# Growth of subsolutions to fully nonlinear equations in halfspaces

NIKLAS L.P. LUNDSTRÖM

Department of Mathematics and Mathematical Statistics,
Umeå University, SE-90187 Umeå, Sweden;
<EMAIL_ADDRESS>
###### Abstract

We characterize lower growth estimates for subsolutions in halfspaces of fully
nonlinear partial differential equations on the form

$F(x,u,Du,D^{2}u)=0$

in terms of solutions to ordinary differential equations built solely upon a
growth assumption on $F$. Using this characterization we derive several sharp
Phragmen–Lindelöf-type theorems for certain classes of well known PDEs. The
equation need not be uniformly elliptic nor homogeneous and we obtain results
both in case the subsolution is bounded or unbounded. Among our results we
retrieve classical estimates in the halfspace for $p$-subharmonic functions
and extend those to more general equations; we prove sharp growth estimates,
in terms of $k$ and the asymptotic behaviour of $\int_{0}^{R}C(s)ds$, for
subsolutions of equations allowing for sublinear growth in the gradient of the
form $C(|x|)|Du|^{k}$ with $k\geq 1$; we establish a Phragmen–Lindelöf theorem
for weak subsolutions of the variable exponent $p$-Laplace equation in
halfspaces, $1<p(x)<\infty$, $p(x)\in C^{1}$, of which we conclude sharpness
by finding the “slowest growing” $p(x)$-harmonic function together with its
corresponding family of $p(x)$-exponents. The paper ends with a discussion of
our results from the point of view of a spatially dependent diffusion problem.

Mathematics Subject Classification: 35B40, 35B50, 35B53, 35D40, 35J25, 35J60,
35J70. Keywords: Phragmen-Lindelöf; general drift; non standard growth;
variable exponent; Laplace; unbounded domain; quasi linear; nonhomogeneous;
sublinear; harmonic.

##  1 Introduction

We consider fully nonlinear nonhomogeneous elliptic partial differential
equations in nondivergence form,

$\displaystyle F(x,u,Du,D^{2}u)=0,$ ($\star$)

in halfspaces in $\mathbb{R}^{n}$ where $n\geq 1$. Here, $Du$ is the gradient,
$D^{2}u$ the hessian,
$F:\mathbb{R}^{n}\times\mathbb{R}\times\mathbb{R}^{n}\times\mathbb{S}^{n}\rightarrow\mathbb{R}$
and $\mathbb{S}^{n}$ is the set of symmetric $n\times n$ matrices equipped
with the positive semi-definite ordering; for $X,Y\in\mathbb{S}^{n}$, we write
$X\leq Y$ if $\langle(X-Y)\xi,\xi\rangle\leq 0$ for all
$\xi\in\mathbb{R}^{n}$. Without loss of generality we fix the halfspace to
$\mathbb{R}^{n}_{+}:=\\{x\in\mathbb{R}^{n}:x_{n}>0\\}$ and assume the
following:

  * Degenerate ellipticity holds, i.e. $F(x,u,p,X)\geq F(x,v,p,Y)$ whenever $u\geq v$, $X\leq Y$, as well as the growth condition

$\displaystyle-F(x,0,p,X)$
$\displaystyle\leq\Phi(|x|,|p|)+\Lambda(x_{n})\text{Tr}(X^{+})-\lambda(x_{n})\text{Tr}(X^{-})$
($\star\star$)

whenever $x,p\in\mathbb{R}^{n},X\in\mathbb{S}^{n}$, $X=X^{+}-X^{-}$,
$X^{+}\geq 0$, $X^{-}\geq 0$ and $X^{+}X^{-}=0$. Here,
$\Phi:[0,\infty)\times[0,\infty)\to(-\infty,\infty)$ is continuous,
nonincreasing in its first argument and
$\lambda,\Lambda:[0,\infty)\to(0,\infty)$ are functions such that ${\lambda}$
is nonincreasing and ${\Lambda}$ is nondecreasing.

Concerning $\Phi$ we will also need the following assumption:

  * Either $\Phi$ is nonnegative and it holds that for all $\epsilon,t>0$ (interpreting $1/0=\infty$)

$\displaystyle\int_{0}^{\epsilon}\frac{ds}{\Phi(t,s)}=\infty$
($\star\star\star$)

or $\Phi$ is nonpositive and $-\Phi(t,s)$ satisfies ($\star\star\star$ ‣ 1)
for all $\epsilon,t>0$.

Under assumptions ($\star\star$ ‣ 1)–($\star\star\star$ ‣ 1) we characterize
the growth of viscosity subsolutions of ($\star$ ‣ 1) in halfspaces in terms
of solutions to ODEs (Theorem 2.1) which are built solely upon functions
$\Phi,\lambda$ and $\Lambda$ in ($\star\star$ ‣ 1). Using this
characterization we are able to derive sharp growth estimates of
Phragmen–Lindelöf type once the solutions to the ODEs are sufficiently
understood. Indeed, to apply Theorem 2.1 one needs to (1) find functions
$\Phi,\lambda$ and $\Lambda$ to ensure assumptions ($\star\star$ ‣ 1) and
($\star\star\star$ ‣ 1), (2) solve the corresponding ODEs given in (2.1) and
(3) find the limit in Theorem 2.1. An estimate is obtained if this limit is
positive. Theorem 2.1 applies both in case the subsolution is bounded or
unbounded, and it can be used to find such border.

In Section 3 we apply Theorem 2.1 to derive sharp estimates for subsolutions
to some well known PDEs of which the corresponding ODEs can be solved
explicitly. For example, we retrieve the classical Phragmen–Lindelöf theorem
in the halfspace for $p$-subharmonic functions by Lindqvist [29] and show in
addition that it holds also for equations of $p$-Laplace type with lower order
terms and vanishing ellipticity. We obtain sharp lower estimates of the
growth, in terms of $k\geq 1$ and the asymptotic behaviour of
$\int_{0}^{R}C(s)\lambda^{-1}(s)ds$, for subsolutions of equations with
sublinear growth in the gradient such as

$\displaystyle-P^{-}_{\lambda,\Lambda}(D^{2}u)+C(|x|)|Du|^{k}=0$

in which $P^{-}_{\lambda,\Lambda}$ is a Pucci operator (definition recalled
below) and $C(t)$ is a nonincreasing function. These results reveal e.g. the
border determining if a subsolution must grow to infinity or not in terms of
$C(t)$ and $k$, see Corollary 3.1 and estimate (3.10). Moreover, Theorem 2.1
applies to nonhomogeneous PDEs including the variable exponent $p$-Laplace
equation

$\nabla\cdot\left(|Du|^{p(x)-2}Du\right)=0$

and we prove a sharp Phragmen–Lindelöf theorem for weak subsolutions of this
equation whenever $1<p(x)<\infty$ is $C^{1}$ regular (Theorem 3.3). It turns
out that the growth estimate heavily depends on whether the subsolution ever
exceeds $x_{n}$ (distance to boundary) or not. We conclude sharpness by
finding the “slowest growing” $p(x)$-harmonic function in the halfspace, for a
given ellipticity bound, together with its corresponding family of $p(x)$
exponents (Remark 3.4). In the geometric setting of halfspaces, this theorem
sharpens some results of Adamowicz [1].

The proof of Theorem 2.1 relies on comparison with certain classical
supersolutions of ($\star$ ‣ 1), which we construct in Lemma 2.2 using
solutions to the aforementioned ODEs. We stress generality by pointing out
that with the validity of Theorem 2.1 at hand, growth estimates for
subsolutions to certain PDEs not considered in Section 3 can be proved mainly
by estimating solutions of first order ODEs and limits.

We end the paper by discussing the problem under investigation from the point
of view of a diffusion problem. Indeed, in Section 4 we briefly explain,
through the application of spatially dependent diffusion, why parts of our
results presented in Theorem 3.3 should hold.

We remark that our main results allow for ellipticity to blow up at infinity
as $\lambda(x_{n})$ may vanish and $\Lambda(x_{n})$ may explode at infinity.
Moreover, the Osgood-type condition in ($\star\star\star$ ‣ 1) is necessary to
ensure that subsolutions must continue to grow. Indeed, for the strong maximum
principle, see Julin [22], Lundström–Olofsson–Toivanen [33] and the remarks
below Theorem 2.1. Furthermore, assumption ($\star\star$ ‣ 1) can be written,
with $\lambda=\lambda(x_{n})$ and $\Lambda=\Lambda(x_{n})$,

$\displaystyle-F(x,0,p,X)\leq\Phi(|x|,|p|)-\mathcal{P}^{-}_{\lambda,\Lambda}(X)\quad\text{whenever}\quad
x,p\in\mathbb{R}^{n},X\in\mathbb{S}^{n},$

where
$\mathcal{P}^{-}_{\lambda,\Lambda}(X)=-\Lambda\text{Tr}(X^{+})+\lambda\text{Tr}(X^{-})$
is the Pucci maximal operator, $X=X^{+}-X^{-}$ with $X^{+}\geq 0$, $X^{-}\geq
0$ and $X^{+}X^{-}=0$. In particular, if $X\in\mathbb{S}^{n}$ has eigenvalues
$e_{1},e_{2},\dots,e_{n}$ the Pucci extremal operators
$\mathcal{P}^{+}_{\lambda,\Lambda}$ and $\mathcal{P}^{-}_{\lambda,\Lambda}$
with ellipticity $0<\lambda\leq\Lambda$ are defined by

$\mathcal{P}^{+}_{\lambda,\Lambda}(X):=-\lambda\sum_{e_{i}\geq
0}e_{i}-\Lambda\sum_{e_{i}<0}e_{i}\quad\text{and}\quad\mathcal{P}^{-}_{\lambda,\Lambda}(X):=-\Lambda\sum_{e_{i}\geq
0}e_{i}-\lambda\sum_{e_{i}<0}e_{i}.$

For properties of the Pucci operators see e.g. Caffarelli–Cabre [9] or
Capuzzo-Dolcetta–Vitolo [10]. We remark also that the above assumption
($\star\star$ ‣ 1) is implied by the standard ellipticity assumption

$\displaystyle\lambda\text{Tr}(Y)\leq
F(x,u,p,X)-F(x,u,p,X+Y)\leq\Lambda\text{Tr}(Y),$ (1.1)

whenever $Y$ is positive semi-definite, together with

$\displaystyle-F(x,0,p,0)\leq\Phi(|x|,|p|)\quad\text{whenever}\quad
x,p\in\mathbb{R}^{n}.$ (1.2)

Observe also that ($\star\star$ ‣ 1) allows for nonlinear degenerate elliptic
operators which do not satisfy (1.1). For example operators of the form

$F(X)=-\Lambda\left(\sum_{i=1}^{n}\Gamma(\mu_{i}^{+})\right)+\lambda\left(\sum_{i=1}^{n}\Psi(\mu_{i}^{-})\right)$

where $\mu_{i},i=1,\dots,n$, are the eigenvalues of the matrix
$X\in\mathbb{S}^{n}$ and $\Gamma,\Psi:[0,\infty)\to[0,\infty)$ are continuous
and nondecreasing functions such that $\Gamma(s)\leq s\leq\Psi(s)$, see
Capuzzo-Dolcetta–Vitolo [10].

The Phragmén-Lindelöf principle and results of Phragmén-Lindelöf type, which
has connections to elasticity theory (Horgan [20], Quintanilla [38],
Leseduarte–Carme–Quintanilla [28]), have been frequently studied during the
last century. To mention a few papers (without giving a complete summary),
Ahlfors [2] extended results from Phragmén–Lindelöf [37] to the upper half
space of $\mathbb{R}^{n}$, Gilbarg [15], Serrin [39] and Herzog [18]
considered more general elliptic equations of second order. Miller [36]
considered uniformly elliptic operators in nondivergence form and unbounded
domains contained in cones. Kurta [27] and Jin–Lancaster [21] estimated growth
of bounded solutions of quasilinear equations, the later used solutions to
boundary value problems, while Vitolo [40] considered elliptic equations in
sectors. Capuzzo-Dolcetta–Vitolo [10] and Armstrong–Sirakov–Smart [3]
considered fully nonlinear equations, the later in certain Lipschitz domains,
and Koike–Nakagawa [26] established Phragmén-Lindelöf theorems for
subsolutions of fully nonlinear elliptic PDEs with unbounded coefficients and
inhomogeneous terms. Adamowicz [1] studied subsolutions of the variable
exponent $p$-Laplace equation, while Bhattacharya [7] and Granlund–Marola [16]
considered infinity-harmonic functions. Lindqvist [29] established Phragmén-
Lindelöf’s theorem for $n$-subharmonic functions when the boundary is an
$m$-dimensional hyperplane in $\mathbb{R}^{n}$, $0\leq m\leq n-1$, which was
extended to $p$-subharmonic functions, $n-m<p\leq\infty$, in Lundström [32].
We also mention that recently, Braga–Moreira [8] showed that nonnegative
solutions to a generalized $p$-Laplace equation in the upper halfplane,
vanishing on $\\{x_{n}=0\\}$, is $u(x)=x_{n}$ (modulo normalization) and
Lundström–Singh [34] proved a similar result for $p$-harmonic functions in
planar sectors as well as a sharp Phragmen–Lindelöf theorem. Lundberg–Weitsman
[30] studied the growth of solutions to the minimal surface equation over
domains containing a halfplane. The spatial behavior of solutions of the
Laplace equation on a semi-infinite cylinder with dynamical nonlinear boundary
conditions was investigated in Leseduarte–Carme–Quintanilla [28]. Finally, we
mention that recently, local estimates such as a sharp Harnack inequality
(Julin [22]), Boundary Harnack inequalities (Avelin–Julin [6]) as well as
strong maximum and minimum principles (Lundström–Olofsson–Toivanen [33]) were
established for fully nonlinear PDEs covered by the class of equations
considered here.

### Preliminaries

For a point $x\in\mathbb{R}^{n}$ we use the notation $x=(x_{1},x_{2},\dots
x_{n-1},x_{n})=(x^{\prime},x_{n})$. By $\Omega$ we denote a domain, that is,
an open connected set. For a set $E\subset\mathbb{R}^{n}$ we let
$\overline{E}$ denote the closure and $\partial E$ the boundary of $E$. By $c$
we denote a positive constant not necessarily the same at each occurrence. We
write $A\precsim B$ if there exists $c$ such that $A\leq cB$.

A function $u:\Omega\to\mathbb{R}$ is a classical subsolution (supersolution)
to ($\star$ ‣ 1) in $\Omega$ if it is twice differentiable in $\Omega$ and
satisfies $F(x,u,Du,D^{2}u)\leq 0$ ($F(x,u,Du,D^{2}u)\geq 0$). If the
inequality holds strict then $u$ is a strict classical subsolution
(supersolution), and if equality holds then it is a classical solution.

We choose to present our main results for viscosity subsolutions, of which we
recall the definition below in case
$F:\mathbb{R}^{n}\times\mathbb{R}\times\mathbb{R}^{n}\times\mathbb{S}^{n}\rightarrow\mathbb{R}$
and $F$ is a continuous function (which is not necessary for our results).

The following definition is from Crandall–Ishii–Lions [12]: An upper
semicontinuous (USC) function $u:\Omega\to\mathbb{R}$ is a viscosity
subsolution if for any $\varphi\in C^{2}(\Omega)$ and any $x_{0}\in\Omega$
such that $u-\varphi$ has a local maximum at $x_{0}$ it holds that

$\displaystyle F(x_{0},u(x_{0}),D\varphi(x_{0}),D^{2}\varphi(x_{0}))\leq 0.$

A lower semicontinuous (LSC) function $u:\Omega\to\mathbb{R}$ is a viscosity
supersolution if for any $\varphi\in C^{2}(\Omega)$ and any $x_{0}\in\Omega$
such that $u-\varphi$ has a local minimum at $x_{0}$ it holds that

$\displaystyle F(x_{0},u(x_{0}),D\varphi(x_{0}),D^{2}\varphi(x_{0}))\geq 0.$

A continuous function is a viscosity solution if it is both a viscosity sub-
and a viscosity supersolution.

Let $u$ be a subsolution and $v$ a supersolution to ($\star$ ‣ 1) and let $a$
and $b$ be constants. As ($\star$ ‣ 1) is not necessarily homogeneous, $a+bu$
and $a+bv$ may fail as sub- and supersolutions. However, degenerate
ellipticity guarantees that $u-c$ is a subsolution, and $u+c$ is a
supersolution whenever $c\geq 0$.

We will not discuss the validity of a general comparison principle for
viscosity solutions of ($\star$ ‣ 1) since we only need the possibility to
compare viscosity subsolutions to classical supersolutions which is possible.
Indeed, let $\Omega$ be a bounded domain, $u$ a viscosity subsolution and $v$
a classical strict supersolution in $\Omega$, $u\leq v$ on $\partial\Omega$
and that $u\geq v$ somewhere in $\Omega$. By USC the function $u-v$ attains a
maximum at some point $x_{0}\in\Omega$. Since $v\in C^{2}(\Omega)$, $u-v$ has
a maximum at $x_{0}$ and $u$ is a viscosity subsolution it follows by
definition of viscosity solutions that

$\displaystyle F(x_{0},u(x_{0}),Dv(x_{0}),D^{2}v(x_{0}))\leq 0.$ (1.3)

But since $v$ is a classical strict supersolution we have
$F(x,v(x),Dv(x),D^{2}v(x))>0$ whenever $x\in\Omega$, and as $u(x_{0})\geq
v(x_{0})$ it follows from degenerate ellipticity that
$F(x_{0},u(x_{0}),Dv(x_{0}),D^{2}v(x_{0}))\geq
F(x_{0},v(x_{0}),Dv(x_{0}),D^{2}v(x_{0}))>0.$ This contradicts (1.3) and hence
we have proved the following simple lemma:

###### Lemma 1.1

Let $\Omega$ be a bounded domain, $u\in USC(\overline{\Omega})$ a viscosity
subsolution and $v\in LSC(\overline{\Omega})$ a viscosity supersolution of
($\star$ ‣ 1) in $\Omega$ satisfying $u\leq v$ on $\partial\Omega$. Assume
degenerate ellipticity. If either $u$ is a strict classical subsolution, or
$v$ is a strict classical supersolution, then $u<v$ in $\Omega$.

Neither the choice of viscosity solutions nor the assumption that $F$ is
continuous are necessary for our results. Any other definition of “weak
solutions” can be considered, whenever more appropriate for the equation, as
long as such weak subsolutions of ($\star$ ‣ 1) can be compared to classical
strict supersolutions of ($\star$ ‣ 1). In particular, our proof relies on
construction of a classical strict supersolution to ($\star$ ‣ 1) and
comparison with this barrier function. What is needed is the validity of
following simple comparison result:

###### Lemma 1.2

Let $\Omega$ be a bounded domain, $u\in USC(\overline{\Omega})$ a subsolution
(in some weak sense) and $v$ a classical strict supersolution to ($\star$ ‣ 1)
in $\Omega$, continuous on $\overline{\Omega}$. If $u\leq v$ on
$\partial\Omega$ then $u\leq v$ in $\Omega$.

##  2 Characterizing growth in terms of solutions to ordinary differential
equations

We will estimate the growth of subsolutions to ($\star$ ‣ 1) in terms of
solutions $f:[0,\infty)\to\mathbb{R}$ to the following initial value problems,
originating from assumption ($\star\star$ ‣ 1): If $\Phi(t,s)\geq 0$ for all
$t,s\in\mathbb{R}_{+}$ then we will make use of solutions to

$\displaystyle\frac{df}{dt}=-\frac{\Phi(t,f(t))}{\lambda(t)}-K(R)\frac{\Lambda(t)}{\lambda(t)}f(t),\quad
t\in(0,R)\quad\text{with}\quad f(0)=\nu,$ (2.1)

where $\nu\geq 0$ and $R>0$. Through the paper, we will by
$f_{\nu,R}=f_{\nu,R}(t)$ denote the solution to (2.1) with
$K(R)=\frac{n}{\gamma(R)}$, in which $\gamma(R)$ appears in the domain defined
in (2.2) below. Further, we denote by $f_{\nu}=f_{\nu}(t)$ the solution of
(2.1) when $K\equiv 0$. If $\Phi\leq 0$ for all $t,s\in\mathbb{R}_{+}$ then we
use instead solutions of (2.1) but with $\lambda(t)$ replaced by $\Lambda(t)$
in the first term on the right hand side of (2.1). We allow ourselves to
simplify notation according to
$\lambda=\lambda(\cdot),\Lambda=\Lambda(\cdot),K=K(R)$ and $\gamma=\gamma(R)$
whenever appropriate.

Since $\Phi(t,0)=0$ and $\nu\in[0,\infty)$ the solutions $f_{\nu,R}$ and
$f_{\nu}$ will be nonegative. Moreover, if $\Phi$ satisfies the Osgood-type
condition ($\star\star\star$ ‣ 1) then the solutions will, for any $\nu>0$,
remain positive. (The only nonpositive solutions are the trivial solutions
$f_{\nu,R}\equiv f_{\nu}\equiv 0$ starting at $\nu=0$.) This plays a role in
our main results, as pointed out in the remarks made below Theorem 2.1. In
Figures 2 and 3 several solutions of (2.1) are plotted for some choices of
$\Phi$.

To proceed we define, for a nondecreasing function $\gamma=\gamma(R)>0$ and
$n\geq 1$, the domain

$\displaystyle
D(R):=\left\\{x\in\mathbb{R}^{n}_{+}:\sum_{i=1}^{n-1}x_{i}^{2}+(x_{n}+\gamma)^{2}<(R+\gamma)^{2}\right\\},$
(2.2)

see Figure 1. Finally, for a subsolution $u$ and for $R>0$ we define

$\displaystyle M(R)=\sup_{\partial D(R)}u$

and

$\displaystyle M^{\prime}(R)=\liminf_{h\to 0^{+}}\frac{M(R)-M(R-h)}{h}.$

The following theorem characterizes a sharp lower growth estimate of
subsolutions to ($\star$ ‣ 1) in terms of solutions $f_{\nu,R}$ and $f_{\nu}$
to the ODE (2.1):

###### Theorem 2.1

Suppose that ($\star\star$ ‣ 1) and ($\star\star\star$ ‣ 1) hold and let $u$
be a subsolution of ($\star$ ‣ 1) in $\mathbb{R}^{n}_{+}$ satisfying

$\displaystyle\limsup_{x\to y}\,u(x)\,\leq\,0\quad\textrm{for all}\quad
y\in\partial\mathbb{R}^{n}_{+}.$

Then either $u\leq 0$ in $\mathbb{R}^{n}_{+}$ or $M(R)$ is increasing and it
holds that

$\displaystyle\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{f_{\nu}(R)}\,\geq\,\liminf_{R\to\infty}\,\frac{f_{\nu,R}(R)}{f_{\nu}(R)},$

where $\nu$ satisfies $u(\bar{x})\geq\int_{0}^{\bar{x}_{n}}f_{\nu}(t)\,dt$ for
some $\bar{x}$ on the $x_{n}$-axis.

Using Theorem 2.1 an “explicit” growth estimate can thus be found by
estimating the limit $f_{\nu,R}(R)/f_{\nu}(R)$ as $R\to\infty$. In Section 3
we will consider certain PDEs for which we can solve the ODE (2.1) explicitly
– calculate the limit – and thereby prove Phragmen–Lindelöf type theorems. Let
us note that if we can prove

$\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{f_{\nu}(R)}>0$

then $M(R)-M(R_{0})\geq c\int_{R_{0}}^{R}f_{\nu}(s)ds$ whenever $R>R_{0}$ for
some small $c$ and thus

$\liminf_{R\to\infty}\,\frac{M(R)}{\int_{0}^{R}f_{\nu}(s)ds}>0.$

Hence, if the integral

$\displaystyle\int_{0}^{\infty}f_{\nu}(s)ds$

converges, then subsolutions may be bounded, but if the integral diverges,
then subsolutions must grow to infinity and the conclusion of Theorem 2.1
takes the form of classical Phrgmen–Lindelöf theorems.

We remark that the assumption “$\bar{x}$ lies on the $x_{n}$-axis” is only for
notational simplicity; we may translate coordinates otherwise. Note also that
Theorem 2.1 holds whenever $\mathbb{R}^{n}_{+}$ is replaced (in the theorem
and in (2.2)) with $\Omega\subset\mathbb{R}^{n}_{+}$, and that Theorem 2.1
gives a growth estimate for any initial condition $\nu\geq 0$ in (2.1) as long
as $u(\bar{x})\geq\int_{0}^{\bar{x}_{n}}f_{\nu}(t)\,dt$ for some $\bar{x}$.
The best estimate corresponds to the largest $\nu$. Moreover, it can be
realized from the proof of Lemma 2.2 that the assumption “${\lambda}$
nonincreasing and ${\Lambda}$ nondecreasing” can be replaced by the slightly
weaker assumption that ${\lambda}/{\Lambda}$ is nonincreasing and $\lambda$ is
nonincreasing ($\Lambda$ is nondecreasing) when $\Phi\geq 0$ ($\Phi\leq 0$).
Finally, it will be clear from the proof that we also have
$M(R)-M(R-h)\geq\int_{R}^{R+h}f_{\nu,R}(t)dt$ for $R>\bar{x}_{n}$ and any
$h>0$. We realize that $M(R)$ must increase as long as $f_{\nu,R}>0$, which
happens whenever $\Phi$ satisfies the Osgood-type condition ($\star\star\star$
‣ 1). Otherwise, the strong maximum principle does not hold and a positive
subsolution to ($\star$ ‣ 1) may stop growing and attain an interior maximum,
see Julin [22] and Lundström-Olofsson-Toivannen [33, Remark 4.3] for a
counterexample.

Concerning sharpness of Theorem 2.1 we consider the function

$\displaystyle u(x)=\int_{0}^{x_{n}}f_{\nu}(t)dt,$ (2.3)

vanishing on $\partial\mathbb{R}_{+}^{n}$, depending only on $x_{n}$ with
derivative $u_{x_{n}}(x)=f_{\nu}(x_{n})$. In case $\Phi(t,s)\geq 0$ it holds
that
$u_{x_{n}x_{n}}(x)=f^{\prime}_{\nu}(x_{n})=-\lambda^{-1}(x_{n})\Phi(x_{n},f_{\nu}(x_{n}))$
and hence we obtain, e.g., that

$\displaystyle-F(x,Du,D^{2}u):=\lambda(x_{n})\Delta u+\phi(f_{\nu}(x_{n}))=0,$
(2.4)

for some function $\phi(s)$ satisfying ($\star\star\star$ ‣ 1). In case
$\Phi(t,s)\leq 0$ the same holds but with $\lambda$ replaced by $\Lambda$.
Thus, the function defined in (2.3) is a classical solution to an equation of
type ($\star$ ‣ 1) satisfying ($\star\star$ ‣ 1) and ($\star\star\star$ ‣ 1).
Moreover, it satisfies $M^{\prime}(R)=f_{\nu}(R)$. In conclusion, when the
limit in Theorem 2.1 is positive, the growth estimate cannot be improved,
ignoring the shape of $D(R)$ and the value of the limit.

Concerning the shape of $D(R)$ we note the following. If $n=1$ then
$D(R)=(0,R)$ independent of $\gamma$, but if $n\geq 2$ then $\gamma=cR$
implies that the spherical segment $D(R)$ preserves its geometric proportions
for all $R>0$. If $\gamma(R)/R$ is increasing then $D(R)$ expands faster in
the $x^{\prime}$-direction, implying slightly weaker estimates since $\partial
D(R)$, on which supremum is taken, becomes larger. Observe that if the problem
is considered in $\Omega\subset\mathbb{R}_{+}^{n}$ this might be of minor
importance, especially if e.g. $\Omega$ is bounded in $x^{\prime}$-directions
or contained in a cone with apex at the origin. There is not much of a gain to
take $\gamma(R)/R$ decreasing since $D(R)$ still expands at rate $R$ in
$x^{\prime}$-directions.

The proof of Theorem 2.1 relies on comparison arguments and the following
construction of a classical strict supersolution to ($\star$ ‣ 1).

###### Lemma 2.2

Suppose that ($\star\star$ ‣ 1) and ($\star\star\star$ ‣ 1) hold, let $R>0$
and put

$\Xi_{R}(x)=\sqrt{\sum_{i=1}^{n-1}x_{i}^{2}+(x_{n}+\gamma)^{2}}-\gamma=|(x^{\prime},x_{n}+\gamma)|-\gamma$

in which $\gamma=\gamma(R)$ is from (2.2). Then the function

$\displaystyle V_{R}(x)=\int_{0}^{\Xi_{R}(x)}f_{\nu,R}(t)dt$

is a strict classical supersolution to ($\star$ ‣ 1) in $D(R)$.

Figure 1: Geometric definitions and constructions.

Proof of Lemma 2.2. For notational simplicity we set $\Xi=\Xi_{R}(x)$,
$V=V_{R}(x)$ and $f(t)=f_{\nu,R}(t)$. Differentiating yields

$\displaystyle\frac{\partial V}{\partial
x_{i}}=\frac{x_{i}}{|(x^{\prime},x_{n}+\gamma)|}f\left(\Xi\right),\quad 1\leq
i\leq n-1,\quad\frac{\partial V}{\partial
x_{n}}=\frac{x_{n}+\gamma}{|(x^{\prime},x_{n}+\gamma)|}f\left(\Xi\right).$

It follows that

$\displaystyle|DV|=f\left(\Xi\right).$ (2.5)

The second derivatives become

$\displaystyle\frac{\partial^{2}V}{\partial
x_{i}^{2}}=\left(\frac{x_{i}}{|(x^{\prime},x_{n}+\gamma)|}\right)^{2}f^{\prime}(\Xi)+\left(\frac{1}{|(x^{\prime},x_{n}+\gamma)|}-\frac{x_{i}^{2}}{|(x^{\prime},x_{n}+\gamma)|^{3}}\right)f\left(\Xi\right),$

for $1\leq i\leq n-1$, and

$\displaystyle\frac{\partial^{2}V}{\partial x_{n}^{2}}$
$\displaystyle=\left(\frac{x_{n}+\gamma}{|(x^{\prime},x_{n}+\gamma)|}\right)^{2}f^{\prime}(\Xi)+\left(\frac{1}{|(x^{\prime},x_{n}+\gamma)|}-\frac{(x_{n}+\gamma)^{2}}{|(x^{\prime},x_{n}+\gamma)|^{3}}\right)f\left(\Xi\right),$

giving

$\displaystyle\text{Tr}(D^{2}V)=f^{\prime}(\Xi)+\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}f\left(\Xi\right).$

By construction we have
$f^{\prime}(t)=-\frac{\Phi(t,f(t))}{\lambda(t)}-K\frac{\Lambda(t)}{\lambda(t)}f(t)$
and hence

$\displaystyle\text{Tr}(D^{2}V)=-\frac{\Phi\left(\Xi,f\left(\Xi\right)\right)}{\lambda(\Xi)}-K\frac{\Lambda(\Xi)}{\lambda(\Xi)}f(\Xi)+\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}f\left(\Xi\right).$

We assume from here on that $\Phi\geq 0$ and decompose
$D^{2}V=\left(D^{2}V\right)^{+}-\left(D^{2}V\right)^{-}$ so that

$\displaystyle\text{Tr}\left(\left(D^{2}V\right)^{+}\right)$
$\displaystyle=\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}f\left(\Xi\right)\quad\textrm{and}$
$\displaystyle\text{Tr}\left(\left(D^{2}V\right)^{-}\right)$
$\displaystyle=\frac{\Phi\left(\Xi,f\left(\Xi\right)\right)}{\lambda(\Xi)}+K\frac{\Lambda(\Xi)}{\lambda(\Xi)}f(\Xi)$

Utilizing the structure assumption ($\star\star$ ‣ 1), the fact that $V\geq 0$
and using (2.5) give

$\displaystyle F(x,V,DV,D^{2}V)\geq$ $\displaystyle F(x,0,DV,D^{2}V)$
$\displaystyle\geq$
$\displaystyle-\Phi\left(|x|,f\left(\Xi\right)\right)-\Lambda(x_{n})\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}f\left(\Xi\right)$
$\displaystyle+\frac{\lambda(x_{n})\Phi\left(\Xi,f\left(\Xi\right)\right)}{\lambda(\Xi)}+K\frac{\lambda(x_{n})\Lambda(\Xi)}{\lambda(\Xi)}f(\Xi)$
$\displaystyle\geq$
$\displaystyle-\Lambda(x_{n})\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}f(\Xi)+K\frac{\lambda(x_{n})\Lambda(\Xi)}{\lambda(\Xi)}f(\Xi)$
(2.6)

since $\lambda(\Xi)\leq\lambda(x_{n})$ and
$\Phi\left(|x|,f\left(\Xi\right)\right)\leq\Phi\left(\Xi,f\left(\Xi\right)\right)$
holds. This last statement follows since $|x|\geq\Xi_{R}(x)\geq x_{n}$ by
geometry, see Figure 1, and functions are nonincreasing by assumption.

To show that $V$ is a strict classical supersolution we need
$F(x,V,DV,D^{2}V)>0$ and by (2) and the fact that $f(\Xi)>0$ it suffices to
ensure

$\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}<K\frac{\lambda(x_{n})\Lambda(\Xi)}{\Lambda(x_{n})\lambda(\Xi)}.$

Observing that
$\frac{\lambda(x_{n})}{\Lambda(x_{n})}\geq\frac{\lambda(\Xi)}{\Lambda(\Xi)}$
holds since $\lambda/\Lambda$ is nonincreasing, it suffices to ensure

$\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}<K.$

We know that $\gamma\leq|(x^{\prime},x_{n}+\gamma)|$ in $\mathbb{R}_{+}^{n}$
so it is enough to have

$\frac{n-1}{\gamma(R)}<K$

which holds since we have $K=\frac{n}{\gamma(R)}$.

If $\Phi\leq 0$ then by construction
$f^{\prime}(t)=-\frac{\Phi\left(t,f\left(t\right)\right)}{\Lambda(t)}-K\frac{\Lambda(t)}{\lambda(t)}f(t)$
and we obtain

$\displaystyle\text{Tr}\left(\left(D^{2}V\right)^{+}\right)$
$\displaystyle=-\frac{\Phi\left(\Xi,f\left(\Xi\right)\right)}{\Lambda(\Xi)}+\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}f(\Xi),$
$\displaystyle\text{Tr}\left(\left(D^{2}V\right)^{-}\right)$
$\displaystyle=K\frac{\Lambda(\Xi)}{\lambda(\Xi)}f(\Xi)$ (2.7)

and thus instead of (2) we end up with

$\displaystyle F(x,V,DV,D^{2}V)\geq$ $\displaystyle F(x,0,DV,D^{2}V)$
$\displaystyle\geq$
$\displaystyle-\Phi\left(|x|,f\left(\Xi\right)\right)+\frac{\Lambda(x_{n})}{\Lambda(\Xi)}\Phi\left(\Xi,f\left(\Xi\right)\right)$
$\displaystyle-\Lambda(x_{n})\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}f(\Xi)+\lambda(x_{n})K\frac{\Lambda(\Xi)}{\lambda(\Xi)}f(\Xi)$
$\displaystyle\geq$
$\displaystyle-\Lambda(x_{n})\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}f(\Xi)+\lambda(x_{n})K\frac{\Lambda(\Xi)}{\lambda(\Xi)}f(\Xi).$
(2.8)

Here, the last inequality holds since $x_{n}\leq\Xi_{R}(x)\leq|x|$, $\Lambda$
is nondecreasing and $-\Phi\geq 0$ is nondecreasing in its first argument so
that

$-\Phi\left(|x|,f\left(\Xi\right)\right)\geq\frac{\Lambda(x_{n})}{\Lambda(\Xi)}\left(-\Phi\left(\Xi,f\left(\Xi\right)\right)\right).$

To ensure that $V$ is a strict supersolution we see from (2) that it remains
to show that

$\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}<K\frac{\lambda(x_{n})}{\Lambda(x_{n})}\frac{\Lambda(\Xi)}{\lambda(\Xi)}$

and we are thus back in the same situation as in the case $\Phi\geq 0$. The
proof of Lemma 2.2 is complete. $\hfill\Box$

Proof of Theorem 2.1. Let $u$ be as in the statement of the theorem and denote
with $\nu_{0}$ the initial condition in (2.1) for which we want to prove the
growth estimate. Let $R>0$ and put $V:=V_{R}(x)$, where $V_{R}(x)$ is the
strict supersolution in $D(R)$ guaranteed by Lemma 2.2.

If $V\geq M(R)$ on $\partial D(R)$ then, if $\Phi$ is nonnegative, it follows
that $f_{\nu}\leq\nu$ and we obtain equality by decreasing $\nu$. If $\Phi$ is
nonpositive we note that
$f^{\prime}_{\nu,R}(t)\leq\widetilde{\Phi}(f_{\nu,R}(t))$ for some
$\widetilde{\Phi}(s)\geq 0$ satisfying ($\star\star\star$ ‣ 1). Thus

$\int_{\nu}^{f_{\nu,R}(t)}\frac{ds}{\widetilde{\Phi}(s)}\leq t$

which implies that $f_{\nu,R}(t)\searrow 0$ as $\nu\searrow 0$ for all
$t\in[0,R]$. Therefore, we obtain equality by decreasing $\nu$ also in this
case. If $V<M(R)$ on $\partial D(R)$ then we increase $\nu$. If this does not
help, (note that we may have $V\leq A$, for all $\nu$ and all $R$, for some
$A>0$), then we lift the supersolution by adding a nonnegative constant.
Indeed, for $C\geq 0$ it follows from degenerate ellipticity that also $V+C$
is a strict supersolution. We conclude that

$V+C\geq C\quad\text{on}\quad\\{x_{n}=0\\}\quad\text{and}\quad
V+C=M(R)\quad\text{on}\quad\partial D(R)\cap\mathbb{R}^{n}_{+}.$

We clarify that if $C>0$ then we have taken $\nu>\nu_{0}$. It follows that

$\displaystyle\limsup_{x\to z}u(x)\leq V(z)+C\quad\textrm{for all}\quad
z\in\partial D(R)$

and the weak comparison principle in Lemma 1.1 implies that $u\leq V+C$ in
$D(R)$.

We next conclude that $\nu\geq\nu_{0}$. In particular, assume $\nu<\nu_{0}$.
By assumption and by the above we have $V(\bar{x})\geq
u(\bar{x})\geq\int_{0}^{\bar{x}_{n}}f_{\nu_{0}}(t)\,dt$ for some $\bar{x}\in
D(R)\cap\\{\bar{x}^{\prime}=0\\}$, but on the other hand

$\displaystyle
V(\bar{x})=\int_{0}^{\Xi_{R}(\bar{x})}f_{\nu,R}\left(t\right)dt=\int_{0}^{\bar{x}_{n}}f_{\nu,R}\left(t\right)dt<\int_{0}^{\bar{x}_{n}}f_{\nu_{0},R}\left(t\right)dt\leq\int_{0}^{\bar{x}_{n}}f_{\nu_{0}}(t)\,dt,$

where the last inequality follows since $f_{\nu,R}\leq f_{\nu}$. Hence, we
have a contradiction and we therefore conclude $\nu\geq\nu_{0}$.

Now let $R>\bar{x}_{n}$, $h\in(0,R)$ and note that by the comparison principle

$\displaystyle M(R)-M(R-h)$ $\displaystyle\geq
V(0,\dots,0,R)-V(0,\dots,0,R-h)$
$\displaystyle=\int_{R-h}^{R}f_{\nu,R}\left(t\right)dt\geq\int_{R-h}^{R}f_{\nu_{0},R}\left(t\right)dt>0.$

Hence $M(R)$ is increasing and

$\displaystyle\frac{M(R)-M(R-h)}{\int_{R-h}^{R}f_{\nu_{0}}\left(t\right)dt}\geq\frac{\int_{R-h}^{R}f_{\nu_{0},R}\left(t\right)dt}{\int_{R-h}^{R}f_{\nu_{0}}\left(t\right)dt}.$
(2.9)

Inequality (2.9) holds for any $R>\bar{x}_{n}$ independent of $h$. Taking the
limit yields

$\displaystyle\frac{M^{\prime}(R)}{f_{\nu_{0}}\left(R\right)}=\liminf_{h\to
0^{+}}\frac{M(R)-M(R-h)}{\int_{R-h}^{R}f_{\nu_{0}}\left(t\right)dt}\geq\frac{f_{\nu_{0},R}\left(R\right)}{f_{\nu_{0}}\left(R\right)}$

and thus

$\liminf_{R\to\infty}\frac{M^{\prime}(R)}{f_{\nu_{0}}\left(R\right)}\geq\liminf_{R\to\infty}\frac{f_{\nu_{0},R}\left(R\right)}{f_{\nu_{0}}\left(R\right)}.$

This completes the proof of the theorem. $\hfill\Box$

##  3 Applications to well known equations

In this section we apply Theorem 2.1 to some well known PDEs for which we can
solve the ODE in (2.1) explicitly – calculate the limit

$\liminf_{R\to\infty}\frac{f_{\nu,R}(R)}{f_{\nu}(R)}$

and conclude explicit growth estimates.

When stating corollaries for specific classes of PDEs we would sometimes
prefer to infer other types of “weak” solutions than viscosity solutions
whenever such are more suitable or more commonly used for such equations in
the literature. As the equivalence of different kinds of ”weak” solutions
often is a nontrivial problem we will in some cases avoid going into these
details, but this should not make things unclear. The reason is that we only
use comparison between “weak” subsolutions and classical strict supersolutions
– i.e. Lemma 1.2.

We begin with the simplest case $\Phi\equiv 0$, including e.g. the famous
$p$-Laplace equation, proceed with PDEs having sublinear growth in the
gradient according to $\Phi(t,s)=C(t)s^{k}$ for $k\geq 1$ and end by the
variable exponent $p$-Laplace equation, which satisfies assumption
($\star\star$ ‣ 1) with $\Phi(s)=C(t)s|\log s|$.

### The case $\Phi(s)\equiv 0$

In this simple case the ODE (2.1) reduces to

$\frac{df}{dt}=-K\frac{\Lambda(t)}{\lambda(t)}f(t)\quad\text{with}\quad
f(0)=\nu$

and hence $f_{\nu,R}(t)=\nu e^{-K\int_{0}^{t}\Lambda(s)\lambda^{-1}(s)ds}$ and
$f_{\nu}(t)\equiv\nu$. We obtain the limit

$\displaystyle\liminf_{R\to\infty}\frac{M^{\prime}(R)}{f_{\nu}(R)}\geq\liminf_{R\to\infty}\frac{f_{\nu,R}(R)}{f_{\nu}(R)}=\lim_{R\to\infty}e^{-K\int_{0}^{R}\Lambda(s)\lambda^{-1}(s)ds}$
(3.1)

which is positive if $K\int_{0}^{R}\Lambda(s)\lambda^{-1}(s)ds\precsim 1.$
Recalling from the definitions in (2.1) that $K=\frac{n}{\gamma(R)}$ this
forces us to chose the function $\gamma(R)$ in the definition of $D(R)$, given
by (2.2),

$\displaystyle\int_{0}^{R}\frac{\Lambda(s)}{\lambda(s)}ds\precsim\gamma(R).$
(3.2)

Following the remark just below Theorem 2.1 we see that (3.1) implies

$\displaystyle\liminf_{R\to\infty}\,\frac{M(R)}{R}\,>\,0,$ (3.3)

and we thus retrieve the classical form of the Phragmen–Lindelöf theorem. If
the PDE is uniformly elliptic, i.e. $\Lambda/\lambda=constant$, then according
to (3.2) we can pick $\gamma(R)=R$ and thereby $D(R)$ preserves its geometric
proportions for all $R>0$, which also agrees with the classical
Phragmen–Lindelöf theorem. If ellipticity blows up at infinity, i.e.
$\Lambda(R)/\lambda(R)\to\infty$, then the loss in estimate (3.3) comes only
in the shape of $D(R)$ – it expands faster in $x^{\prime}$-direction since we
need to take a larger $\gamma(R)$ according to (3.2).

Concerning sharpness of (3.3) we note that with $\phi\equiv 0$ the solution of
(2.3) yields

$\int_{0}^{x_{n}}\,f_{\nu}(s)\,ds=\nu\,x_{n},$

which clearly hits the bottom of (3.3).

Following ($\star$ ‣ 1) and ($\star\star$ ‣ 1) we see that (3.3) holds, e.g.,
for subsolutions of the quasilinear equations

$\displaystyle-\sum_{i,j=1}^{n}A_{ij}(x)\frac{\partial^{2}u}{\partial
x_{i}\partial x_{j}}+f(x,u,Du)=0,$ (3.4)

corresponding to $F(x,r,p,X)=-\text{Tr}\left(A(x)X\right)+f(x,u,Du)$, where
$A(x)\in\mathbb{S}^{n}$ satisfies
$\lambda(x_{n})\text{Tr}\left(Y\right)\leq\text{Tr}\left(A(x)Y\right)\leq\Lambda(x_{n})\text{Tr}\left(Y\right)$
for all $Y\geq 0$, and

$\displaystyle P^{-}_{\lambda,\Lambda}(D^{2}u)+f(x,u,Du)=0,$ (3.5)

whenever $f(x,u,Du)\geq 0$ is nondecreasing in $u$. One such PDE is the
following $p$-Laplace equation, $p\in(1,\infty)$, with lower order terms

$\displaystyle-\nabla\cdot\left(|Du|^{p-2}Du\right)+f(x,u,Du)=0.$ (3.6)

Indeed, with

$\displaystyle-F(x,u,Du,D^{2}u)=\Delta
u+(p-2)\Delta_{\infty}u-\frac{f(x,u,Du)}{|Du|^{p-2}},$

where $\Delta_{\infty}u=\langle D^{2}u\frac{Du}{|Du|},\frac{Du}{|Du|}\rangle$
denotes the infinity Laplace operator, we see that $F$ satisfies (1.2) with
$\phi\equiv 0$ and

$\min\left\\{1,p-1\right\\}\text{Tr}(Y)\leq
F(x,u,Du,X)-F(x,u,Du,X+Y)\leq\max\left\\{1,p-1\right\\}\text{Tr}(Y),$

whenever $Y\geq 0$. Hence $F$ satisfies (1.1) with
$\lambda=\min\left\\{1,p-1\right\\}$ and $\Lambda=\max\left\\{1,p-1\right\\}$
and Theorem 2.1 applies.

Recalling that Lemma 1.2 holds for weak solutions (defined in the usual way)
to $p$-Laplace type problems or that viscosity solutions and weak solutions
are equivalent for some $p$-Laplace type equations , see e.g.
Juutinen–Lindqvist–Manfredi [24], Julin–Juutinen [23] and Medina–Ochoa [35],
we retrieve the well known Phragmen–Lindelöf results in Lindqvist[29] in the
setting of halfspaces.

### The case $\Phi(t,s)=C(t)s^{k}$

We now consider equations satisfying ($\star\star$ ‣ 1) with
$\Phi(t,s)=C(t)s^{k}$ for some $k\geq 1$. Note that such $\Phi$ satisfies
($\star\star\star$ ‣ 1) and hence we expect that Theorem 2.1 implies that
subsolutions must be increasing on the boundary of $D(R)$. The ODE in (2.1)
yields

$\frac{df}{dt}=-\frac{C(t)f^{k}}{\lambda(t)}-K\frac{\Lambda(t)}{\lambda(t)}f,\quad
t\in(0,R),\quad\text{with}\quad f(0)=\nu.$

As $C(t)$ is nonincreasing (by assumptions on $\Phi$) and $\Lambda$ is
nondecreasing we can replace the above equation with the separable ODE

$\frac{df}{dt}=-A(t)\left(f^{k}+\widetilde{K}f\right),$

where $A(t):=\lambda^{-1}(t)C(t)$ and
$\widetilde{K}:=\frac{K\Lambda(R)}{C(R)}$. This is possible since solutions of
this ODE will approach the origin faster when $t\in(0,R)$, and hence it
creates a lower bound on the limit in Theorem 2.1. To find the solution for
$k>1$ we observe that

$\frac{1}{\widetilde{K}}\int_{\nu}^{f(t)}\left(\frac{1}{y}-\frac{y^{k-2}}{y^{k-1}+\widetilde{K}}\right)dy=-\int_{0}^{t}A(s)ds$

and

$\frac{1}{k-1}\left[\log
y^{k-1}-\log\left(y^{k-1}+\widetilde{K}\right)\right]_{\nu}^{f(t)}=-\widetilde{K}\int_{0}^{t}A(s)ds.$

Thus

$f_{\nu,R}(t)=\left\\{\begin{array}[]{ll}\nu
e^{-\left(1+\widetilde{K}\right)\int_{0}^{t}A(s)ds}&\text{if $k=1$},\\\
\widetilde{K}^{\frac{1}{k-1}}\left({e^{(k-1)\widetilde{K}\int_{0}^{t}A(s)ds}\left(\frac{\widetilde{K}}{\nu^{k-1}}+1\right)-1}\right)^{\frac{1}{1-k}}&\text{if
$k>1$},\end{array}\right.$

and by solving $\frac{df}{dt}=-\lambda(t)^{-1}C(t)f^{k}$ with $f(0)=\nu$ we
also obtain

$\displaystyle f_{\nu}(t)=\left\\{\begin{array}[]{ll}\nu
e^{-\int_{0}^{t}A(s)ds}&\text{if $k=1$},\\\
\left((k-1)\int_{0}^{t}A(s)ds+\nu^{1-k}\right)^{\frac{1}{1-k}}&\text{if
$k>1$}.\end{array}\right.$ (3.9)

The limit in Theorem 2.1 becomes, for $k=1$,

$\displaystyle\liminf_{R\to\infty}\frac{M^{\prime}(R)}{f_{\nu}(R)}\geq\liminf_{R\to\infty}\frac{f_{\nu,R}(R)}{f_{\nu}(R)}=\lim_{R\to\infty}e^{-\widetilde{K}\int_{0}^{R}A(s)ds},$

and for $k>1$,

$\displaystyle\liminf_{R\to\infty}\frac{M^{\prime}(R)}{f_{\nu}(R)}\geq\liminf_{R\to\infty}\frac{f_{\nu,R}(R)}{f_{\nu}(R)}=\lim_{R\to\infty}\left(\frac{\widetilde{K}(k-1)\int_{0}^{R}A(s)ds+\widetilde{K}\nu^{1-k}}{e^{\widetilde{K}(k-1)\int_{0}^{R}A(s)ds}\left(\widetilde{K}\nu^{1-k}+1\right)-1}\right)^{\frac{1}{k-1}}.$

Let’s observe that if $\widetilde{K}(R)\int_{0}^{R}A(s)ds\precsim 1,$ which is
obtained by taking
$\frac{\Lambda(R)}{C(R)}\int_{0}^{R}A(s)ds\precsim\gamma(R)$, then the limits
are positive and we obtain

$\displaystyle\liminf_{R\to\infty}\frac{M^{\prime}(R)}{f_{\nu}(R)}>0.$ (3.10)

Thus, we may derive several Phragmen–Lindelöf type results using Theorem 2.1,
whose form will depend on the exponent $k$ and the functions $C$, $\lambda$
and $\Lambda$. For example, using (3.9)-(3.10) we have proved:

###### Corollary 3.1

Suppose that ($\star\star$ ‣ 1) holds with $\Phi(t,s)=C(t)s^{k}$, $k\geq 1$.
Let $u$ be a subsolution of ($\star$ ‣ 1) in $\mathbb{R}^{n}_{+}$ satisfying

$\displaystyle\limsup_{x\to y}\,u(x)\,\leq\,0\quad\textrm{for all}\quad
y\in\partial\mathbb{R}^{n}_{+}.$

Assume also that $u(\bar{x})>0$ for some $\bar{x}$ on the $x_{n}$-axis. Then
the following is true, with $A(t)=C(t)\lambda(t)^{-1}$:

  * $(i)$

If $\int_{0}^{R}A(s)ds\precsim R^{\alpha(k-1)}$ for $\alpha\geq 0,k>1$ and
$\frac{\Lambda(R)}{\gamma(R)C(R)}\precsim R^{-\alpha(k-1)}$ then

$\displaystyle\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{R^{-\alpha}}\,>\,0\quad\text{implying}\quad\liminf_{R\to\infty}\,\frac{M(R)}{R^{1-\alpha}}\,>\,0.$

  * $(ii)$

If $\int_{0}^{R}A(s)ds\precsim R$ and
$\frac{\Lambda(R)}{\gamma(R)C(R)}\precsim\frac{1}{R}$ then

$\displaystyle\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{e^{-R}}\,>\,0\quad\text{if
$k=1$},\quad\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{R^{\frac{1}{1-k}}}\,>\,0\quad\text{if
$k\in(1,2)$},$
$\displaystyle\liminf_{R\to\infty}\,\frac{M(R)}{\log(R)}\,>\,0\quad\text{if
$k=2$}\quad\text{and}\quad\liminf_{R\to\infty}\,\frac{M(R)}{R^{\frac{2-k}{1-k}}}\,>\,0\quad\text{if
$2<k$}.$

  * $(iii)$

If $\int_{0}^{R}A(s)ds\precsim\log(R)$,
$\frac{\Lambda(R)}{\gamma(R)C(R)}\precsim\frac{1}{\log(R)}$ and $k=1$ then

$\displaystyle\liminf_{R\to\infty}\,\frac{M(R)}{\log(R)}\,>\,0.$

We remark that in Corollary 3.1 we only summarize some examples of growth
estimates that take simple forms – the reader may return to conclusion (3.10)
for the general case. Note also that all conclusions in Corollary 3.1 are
independent of $\nu$, meaning that we only have to use arbitrary small $\nu>0$
to prove them. Therefore, since $f_{\nu}$ and $f_{\nu,R}$ are nonincreasing
functions in this case, we only need that the assumption $\Phi(t,s)=C(t)s^{k}$
holds for arbitrary small $s$.

Conclusion $(i)$ takes the form of the classical Phragmen–Lindelöf theorem and
when $\alpha=0$ it applies e.g. when

$\displaystyle\Phi(|x|,s)=C(|x|)s^{k}=\frac{c}{(1+|x|)^{a}}s^{k},$ (3.11)

$k\geq 1$, $a>1$, ellipticity $\lambda(R)=constant$, $\Lambda(R)=constant$ and
$\gamma(R)\succsim R^{a}$. Conclusion $(ii)$ holds e.g. when
$A=C/\lambda=constant$, $\Lambda/C=constant$ and $\gamma(R)\equiv R$. We
observe that the exponent $k$ in $\Phi(t,s)=C(t)s^{k}$ has a borderline value
at $k=2$. Namely, if $k\in[1,2)$ then subsolutions may be bounded, but if
$k\in[2,\infty)$ then any subsolution must grow to infinity. As already
mentioned in Section 2 such border is, beyond the assumptions in Corollary
3.1, characterized by convergence/divergence of

$\displaystyle\int_{0}^{\infty}f_{\nu}(s)ds.$

Conclusion $(iii)$ holds e.g. when $a=1$ in (3.11), $\gamma(R)=R\log(R)$ and
$\Lambda=constant$.

We further remark that upper bounds on $\int_{0}^{R}A(s)ds$ have played an
important role for related results in the literature, see e.g. Gilbarg [15],
Hopf [19] and Vitolo [40], and that Phragmen–Lindelöf theorems for similar
equations in more general domains but with $k=1$ and $k=2$ are proved by
Capuzzo-Dolcetta–Vitolo [10] and Koike–Nakagawa [26].

As in the case $\Phi\equiv 0$ the results in this subsection apply to PDEs of
type (3.4) and (3.5) but now with relaxed assumption on $f$, namely
$f(x,u,Du)\geq-C(|x|)|Du|^{k}.$ In case of the $p$-Laplace type equation
(3.6), $1<p<\infty$, the growth condition on the lower order terms becomes
$f(x,u,Du)\geq-C(|x|)|Du|^{\,k+p-2}$.

When $A(s)=C(s)/\lambda(s)=constant$ we can explicitly find the classical
solution of (2.4), ensuring sharpness, in case of $\Phi(t,s)=C(t)s^{k}$:

$\displaystyle
u(x_{n})=\int_{0}^{x_{n}}f_{\nu}(t)dt=\frac{1}{A}\left\\{\begin{array}[]{ll}\nu\left(1-e^{-Ax_{n}}\right)&\text{if
$k=1$},\\\ \log\left(Ax_{n}\nu+1\right)&\text{if $k=2$},\\\
\frac{\nu^{2-k}}{2-k}\left(1-\left((k-1)Ax_{n}\nu^{k-1}+1\right)^{\frac{2-k}{1-k}}\right)&\text{otherwise}.\end{array}\right.$

Figure 2 shows the solutions $\int_{0}^{x_{n}}f_{\nu}(t)dt$ for some values of
$k$, $\nu$ and different functions $A(t)$.

Figure 2: The derivative $f_{\nu}$ in (3.9) and the solution
$\int_{0}^{x}f_{\nu}(t)dt$. Solid curves (for $u$) approach infinity at speed
$\log(x)$ and dashed-dot at speed $x^{8/9}$, while dashed curve is bounded.
Corresponding growth estimates are established in Corollary 3.1. In all
simulations, $\nu=5.$

### The case $\Phi(t,s)=C(t)s|\log s|$ : variable exponent $p$-Laplace
equation

We set $\Phi(t,s)=C(t)s|\log s|$ and obtain the ODE

$\displaystyle\frac{df}{dt}=-\frac{C(t)f|\log
f|}{\lambda(t)}-K\frac{\Lambda(t)}{\lambda(t)}f,\quad
t\in(0,R),\quad\text{with}\quad f(0)=\nu.$

By the same argument as in the case $\Phi=C(t)s^{k}$ we replace the above ODE
with

$\frac{df}{dt}=-A(t)\left(f|\log f|-\widetilde{K}f\right),$

where $A(t)=\lambda^{-1}(t)C(t)$ and $\widetilde{K}=\frac{K\Lambda(R)}{C(R)}$.
This equation separates, when $0<f\leq 1$, to

$\log\left(\widetilde{K}-\log\nu\right)-\log\left(\widetilde{K}-\log
f\right)=-\int_{0}^{t}A(s)ds.$

Thus

$f_{\nu,R}(t)=e^{\widetilde{K}\left(1-e^{\int_{0}^{t}A(s)ds}\right)}\nu^{e^{\int_{0}^{t}A(s)ds}}$

and since $f$ must be nonincreasing this holds for $0<\nu\leq 1$. If $f>1$ the
solution takes a similar form. In particular

$\displaystyle
f_{\nu,R}(t)=\left\\{\begin{array}[]{ll}e^{\widetilde{K}\left(1-e^{\int_{0}^{t}A(s)ds}\right)}\nu^{e^{\int_{0}^{t}A(s)ds}}&\text{if
$0<\nu\leq 1$},\\\
e^{\widetilde{K}\left(e^{-\int_{0}^{t}A(s)ds}-1\right)}\nu^{e^{-\int_{0}^{t}A(s)ds}}&\text{if
$1<f(t)$}\end{array}\right.$ (3.14)

and

$\displaystyle
f_{\nu}(t)=\left\\{\begin{array}[]{ll}\nu^{e^{\int_{0}^{t}A(s)ds}}&\text{if
$0<\nu\leq 1$},\\\ \nu^{e^{-\int_{0}^{t}A(s)ds}}&\text{if
$1<\nu$}.\end{array}\right.$ (3.17)

The limit in Theorem 2.1 becomes, for $0<\nu\leq 1$,

$\displaystyle\liminf_{R\to\infty}\frac{M^{\prime}(R)}{f_{\nu}(R)}\geq\liminf_{R\to\infty}\frac{f_{\nu,R}(R)}{f_{\nu}(R)}=\lim_{R\to\infty}e^{\widetilde{K}\left(1-e^{\int_{0}^{R}A(s)ds}\right)}$
(3.18)

which is positive if $\widetilde{K}e^{\int_{0}^{R}A(s)ds}\precsim 1$.
Therefore, we have to pick
$\gamma(R)\succsim\frac{\Lambda(R)}{C(R)}e^{\int_{0}^{R}A(s)ds}$ to achieve a
growth estimate.

When $\nu>1$ we know that $f_{\nu,R}$ in (3.14) stays above 1 if

$\left(\widetilde{K}+\log\nu\right)e^{-\int_{0}^{t}A(s)ds}>\widetilde{K}$

which forces us to take
$\gamma(R)\succsim\frac{\Lambda(R)}{C(R)}e^{\int_{0}^{R}A(s)ds}$. Then

$\displaystyle\liminf_{R\to\infty}\frac{M^{\prime}(R)}{f_{\nu}(R)}\geq\liminf_{R\to\infty}\frac{f_{\nu,R}(R)}{f_{\nu}(R)}=\lim_{R\to\infty}e^{\widetilde{K}\left(e^{-\int_{0}^{R}A(s)ds}-1\right)}\geq\lim_{R\to\infty}e^{-\widetilde{K}}>0$

as $\gamma(R)\succsim\frac{\Lambda(R)}{C(R)}$. If the solution $f_{\nu,R}$
decreases to 1 then the limit can be estimated as in (3.18) since $f_{\nu,R}$
follows the expression for $\nu\in(0,1]$ with $\nu=1$.

In summary, since $\Phi(t,s)=C(t)s|\log s|$ satisfies ($\star\star\star$ ‣ 1)
we can conclude that for a subsolution $u$ satisfying the assumptions in
Theorem 2.1 with $\Phi(t,s)=C(t)s|\log s|$, the following is true when
$\gamma(R)\succsim\frac{\Lambda(R)}{C(R)}e^{\int_{0}^{R}A(s)ds}$:

  * •

If $u(\bar{x})\geq\check{u}$ for some $\bar{x}$ on the $x_{n}$-axis and
$\nu\in(0,1]$, then $M(R)$ may be bounded but

$\displaystyle\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{\nu^{e^{\int_{0}^{R}A(s)ds}}}>0.$
(3.19)

  * •

If $u(\bar{x})\geq\check{u}$ for some $\bar{x}$ on the $x_{n}$-axis and
$\nu>1$, then $M(R)$ approaches infinity according to

$\displaystyle\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{\nu^{e^{-\int_{0}^{R}A(s)ds}}}>0\quad\text{implying}\quad\liminf_{R\to\infty}\,\frac{M(R)}{R}>0.$
(3.20)

Thus we retrieve the classical form of a Phragmen–Lindelöf theorem if the
subsolution exceeds $\check{u}$ in (3.25) with $\nu\geq 1$, but if the
subsolution only exceeds $\check{u}$ with $\nu<1$, it may grow very slowly and
need not approach infinity. The border at $\nu=1$ originates from the fact
that $\Phi(1)=0$ and thus $f_{1}\equiv 1$, while $\Phi(s)>0$ for all other
positive $s$ implying $f_{\nu}\to 0$ if $\nu\in(0,1)$ and $f_{\nu}\to 1$ if
$\nu>1$.

The solution of (2.4) with $\Phi(t,s)=C(t)s|\log s|$, i.e.

$\displaystyle C(t)|Du||\log|Du||+\lambda(x_{n})\Delta u=0,$ (3.21)

ensuring sharpness, can be calculated analytically when
$A(t)=C(t)/\lambda(t)=constant$ and then yields

$\displaystyle\check{u}(x)=\int_{0}^{x_{n}}f_{\nu}(s)ds=\frac{1}{A}\left\\{\begin{array}[]{ll}-E_{i}\left(\log\nu\right)+E_{i}\left(e^{At}\log\nu\right)&\text{if
$0<\nu<1$},\\\ x_{n}&\text{if $\nu=1$},\\\
E_{i}\left(\log\nu\right)-E_{i}\left(e^{-At}\log\nu\right)&\text{if
$1<\nu$}\end{array}\right.$ (3.25)

where $E_{i}$ is the Exponential integral. See Figure 3 (upper row) for some
illustrations of the functions $f_{\nu}$ in (3.17) and $\check{u}$ in (3.25).

#### Variable exponent $p$-Laplace equation

The $p(x)$-Laplace equation, which often serves as a model example for PDEs
with nonstandard growth, yields

$\nabla\cdot(|Du|^{p(x)-2}Du)\,=\,0.$ (3.26)

The function $p:\Omega\to(1,\infty)$ is usually called a variable exponent. If
$p=constant$, then this equation is the classical $p$-Laplace equation and if
$p=2$ it’s the famous Laplace equation. Apart from interesting theoretical
considerations such equations arise in the applied sciences, for instance in
fluid dynamics, see e.g. Diening–Růžička [13], in the study of image
processing, see e.g. Chen-Levine-Rao [11] and for electro-rheological fluids,
we refer the reader to Harjulehto–Hästö–Lê–Nuortio [17] for a recent survey
and further references.

We recall the following standard definition of weak solutions of (3.26): A
function $u\in W^{1,p(x)}_{loc}(\Omega)$ is a weak (sub)solution of (3.26) if

$\int_{\Omega}|Du|^{p(x)-2}Du\cdot D\psi\,dx(\leq)=0$

for all (nonnegative) $\psi\in C_{0}^{\infty}(\Omega)$. Similarly, $u$ is a
weak supersolution if $-u$ is a weak subsolution. A function which is both a
weak subsolution and a weak supersolution is called a weak solution. An
(USC/LSC) weak (sub/super)solution is called a $p(x)$-(sub/super)harmonic
function. We also note that $u\in W^{1,1}_{loc}(\Omega)$ is $p(x)$-harmonic in
$\Omega$ if it is a local minimizer of the energy

$\int_{\Omega}\frac{1}{p(x)}|Du|^{p(x)}dx,$

where $1<p(x)<\infty$.

To proceed we define the operator

$\displaystyle\Delta_{p(x)}u:=\Delta
u+(p(x)-2)\Delta_{\infty}u+\log|Du|\langle Dp(x),Du\rangle,$

where $\Delta_{\infty}u=\langle D^{2}u\frac{Du}{|Du|},\frac{Du}{|Du|}\rangle$
denotes the infinity Laplace operator. We note that $\Delta_{p(x)}u\geq 0$
implies

$\displaystyle-\widehat{F}(x,Du,D^{2}u):=\Lambda(x)\text{Tr}(D^{2}u^{+})-\lambda(x)\text{Tr}(D^{2}u^{-})+|Dp(x)||Du||\log|Du||\geq
0$ (3.27)

with $\lambda(x)=\min\\{1,p(x)-1\\}$ and $\Lambda(x)=\max\\{1,p(x)-1\\}$. This
suggests that $p(x)$-subharmonic functions should be viscosity subsolutions to
$\widehat{F}=0$, which is the case. Indeed, following the proof in Julin [22],
which expands on Juutinen–Lukkari–Parviainen [25], we can conclude the
following slightly generalized version of [22, Lemma 5.2]:

###### Lemma 3.2

Suppose that $p(x)$ is $C^{1}(\mathbb{R}_{+})$, $1<p(x)<\infty$,
$\lambda(x)=\min\\{1,p(x)-1\\}$ and $\Lambda(x)=\max\\{1,p(x)-1\\}$. If $u$ is
$p(x)$-subharmonic in a domain $\Omega\in\mathbb{R}^{n}$, $\varphi\in
C^{2}(\Omega)$ is such that $\varphi(x_{0})=u(x_{0})$ at $x_{0}\in\Omega$ and
$\varphi\geq u$ then

$\widehat{F}(x_{0},D\varphi(x_{0}),D^{2}(\varphi(x_{0})))\leq-\Delta_{p(x)}\varphi(x_{0})\leq
0.$

To obtain a PDE satisfying the required assumptions we redefine $\widehat{F}$
by replacing ellipticity with $\lambda(x_{n})\leq\min\\{1,p(x)-1\\}$
nonincreasing, $\Lambda(x_{n})\geq\max\\{1,p(x)-1\\}$ nondecreasing and also
replacing the nonhomogeneous term with $\Phi(|x|,s)\geq|Dp(x)|s|\log s|$,
where $\Phi(|x|,s)$ is nonincreasing in $|x|$. In particular, we can take

$\displaystyle\lambda(t)=$ $\displaystyle p_{\lambda}:=\inf_{y:y_{n}\leq
t}\min\\{1,p(y)-1\\},\quad\Lambda(t)=p_{\Lambda}:=\sup_{y:y_{n}\leq
t}\max\\{1,p(y)-1\\}\quad\text{and}$ $\displaystyle\Phi(t,s)$
$\displaystyle=||Dp||_{\infty,t}\,s|\log
s|,\quad\text{where}\quad||f||_{\infty,t}=\sup_{y:|y|\geq t}|f(y)|.$ (3.28)

In conclusion, a weak USC subsolution (a $p(x)$-subharmonic function) to the
variable exponent $p$-Laplace equation is a viscosity subsolution of a PDE of
type ($\star$ ‣ 1) satisfying ($\star\star$ ‣ 1) and ($\star\star\star$ ‣ 1).
We can therefore conclude that deductions (3.19) and (3.20) hold for
$p(x)$-subharmonic functions whenever $p(x)$ is $C^{1}(\mathbb{R}_{+})$ and
$1<p(x)<\infty$.

We summarize our findings in the following theorem yielding Phragmen–Lindelöf-
type results, of which some are sharp, for weak solutions to the variable
exponent $p$-Laplace equation:

###### Theorem 3.3

Suppose that $p(x)$ is $C^{1}(\mathbb{R}^{n}_{+})$, $1<p(x)<\infty$, and let
$u$ be $p(x)$-subharmonic in $\mathbb{R}^{n}_{+}$ satisfying

$\displaystyle\limsup_{x\to y}\,u(x)\,\leq\,0\quad\textrm{for all}\quad
y\in\partial\mathbb{R}^{n}_{+}.$

Then $u$ is a viscosity subsolution of an equation of type ($\star$ ‣ 1)
satisfying ($\star\star$ ‣ 1) and ($\star\star\star$ ‣ 1) with
$\Phi,\lambda=p_{\lambda}$ and $\Lambda=p_{\Lambda}$ as in (3). Moreover, if
$\frac{p_{\Lambda}(R)}{||Dp||_{\infty,R}}{\exp\left({\int_{0}^{R}\frac{||Dp||_{\infty,s}}{p_{\lambda}(s)}ds}\right)}\precsim\gamma(R)$
and $\check{u}(x)=\int_{0}^{x_{n}}f_{\nu}(s)ds$ with $f_{\nu}$ from (3.17)
then the following is true:

  * •

If $u\geq\check{u}$ somewhere on the $x_{n}$-axis for $\nu\in(0,1]$ then

$\displaystyle\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{\nu^{\exp\left({\int_{0}^{R}\frac{||Dp||_{\infty,s}}{p_{\lambda}(s)}ds}\right)}}>0.$

  * •

If $u\geq\check{u}$ somewhere on the $x_{n}$-axis for $\nu>1$ then

$\displaystyle\liminf_{R\to\infty}\,\frac{M(R)}{R}>0.$

We thus retrieve the classical form of a Phragmen–Lindelöf theorem if the
subsolution exceeds $\check{u}$ with $\nu\geq 1$, in particular if it exceeds
$x_{n}$. On the other hand, if the subsolution only exceeds $\check{u}$ with
$\nu<1$, then Theorem 3.3 states that it may grow very slowly and be bounded.
The sharpness in the case $\nu\geq 1$ follows by observing, e.g., that

$\displaystyle u(x)=c\,x_{n}\quad\text{is $p(x)$-harmonic with}\quad
p(x)=M_{0}+\sum_{i=1}^{n-1}M_{i}x_{i}^{2},$ (3.29)

whenever $c\geq 1,M_{0}>1$ and $M_{i}$, for $i\in[1,n-1]$, are constants. It
is worth observing that the conclusion

$\displaystyle\liminf_{R\to\infty}\,\frac{M(R)}{R}>0$

follows also in the case $\nu\in(0,1]$ if
${\int_{0}^{R}\frac{||Dp||_{\infty,s}}{p_{\lambda}(s)}ds}\precsim 1$ since
then $\liminf_{R\to\infty}\,M^{\prime}(R)>0$. This holds e.g. if the exponent
satisfies $p^{-}<p(x)$ and $||Dp||_{\infty,s}\precsim s^{-k}$ for some
constants $p^{-},k>1$; a natural conclusion since these assumptions force the
equation toward the constant exponent $p$-Laplace equation far away from the
origin.

Versions of Theorem 3.3 are possible to derive from (3.19) and (3.20); e.g.,
it may be useful to redefine the norm in (3) as
$||f||_{\infty,t}=\sup_{y:y_{n}=t}|f(y)|$. Then, if
$||Dp(x)||_{\infty,x_{n}}\neq 0$ for all $x_{n}>0$ we may divide (3.27) by
$||Dp(x)||_{\infty,x_{n}}$ and conclude, for
$\lambda(x_{n})\leq\frac{\min\\{1,p(x)-1\\}}{||Dp(x)||_{\infty,x_{n}}}$
nonincreasing and
$\Lambda(x_{n})\geq\frac{\max\\{1,p(x)-1\\}}{||Dp(x)||_{\infty,x_{n}}}$
nondecreasing, that Theorem 3.3 holds with
$\frac{p_{\Lambda}(R)}{||Dp||_{\infty,R}}$ replaced by $\Lambda(R)$ and
$\frac{||Dp||_{\infty,s}}{p_{\lambda}(s)}$ replaced by $\lambda^{-1}(s)$. In
particular, in the case $\nu\in(0,1]$ the conclusion then reads

$\displaystyle\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{\nu^{e^{\int_{0}^{R}\lambda^{-1}(s)ds}}}>0.$
(3.30)

We build sharpness of this result in Remark 3.4 below in which we find a
family of exponents for which the solution in (3.25), which satisfies
$M^{\prime}(R)=\nu^{e^{\int_{0}^{R}\lambda^{-1}(s)ds}}$, is $p$-harmonic.

We further remark that Theorem 3.3 sharpens some results of Adamowicz [1] in
the geometric setting of halfspaces, and the $C^{1}$-assumption on $p(x)$
should be replaceable with locally Lipschitz continuity by approximation
arguments. Furthermore, the reader may recall the remarks made below
deductions (3.19) and (3.20) and also note that contrary to the results in the
former subsection, for $\Phi(t,s)=C(t)s^{k}$, the growth estimates here depend
on $\nu$.

Our estimates may not be optimal when $\log|Du|\langle Dp(x),Du\rangle$ is
negative since then we lose information by our choice of $\Phi$. We can
improve by taking $\phi(s)\equiv 0$, but we still lose information when
subsolutions gradients are not “close to perpendicular” to $Dp(x)$. This
motivates us to derive better estimates under assumptions excluding e.g. the
solution in (3.29). We do so by studying a nonpositive $\Phi$; the case
$\Phi(s)=-s|\log s|$, in the next section.

We proceed by proving the following result, in which we find the “slowest
growing” $p(x)$-harmonic function, for a given ellipticity bound, and the
corresponding family of exponents.

###### Remark 3.4

The function $\check{u}(x)=\int_{0}^{x_{n}}f_{\nu}(s)ds$ with $f_{\nu}$ from
(3.17) is $p(x)$-harmonic with exponent

$\check{p}(x)=1+Me^{-\int_{0}^{x_{n}}A(s)ds}\quad\text{if $\nu\in(0,1]$
and}\quad\check{p}(x)=1+Me^{\int_{0}^{x_{n}}A(s)ds}\quad\text{if $\nu>1$},$

whenever $M\in\mathbb{R}$ is a constant.

Suppose that $\nu\in(0,1]$ and $A^{-1}(s)=\lambda(s)$ is nonincreasing. Then
$\check{u}(x)$ is the slowest growing $p(x)$-harmonic function in the sense of
version (3.30) of Theorem 3.3. In particular, any $p(x)$-subharmonic function
with exponent $p(x)\in C^{1}(\mathbb{R}^{n}_{+})$, $1<p(x)<\infty$,
$\lambda(x_{n})\leq\frac{\min\\{1,p(x)-1\\}}{||Dp||_{\infty,x}}$ and
$\frac{\max\\{1,p(x)-1\\}}{||Dp||_{\infty,x}}\leq\Lambda(x_{n})$, satisfying

$\displaystyle\limsup_{x\to y}\,u(x)\,\leq\,0\quad\textrm{for all}\quad
y\in\partial\mathbb{R}^{n}_{+}$

that exceeds $\check{u}(x)$ somewhere on the $x_{n}$-axis satisfies

$\displaystyle\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{\check{u}(Re_{n})}=\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{\nu^{e^{\int_{0}^{R}\lambda^{-1}(s)ds}}}>0.$

Finally, if $\lambda$ is constant then

$\displaystyle\check{u}(x)=\lambda\left\\{\begin{array}[]{ll}-E_{i}\left(\log\nu\right)+E_{i}\left(e^{\lambda^{-1}x_{n}}\log\nu\right)&\text{if
$0<\nu<1$},\\\ x_{n}&\text{if $\nu=1$}\end{array}\right.$

where $E_{i}$ is the Exponential integral.

Proof. Since $\check{u}$ depends only on $x_{n}$ and solves (3.21) the first
statement follows if we prove that the variable exponent $p(x)$-Laplace
equation, with exponent $\check{p}(x)=1+Me^{\mp\int_{0}^{x_{n}}A(s)ds}$,
reduces to the PDE (3.21) in one dimension. Without derivatives in
$x^{\prime}$-direction we have

$\displaystyle\Delta_{p(x)}u(x)=(p(x)-1)u^{\prime\prime}_{x_{n}x_{n}}(x)+\log|u_{x_{n}}^{\prime}(x)|p^{\prime}_{x_{n}}(x)u_{x_{n}}^{\prime}(x)=0.$
(3.31)

Observe that exponent $\check{p}(x)$ is the unique family of
$C^{1}(\mathbb{R}_{+}^{n})$ solutions to the ODE

$p_{x_{n}}^{\prime}(x)=\mp(p(x)-1)A(x_{n})$

and substituting this equality into (3.31) yields
$u^{\prime\prime}_{x_{n}x_{n}}(x)\mp
A(x_{n})\log|u_{x_{n}}^{\prime}(x)|u_{x_{n}}^{\prime}(x)=0$ where the “$-$”
sign is for $\nu\in(0,1]$ when $\log|\check{u}_{x_{n}}^{\prime}(x)|<0$. Thus

$\displaystyle
u^{\prime\prime}_{x_{n}x_{n}}(x)+A(x_{n})|\log|u_{x_{n}}^{\prime}(x)||u_{x_{n}}^{\prime}(x)=0$

which is (3.21) in one dimension.

To prove the second statement we need to ensure that a weak subsolution of the
variable exponent $\check{p}$-Laplace equation,
$\check{p}(x)=1+Me^{-\int_{0}^{x_{n}}\lambda(s)^{-1}ds}$ for some $M$, of
which $\check{u}$ is a solution, is a viscosity subsolution of ($\star$ ‣ 1)
where ($\star\star$ ‣ 1) holds with the same $\Phi(s)$ and $\lambda(t)$ as in
version (3.30) of Theorem 3.3. To do so we observe that, recalling Lemma 3.2,
any $p(x)$-subharmonic function is viscosity solution of

$\displaystyle\Delta_{p(x)}u=\Delta u+(p(x)-2)\Delta_{\infty}u+\log|Du|\langle
Dp(x),Du\rangle\geq 0,$

and hence of

$\displaystyle|Dp||Du||\log|Du||+\max\\{1,p(x)-1\\}Tr((D^{2}u)^{+})-\min\\{1,p(x)-1\\}Tr((D^{2}u)^{-})\geq
0.$

Inserting $\check{p}(x)=1+Me^{-\int_{0}^{x_{n}}\lambda(s)^{-1}ds}$,
$|D\check{p}|=(\check{p}(x)-1)\lambda^{-1}(x_{n})$ and assuming that
$1<\check{p}(x)\leq 2$, which we may by taking $M\in(0,1]$, we see that

$\displaystyle|Du||\log|Du||+\frac{\lambda(x_{n})}{Me^{-\int_{0}^{x_{n}}\lambda^{-1}(s)ds}}Tr((D^{2}u)^{+})-\lambda(x_{n})Tr((D^{2}u)^{-})\geq
0.$

This is a PDE satisfying ($\star\star$ ‣ 1) with $\Phi(s)=s|\log s|$ and
$\lambda(t)$ as in version (3.30) of Theorem 3.3.

It remains to show that $\check{p}$ satisfies

$\lambda(x_{n})\leq\frac{\min\\{1,\check{p}(x)-1\\}}{||D\check{p}||_{\infty,x}}.$

This holds with equality since

$\check{p}(x)-1=Me^{-\int_{0}^{x_{n}}\lambda^{-1}(s)ds},\qquad||D\check{p}||_{\infty,x}=|\check{p}^{\prime}(x)|=-\lambda(x_{n})^{-1}Me^{-\int_{0}^{x_{n}}\lambda^{-1}(s)ds}$

and we have assumed $M\in(0,1]$. The proof is complete. $\hfill\Box$

### The case $\Phi(s)=-s|\log s|$

In this case the ODE (2.1) becomes (we skip the $t$-dependence in $\Phi$ only
for simplicity)

$\frac{df}{dt}=\frac{f|\log
f|}{\Lambda(t)}-K\frac{\Lambda(t)}{\lambda(t)}f,\quad
t\in(0,R),\quad\text{with}\quad f(0)=\nu.$

By the same argument as in the former cases we replace this ODE by

$\frac{df}{dt}=\Lambda^{-1}(t)\left(f|\log f|-\widehat{K}f\right),$

which separates, and we obtain, with
$\widehat{K}=K\frac{\Lambda^{2}(R)}{\lambda(R)}$

$\displaystyle
f_{\nu,R}(t)=\left\\{\begin{array}[]{ll}e^{-\widehat{K}(1-e^{-\int_{0}^{t}\Lambda^{-1}(s)ds})}\nu^{e^{-\int_{0}^{t}\Lambda^{-1}(s)ds}}&\text{if
$0<\nu\leq 1$},\\\
e^{-\widehat{K}\left(e^{\int_{0}^{t}\Lambda^{-1}(s)ds}-1\right)}\nu^{e^{\int_{0}^{t}\Lambda^{-1}(s)ds}}&\text{if
$1<f(t)$}\end{array}\right.$ (3.34)

and

$\displaystyle
f_{\nu}(t)=\left\\{\begin{array}[]{ll}\nu^{e^{-\int_{0}^{t}\Lambda^{-1}(s)ds}}&\text{if
$0<\nu\leq 1$},\\\ \nu^{e^{\int_{0}^{t}\Lambda^{-1}(s)ds}}&\text{if
$1<\nu$}.\end{array}\right.$ (3.37)

The limits become, for $0<\nu\leq 1$,

$\displaystyle\liminf_{R\to\infty}\frac{M^{\prime}(R)}{f_{\nu}(R)}\geq\liminf_{R\to\infty}\frac{f_{\nu,R}(R)}{f_{\nu}(R)}\geq\lim_{R\to\infty}e^{-\widehat{K}}$

and we only need $\widehat{K}\precsim 1$. When $\nu>1$ we know that
$f_{\nu,R}$ in (3.34) stays above 1 if

$\left(\log\nu-\widehat{K}\right)e^{\int_{0}^{t}\Lambda^{-1}(s)ds}>-\widehat{K}$

which forces us to take $\widehat{K}<\log\nu$. Then

$\displaystyle\liminf_{R\to\infty}\frac{M^{\prime}(R)}{f_{\nu}(R)}\geq\liminf_{R\to\infty}\frac{f_{\nu,R}(R)}{f_{\nu}(R)}=\lim_{R\to\infty}e^{-\widehat{K}\left(e^{\int_{0}^{R}\Lambda^{-1}(s)ds}-1\right)}$

and we need also $\widehat{K}e^{\int_{0}^{R}\Lambda^{-1}(s)ds}\precsim 1$ to
achieve a growth estimate.

We have defined $\widehat{K}(R)=\frac{n\Lambda(R)^{2}}{\lambda(R)\gamma(R)}$
in this case. However, from (3.34) it can be shown that $f_{\nu,R}(t)$ is
nondecreasing if $\widehat{K}\leq|\log(\nu)|$. This means that

$\frac{df}{dt}=\Lambda^{-1}(t)\left(f|\log f|-\widehat{K}f\right)\geq 0,\qquad
t\in(0,R).$

Now, we let $f_{\nu,R}$ solve this ODE in place of (2.1) and in the proof of
Lemma 2.2 we replace (2) with

$\displaystyle\text{Tr}\left(\left(D^{2}V\right)^{+}\right)$
$\displaystyle=-\Lambda(\Xi)^{-1}\Phi\left(\Xi,f\left(\Xi\right)\right)-\Lambda(\Xi)^{-1}\widehat{K}f(\Xi)+\frac{n-1}{|(x^{\prime},x_{n}+\gamma)|}f(\Xi),$
$\displaystyle\text{Tr}\left(\left(D^{2}V\right)^{-}\right)$
$\displaystyle=0.$

By tracing the remaining part of the proof of Lemma 2.2 we realize that it is
enough to pick $\widehat{K}(R)=\frac{n\Lambda(R)}{\gamma(R)}$.

As in the former situation the solution of (2.4) with $\Phi(s)=-s|\log s|$ can
be calculated analytically when $\Lambda(t)=constant$:

$\displaystyle\check{u}(x)=\int_{0}^{x_{n}}f_{\nu}(s)ds=\Lambda\left\\{\begin{array}[]{ll}E_{i}\left(\log\nu\right)-E_{i}\left(e^{-\Lambda^{-1}t}\log\nu\right)&\text{if
$0<\nu<1$},\\\ x_{n}&\text{if $\nu=1$},\\\
-E_{i}\left(\log\nu\right)+E_{i}\left(e^{\Lambda^{-1}t}\log\nu\right)&\text{if
$1<\nu$}.\end{array}\right.$ (3.41)

See Figure 3 (lower row) for functions $f_{\nu}$ in (3.37) and $\check{u}$ in
(3.41).

Now, using the calculations above (3.27) we see that $\Delta_{p(x)u\geq 0}$
implies

$\displaystyle\max\\{1,p(x)-1\\}\text{Tr}(D^{2}u^{+})-\min\\{1,p(x)-1\\}\text{Tr}(D^{2}u^{-})+|Dp||Du|\cos\theta\log|Du|\geq
0$

where $\theta=\theta(x)$ is the angle between $Du$ and $Dp$. Assume
$|Dp||\cos\theta|>0$ and divide the PDE with this factor to obtain

$\displaystyle\Lambda(x_{n})\text{Tr}(D^{2}u^{+})-\lambda(x_{n})\text{Tr}(D^{2}u^{-})+\frac{\cos\theta}{|\cos\theta|}|Du|\log|Du|\geq
0$

where $\lambda(x_{n})\leq\frac{\min\\{1,p(x)-1\\}}{|Dp||\cos\theta|}$ and
$\Lambda(x_{n})\geq\frac{\max\\{1,p(x)-1\\}}{|Dp||\cos\theta|}$ for some
nonincreasing function $\lambda$ and nondecreasing function $\Lambda$.
Assuming $\cos\theta\log|Du|\leq 0$ leads to

$\displaystyle\Lambda(x_{n})\text{Tr}(D^{2}u^{+})-\lambda(x_{n})\text{Tr}(D^{2}u^{-})-|Du||\log|Du||\geq
0$

and we can apply the results from (3.37)-(3.41) and Lemma 3.2 to obtain:

###### Corollary 3.5

Suppose that $p(x)$ and $u$ are as in Theorem 3.3. Let $\theta=\theta(x)$ be
the angel between $Dp$ and $Du$ and assume that

$|Dp||\cos\theta|>0\qquad\text{and}\qquad\cos\theta\log|Du|\leq 0$

holds in $\mathbb{R}^{n}_{+}$ (in a suitable weak sense if $u$ is not $C^{1}$
with $|Du|\neq 0$). Then $u$ is a subsolution of an equation of type ($\star$
‣ 1) satisfying ($\star\star$ ‣ 1) with $\Phi(s)=-s|\log s|$,
$\lambda(x_{n})\leq\frac{\min\\{1,p(x)-1\\}}{|Dp||\cos\theta|}$ and
$\Lambda(x_{n})\geq\frac{\max\\{1,p(x)-1\\}}{|Dp||\cos\theta|}$ for some
nonincreasing function $\lambda$ and nondecreasing function $\Lambda$. If
$\frac{n\Lambda(R)}{|\log\nu|}<\gamma(R)$ and
$\check{u}(x)=\int_{0}^{x_{n}}f_{\nu}(s)ds$ with $f_{\nu}$ from (3.37) then
the following is true:

  * •

If $u\geq\check{u}$ somewhere on the $x_{n}$-axis, $\nu\in(0,1)$ then

$\displaystyle\liminf_{R\to\infty}\,\frac{M(R)}{R}>0.$

  * •

If $u\geq\check{u}$ somewhere on the $x_{n}$-axis, $\nu>1$ and
$\Lambda(R)e^{\int_{0}^{R}\Lambda^{-1}(s)ds}\precsim\gamma(R)$, then

$\displaystyle\liminf_{R\to\infty}\,\frac{M^{\prime}(R)}{\nu^{e^{\int_{0}^{R}\Lambda^{-1}(s)ds}}}>0$

which, if $\Lambda=constant$ and $E_{i}$ is the Exponential integral, implies

$\displaystyle\liminf_{R\to\infty}\,\frac{M(R)}{E_{i}\left(e^{\Lambda^{-1}R}\log\nu\right)-E_{i}\left(\log\nu\right)}>0.$

In the one dimensional case Corollary 3.5 shows that if we know that the
exponent $p(x)$ is increasing, $p^{\prime}>0$, and that the subsolution
satisfies $0<u^{\prime}<1$, then $\liminf_{R\to\infty}u(R)/R>0$, which is much
stronger than the growth estimates that can be derived from Theorem 3.3.
Similarly, if we know that $p^{\prime}<0$ and $1<u^{\prime}$ then
$\liminf_{R\to\infty}u(R)/{\nu^{e^{\int_{0}^{R}\Lambda^{-1}(s)ds}}}>0$. These
improvements can be visualized by comparing the right panels in Figure ; the
upper right panel corresponds to Theorem 3.3 while the lower right panel
corresponds to the results in Corollary 3.5.

.  .

Figure 3: Functions $f_{\nu}$ in (3.17) and $\check{u}$ in (3.25) (upper row),
$f_{\nu}$ in (3.37) and $\check{u}$ in (3.41) (lower row). In all simulations,
$A=\Lambda=1$.

##  4 Connections with nonlinear diffusion problems

We follow the presentation in Lundström [31] and let $u$ denote the density of
some quantity in equilibrium, let $\Omega$ be a domain and $E\subset\Omega$ be
a $C^{1}$-domain so that the divergence theorem can be applied. Due to the
equilibrium, the net flux of $u$ through $\partial E$ is zero, that is

$\displaystyle\oint\limits_{\partial
E}\langle\boldsymbol{F},\boldsymbol{n}\rangle\,ds\,=\,0,$

where $\boldsymbol{F}$ denotes the flux density, $\boldsymbol{n}$ the normal
to $\partial E$ and $ds$ is the surface measure. The divergence theorem gives

$\displaystyle\int\limits_{E}\nabla\cdot\boldsymbol{F}\,dx\,=\,\oint\limits_{\partial
E}\langle\boldsymbol{F},\boldsymbol{n}\rangle\,ds\,=\,0.$

Since $E$ was arbitrary, we conclude

$\displaystyle\nabla\cdot\boldsymbol{F}\,=\,0\quad\mbox{in}\quad\Omega.$ (4.1)

In many situations it is physically reasonable to assume that the flux vector
$\boldsymbol{F}$ and the gradient $\nabla u$ are related by a power-law of the
form

$\displaystyle\boldsymbol{F}\,=\,-c\,|Du|^{\,q}\,Du,$ (4.2)

for some factor $c$ and exponent $q$, which may depend on space as well. One
reason is that flow is usually from regions of higher concentration to regions
of lower concentration. From this assumption, with $q=p-2$, and from (4.1), we
obtain the $p$-Laplace equation

$\displaystyle\nabla\cdot(|Du|^{p-2}Du)\,=\,0\quad\mbox{in}\quad\Omega.$

The linear case $p=2$ in (4.2) arises as a physical law in the following: If
$u$ denotes a chemical concentration, then it is the well known Fick’s law of
diffusion, if $u$ denotes a temperature, then it is Fourier’s law of heat
conduction, if $u$ denotes electrostatic potential, it is Ohm’s law of
electrical conduction, and if $u$ denotes pressure, then it is Darcy’s law of
fluid flow through a porous media. A problem involving the nonlinear case
$p\neq 2$ is fast/slow diffusion of sandpiles, see Aronsson–Evans–Wu [4]. In
that case $p$ is very large and $u$ models the height of a growing sandpile.
If $|Du|>1+\delta$ for some $\delta>0$, then $|Du|^{p-2}$ is very large, and
hence the transport of sand is also large, and if $|Du|<1-\delta$, then
$|Du|^{p-2}$ is very small. Therefore, when adding sand particles to a
sandpile, they accumulate as long as the slope of the pile does not exceed
one. If the slope exceeds one, then the sand becomes unstable and instantly
slides. Other application in which (4.2) arises with $p\neq 2$ is Hele-Shaw
flow of power-law fluids (Aronsson–Janfalk [5], Fabricius–Manjate–Wall [14])
and electro-rheological fluids (Harjulehto–Hästö–Lê–Nuortio [17]). When
properties of the quantity under investigation depends on space we may model
it by a variable exponent $p=p(x)$ in (4.2) and thus enter equations of type
(3.26) studied in Section 3.

We will now discuss the problem under investigation from the point of a
diffusion problem. Indeed, we will briefly explain, through spatially
dependent diffusion, why parts of our results presented in Theorem 3.3 holds
true. Suppose that $u$ denotes the density of some quantity at equilibrium in
the $n$-dimensional halfspace $\\{x_{n}>0\\}$ and that (4.2) holds with a
variable exponent $p(x_{n}),1<p(x_{n})<\infty$. Assume also that $u=0$ on the
boundary $x_{n}=0$, and at some $x_{n}=a>0$ we assume that $u(x)>0$. We
conclude that then $u$ satisfies the $p(x)$-Laplace equation (3.26) in the
halfspace and that our results applies. We simplify by further assuming that
concentration $u(x)$ is independent of $x^{\prime}$-directions. Since $|Du|$
must be positive there is a flux of $u$, independent of $x^{\prime}$, flowing
perpendicular through the plane at $x_{n}=a$ toward the boundary $x_{n}=0$.
Due to the equilibrium, the flux must be independent also of $x_{n}$ and is
therefore constant through the halfspace. Since the problem is herefrom
independent of $x^{\prime}$, we drop the index and write in the following
$x=x_{n}$.

Figure 4: Examples of how the concentration $u$ may depend on $x$ for
decreasing exponents (left) and increasing exponent (right). The slope
explodes or vanish as $p(x)\to 1$: The green dashed curves correspond to an
exponent $p(x)$ that approaches 1 as $x$ increases (left) and as $x\to 0$
(right). The slope approaches 1 as $p(x)\to\infty$, i.e. fast/slow diffusion:
The red solid curves correspond to an exponent $p(x)$ that becomes very large
as $x\to 0$ (left) and as $x$ increases (right).

Suppose that $p(x)$ is decreasing. As the flux of $u$, given by assumption
(4.2), is constant, the concentration $u$ must be convex (upwards) if
$|Du|=u^{\prime}>1$. Indeed, if $u^{\prime}>1$ near the boundary we locally
have that (4.2) yields flux ${\bf F}=-c\,(u^{\prime})^{p(x)-1}$ and since
$p(x)-1>0$ is decreasing it follows that $u^{\prime}$ must be increasing. A
similar reasoning explains that if $u^{\prime}=1$ somewhere then the flux
${\bf F}=c$ implying $u(x)=x$, and if $u^{\prime}<1$ then $u$ must be concave.
Figure 4(left) shows examples of how the concentration $u(x)$ may depend on
$x$ for two different decreasing exponents. We remark that if $p(x)$ becomes
very large near the boundary then $u^{\prime}$ must be very close 1 there,
otherwise the flux becomes zero or infinity – that is fast/slow diffusion (red
solid curve). Similarly, if $p(x)$ comes close to 1 as we move into the domain
then $u^{\prime}$ must grow fast if $u^{\prime}$ ever was larger than 1 along
the curve in order to keep the flux constant (green dashed curve). Finally, we
realize that if $p(x)$ becomes constant then $u^{\prime}$ becomes constant
(recall that $u(x)=cx$ is $p$-harmonic when $p=constant$).

Suppose now instead that $p(x)$ is decreasing. Reasoning as in the former case
we realize that we may switch our conclusions made near the boundary in the
former case with those made further away into the domain. Thus fast/slow
diffusion may occur away from the boundary and in such a case the slope of
$u(x)$ must approach 1. If $p(x)$ approaches 1 near the boundary then
$u^{\prime}$ must explode there, see Figure 4(right).

Returning to (3.25), (3.41) and (3.31) in Section 3 we find that the one-
dimensional $p(x)$-Laplace equation yields

$\displaystyle\Delta_{p(x)}u(x)=(p(x)-1)u^{\prime\prime}(x)+\log|u^{\prime}(x)|p^{\prime}(x)u^{\prime}(x)=0,$

and with the decreasing exponent

$p(x)=1+Me^{-Ax},$

where $M>0,A>0$ are a constants, the solution yields

$\displaystyle
u(x)=\frac{1}{A}\left\\{\begin{array}[]{ll}-E_{i}\left(\log\nu\right)+E_{i}\left(e^{Ax}\log\nu\right)&\text{if
$\nu\neq 1$},\\\ x&\text{if $\nu=1$}.\end{array}\right.$ (4.5)

Similarly, with the increasing exponent

$p(x)=1+Me^{Ax}$

the solution yields

$\displaystyle
u(x)=\frac{1}{A}\left\\{\begin{array}[]{ll}E_{i}\left(\log\nu\right)-E_{i}\left(e^{-Ax}\log\nu\right)&\text{if
$\nu\neq 1$},\\\ x&\text{if $\nu=1$}.\end{array}\right.$ (4.8)

With $A=\lambda=\Lambda=1$, solution curves for decreasing exponent in (4.5)
are plotted in Figure 3(upper right) (below line $u=x$) and (lower right)
(above line $u=x$), and solution curves for increasing exponent in (4.8) are
plotted in Figure 3(upper right) (above line $u=x$) and (lower right) (below
line $u=x$). Compare the structure of these curves to those in Figure 4 with
properties of the exponent $p(x)$ in mind.

Acknowledgement. This work was partially supported by the Swedish research
council grant 2018-03743.

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