Existence and multiplicity of positive solutions for one-dimensional prescribed mean curvature equations

In this work, we establish the existence and multiplicity results of positive solutions for one-dimensional prescribed mean curvature equations. Our approach is based on fixed point index theory for completely continuous operators which leave invariant a suitable cone in a Banach space of continuous functions.

MSC:34B10, 34B18.

The prescribed mean curvature problems like

have attracted much attention in recent years, see [1–4] and the references therein. Since the problem is quasilinear non-uniformly elliptic, it is more difficult to study the existence of classical solutions. The greatest obstacle is the lack of gradient estimate, such kind of estimate does not hold in general and boundary gradient blow-up may occur. This leads to some new phenomena very different from those in semilinear problems. Many well-known results of semilinear problems have to be reconsidered for this quasilinear problem. Motivated by the search for solutions of the above problem, many authors (see [5–13]) studied the existence of (positive) solutions for one-dimensional prescribed mean curvature equations with Dirichlet boundary conditions

where \kappa >0 is a constant, f\in C([0,\mathrm{\infty }),[0,\mathrm{\infty })) and f(u)>0 for u>0 and x\in [0,1].

Note that if \kappa =0, problem (1.1) is degenerate to the second-order ordinary differential equation boundary value problems

The existence of (positive) solutions of (1.1) has been well known with various qualitative assumptions of the nonlinearity f, see [14, 15] and the references therein.

If \kappa =1, Bonheure et al. [6], Habets and Omari [9], Kusahara and Usami [10], Pan and Xing [12, 13] studied the existence of (positive) solutions of (1.2) by using the variational method, lower and upper solutions method and time mapping method, respectively.

However, to the best of our knowledge, the existence and multiplicity of positive solutions for (1.1) are relatively few by the fixed point index theory. In this paper, based on the fixed point index theory, we shall investigate the existence and multiplicity of positive solution of (1.1) when f is ϕ-superlinear and ϕ-sublinear at 0 and ∞, respectively, here \varphi (s)=\frac{s}{\sqrt{1+\kappa s^2}}.

Let \varphi (s)=\frac{s}{\sqrt{1+\kappa s^2}}. Then (1.1) can be rewritten as

Obviously, \varphi :\mathbb{R}\to (-\frac{1}{\sqrt{\kappa }},\frac{1}{\sqrt{\kappa }}) is an odd, increasing homeomorphism with \varphi (0)=0.

For convenience, we introduce some notations

We will also need the function f^{\ast }(u)={max}_{0\le t\le u}\{f(t)\}, and let f_0^{\ast }={lim}_{u\to 0}\frac{f^{\ast }(u)}{\varphi (u)}. By a similar method in [[14], Lemma 2.8], it is not difficult to verify that f_0^{\ast }=f_0.

In the rest of this paper, we shall study the existence of positive solutions of (1.3) by using the fixed point index theory to give a brief and clear proof for the existence of positive solutions of (1.1). More concretely, we shall prove the following.

Theorem 1.1 Assume that f\in C([0,\mathrm{\infty }),[0,\mathrm{\infty })) and f(u)>0 for u>0.

1. (i)

   If f_0=0, then there exists such that (1.1) has a positive solution for .

2. (ii)

   If f_0=\mathrm{\infty }, then there exists {\lambda }_0>0 such that (1.1) has a positive solution for .

3. (iii)

   If f_0=f_{\mathrm{\infty }}=0, then there exists such that (1.1) has at least two positive solutions for .

4. (iv)

   If f_0=f_{\mathrm{\infty }}=\mathrm{\infty }, then there exists such that (1.1) has at least two positive solutions for .

Corollary 1.2 Assume that f\in C([0,\mathrm{\infty }),[0,\mathrm{\infty })) and f(u)>0 for u>0.

1. (a)

   If f_0=0 and f_{\mathrm{\infty }}=\mathrm{\infty }, then there exists such that (1.1) has a positive solution for .

2. (b)

   If f_0=\mathrm{\infty } and f_{\mathrm{\infty }}=0, then there exists {\lambda }_0>0 such that (1.1) has a positive solution for .

Remark 1.1 The results of Theorem 1.1 and Corollary 1.2 are different with the case \kappa =0 which is the classical Dirichlet boundary value problem (1.2). This phenomenon is a striking feature of problem (1.1), which is just the reason why we study the existence of positive solutions of problem (1.1). It is pointed out that in equation of (1.1), having replaced f(u) with f(x,u), Theorem 1.1 and Corollary 1.2 also hold as well as all of the proofs with obvious changes.

Throughout the paper |\cdot | will denote absolute value, and let

Then E is a Banach space endowed with the norm {\parallel u\parallel }_{\mathrm{\infty }}={max}_{x\in [0,1]}|u(x)|.

We first establish some preliminary results to prove our main result. An easy but useful property of ϕ and {\varphi }^{-1} is the following one.

Lemma 2.1 Let \varphi (s)=\frac{s}{\sqrt{1+\kappa s^2}}. Then \varphi :\mathbb{R}\to (-\frac{1}{\sqrt{\kappa }},\frac{1}{\sqrt{\kappa }}) is an odd, increasing homeomorphism with \varphi (0)=0. Moreover, ϕ has the following properties:

1. (i)

   ϕ is convex up on [0,\mathrm{\infty }) and {\varphi }^{-1} is concave up on (0,\frac{1}{\sqrt{\kappa }}).

2. (ii)

   For each c\ge 1, there exists {\hat{A}}_c\ge c such that \varphi (cs)\le {\hat{A}}_c\varphi (s), \mathrm{\forall }s\ge 0 with {lim}_{c\to \mathrm{\infty }}{\hat{A}}_c=\mathrm{\infty } and for each , there exists such that \varphi (cs)\ge {\hat{B}}_c\varphi (s), \mathrm{\forall }s\ge 0 with {lim}_{c\to 0}{\hat{B}}_c=0.

3. (iii)

   For each , there exists B_c\ge c such that {\varphi }^{-1}(cs)\le B_c{\varphi }^{-1}(s), \mathrm{\forall }s\in (0,\frac{1}{\sqrt{\kappa }}). For each c\ge 1 with , there exists A_c\le c such that {\varphi }^{-1}(cs)\ge A_c{\varphi }^{-1}(s), \mathrm{\forall }s\in (0,\frac{1}{\sqrt{\kappa }}).

Proof By a simple computation, it follows that \varphi (-s)=-\varphi (s) and {\varphi }^{\prime }(s)=\frac{1}{\sqrt{{(1+\kappa s^2)}^3}}>0. So ϕ is an odd, increasing homeomorphism with \varphi (0)=0. Moreover, from {\varphi }^{″}(s)=\frac{-3\kappa s}{\sqrt{{(1+\kappa s^2)}^5}}, we get that ϕ is convex up on [0,\mathrm{\infty }). Notice that {\varphi }^{-1}(s)=\frac{s}{\sqrt{1-\kappa s^2}} is also an odd, increasing homeomorphism with \varphi (0)=0. It is easy to verify that {\varphi }^{-1} is concave up on (0,\frac{1}{\sqrt{\kappa }}).

1. (ii)

   For each , there exists {\hat{B}}_c\le c such that

   \varphi (cs)=\frac{cs}{\sqrt{1+\kappa c^2{s}^2}}\ge {\hat{B}}_c\frac{s}{\sqrt{1+\kappa s^2}},\mathrm{\forall }s\ge 0;

and for each c\ge 1, there exists {\hat{A}}_c\ge c such that

1. (iii)

   By a similar argument, it is not difficult to compute that for each , there exists B_c\ge c such that {\varphi }^{-1}(cs)\le B_c{\varphi }^{-1}(s), \mathrm{\forall }s\in (0,\frac{1}{\sqrt{\kappa }}). For each c\ge 1 with , there exists A_c\le c such that {\varphi }^{-1}(cs)\ge A_c{\varphi }^{-1}(s), \mathrm{\forall }s\in (0,\frac{1}{\sqrt{\kappa }}). □

Lemma 2.2 Let h\in C([0,1],[0,\mathrm{\infty })) with h\not\equiv 0. Assume that w is the solution of

Then w>0 on (0,1) and {\parallel w^{\prime }\parallel }_{\mathrm{\infty }}\le {\varphi }^{-1}(M), where M=min\{\frac{1}{\sqrt{\kappa }},{sup}_{x\in [0,1]}|h(x)|\}.

Proof By integrating, it follows that (2.1) has the unique solution given by

where C is such that w(1)=0. Hence we must have . Further, since \varphi (u^{\prime })\in (-\frac{1}{\sqrt{\kappa }},\frac{1}{\sqrt{\kappa }}), by using \varphi (w^{\prime }(x))=C-{\int }_0^{x}h(t)dt, we obtain -M\le \varphi (w^{\prime }(x))\le M, \mathrm{\forall }x\in [0,1] and {\parallel w^{\prime }\parallel }_{\mathrm{\infty }}\le {\varphi }^{-1}(M) follows, here M=min\{\frac{1}{\sqrt{\kappa }},{sup}_{x\in [0,1]}|h(x)|\}.

Since w(0)=w(1)=0, there exists x_0\in (0,1) such that w^{\prime }(x_0)=0 and it follows from -{(\varphi (w^{\prime }))}^{\prime }\ge 0 that \varphi (w^{\prime }) is decreasing on (0,1). Then w^{\prime }(x)>0 for and for x>x_0. Hence, w>0 on (0,1). □

Note that from Lemma 2.2, there exists {\tau }_i\in (0,\frac{1}{2}), i=1,2, such that {min}_{x\in [{\tau }_1,1-{\tau }_2]}w(x)\ge \sigma {\parallel w\parallel }_{\mathrm{\infty }} with depending on {\tau }_i. Define the cone P in E by

and for r>0, let .

Lemma 2.3 ([[11], Lemma 4.1 and Lemma 4.2])

For each h\in C[0,1], (2.1) has a unique solution given by

where C is such that u(1)=0 with . Moreover, the operator T_h:E\to E is continuous and sends equicontinuous sets in C[0,1] into a relatively compact set in E.

We next state the fixed point index theorem which will be used to prove our results.

Lemma 2.4 ([[16], Chapter 6])

Let E be a Banach space and P be a cone in E. Assume that Ω is a bounded open subset of E with 0\in \mathrm{\Omega }, and let T:P\cap \overline{\mathrm{\Omega }}\to P be a completely continuous operator such that Tu\ne u, u\in \partial \mathrm{\Omega }\cap P.

1. (i)

   If \parallel Tu\parallel \le \parallel u\parallel, u\in \partial \mathrm{\Omega }\cap P, then i(A,\mathrm{\Omega }\cap P,P)=1.

2. (ii)

   If \parallel Tu\parallel \ge \parallel u\parallel, u\in \partial \mathrm{\Omega }\cap P, then i(A,\mathrm{\Omega }\cap P,P)=0.

From Lemma 2.2, problem (1.3) is equivalent to the fixed point problem

in the space E, where C_0 is such that u(1)=0 with , since otherwise,

which is a contradiction. This together with Lemma 2.3 implies that T_{\lambda }:E\to E is a completely continuous mapping. Moreover, for any fixed u\in P, we have

and T_{\lambda }(u)(0)=T_{\lambda }(u)(1)=0. In addition, from Lemma 2.2, it follows that T_{\lambda }(u)(x)>0 on (0,1) and there exist {\tau }_i\in (0,\frac{1}{2}), i=1,2, such that {min}_{x\in [{\tau }_1,1-{\tau }_2]}{T}_{\lambda }u\ge \sigma {\parallel T_{\lambda }u\parallel }_{\mathrm{\infty }}. So T_{\lambda }:P\to P is a completely continuous operator.

Lemma 2.5 Let r>0 be given. If there exists \epsilon >0 small enough with such that f^{\ast }(r)\le \epsilon \varphi (r), then

where B_{\lambda \epsilon } is defined as in Lemma  2.1(iii).

Proof From the definition of T_{\lambda }, for any u\in \partial {\mathrm{\Omega }}_r, we have

 □

Lemma 2.6 Let \eta >0 be given. If u\in P and f(u(x))\ge \eta \varphi (u(x)) for x\in [0,1], then

where x_{\ast }=min\{x_0,1-x_0\} and u(x_0)={max}_{x\in [0,1]}u(x)={\parallel u\parallel }_{\mathrm{\infty }}.

Proof From problem (1.3), since u(0)=u(1)=0, it follows that there exists x_0\in (0,1) such that u^{\prime }(x_0)=0. Let {\parallel u\parallel }_{\mathrm{\infty }}=u(x_0). Then u satisfies the following boundary value problem:

Let v be the solution of the problem

Then we have

and by a comparison argument, we get that u>v on (0,x_0). In fact, from (u-v)(0)=0, (u-v)(x_0)=0, there exists \hat{x}\in (0,x_0) such that {(u-v)}^{\prime }(\hat{x})=0, i.e., u^{\prime }(\hat{x})=v^{\prime }(\hat{x}). Thus, by a simple computation, we have that

and

This together with ϕ is an increasing homeomorphism implies that

Integrating from 0 to x for (2.6) and integrating from x to x_0 for (2.7), respectively, we have that u(x)\ge v(x) for x\in [0,x_0].

Note that

where C_1 is such that v(0)=0, and hence {\parallel u\parallel }_{\mathrm{\infty }}={\int }_0^{x_0}{\varphi }^{-1}(C_1)ds. If C_1>\varphi (\frac{{\parallel u\parallel }_{\mathrm{\infty }}}{x_0}), then it follows that

which is a contradiction. Thus, 0\le C_1\le \varphi (\frac{{\parallel u\parallel }_{\mathrm{\infty }}}{x_0}). Moreover, we have

where x_1=\sigma x_0. Consequently, u(x)\ge v(x)\ge \sigma {\parallel u\parallel }_{\mathrm{\infty }} for x\in [x_1,x_0]. Obviously,

where C_2 satisfies u^{\prime }(x_0)=0. It follows from {\varphi }^{-1}(C_2-\lambda {\int }_0^{x_0}f(u(t))dt)=0 that C_2=\lambda {\int }_0^{x_0}f(u(t))dt. Therefore,

If x_0\ge \frac{1}{2}, then

If x_0\le \frac{1}{2}, let w be the solution of

Then

where satisfies w(1)=0, i.e., {\parallel u\parallel }_{\mathrm{\infty }}=-{\int }_{x_0}^1{\varphi }^{-1}(C_3)ds. If , then

which is contradiction. Hence -\varphi (\frac{{\parallel u\parallel }_{\mathrm{\infty }}}{1-x_0})\le C_3\le 0.

By a similar argument as before, it follows that u\ge w on (x_0,1). Moreover,

where x_2=1-\sigma (1-x_0). So u(x)\ge w(x)\ge \sigma {\parallel u\parallel }_{\mathrm{\infty }}, x\in [x_0,x_2]. Therefore, we have

and subsequently,

Let x_{\ast }=min\{x_0,1-x_0\}. Then

 □

Lemma 2.7 Let r>0 be given. If u\in \partial {\mathrm{\Omega }}_r, then

where M_r=1+{max}_{0\le u\le r}\{f(u)\}>0.

Proof Obviously, for any u\in \partial {\mathrm{\Omega }}_r, it follows that f(u(x))\le M_r for x\in [0,1]. So we have

 □

Lemma 2.8 Let r>0 be given. If u\in \partial {\mathrm{\Omega }}_r, then

where m_r={min}_{\sigma r\le u\le r}\{f(u)\}>0 and x_{\ast }=min\{x_0,1-x_0\}.

Proof By using a similar argument of the proof of Lemma 2.6, we have that u(x)\ge \sigma {\parallel u\parallel }_{\mathrm{\infty }} for x\in [x_1,x_2]. Meanwhile, (2.8) is true. If x_0\ge \frac{1}{2}, then we can get that

If x_0\le \frac{1}{2}, then (2.10) holds and it follows that

Let x_{\ast }=min\{x_0,(1-x_0)\}, then

 □

Proof of Theorem 1.1 (i) Choose a suitable number r_1>0. By Lemma 2.8, we have

where

If f_0=0, then f_0^{\ast }=0, and so we can choose r_2\in (0,r_1) such that f^{\ast }(r_2)\le \epsilon \varphi (r_2), where \epsilon >0 small enough satisfies

Then Lemma 2.5 implies that

From Lemma 2.4, it follows that i(T_{\lambda },{\mathrm{\Omega }}_{r_1},P)=0 and i(T_{\lambda },{\mathrm{\Omega }}_{r_2},P)=1. By using the additivity-excision property of the fixed point index [16], we have that

Therefore, T_{\lambda } has a fixed point in {\overline{\mathrm{\Omega }}}_{r_1}\mathrm{\setminus }{\mathrm{\Omega }}_{r_2}. Consequently, (1.1) has a positive solution for .

1. (ii)

   Choose a suitable number r_1>0. By Lemma 2.7, there exists

such that for u\in \partial {\mathrm{\Omega }}_{r_1} and . That is, i(T_{\lambda },{\mathrm{\Omega }}_{r_1},P)=1.

If f_0=\mathrm{\infty }, then there exists r_2\in (0,r_1) such that f(u)\ge \eta \varphi (u) for 0\le u\le r_2, where \eta >0 is chosen large enough so that

Clearly, f(u)\ge \eta \varphi (u(x)) for u\in \partial {\mathrm{\Omega }}_{r_2}, x\in [0,1]. From Lemma 2.6, we get that

This together with Lemma 2.4 implies i(T_{\lambda },{\mathrm{\Omega }}_{r_2},P)=0. By using the additivity-excision property of the fixed point index [16], we have

Therefore, T_{\lambda } has a fixed point in {\overline{\mathrm{\Omega }}}_{r_1}\mathrm{\setminus }{\mathrm{\Omega }}_{r_2}. Consequently, (1.1) has a positive solution for .

1. (iii)

   Since ϕ is a bounded operator, multiplying (1.3) by u^{\prime } and integrating from 0 to x_0, we get that

   \lambda {\int }_0^{u(x_0)}f(u)du=\frac{1}{\kappa }[1-\frac{1}{\sqrt{1+\kappa {(u^{\prime }(0))}^2}}]\to \frac{1}{\kappa }\text{as }{u}^{\prime }(0)\to \mathrm{\infty }.

Let {\lambda }^{\ast }>0 be the solution of \lambda {\int }_0^{r^{\ast }}f(u)du=\frac{1}{\kappa }. Then there exists \overline{r}:=r^{\ast }-ϵ>0 such that

with \overline{r}=u(x_0)={\parallel u\parallel }_{\mathrm{\infty }}.

Choose two numbers satisfying

By Lemma 2.8, there exist

and

such that for , we have

This together with Lemma 2.4 implies i(T_{\lambda },{\mathrm{\Omega }}_{r_i},P)=0, i=3,4.

Since f_0=0, from the proof of the case (i), it follows that we can choose r_1\in (0,\frac{r_3}{2}) such that i(T_{\lambda },{\mathrm{\Omega }}_{r_1},P)=1. Subsequently,

On the other hand, f_{\mathrm{\infty }}=0, and (3.3) together with Lemma 2.7 implies

That is, i(T_{\lambda },{\mathrm{\Omega }}_{\overline{r}},P)=1. Subsequently,

Therefore, T_{\lambda } has two fixed points u_1 and u_2 such that u_1\in {\overline{\mathrm{\Omega }}}_{r_3}\mathrm{\setminus }{\mathrm{\Omega }}_{r_1} and u_2\in {\overline{\mathrm{\Omega }}}_{\overline{r}}\mathrm{\setminus }{\mathrm{\Omega }}_{r_4}. These are the desired distinct positive solutions of (1.1) for satisfying

1. (iv)

   Choose two numbers satisfying

   \varphi (r_3)>\frac{M_{r_3}}{(1-\sigma )x_{\ast }{m}_{\overline{r}}}\varphi \left(\frac{\overline{r}}{\sigma x_{\ast }}\right)\text{and}\varphi (r_4)>\frac{M_{r_4}}{(1-\sigma )x_{\ast }{m}_{\overline{r}}}\varphi \left(\frac{\overline{r}}{\sigma x_{\ast }}\right).

By Lemma 2.7, there exists such that for , we have

That is, i(T_{\lambda },{\mathrm{\Omega }}_{r_3},P)=1 and i(T_{\lambda },{\mathrm{\Omega }}_{r_4},P)=1.

Since f_0=\mathrm{\infty }, from the proof of the case (ii), choose r_1\in (0,\frac{r_3}{2}) such that i(T_{\lambda },{\mathrm{\Omega }}_{r_1},P)=0. Consequently,

On the other hand, f_{\mathrm{\infty }}=\mathrm{\infty } and (3.3) together with Lemma 2.8 implies that

Let {\lambda }_{\ast }=\frac{\varphi (\frac{\overline{r}}{\sigma x_{\ast }})}{(1-\sigma )x_{\ast }{m}_{\overline{r}}} and {\lambda }^{\ast }=min\{{\lambda }_0,\frac{1}{\sqrt{\kappa }(1-\sigma )x_{\ast }{m}_{\overline{r}}}\}. Then, for any , we have

That is, i(T_{\lambda },{\mathrm{\Omega }}_{\overline{r}},P)=0. Subsequently,

Therefore, T_{\lambda } has two fixed points u_1 and u_2 such that u_1\in {\overline{\mathrm{\Omega }}}_{r_3}\mathrm{\setminus }{\mathrm{\Omega }}_{r_1} and u_2\in {\overline{\mathrm{\Omega }}}_{\overline{r}}\mathrm{\setminus }{\mathrm{\Omega }}_{r_4}. These are the desired distinct positive solutions of (1.1) for satisfying (3.4). □

Proof of Corollary 1.2 It is easy to show by the result of Theorem 1.1(i) and (ii). □

Example 3.1 Let us consider the following problem:

Obviously, f_0={lim}_{u\to 0}\frac{e^u}{\varphi (u)}=\mathrm{\infty } and f_{\mathrm{\infty }}={lim}_{u\to \mathrm{\infty }}\frac{e^u}{\varphi (u)}=\mathrm{\infty }. From Theorem 1.1(iv), there exists such that (3.5) has at least two positive solutions for .