A uniqueness result for a class of infinite semipositone problems with nonlinear boundary conditions

We study positive solutions to the two-point boundary value problem:

where \(\lambda >0\) is a parameter, \(h \in C^1((0,1],(0,\infty ))\) is a decreasing function, \(f \in C^1((0,\infty ),\mathbb {R}) \) is an increasing concave function such that \(\lim \limits _{s \rightarrow \infty }f(s)=\infty \), \(\lim \limits _{s \rightarrow \infty }\frac{f(s)}{s}=0\), \(\lim \limits _{s \rightarrow 0^+}f(s)=-\infty \) (infinite semipositone) and \(c \in C([0,\infty ),(0,\infty ))\) is an increasing function. For classes of such h and f, we establish the uniqueness of positive solutions for \(\lambda \gg 1\).

Authors and Affiliations

1. Department of Mathematics and Statistics, Mississippi State University, Mississippi State, MS, 39762, USA

   D. D. Hai

2. Department of Mathematics and Statistics, University of North Carolina at Greensboro, Greensboro, NC, 27412, USA

   A. Muthunayake & R. Shivaji

Corresponding author

Correspondence to R. Shivaji.

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4 Appendix

Throughout this appendix, assume c satisfies \((H_3)\), and \(c(z)=c(0)~;~z<0\). First we consider the boundary value problem:

We establish:

Lemma A.1

Let \(\bar{h}\in L^1(0,1).\) Then (A.1) has a unique solution. Further, \(T:L^1(0,1)\rightarrow C[0,1]\) such that \(u=T(\bar{h})\) is the solution of (A.1), is a completely continuous operator.

Proof

By integrating we obtain

where \(A\in \mathbb {R}\) is such that \(u'(1)+c(u(1))u(1)=0\), i.e., \(G(A)=0\) where

Since \(G:\mathbb {R}\rightarrow \mathbb {R}\) is increasing and \(\lim \limits _{z \rightarrow \infty }G(z)=\infty \), \(\lim \limits _{z \rightarrow -\infty }G(z)=-\infty \), \(\exists _1\) \(A\in \mathbb {R}\) such that \(G(A)=0\). Thus (A.1) has a unique solution \(u\equiv T(\bar{h})\). Note that \(|A|\le \int _0^1|\bar{h}|ds\). (If \(z>\int _0^1|\bar{h}|ds\) then \(G(z)>0\) while if \(z<-\int _0^1|\bar{h}|ds\) then \(G(z)<0\).) Hence (A.2) and (A.3) give:

So T maps bounded sets in \(L^1(0,1)\) into relatively compact subsets of C[0, 1]. It remains to show that T is continuous. Let \({\bar{h}_n}\subset L^1(0,1)\) be such that \(\bar{h}_n\rightarrow \bar{h}\) in \(L^1(0,1)\) and let \(u_n=T(\bar{h}_n)\), \(u=T( \bar{h})\). Then \(-(u''_n-u'')=\bar{h}_n-\bar{h}.\) So (A.4) gives \(|u_n-u|_{C^1}\le 2\Vert \bar{h}_n-\bar{h}\Vert _{L^1}\rightarrow 0\). Thus Lemma A.1 is proven. \(\square \)

Next, we consider the boundary value problem:

when c satisfies \((H_3)\) and the following hypotheses hold:

\((G_1)\):\(~~g:(0,1)\times (0,\infty )\rightarrow \mathbb {R}\) is continuous and \(\exists ~\Phi ,\Psi \in C^2(0,1)\cap C^1[0,1]\) with

\(~~~~~~~~~\Psi \le \Phi ;~(0,1)\), \(\inf \limits _{(0,1)}\frac{\Psi }{p}>0\) where \(p(t)=\min (t,1-t)\) such that

and

\((G_2)\):\(~~\exists ~\Gamma \in L^1(0,1)\) such that for all \(\zeta \in [\Psi ,\Phi ]\), \(|g(t,\zeta (t))|\le \Gamma (t);~(0,1).\)

We establish:

Lemma A.2

Let \((G_1)-(G_2)\) hold. Then (A.5) has a positive solution \(u\in [\Psi ,\Phi ]~;~[0,1]\).

Proof

For \(v\in C[0,1]\) define \(u=S(v)\) to be the unique solution of:

where \(\tilde{v}(s)=\min \big (\max (v(s),\Psi (s)),\Phi (s)\big ).\) Note that by \((G_2)\), \(\exists \) \(\Gamma \in L^1(0,1)\) such that \(|g(t,\tilde{v})|\le \Gamma (t))\) and so (A.6) has a unique solution by Lemma A.1. Since \(\tilde{T}:C[0,1]\rightarrow L^1(0,1)\) defined \(\tilde{T}(v)=\tilde{v}\) is continuous and \(S=T\circ \tilde{T}\) (here T is as defined in Lemma A.1), it follows that \(S:C[0,1]\rightarrow C[0,1]\) is completely continuous. Since S(C[0, 1]) is bounded, S has a fixed point u by the Schauder Fixed Point Theorem. We now claim that \(u\in [\Psi ,\Phi ]\). Suppose there exists \(t_0\in (0,1)\) such that \(u(t_0)<\Psi (t_0).\) Let \((\alpha ,\beta )\) be the largest open interval in (0, 1] containing \(t_0\) such that \(u<\Psi \) on \((\alpha ,\beta )\). Then \(u(\alpha )=\Psi (\alpha )\). If \(\beta <1\) then \(u(\beta )=\Psi (\beta )\), while if \(\beta =1\) then \(u(\beta )\le \Psi (\beta )\) and \(u'(\beta )\ge \Psi '(\beta )\) in view of the boundary conditions of u and \(\Psi \) at 1. Thus in either case we have

Further, since \(\tilde{u}=\Psi \) on \((\alpha ,\beta )\) we have

Multiplying (A.8) by \(u-\Psi \) and integrating we obtain

Hence using (A.7), we obtain \(u'=\Psi '~;~(\alpha ,\beta )\), i.e., \(u=\Psi ~;~(\alpha ,\beta )\) (since \(u(\alpha )=\Psi (\alpha )\)), a contradiction. Hence \(u(t)\ge \Psi (t)~;~(0,1)\). Similarly, we obtain \(u(t)\le \Phi (t)~;~(0,1)\). Thus \(u\in [\Psi ,\Phi ]~;~[0,1]\) and so \(\tilde{u}=u\), i.e., u is a solution of (A.5). Hence Lemma A.2 is proven. \(\square \)

Finally, we consider (1.1) when for some \(l>0, \alpha>0, \gamma >0\) such that \(\alpha +\gamma <1\), h and f satisfy:

\((\tilde{H}_1)\):

:  \(h:(0,1]\rightarrow (0,\infty )\) is \(L^1(0,1)\) and \(\limsup \limits _{s \rightarrow 0^+}s^\gamma h(s)<\infty .\)

\((\tilde{H}_2)\):

:  \(f:(0,\infty )\rightarrow \mathbb {R}\) is \(C^1\), \(\lim \limits _{s \rightarrow \infty }f(s)=\infty \), \(\lim \limits _{s \rightarrow \infty }\frac{f(s)}{s}=0\), \(\lim \limits _{s \rightarrow 0^+}f(s)=-\infty \) and \(~~~~~~~~~\lim \limits _{s \rightarrow 0^+}s^{1+\alpha }f'(s)=l\).

We establish:

Lemma A.3

Let \((\tilde{H}_1)-(\tilde{H}_2)\) hold. Then (1.1) has a maximal positive solution for \(\lambda \gg 1\).

Proof

By Theorem 1.3 in [19], the Dirichlet boundary value problem:

has a positive solution w for \(\lambda \gg 1\). Clearly this w satisfies the role of \(\Psi \) in \((G_1)\) since \(w'(1)\le 0\). Let e be the unique positive solution of

and choose \(z(t)=M(\lambda )e(t)~;~[0,1]\) where \(M(\lambda )\gg 1\) so that

where \(\tilde{f}(s)=\max \limits _{(0,s]}f(t)\) (which is possible since \(\lim \limits _{s \rightarrow \infty }\frac{f(s)}{s}=0\)). Then

Also \(z(0)=0\) and

since c is increasing. Now choose \(M(\lambda )\gg 1\) so that \(z\ge \Psi (=w)\). Then z satisfies the role of \(\Phi \) in \((G_1)\) and hence \((G_1)\) hold. Also, for \(\zeta \in [\Psi ,\Phi ]\), \(\exists ~C>0\) such that

Note that \(\Gamma \in L^1(0,1)\) since \(\inf \limits _{(0,1)}\frac{\Psi }{p}>0\) and \(\limsup \limits _{s\rightarrow 0^+}s^\gamma h(s)<\infty \). So \((G_2)\) holds and by Lemma A.2, (1.1) has a positive solution for \(\lambda \gg 1\). To show the existence of a maximal solution, let u be a positive solution of (1.1). Then

This implies

Since \(\lim \limits _{s\rightarrow \infty }\frac{f(s)}{s}=0=\lim \limits _{s\rightarrow \infty }\frac{\tilde{f}(s)}{s}\) , \(\exists ~c_\lambda >0\) such that

and hence

This implies u is bounded by a supersolution of the form \(M(\lambda ) e(t)\) for \(M(\lambda )\gg 1\), and the existence of the maximal positive solution follows. \(\square \)

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