A three solution theorem for a singular differential equation with nonlinear boundary conditions

Rajendran Dhanya, Ratnasingham Shivaji, Byungjae Son


We study positive solutions to singular boundary value problems of the form: \begin{equation*} \begin{cases} -u'' = h(t) \dfrac{f(u)}{u^\alpha} &\text{for } t \in (0,1), \\ u(0) = 0, \\ u'(1) + c(u(1)) u(1) = 0,\hidewidth \end{cases} \end{equation*} where $0< \alpha< 1$, $h\colon(0,1]\rightarrow(0,\infty)$ is continuous such that $h(t)\leq {d}/{t^\beta}$ for some $d> 0$ and $\beta\in[0,1-\alpha)$ and $c\colon [0,\infty)\rightarrow [0,\infty)$ is continuous such that $c(s)s$ is nondecreasing. We assume that $f\colon[0,\infty)\rightarrow(0,\infty)$ is continuously differentiable such that $[(f(s)-f(0))/s^\alpha ]+\tau s$ is strictly increasing for some $\tau\geq 0$ for $s\in(0,\infty)$. When there exists a pair of sub-supersolutions $(\psi,\phi)$ such that $0\leq \psi\leq\phi$, we first establish a minimal solution $\underline u$ and a maximal solution $\overline u$ in $[\psi,\phi]$. When there exist two pairs of sub-supersolutions $(\psi_1,\phi_1)$ and $(\psi_2,\phi_2)$ where $0\leq \psi_1 \leq \psi_2 \leq \phi_1$, $\psi_1 \leq \phi_2 \leq \phi_1$ with $\psi_2\not \leq \phi_2$, and $\psi_2$, $\phi_2$ are not solutions, we next establish the existence of at least three solutions $u_1$, $u_2$ and $u_3$ satisfying $u_1\in [\psi_1,\phi_2], u_2\in [\psi_2,\phi_1]$ and $u_3\in [\psi_1,\phi_1]\setminus ([\psi_1,\phi_2]\cup [\psi_2,\phi_1])$.


Singular boundary value problem; nonlinear boundary conditions; three solution theorem

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