On the Existence of Radially Symmetric Solutions for the $p$-Laplace Equation with Strong Gradient Nonlinearities

Abstract

We consider the Dirichlet problem for the $p$-Laplace equation in presence of a gradient not satisfying the Bernstein–Nagumo type condition. We define some class of gradient nonlinearities, for which we prove the existence of a radially symmetric solution with a Hölder continuous derivative.

1. Introduction and the Main Results

Consider the Dirichlet problem for the $p$-Laplace equation

$$ -\operatorname{div}(|\nabla u|^{p-2}\nabla u)=F(x,u,\nabla u)\quad\text{in}\quad B_{R}\subset{𝕉}^{n}, $$

(1.1)

$$ u|_{\partial{B_{R}}}=0, $$

(1.2)

where $B_{R}$ is a ball of radius $R$ and $\partial{B_{R}}$ is the boundary of $B_{R}$. Regarding the history of the issue and the first results on the existence and regularity of solutions to equations with $p$-Laplacian, we can refer to the monograph [1] by Ladyzhenskaya and Uraltseva. Variational, approximation, and topological methods are of use in studying boundary value problems for (1.1). Involving the method of the calculus of variations is due to the variational structure of the main part of (1.1). However, the presence of gradient terms in the equation essentially complicates the use of these methods. In this case, to prove the solvability of the boundary value problem of wide use are the topological methods based on a priori estimates, as well as various approximation methods.

In this article we will specifically focus on the dependence of $F(x,u,\nabla u)$ on the gradient of $u$. In this regard, we note the following works in which the study of boundary value problems was carried out in presence of gradient terms.

In [2,3,4,5], the existence of weak solutions to problem (1.1), (1.2) is proved on using approximation methods. In the works [6,7,8,9,10] similar results were proven on using the various topological methods based on theorems of Liouville type, the method of sub/supersolutions, and application of Krasnoselsky’s Theorem. In [11], the results on the existence of solutions were obtained on using an iterative method that bases on the mountain pass theorem. In [12], the connection is studied between viscosity solutions and Sobolev solutions in equations with low order terms containing derivatives of the solution. The authors of [13, 14] apply the Leray–Schauder Fixed Point Theorem to obtaining the existence of solutions by linearization methods, a priori estimates with weights, and Comparison Theorems.

Note that in all of the above works, the low order terms in the equation satisfy the Bernstein–Nagumo condition which in the case of equation (1.1) takes the form

$$ |F(x,u,q)|\leq c\left(1+|q|^{p}\right)\quad\text{for}\quad(x,u,q)\in\overline{\Omega}\times[-M,M]\times 𝕉^{n}, $$

(1.3)

with some constant $c$, provided that the solution satisfies $\max|u|\leq M$ with some constant $M$. The novelty of our results lies in the proof of the existence of a solution in the case that the Bernstein–Nagumo condition is violated. Note however that, unlike the works mentioned above, we confine our research to radially symmetric solutions.

In [15, 16] the existence of radially symmetric solutions to problem (1.1), (1.2) was proven when condition (1.3) is violated. Sufficient conditions for solvability in the specified class of functions were given, connecting the behavior of the nonlinear source and the convective term.

The novelty of the results of this article in comparison with [15, 16] lies in the proof of the existence of a solution in the absence of the Bernstein–Nagumo condition for a significantly wider class of gradient nonlinearities.

We are interested in the existence of bounded radially symmetric solutions to problem (1.1), (1.2). We will assume that $F(x,u,\nabla u)$ can be represented in the form $F(r,u,u_{r})$ with the change of variables $r=|x|$. Some examples of these functions are as follows:

$$ F(|x|,u,|\nabla u|),\quad F(|x|,u,x\cdot\nabla u), $$

where

$$ x\cdot\nabla u=\sum\limits^{n}_{i=1}x_{i}u_{x_{i}}. $$

In the sequel, we will simply denote the derivative of a function $u$ with respect to $r$ as $u^{\prime}$. It is well known that the radially symmetric solution of (1.1) satisfies the equation

$$ -(|u^{\prime}|^{p-2}u^{\prime})^{\prime}-\frac{n-1}{r}|u^{\prime}|^{p-2}u^{\prime}=F(r,u,u^{\prime}),\quad r\in(0,R), $$

(1.4)

and boundary conditions

$$ u^{\prime}(0)=0,\quad u(R)=0. $$

(1.5)

Owing to the degeneracy (singularity) of equation (1.4), its solutions can fail to belong to the space of twice continuously differentiable functions. In this regard, let us define what we will mean by a solution of problem (1.4), (1.5).

Definition 1.1

Say that $u(r)$ is a weak solution of problem (1.4), (1.5), if $u^{\prime}(r)$ is Hölder continuous on $[0,R]$, satisfies (1.5) and the integral identity

$$ \int\limits^{R}_{0}|u^{\prime}(r)|^{p-2}u^{\prime}(r)\phi^{\prime}(r)dr $$

$$ =\int\limits^{R}_{0}\frac{n-1}{r}|u^{\prime}(r)|^{p-2}u^{\prime}(r)\phi(r)dr+\int\limits^{R}_{0}F(r,u(r),u^{\prime}(r))\phi(r)dr $$

$$ \quad\text{for all }\phi(r)\in 𝔺^{\infty}_{0}(0,R). $$

Owing to the smoothness of the so-defined solution, we will understood (1.5) in the usual sense.

Let us make a few more comments related directly to the goal of this article. These remarks concern $F(r,u,u^{\prime})$. We will construct a weak solution in the sense of Definition 1.1 by passing to the limit in a sequence of classical solutions of the corresponding regularized problem. The proof of the existence of classical solutions is carried out on the basis of a priori estimates and the Fixed Point Theorem. Obtaining a priori estimates begins with an a priori estimate of $\max|u|$. The difficulty in obtaining such estimate arises in the case when the maximum principle cannot be directly applied to problem (1.4), (1.5). Obtaining an a priori estimate of $\max|u|$ for a regularized problem can be found in [15, 16]. To simplify the presentation and focus attention on the result of this work, which constitutes its novelty, we will assume that $F$ has the form

$$ F(r,u,u^{\prime})=-u+g(r,u,u^{\prime})+f(r),\quad g(r,u,0)=0. $$

(1.6)

In [16] we assume that the gradient term $g(r,u,u^{\prime})$ is a continuous function satisfying the conditions

$$ g(s,u_{1},q)-g(r,u_{2},q)\leq 0,\quad q>0, $$

(1.7)

$$ g(r,u_{1},-q)-g(s,u_{2},-q)\leq 0,\quad q>0, $$

(1.8)

for $r>s$ and $u_{1}-u_{2}>0$. In this article we will show that, owing to Lipschitz continuity in the spatial variable and strict monotonicity in $u$, it is possible in fact to remove the structural constraints on the gradient term associated with the behavior in the spatial variable to obtain the existence of a solution to problem (1.4), (1.5).

To formulate the main result we need some notations. Put $f_{0}=\max_{r\in[0,R]}|f(r)|$, and let $\widetilde{M}$ be a constant satisfying the inequality

$$ \left(\frac{Rf_{0}}{n-1}\right)^{\frac{1}{p-1}}<\widetilde{M}. $$

(1.9)

For the boundary estimate of the gradient of a solution we need the condition on $g(r,u,u^{\prime})$ which follows from Lemma 1 of [16]. Assume that $g(r,u,u^{\prime})$ satisfies

$$ g(r,u,-\widetilde{M})\leq 0,\quad r\in(0,R),\quad u>0, $$

(1.10)

$$ g(r,u,\widetilde{M})\geq 0,\quad r\in(0,R),\quad u<0. $$

(1.11)

Let us introduce some more notations

$$ M_{1}=\min\left\{-f_{0},0\right\},\quad M_{2}=\max\left\{f_{0},0\right\},\quad M_{*}=\widetilde{M}R,\quad M=\min\{M_{*},\max\{|M_{1}|,M_{2}\}\} $$

$$ C^{p-1}\geq\max\left\{\frac{M+f_{0}}{(p-1)},\left(\frac{M_{*}}{R}\right)^{p-1}\right\},\quad p\geq 2, $$

(1.12)

$$ C^{p-1}\geq\max\left\{\frac{M+f_{0}}{2(p-1)(1+R)^{p-2}},\left(\frac{M_{*}}{R}\right)^{p-1}\right\},\quad 1<p<2. $$

(1.13)

Replace conditions (1.7) and (1.8) with

$$ |g(r,u,q)-g(s,u,q)|\leq K(r,s,u,q)(r-s) $$

(1.14)

for $r,s\in(0,R)$, $0<r-s$, $|u|\leq M$, and $q\in[-(1+R)C,-C]\cup[C,(1+R)C]$, where $K\geq 0$;

$$ g(r,u_{2},q)-g(r,u_{1},q)\geq\gamma(r,u_{1},u_{2},q)(u_{1}-u_{2}) $$

(1.15)

for $r\in(0,R)$, $|u_{1}|,|u_{2}|\leq M$, $u_{1}>u_{2}$, and

$$ q\in[-(1+R)C,-C]\cup[C,(1+R)C], $$

where $\gamma(r,u_{1},u_{2},q)\geq\gamma_{0}>0$. Introduce ${\mathbf{V}}$ as follows:

$$ \begin{gathered}\displaystyle{\mathbf{V}}=\{r,s)\in(0,R),\ 0<r-s,\ |u_{1}|,\ |u_{2}|\leq M,\ u_{1}>u_{2},\\ \displaystyle q\in[-(1+R)C,-C]\cup[C,(1+R)C]\}.\end{gathered} $$

Suppose that

$$ \max\limits_{\mathbf{V}}\frac{K(r,s,u_{1},q)}{\gamma(r,u_{1},u_{2},q)}\leq C_{0}|q|^{\mu},\quad\mu<1, $$

(1.16)

for $C$ sufficiently large, where $C_{0}$ is a positive constant.

Remark

In [16], there are given some conditions similar to (1.14) and (1.15) in the case that $K$ and $\gamma$ are constants. In this article we abandon the constancy of $K$ and $\gamma$ and add (1.16).

Theorem 1.2

Assume that $F(r,u,u^{\prime})$ is jointly continuous and satisfies (1.6). Under conditions (1.9)–(1.16), there is a weak solution of (1.4), (1.5) satisfying

$$ |u|\leq M,\quad|u^{\prime}|\leq(1+R)C. $$

Remark

If (1.10) and (1.11) take place for an arbitrary $\widetilde{M}$ satisfying (1.9), then we can choose $M$ in Theorem 1.2 to be equal to

$$ M=\min\left\{R^{\frac{p}{p-1}}\left(\frac{f_{0}}{n-1}\right)^{\frac{1}{p-1}},\max\{|M_{1}|,M_{2}\}\right\}. $$

Let us give several simple examples of the problems of the form (1.4), (1.5) for which Theorem 1.2 holds. In all examples below, $g(r,u,u^{\prime})$ does not satisfy (1.7) and (1.8).

Example 1.3

Consider in $B_{R}$ the equation

$$ -\operatorname{div}(|\nabla u|^{p-2}\nabla u)=-u+\frac{1}{|x|^{2k}}\left(\sum\limits^{n}_{i=1}x_{i}u_{x_{i}}\right)^{2k+1}-u|\nabla u|^{\nu}+f(|x|), $$

(1.17)

where $\nu>0$ and $k\geq 0$ is an integer. The radially symmetric solution of (1.17) satisfies the equation

$$ -(|u^{\prime}|^{p-2}u^{\prime})^{\prime}-\frac{n-1}{r}|u^{\prime}|^{p-2}u^{\prime}=-u+ru^{\prime}{}^{2k+1}-u|u^{\prime}|^{\nu}+f(r),\quad r\in(0,R). $$

(1.18)

The function $g=ru^{\prime}{}^{2k+1}-u|u^{\prime}|^{\nu}$ satisfies (1.10), (1.11), and (1.14)–(1.16) for $2k<\nu$. Thus, problem (1.17), (1.5) has a weak radially symmetric solution in the sense of Definition 1.1 for $2k<\nu$.

Example 1.4

Consider in $B_{R}$ the equation

$$ -\operatorname{div}(|\nabla u|^{p-2}\nabla u)=-u-ue^{|x|^{1-\mu}\left(\sum\limits^{n}_{i=1}x_{i}u_{x_{i}}\right)^{\mu}}+f(|x|). $$

(1.19)

The radially symmetric solution of (1.19) satisfies the equation

$$ -(|u^{\prime}|^{p-2}u^{\prime})^{\prime}-\frac{n-1}{r}|u^{\prime}|^{p-2}u^{\prime}=-u-ue^{ru^{\prime}{}^{\mu}}+f(r),\quad r\in(0,R). $$

(1.20)

The function $g=-ue^{ru^{\prime}{}^{\mu}}$ satisfies (1.10), (1.11), and (1.14)–(1.16) for $\mu<1$; consequently, (1.19), (1.5) has a weak radially symmetric solution in the sense of Definition 1.1 with a specified $\mu$.

Example 1.5

Consider in $B_{R}$ the equation

$$ -\operatorname{div}(|\nabla u|^{p-2}\nabla u)=-u+\frac{1}{|x|^{2k}}\left(\sum\limits^{n}_{i=1}x_{i}u_{x_{i}}\right)^{2k+1}-ue^{|\nabla u|}+f(|x|). $$

(1.21)

The radially symmetric solution of (1.21) satisfies the equation

$$ -(|u^{\prime}|^{p-2}u^{\prime})^{\prime}-\frac{n-1}{r}|u^{\prime}|^{p-2}u^{\prime}=-u+ru^{\prime}{}^{2k+1}-ue^{|u^{\prime}|}+f(r),\quad r\in(0,R). $$

(1.22)

The function $g=ru^{\prime}{}^{2k+1}-ue^{|u^{\prime}|}$ satisfies (1.10), (1.11), (1.14), and (1.15). It is easy to see that $\frac{K}{\gamma}\to 0$ as $C\to\infty$, and therefore $g$ also satisfies condition (1.16) for $C$ sufficiently large. As a consequence, problem (1.21), (1.5) has a weak radially symmetric solution in the sense of Definition 1.1.

In order to prove Theorem 1.2, we regularize equation (1.4) and prove the classical solvability of the regularized problem, on using the technique of [17] and the fixed point principle. Next, we pass to the limit to obtain a weak solution of problem (1.4), (1.5). The article is organized as follows. In Section 2 we obtain an a priori estimate of the classical solution of the regularized problem and its derivative. Section 3 contains a proof of the existence of a classical solution to the regularized problem (see Theorem 3.2), as well as the existence of a weak solution to problem (1.4), (1.5) in the sense of Definition 1.1 (see Theorem 1.2).

2. A Priori Estimates of a Solution of the Regularized Problem and Its Derivative

Instead of equation (1.4), we will consider its regularization where instead of $F$ we will write its representation (1.6)

$$ -((u^{\prime}{}^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime})^{\prime}-\frac{n-1}{r}(u^{\prime}{}^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime}=-u+g(r,u,u^{\prime})+f(r). $$

(2.1)

The constant $\alpha\in(0,1)$ is such that $(u^{\prime}{}^{\alpha})^{\frac{p-2}{\alpha}}=|u^{\prime}|^{p-2}$, $\varepsilon>0$. As $\alpha$ we may take $\alpha=\frac{m}{k}$, where $m$ is even and $k$ is a positive integer. Moreover, we will assume that $\alpha>p-1$ for $1<p<2$. Let us rewrite (2.1) in nondivergent form

$$ -a_{\varepsilon}(u^{\prime})u^{\prime\prime}-\frac{n-1}{r}(u^{\prime}{}^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime}=-u+g(r,u,u^{\prime})+f(r), $$

(2.2)

where

$$ a_{\varepsilon}(z)=(z^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}-1}((p-1)z^{\alpha}+\varepsilon). $$

Obviously $a_{\varepsilon}(z)=a_{\varepsilon}(-z)$. We will study the existence of a classical solution to problem (2.1), (2.2). To this end, let us give a definition of this notion.

Definition 2.1

Say that $u(r)\in 𝔺^{2}(0,R)\cap 𝔺^{1}([0,R])$ is a classical solution of (2.2), (1.5), if it satisfies (2.2) pointwise in $(0,R)$ as well as the boundary conditions (1.5) understood in the usual sense.

Our goal in this section is to obtain some a priori estimates of a solution and its derivative that do not depend on the regularization parameter. The next lemma follows easily from the classical maximum principle.

Lemma 2.2

For every classical solution to problem (2.2), (1.5) we have the estimate

$$ M_{1}\leq u(r)\leq M_{2}. $$

Note that such estimate of the maximum of a solution is not sufficient for obtaining an a priori estimate of its derivative near the boundary point $r=R$. Consider the function $H(r)=\widetilde{M}(R-r)$, where $\widetilde{M}$ is defined in (1.9). The following lemma is a special case of Lemma 1 of [16]:

Lemma 2.3

Suppose that (1.9)–(1.11) take place. Assume that if $1<p<2$ then $\varepsilon<\varepsilon_{0}=(1+\alpha-p)\big{(}\widetilde{M}\big{)}^{\alpha}$. Moreover, for each classical solution of (2.2), (1.5) we have

$$ |u(r)|\leq H(r)\leq H(0)=\widetilde{M}R. $$

Let us proceed with the estimate of the derivative of a classical solution of the regularized problem. Put

$$ \Phi(\tau)=-C\frac{\tau^{2}}{2}+(1+R)C\tau,\quad\tau\in[0,R], $$

where $C$ is defined by (1.12), (1.13). It is easy to see that

$$ C\leq\Phi^{\prime}=-C\tau+(1+R)C\leq(1+R)C,\quad\Phi^{\prime\prime}=-C. $$

Lemma 2.4

Let the conditions of Lemma 2.3 be satisfied. Assume conditions (1.12)–(1.16). Then

$$ |u^{\prime}|\leq\Phi^{\prime}(0)=(1+R)C $$

for every classical solution of (2.2), (1.5).

Proof

Write (2.2) in the two different points $r=x$ and $r=y$

$$ -a_{\varepsilon}(u_{x}(x))u_{xx}(x)-\frac{n-1}{x}\left(u_{x}^{\alpha}(x)+\varepsilon\right)^{\frac{p-2}{\alpha}}u_{x}(x)=-u(x)+g(x,u,u_{x})+f(x), $$

(2.3)

$$ -a_{\varepsilon}(u_{y}(y))u_{yy}(y)-\frac{n-1}{y}(u_{y}^{\alpha}(y)+\varepsilon)^{\frac{p-2}{\alpha}}u_{y}(y)=-u(y)+g(y,u,u_{y})+f(y), $$

(2.4)

where $x,y\in(0,R)$. Subtracting (2.4) from (2.3), we obtain

$$ \begin{gathered}\displaystyle-a_{\varepsilon}(u_{x}(x))u_{xx}(x)+a_{\varepsilon}(u_{y}(y))u_{yy}(y)\\ \displaystyle-\frac{n-1}{x}(u_{x}^{\alpha}(x)+\varepsilon)^{\frac{p-2}{\alpha}}u_{x}(x)+\frac{n-1}{y}\bigl{(}u_{y}^{\alpha}(y)+\varepsilon\bigr{)}^{\frac{p-2}{\alpha}}u_{y}(y)\\ \displaystyle=-u(x)+u(y)+g(x,u,u_{x})-g(y,u,u_{y})+f(x)-f(y).\end{gathered} $$

(2.5)

Put $V(x,y)=u(x)-u(y)$. Considering that $V_{xx}=u_{xx}$ and $V_{yy}=-u_{yy}$, write (2.5) as

$$ \begin{gathered}\displaystyle-a_{\varepsilon}(u_{x})V_{xx}-a_{\varepsilon}(u_{y})V_{yy}=\frac{n-1}{x}(u_{x}^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u_{x}-\frac{n-1}{y}(u_{y}^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u_{y}\\ \displaystyle-u(x)+u(y)+g(x,u,u_{x})-g(y,u,u_{y})+f(x)-f(y).\end{gathered} $$

Define the operator

$$ \widetilde{L}z\equiv-a_{\varepsilon}(u_{x})z_{xx}-a_{\varepsilon}(u_{y})z_{yy}. $$

The function $\Phi(x-y)$ satisfies the inequality

$$ \widetilde{L}\Phi=-a_{\varepsilon}(u_{x})\Phi_{xx}-a_{\varepsilon}(u_{y})\Phi_{yy}=(a_{\varepsilon}(u_{x})+a_{\varepsilon}(u_{y}))C. $$

For $W(x,y)=V(x,y)-\Phi(x-y)$ we have

$$ \begin{gathered}\displaystyle\widetilde{L}W=-a_{\varepsilon}(u_{x})W_{xx}-a_{\varepsilon}(u_{y})W_{yy}=\widetilde{L}V-\widetilde{L}\Phi\\ \displaystyle=\frac{n-1}{x}(u_{x}^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u_{x}-\frac{n-1}{y}(u_{y}^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u_{y}-u(x)+u(y)+g(x,u,u_{x})-g(y,u,u_{y})\\ \displaystyle+f(x)-f(y)-(a_{\varepsilon}(u_{x})+a_{\varepsilon}(u_{y}))C.\end{gathered} $$

(2.6)

Consider (2.6) in the domain $P=\{(x,y):x\in(0,R),y\in(0,R),x>y\}$. Suppose that the function $W(x,y)$ attains its positive maximum at some $Q_{0}=(x_{0},y_{0})\in P$. Then $W_{xx}(x_{0},y_{0})\leq 0$, $W_{yy}(x_{0},y_{0})\leq 0$, and we conclude that

$$ \widetilde{L}W\Big{|}_{Q_{0}}\geq 0. $$

(2.7)

At the same time, from the representation of $W$ we have

$$ W_{x}(x_{0},y_{0})=W_{y}(x_{0},y_{0})=0,\quad u_{x}(x_{0})=u_{y}(y_{0})=\Phi^{\prime}(x_{0}-y_{0}),\quad u(x_{0})>u(y_{0}). $$

(2.8)

From (2.6) and (2.8) it follows that

$$ \begin{gathered}\displaystyle\widetilde{L}W|_{Q_{0}}\leq\left[\frac{n-1}{x}-\frac{n-1}{y}\right]((\Phi^{\prime})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}\Phi^{\prime}|_{Q_{0}}\\ \displaystyle+g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))+2(M+f_{0})-2a_{\varepsilon}(\Phi^{\prime}(x_{0}-y_{0}))C\\ \displaystyle<g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))+2(M+f_{0})-2a_{\varepsilon}(\Phi^{\prime}(x_{0}-y_{0}))C,\end{gathered} $$

(2.9)

since

$$ \left[\frac{n-1}{x}-\frac{n-1}{y}\right]((\Phi^{\prime})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}\Phi^{\prime}\Big{|}_{Q_{0}}<0. $$

(2.10)

We will show now that the function

$$ a_{\varepsilon}(\Phi^{\prime})=((\Phi^{\prime})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}-1}((p-1)(\Phi^{\prime})^{\alpha}+\varepsilon) $$

is increasing in $\varepsilon$ for $p\neq 2$. Obviously for $p=2$ we have $a_{\varepsilon}(\Phi^{\prime})\equiv 1$.

Indeed,

$$ \begin{gathered}\displaystyle\frac{d}{d\varepsilon}a_{\varepsilon}(\Phi^{\prime})=\left(\frac{p-2}{\alpha}-1\right)((\Phi^{\prime})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}-2}((p-1)(\Phi^{\prime})^{\alpha}+\varepsilon)+((\Phi^{\prime})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}-1}\\ \displaystyle=((\Phi^{\prime})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}-2}\left[\left(\frac{p-2}{\alpha}-1\right)((p-1)(\Phi^{\prime})^{\alpha}+\varepsilon)+(\Phi^{\prime})^{\alpha}+\varepsilon\right]\\ \displaystyle=\frac{1}{\alpha}((\Phi^{\prime})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}-2}\left[(p-2)(p-1-\alpha)(\Phi^{\prime})^{\alpha}+(p-2)\varepsilon\right].\end{gathered} $$

(2.11)

From the definition of $\alpha$ it follows that $(\Phi^{\prime})^{\alpha}>0$. Considering that $p-1-\alpha>0$ for $p>2$, we obtain

$$ (p-2)(p-1-\alpha)(\Phi^{\prime})^{\alpha}+(p-2)\varepsilon>0. $$

(2.12)

Moreover, one may easily notice that (2.12) remains valid for $1<p<2$, if we choose $\varepsilon\leq(1+\alpha-p)C^{\alpha}$. The last inequality takes place since from (1.13) it follows that $C\geq\widetilde{M}$ and as a consequence $\varepsilon$ satisfies the conditions of Lemma 2.3. Remind that $\alpha>p-1$ for $1<p<2$. Relation (2.12) implies that $a_{\varepsilon}(\Phi^{\prime})$ increases in $\varepsilon$ for $p>1$ and $p\neq 2$. Thus,

$$ a_{\varepsilon}(\Phi^{\prime})=((\Phi^{\prime})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}-1}((p-1)(\Phi^{\prime})^{\alpha}+\varepsilon)\geq(p-1)(\Phi^{\prime})^{p-2}. $$

(2.13)

From $C\leq\Phi^{\prime}\leq(1+R)C$ it follows that $(\Phi^{\prime})^{p-2}\geq C^{p-2}$ for $p\geq 2$ and $(\Phi^{\prime})^{p-2}\geq\left((1+R)C\right)^{p-2}$ for $1<p<2$. Consequently, from (2.13) we obtain that

$$ a_{\varepsilon}(\Phi^{\prime})C\geq(p-1)C^{p-1}\quad\text{for}\quad p\geq 2, $$

(2.14)

$$ a_{\varepsilon}(\Phi^{\prime})C\geq(p-1)(1+R)^{p-2}C^{p-1}\quad\text{for}\quad 1<p<2. $$

(2.15)

Finally, (1.12), (1.13), (2.9), (2.14), and (2.15) imply that

$$ \begin{gathered}\displaystyle\widetilde{L}W<g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))+2(M+f_{0})-2(p-1)C^{p-1}\\ \displaystyle\leq g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))\quad\text{for}\quad p\geq 2,\end{gathered} $$

(2.16)

$$ \begin{gathered}\displaystyle\widetilde{L}W<g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))\\ \displaystyle+2(M+f_{0})-2(p-1)(1+R)^{p-2}C^{p-1}\\ \displaystyle\leq g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))\quad\text{for}\ 1<p<2.\end{gathered} $$

(2.17)

To obtain a contradiction with (2.7), one has to show that

$$ g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))\leq 0. $$

(2.18)

Represent (2.18) as

$$ \begin{gathered}\displaystyle g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))\\ \displaystyle=[g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0})))-g(y_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))]\\ \displaystyle+[g(y_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))].\end{gathered} $$

(2.19)

Using (1.14) and (1.15), from (2.19) we find that

$$ g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0})) $$

$$ \leq\Big{[}K\Big{(}x_{0},y_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))\Big{)}(x_{0}-y_{0}) $$

$$ -\gamma\Big{(}x_{0},u(x_{0}),u(y_{0}),\Phi^{\prime}(x_{0}-y_{0})\Big{)}(u(x_{0})-u(y_{0}))\Big{]}. $$

(2.20)

At the maximum point we have

$$ u(x_{0})-u(y_{0})>\Phi(x_{0}-y_{0})=\Phi(x_{0}-y_{0})-\Phi(0)=\Phi^{\prime}(\xi)(x_{0}-y_{0}), $$

(2.21)

where $\xi\in(0,x_{0}-y_{0})$. The function $\Phi^{\prime}$ satisfies $C\leq\Phi^{\prime}\leq(1+R)C$. Using (1.16) and (2.21), rewrite (2.20) as

$$ \begin{gathered}\displaystyle g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))\\ \displaystyle\leq[K(x_{0},y_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))\\ \displaystyle-\gamma(x_{0},u(x_{0}),u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))\Phi^{\prime}(\xi)](x_{0}-y_{0})\\ \displaystyle\leq[\gamma(x_{0},u(x_{0}),u(y_{0}),\Phi^{\prime}(x_{0}-y_{0})C_{0}\Phi^{\prime}{}^{\mu}(x_{0}-y_{0})\\ \displaystyle-\gamma(x_{0},u(x_{0}),u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))\Phi^{\prime}(\xi)](x_{0}-y_{0})\\ \displaystyle\leq\gamma(x_{0},u(x_{0}),u(y_{0}),\Phi^{\prime}(x_{0}-y_{0})[C_{0}((1+R)C)^{\mu}-C](x_{0}-y_{0})\leq 0\end{gathered} $$

(2.22)

under the condition

$$ C_{0}((1+R)C)^{\mu}-C\leq 0. $$

(2.23)

Obviously for $\mu<1$ and all $C_{0}$ and $R$ one may choose a sufficiently large $C$ such that (2.23) holds. Thus, $\widetilde{L}W\big{|}_{Q_{0}}<0$, which contradicts (2.7). So, $W$ cannot attain a positive maximum inside $P$.

Consider $W$ on the boundary of $P$. If $x=R$ and $y\in[0,R]$ then

$$ W(R,y)=u(R)-u(y)-\Phi(R-y)=-u(y)-\Phi(R-y). $$

Prove that $W(R,y)\leq 0$. From Lemma 2.3 it follows that $-u(y)-H(y)\leq 0$. So, it remains to show that $\Phi(R-y)\geq H(y)$. Indeed, using (1.12) and (1.13), we see that

$$ \begin{gathered}\displaystyle\Phi(R-y)-H(y)=-C\frac{(R-y)^{2}}{2}+C(1+R)(R-y)-\frac{M_{*}}{R}(R-y)\\ \displaystyle=\frac{C}{2}(R^{2}-y^{2})+C(R-y)-\frac{M_{*}}{R}(R-y)=\frac{C}{2}(R^{2}-y^{2})+\left(C-\frac{M_{*}}{R}\right)(R-y)\geq 0\end{gathered} $$

for $C\geq\frac{M_{*}}{R}$. Thus $W(R,y)\leq 0$. On the part of $\partial P$ where $x=y$ we obtain

$$ W(x,y)\Big{|}_{x=y}=u(x)-u(y)-\Phi(x-y)\Big{|}_{x=y}=-\Phi(0)=0. $$

Finally,

$$ W_{y}(x,0)=-u_{y}(0)+\Phi^{\prime}(x)=\Phi^{\prime}(x)\geq C>0 $$

for $y=0$ and $x\in[0,R]$. This implies that $W$ cannot attain its maximum on this part of the boundary. So, $W(x,y)\leq 0$ implying that

$$ u(x)-u(y)\leq\Phi(x-y). $$

(2.24)

Let us estimate the difference $u(x)-u(y)$ from below. Put $\widetilde{W}=u(y)-u(x)-\Phi(x-y)$. Subtracting (2.3) from (2.4), by analogy to (2.6), for $\widetilde{W}$ we have

$$ \begin{gathered}\displaystyle\widetilde{L}\widetilde{W}=-a_{\varepsilon}(u_{x})\widetilde{W}_{xx}-a_{\varepsilon}(u_{y})\widetilde{W}_{yy}\\ \displaystyle=\frac{n-1}{y}\bigl{(}u_{y}^{\alpha}+\varepsilon\bigr{)}^{\frac{p-2}{\alpha}}u_{y}-\frac{n-1}{x}\bigl{(}u_{x}^{\alpha}+\varepsilon\bigr{)}^{\frac{p-2}{\alpha}}u_{x}-u(y)+u(x)\\ \displaystyle+g(y,u,u_{y})-g(x,u,u_{x})+f(y)-f(x)-(a_{\varepsilon}(u_{x})+a_{\varepsilon}(u_{y}))C.\end{gathered} $$

(2.25)

Suppose that $\widetilde{W}(x,y)$ attains its positive maximum at some $Q_{1}=(x_{1},y_{1})\in P$. Then the following relations take place at $Q_{1}$:

$$ \widetilde{W}_{x}=\widetilde{W}_{y}=0,\quad\widetilde{W}_{xx}\leq 0,\quad\widetilde{W}_{yy}\leq 0,\quad u_{x}=u_{y}=-\Phi^{\prime},\quad\widetilde{L}\widetilde{W}\geq 0,\quad u(y_{1})>u(x_{1}). $$

(2.26)

On the other hand, (1.12), (1.13), (2.10), (2.14), (2.15), (2.25), (2.26), the fact that $a_{\varepsilon}$ is nondecreasing in $\varepsilon$, and $a_{\varepsilon}(-\Phi^{\prime})=a_{\varepsilon}(\Phi^{\prime})$ imply by analogy with (2.16) and (2.17) that

$$ \begin{gathered}\displaystyle\widetilde{L}\widetilde{W}|_{Q_{1}}<g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))\\ \displaystyle+2(M+f_{0})-2(p-1)C^{p-1}\\ \displaystyle\leq g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))\quad\text{for}\ p\geq 2,\end{gathered} $$

(2.27)

$$ \begin{gathered}\displaystyle\widetilde{L}\widetilde{W}|_{Q_{1}}<g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))\\ \displaystyle+2(M+f_{0})-2(p-1)(1+R)^{p-2}C^{p-1}\\ \displaystyle\leq g(y_{0},u(y_{0}),\Phi^{\prime}(x_{0}-y_{0}))-g(x_{0},u(x_{0}),\Phi^{\prime}(x_{0}-y_{0}))\quad\text{for}\ 1<p<2.\end{gathered} $$

(2.28)

The inequality similar to (2.21) looks like

$$ u(y_{1})-u(x_{1})>\Phi(x_{1}-y_{1})=\Phi(x_{1}-y_{1})-\Phi(0)=\Phi^{\prime}(\xi)(x_{1}-y_{1}),\quad\xi\in(0,x_{1}-y_{1}). $$

(2.29)

Using (1.14)– (1.16), (2.29) and some calculations similar to (2.22), in the case of $\widetilde{W}$ we obtain $\widetilde{L}\widetilde{W}\big{|}_{Q_{1}}<0$, which contradicts the assumption that $\widetilde{W}$ attains its positive maximum inside $P$.

Consider $\widetilde{W}$ on the boundary of $P$. If $x=R$ and $y\in[0,R]$ then

$$ \widetilde{W}(R,y)=u(y)-u(R)-\Phi(R-y)=u(y)-\Phi(R-y)\leq 0; $$

since $u(y)-H(y)\leq 0$ and $\Phi(R-y)\geq H(y)$ for $C\geq\frac{M_{*}}{R}$. On the part of $\partial P$, where $x=y$ we obtain

$$ \widetilde{W}(x,y)\Big{|}_{x=y}=u(y)-u(x)-\Phi(x-y)\Big{|}_{x=y}=-\Phi(0)=0. $$

Finally, if $y=0$ and $x\in[0,R]$ then

$$ \widetilde{W}_{y}(x,0)=u_{y}(0)+\Phi^{\prime}(x)=\Phi^{\prime}(x)\geq C>0 $$

and, as a consequence, $\widetilde{W}$ cannot attain its maximum on this part of the boundary. Consequently, $\widetilde{W}(x,y)\leq 0$, which entails

$$ u(y)-u(x)\leq\Phi(x-y). $$

(2.30)

Furthermore, (2.24) and (2.30) imply

$$ |u(x)-u(y)|\leq\Phi(x-y). $$

(2.31)

Owing to the symmetry between $x$ and $y$, we have $|u(x)-u(y)|\leq\Phi(y-x)$ and so

$$ |u(x)-u(y)|\leq\Phi(|x-y|). $$

(2.32)

Since $\Phi(0)=0$, (2.32) may be represented as

$$ \frac{|u(x)-u(y)|}{|x-y|}\leq\frac{\Phi(|x-y|)-\Phi(0)}{|x-y|}, $$

(2.33)

which implies the needed estimate

$$ |u_{x}|\leq\Phi^{\prime}(0)=(1+R)C, $$

(2.34)

where $C$ is defined by (1.12) ((1.13)).  ☐

3. Proof of Existence Theorems

Consider a solution $u_{\varepsilon}$ of regularized equation (2.2)

$$ -((u^{\prime}_{\varepsilon})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime}_{\varepsilon})^{\prime}-\frac{n-1}{r}((u^{\prime}_{\varepsilon})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime}_{\varepsilon}=-u_{\varepsilon}+g(r,u_{\varepsilon},u^{\prime}_{\varepsilon})+f(r), $$

(3.1)

together with boundary conditions (1.5) which we rewrite for $u_{\varepsilon}$ as follows:

$$ u^{\prime}_{\varepsilon}(0)=0,\quad u_{\varepsilon}(R)=0. $$

(3.2)

In order to prove the existence of a classical solution of problem (3.1), (3.2) we have to show that

$$ \frac{n-1}{r}((u^{\prime}_{\varepsilon})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime}_{\varepsilon} $$

is bounded as $r\to 0$. Put

$$ Z(r)=((u^{\prime}_{\varepsilon})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime}_{\varepsilon}. $$

For completeness, we present a proof of this technical lemma [18].

Lemma 3.1

If $u_{\varepsilon}$ is a classical solution of (3.1), (3.2), then $Z(r)\in 𝔺^{1}[0,R]$ and

$$ Z^{\prime}(0)=\frac{u_{\varepsilon}(0)-f(0)}{n}. $$

Proof

Rewrite (3.1) as

$$ Z^{\prime}+\frac{n-1}{r}Z-u_{\varepsilon}+g(r,u_{\varepsilon},u^{\prime}_{\varepsilon})+f(r)=0. $$

(3.3)

Represent (3.3) as

$$ (r^{n-1}Z)^{\prime}+r^{n-1}(-u_{\varepsilon}+g(r,u_{\varepsilon},u^{\prime}_{\varepsilon})+f(r))=0. $$

(3.4)

From (3.4) it is immediate that

$$ r^{n-1}Z=-\int\limits^{r}_{0}t^{n-1}(-u_{\varepsilon}(t)+g(t,u_{\varepsilon}(t),u^{\prime}_{\varepsilon}(t))+f(t))dt. $$

(3.5)

Equalities (3.3) and (3.5) imply the presentation for $Z^{\prime}$:

$$ \begin{gathered}\displaystyle Z^{\prime}=u_{\varepsilon}-g(r,u_{\varepsilon},u^{\prime}_{\varepsilon})-f(r)\\ \displaystyle+\frac{n-1}{r^{n}}\int\limits^{r}_{0}t^{n-1}(-u_{\varepsilon}(t)+g(t,u_{\varepsilon}(t),u^{\prime}_{\varepsilon}(t))+f(t))\,dt.\end{gathered} $$

(3.6)

Notice that

$$ \begin{aligned} \lim_{r\to 0}Z^{\prime}&=\lim_{r\to 0}\frac{Z(r)-Z(0)}{r}=\lim_{r\to 0}\frac{Z(r)}{r}\\ &=\lim_{r\to 0}\frac{1}{r}(u^{\prime}_{\varepsilon})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime}_{\varepsilon}=\frac{u_{\varepsilon}(0)-f(0)}{n}\end{aligned} $$

(3.7)

for all $\varepsilon>0$, as (3.7) is a consequence of applying the L’Hôspital rule to the right-hand side of (3.6).  ☐

Notice that Lemma 3.1 implies the boundedness of $Z_{r}$ on $[0,R]$ for all $\varepsilon\geq 0$ which in turns implies the Hölder continuity of $u^{\prime}_{\varepsilon}$ for $p>2$

$$ |u^{\prime}_{\varepsilon}(r_{1})-u^{\prime}_{\varepsilon}(r_{2})|\leq c_{1}|r_{1}-r_{2}|^{\frac{1}{p-1}} $$

(3.8)

and the Lipschitz continuity of $u^{\prime}_{\varepsilon}$ for $1<p\leq 2$

$$ |u^{\prime}_{\varepsilon}(r_{1})-u^{\prime}_{\varepsilon}(r_{2})|\leq c_{2}|r_{1}-r_{2}|, $$

(3.9)

where constants $c_{1}$ and $c_{2}$ do not depend on $\varepsilon$. Having the a priori estimates of the classical solution in the class $𝔺^{1,\beta}$ with the above $\beta$, we can apply the Fixed Point Theorem [19, 20] to prove the existence of a solution to problem (3.1), (3.2). In this article we will provide only the corresponding formulation. The proof of this result is exactly the same as in the proof of Theorem 3 in [16].

Theorem 3.2

Assume that $f(r)\in 𝔺[0,R]$, $g(r,u_{\varepsilon},u^{\prime}_{\varepsilon})\in 𝔺\left([0,R]\times 𝕉\times 𝕉\right)$, and conditions (1.9)–(1.16) take place. Then there exists a classical solution $u_{\varepsilon}\in 𝔺^{2}(0,R)\cap 𝔺^{1,\beta}[0,R]$ of (3.1), (3.2) satisfying

$$ |u_{\varepsilon}|\leq M,\quad|u_{\varepsilon r}|\leq(1+R)C. $$

Proof of Theorem 1.2

Consider (3.1). Multiplying (3.1) by $\phi\in 𝔺^{\infty}_{0}(0,R)$ and integrating by parts, we obtain

$$ \int\limits^{R}_{0}((u^{\prime}_{\varepsilon})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime}_{\varepsilon}\phi^{\prime}\,dr-\int\limits^{R}_{0}\frac{n-1}{r}((u^{\prime}_{\varepsilon})^{\alpha}+\varepsilon)^{\frac{p-2}{\alpha}}u^{\prime}_{\varepsilon}\phi(r)dr $$

$$ =\int\limits^{R}_{0}(-u_{\varepsilon}+g(r,u_{\varepsilon},u^{\prime}_{\varepsilon})+f(r))\phi(r)dr. $$

(3.10)

From (3.8) and(3.9) we conclude that there exists a subsequence $\varepsilon_{n}\to 0$ such that

$$ u_{\varepsilon_{n}}(r)\rightarrow u(r),\quad u^{\prime}_{\varepsilon_{n}}(r)\rightarrow u^{\prime}(r)\quad\text{in }\ 𝔺[0,R], $$

(3.11)

whence, due to the joint continuity of $g$, it is immediate that

$$ g(r,u_{\varepsilon},u^{\prime}_{\varepsilon})\rightarrow g(r,u,u^{\prime})\quad\text{in }\ 𝔺[0,R]. $$

Also from (3.11) we have

$$ \left(u^{\prime}_{\varepsilon_{n}}\right)^{\alpha}+\varepsilon_{n}\rightarrow\left(u^{\prime}\right)^{\alpha}\quad\text{in }\ 𝔺[0,R], $$

$$ \left(\left(u^{\prime}_{\varepsilon_{n}}\right)^{\alpha}+\varepsilon_{n}\right)^{\frac{p-2}{\alpha}}u^{\prime}_{\varepsilon_{n}}\rightarrow\left|u^{\prime}\right|^{p-2}u^{\prime}\quad\text{in }\ 𝔺[0,R]. $$

(3.12)

From (3.12) and the fact that $\frac{\phi(r)}{r}$ is continuous $[0,R]$ it follows that

$$ \int\limits^{R}_{0}\frac{n-1}{r}\left(\left(u^{\prime}_{\varepsilon_{n}}\right)^{\alpha}+\varepsilon_{n}\right)^{\frac{p-2}{\alpha}}u^{\prime}_{\varepsilon_{n}}\phi(r)dr\rightarrow\int\limits^{R}_{0}\frac{n-1}{r}\left|u^{\prime}\right|^{p-2}u^{\prime}\phi(r)dr. $$

Passing to the limit in (3.10) as $\varepsilon\to 0$, we find that $u=\lim_{\varepsilon\to 0}u_{\varepsilon}$ is the sought weak radially symmetric solution of problem (1.4), (1.5).  ☐

Remark

Note that $u^{\prime}_{\varepsilon}$ is Lipschitz continuous for $1<p\leq 2$, where the Lipschitz constant does not depend on $\varepsilon$. Thus, in this case the solution $u(r)$ to problem (1.4), (1.5) will have the same smoothness. This means that $u$, being a weak solution to problem (1.4), (1.5), satisfies (1.4) in the classical sense almost everywhere.