and is increasing. Moreover, since d is bounded on .
Corollary 2.5.
Let assumption (1.7) hold. Then
and for each satisfying the inequality
there exists a unique such that
Proof.
The equalities for and imply that . Since the function defined by (2.23) is continuous and increasing on , it follows that for ; see (2.26). Let us choose an arbitrary satisfying (2.27). Then . Now, the properties of guarantee that equation has a unique solution . This means that (2.28) holds for a unique .
2.2. Dependence of Solutions on the Parameter
The following two lemmas characterize the dependence of positive and dead core solutions of problem (1.6a)(1.6b) on the parameter .
Lemma 2.6.
Let assumption (1.7) hold and let be a positive solution of problem (1.6a)(1.6b) for some . Also, let , and . Then , ,
where the function is given by (2.2).
Proof.
Since and for , we conclude that on and , . By integrating the equality over , we obtain
and consequently, since on ,
Finally, integrating
over yields (2.30). Now we set in (2.30) and obtain (2.29). Equality (2.31) follows from and from
Remark 2.7.
Let (1.7) hold and let be a pseudo dead core solution of problem (1.6a)(1.6b). Then, by the definition of pseudo dead core solutions, . We can proceed analogously to the proof of Lemma 2.6 in order to show that
where , and
From (2.38), we finally have . Consequently, .
Remark 2.8.
If , then , is the unique solution of problem (1.6a)(1.6b). This solution is positive.
Lemma 2.9.
Let assumption (1.7) hold and let be a dead core solution of problem (1.6a)(1.6b) for some . Moreover, let . Then and there exists a point such that for ,
where the function is given by (2.3). Furthermore, is the unique dead core solution of problem (1.6a)(1.6b) with .
Proof.
Since is a dead core solution of problem (1.6a)(1.6b) with , there exists by definition, a point such that , for and on . Consequently, on , and . We can now proceed analogously to the proof of Lemma 2.6 to show that
and (2.39) holds. Setting in (2.39), we obtain (2.40). Also, from (1.6b), ,
equality (2.41) follows.
It remains to verify that is the unique dead core solution of problem (1.6a)(1.6b) with . Let us suppose that is another dead core solution of the above problem. Let for and on for some . Then for , and consequently on and . Hence, compare (2.40) and (2.41),
Since
by (2.41), and the function is increasing and continuous on , we deduce from (2.45) and (2.46) that . Then (2.40) and (2.44) yield . Therefore, d for . Finally, since for and since by Lemma 2.1 the function is increasing on , follows. This completes the proof.
2.3. Main Results
Let the function be given by (2.5) and let us denote by the range of the function restricted to the interval ,
Since by Lemma 2.2(iii), for and , we can have either (i) for , or (ii) for some . For (i), we have , while in case of (ii), with
holds. Clearly, .
Positive solutions of problem (1.6a)(1.6b) are analyzed in the following theorem.
Theorem 2.10.
Let assumption (1.7) hold. Then problem (1.6a)(1.6b) has a positive solution if and only if . Additionally, for each , problem (1.6a)(1.6b) with has a unique positive solution such that and .
Proof.
Let be a positive solution of problem (1.6a)(1.6b) for . By Lemma 2.6, (2.31) holds with and . Furthermore, by Lemmas 2.2(ii) and 2.6, , which together with (2.29) implies that . Consequently, . For , problem (1.6a)(1.6b) has the unique positive solution ; see Remark 2.8. Since , . Consequently, if problem (1.6a)(1.6b) has a positive solution, then .
We now show that for each , problem (1.6a)(1.6b) has a positive solution, and if for some , then problem (1.6a)(1.6b) has a unique positive solution such that and . Let us choose . Then for some . If , then . Consequently, and is the unique solution of problem (1.6a)(1.6b). Clearly, and since . Let us suppose that . If is a positive solution of problem (1.6a)(1.6b) and , then, by Lemma 2.6; see (2.30), the equality holds for , where is given by (2.1). Hence, in order to prove that for problem (1.6a)(1.6b) has a unique positive solution such that and , we have to show that the equation
has a unique solution ; this solution is a positive solution of problem (1.6a)(1.6b), and , . Since , is increasing by Lemma 2.1, and , (2.49) has a unique solution . It follows from and that and . In addition,
Hence, and . In order to show that is continuous at , we set . Then, compare (2.49),
and therefore,
Consequently, , and so is continuous at , or equivalently, . Now (2.50) indicates that and
Moreover, by the de L'Hospital rule,
As a result and for . Since and, by (2.50), , we have
by Lemma 2.2(ii). Thus, satisfies (1.6b), and therefore is a unique positive solution of problem (1.6a)(1.6b) such that and .
The following theorem deals with multiple positive solutions of problem (1.6a)(1.6b).
Theorem 2.11.
Let assumption (1.8) hold. Then , with given by (2.48), and for each , there exist multiple positive solutions of problem (1.6a)(1.6b).
Proof.
By Lemmas 2.2(iii) and 2.3, , , and in a right neighbourhood of . Hence, . Let us choose . Then there exist such that for . Now Theorem 2.10 guarantees that problem (1.6a)(1.6b) has positive solutions and such that , . Since , we have and therefore, for each , problem (1.6a)(1.6b) has multiple positive solutions.
Next, we present results for pseudo dead core solutions of problem (1.6a)(1.6b). Note that here .
Theorem 2.12.
Let assumption (1.7) hold. Then problem (1.6a)(1.6b) has a pseudo dead core solution if and only if
Moreover, for given by (2.56), problem (1.6a)(1.6b) has a unique pseudo dead core solution such that .
Proof.
Let us assume that is a pseudo dead core solution of problem (1.6a)(1.6b) and let . Then, by Remark 2.7, equalities (2.36), (2.38) hold, and . Also, (2.37) implies that is a solution of the equation
where and are given by (2.1) and (2.56), respectively. The result follows by showing that equation (2.57) has a unique solution and that this solution is a pseudo dead core solution of problem (1.6a)(1.6b). We verify these facts for solutions of (2.57) arguing as in the proof of Theorem 2.10, with replaced by 0.
In the final theorem below, we deal with dead core solutions of problem (1.6a)(1.6b).
Theorem 2.13.
Let assumption (1.7) hold and let be the function defined in Lemma 2.4. Then the following statements hold.

(i)
Problem (1.6a)(1.6b) has a dead core solution if and only if

(ii)
For each satisfying (2.58), problem (1.6a)(1.6b) has a unique dead core solution.

(iii)
If the subinterval is the dead core of a dead core solution of problem (1.6a)(1.6b), then and
Proof.

(i)
Let be a dead core solution of problem (1.6a)(1.6b) for some and let . Then there exists a point such that for , and equalities (2.39), (2.40), and (2.41) are satisfied by Lemma 2.9. We deduce from (2.41) and from Lemma 2.4 that . Therefore, compare (2.40),
Since
by Corollary 2.5, we have
Hence, if problem (1.6a)(1.6b) has a dead core solution, then satisfies inequality (2.58).
We now prove that for each satisfying (2.58), problem (1.6a)(1.6b) has a dead core solution. Let us choose satisfying (2.58). Then, by Corollary 2.5, there exists a unique such that
Let us now consider, compare (2.39),
where is given by (2.1). Since and is increasing on by Lemma 2.1, , and, by (2.63), , there exists a unique solution of (2.64) and , . In addition,
and consequently, and . Since
by the Mean Value Theorem for integrals, where , we have
Therefore,
since . Hence, is continuous at , and . Furthermore,
Let
Then , for , , , and
Thus,
where is given by (2.3). Since by Lemma 2.4, satisfies the boundary conditions (1.6b). Consequently, is a dead core solution of problem (1.6a)(1.6b).

(ii)
Let us choose an arbitrary satisfying (2.58). By (i), problem (1.6a)(1.6b) has a dead core solution which is unique by Lemma 2.9.

(iii)
Let the subinterval be the dead core of a dead core solution of problem (1.6a)(1.6b). Then, by Lemma 2.9, equalities (2.40) and (2.41) hold with replaced by and . Since by the definition of the function , we have . Equality (2.59) now follows from (2.40) with and replaced by and , respectively.
Example 2.14.
We now turn to the case study of the boundary value problem (1.9a)(1.9b),
Note that (1.9a)(1.9b) is a special case of (1.6a)(1.6b) with satisfying (1.8). Since
we have
for , and for . By Lemma 2.2, the equation has a unique solution for , , for , and . Let
Then , , and . In order to show that is increasing on it is sufficient to verify that is injective. Let us assume that this is not the case, then there exist , , such that . From , , or equivalently, from
it follows that , and which is a contradiction. Hence, is increasing on and therefore, there exists the inverse function mapping onto . Since
and for , we have
Consequently,
In order to discuss the range of the function and the value of for , we first consider properties of the function
defined on . Let
Then
where . The function vanishes only at point
in the interval , and , because and . Since , on , on and
for , we have on and on . Let us define . Then , and it follows from that on and on . Consequently, is increasing on and decreasing on . It follows from the equality for and from the properties of the functions and that is increasing on and decreasing on . Hence, , where . Also,
Using properties of the function and the results of Theorems 2.10–2.13, we can now characterize the structure of the solution .

(i)
For each , there exists only a unique dead core solution of problem (1.9a)(1.9b).

(ii)
For , there exist a unique dead core solution and a unique positive solution of problem (1.9a)(1.9b).

(iii)
For each , there exist a unique dead core solution and exactly two positive solutions of problem (1.9a)(1.9b).

(iv)
For , there exist the unique pseudo dead core solution and a unique positive solution of problem (1.9a)(1.9b).

(v)
For each , there exist only a unique positive solution of problem (1.9a)(1.9b).
Using Theorem 2.10, Lemma 2.6, and the properties of the function , we can specify further properties of positive solutions of problem (1.9a)(1.9b).

(i)
If is the (unique) positive solution of problem (1.9a)(1.9b) with , then , where is the root of the equation .

(ii)
If are the (unique) positive solutions of problem (1.9a)(1.9b) with , then , , where , are the roots of the equation .
We are also able to give some more information on the dead core solutions of problem (1.9a)(1.9b). Since
the function , , is the solution of the equation . Let us choose an arbitrary . By Corollary 2.5, the equation; see (2.28),
has a unique solution . Consequently,
One can easily show that the function
is the unique dead core solution of problem (1.9a)(1.9b). Additionally, it follows from Theorem 2.13(iii) that since .