On the localization and numerical computation of positive radial solutions for φ-Laplace equations in the annulus

Abstract


The paper deals with the existence and localization of positive radial solutions for stationary partial differential equations involving a general ϕ-Laplace operator in the annulus. Three sets of boundary conditions are considered: Dirichlet–Neumann, Neumann–Dirichlet and Dirichlet–Dirichlet. The results are based on the homotopy version of Krasnosel’skiıs fixed point theorem and Harnack type inequalities, first established for each one of the boundary conditions. As a consequence, the problem of multiple solutions is solved in a natural way. Numerical experiments confirming the theory, one for each of the three sets of boundary conditions, are performed by using the MATLAB object-oriented package Chebfun.

Authors

Jorge Rodríguez-López
CITMAga & Departamento de Estatística, Análise Matemática e Optimización, Universidade de
Santiago de Compostela, 15782, Facultade de Matemáticas, Campus Vida, Santiago, Spain

Radu Precup
Faculty of Mathematics and Computer Science and Institute of Advanced Studies in Science and Technology, Babes-Bolyai University, Cluj-Napoca, Romania
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy

Călin-Ioan Gheorghiu
Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy

Keywords

ϕ-Laplace operator; radial solution; positive solution; fixed point index; Harnack type inequality; numerical solution.

Paper coordinates

J. Rodriguez-Lopez, R. Precup, C.-I. Gheorghiu, On the localization and numerical computation of positive radial solutions for φ-Laplace equations in the annulus, Electronic Journal of Qualitative Theory of Differential Equations, 2022, no. 47, pp. 1-22, doi.org/10.14232/ejqtde.2022.1.47

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Electronic Journal of Qualitative Theory of Differential Equations

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University of Szeged

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ISSN 14173875

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On the localization and numerical computation of positive radial solutions for ϕ-Laplace equations in the annulus

On the localization and numerical computation of positive radial solutions for ϕ-Laplace equations in the annulus

Jorge Rodríguez-López J. Rodríguez-López, Departamento de Estatística, Análise Matemática e Optimización, Instituto de Matemáticas, Universidade de Santiago de Compostela, 15782, Facultade de Matemáticas, Campus Vida, Santiago, Spain jorgerodriguez.lopez@usc.es Radu Precup R. Precup, Faculty of Mathematics and Computer Science and Institute of Advanced Studies in Science and Technology, Babeş-Bolyai University, 400084 Cluj-Napoca, Romania & Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, P.O. Box 68-1, 400110 Cluj-Napoca, Romania r.precup@math.ubbcluj.ro  and  Călin-Ioan Gheorghiu C.-I. Gheorghiu, Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy, P.O. Box 68-1, 400110 Cluj-Napoca, Romania ghcalin@ictp.acad.ro
Abstract.

The paper deals with the existence and localization of positive radial solutions for stationary partial differential equations involving a general ϕ-Laplace operator in the annulus. Three sets of boundary conditions are considered: Dirichlet-Neumann, Neumann-Dirichlet and Dirichlet-Dirichlet. The results are based on the homotopy version of Krasnosel’skiĭ’s fixed point theorem and Harnack type inequalities, first established for each one of the boundary conditions. As a consequence, the problem of multiple solutions is solved in a natural way. Numerical experiments confirming the theory, one for each of the three sets of boundary conditions, are performed by using the MATLAB object-oriented package Chebfun.

Key words and phrases:
ϕ-Laplace operator, radial solution, positive solution, fixed point index, Harnack type inequality, numerical solution
1991 Mathematics Subject Classification:
35J60, 34B18

1. Introduction

In the short but clever paper [19], Hayan Wang solved the problem of existence of positive radial solutions for the semilinear elliptic equation

Δw+g(|x|)f(w)=0,R1<|x|<R2,xN,N2,

with one of the following sets of boundary conditions,

(1.2) w=0on |x|=R1and|x|=R2,
(1.4) w=0on |x|=R1andw/r=0on |x|=R2,
(1.6) w/r=0on |x|=R1andw=0on |x|=R2,

where r=|x| and w/r denotes differentiation in the radial direction, and 0<R1<R2<+.

The tools were a Krasnosel’skiĭ type fixed point theorem in cones and the property of bilateral boundedness of the corresponding Green functions. The first one is based on the fixed point index theory, while the second, as shown in [13], on Harnack type inequalities. Since then, many authors have considered the problem of radial solutions for equations and systems involving the Laplacian or some of its generalizations, various boundary conditions and domains, by using different topological or variational methods. We refer the interested reader to some of these contributions [1, 2, 3, 6, 5, 8] and the references therein.

Most of the works that followed deviated from the spirit of the original ideas. On their line, we mention our recent papers [14], [15] and [16]. It is the scope of the present paper to complement them, as close as possible to paper [19], for the case of equations with a general ϕ-Laplacian. Here in the absence of a Green function we are forced to produce Harnack type inequalities for each set of boundary conditions.

More exactly, in this paper, we deal with the existence, localization and multiplicity of positive radial solutions to the Dirichlet problem involving ϕ-Laplacian operators:

(1.7) div(ψ(|w|)w)=g(|x|)f(u) R1<|x|<R2,xN,N2,

where 0<R1<R2<+, the functions g:[R1,R2]+,f:++ are continuous and ψ:(a,a) is such that ϕ(s):=sψ(s) is an increasing homeomorphism between two intervals (a,a) and (b,b) (0<a,b+).

The following particular cases are of much interest due to their corresponding models arising from physics:

(a) ϕ:,ϕ(s)=|s|p2s, where p>1(here a=b=+), when the left side L0w in (1.7) is

L0w=div(|w|p2w)(p-Laplace operator),

involved in a nonlinear Darcy law for flows through porous media;

(b) (singular homeomorphism) ϕ:(a,a),ϕ(s)=sa2s2 (here 0<a<+ and b=+), when

L0w=div(wa2|w|2)(Minkowski mean curvature operator),

arised from the relativistic mechanics;

(c) (bounded homeomorphism) ϕ:(b,b),ϕ(s)=bs1+s2 (here a=+ and 0<b<+), when

L0w=bdiv(w1+|w|2)(Euclidian mean curvature operator),

associated to capillarity problems.

Looking for radial solutions of (1.7), that is, functions of the form w(x)=v(r) with r=|x|, (1.7) reduces to the ordinary differential equation

(1.8) L(v):=r1N(rN1ϕ(v))=g(r)f(v),R1<r<R2,

while boundary conditions (1.2) become

(1.10) v(R1)=0andv(R2)=0,
(1.12) v(R1)=0andv(R2)=0,
(1.14) v(R1)=0andv(R2)=0.

2. Harnack type inequalities

Originally, Harnack’s inequality was introduced in order to give estimates from above and from below for the ratio u(x)/u(y) of two values of a positive harmonic function. It can be generically put under the form

minxωu(x)kmaxxωu(x),

where k is a positive constant depending on the subdomain ω. Next it was generalized to nonnegative solutions or supersolutions of a wide class of linear elliptic equations. For the origin of the notion and many references, we refer the reader to [11].

More general, we speak about a Harnack type inequality whenever for a given operator L acting on a space of functions defined on a set Ω and endowed with a norm , there is a subdomain ωΩ and a constant k>0 such that

minxωu(x)ku

for all nonnegative functions u satisfying L(u)0 and eventually some additional behavior properties. In [12], Harnack inequalities have been put in connexion with the compression-expansion method of Krasnosel’skiĭ for the localization of positive solutions of nonlinear problems. In case of boundary value problems for ordinary differential equations, when a Green function is known, a Harnack inequality immediately can be derived using the bilateral estimates of the Green function. However, Harnack inequalities can be obtained even for differential operators for which a Green function does not exist. This is the case of the ϕ-Laplace operators. Deduction of such inequalities requires a fine analysis and makes use of priority properties of solutions such as monotony and concavity (see, e.g., [9] and [10]). The analysis is even more difficult in the case of radial solutions. It is the goal of this section to obtain Harnack inequalities for ϕ-Laplace operators subject to each of the three boundary conditions (1.10), (1.12), (1.14).

2.1. Case of the boundary conditions (1.12)

Theorem 2.1.

If vC1[R1,R2] is such that (1.12) are satisfied, rN1ϕ(v) is differentiable and L(v)(r)[0,b(R1/R2)N1) for all r[R1,R2], then v is nonnegative, increasing and concave. In addition, for any c(R1,R2), one has

(2.1) v(c)cR1R2R1v(R2).
Proof.

Let h:=L(v). Integrating from r to R2 and taking into account that v(R2)=0 yields

v(r)=ϕ1(r1NrR2τN1h(τ)𝑑τ).

A new integration, this time from R1 to r gives the expression of the solution, namely

v(r)=R1rϕ1(s1NsR2τN1h(τ)𝑑τ)𝑑s

and of the associated solution operator

S(h)(r)=R1rϕ1(s1NsR2τN1h(τ)𝑑τ)𝑑s.

Since h0, these formulas show that v is nonnegative and increasing. Also v is decreasing, i.e., v is concave. Finally, the concavity implies that the graph of v is over the line joining the point (R1,0) and (R2,v(R2)), whose equation is y=v(R2)R2R1(xR1). Taking x=c gives (2.1). ∎

Note that under the assumptions of Theorem 2.1, in (2.1), one has v(c)=minr[c,R2]v(r) and v(R2)=maxr[R1,R2]v(r)=v. Hence,

minr[c,R2]v(r)k1v,

with k1=(cR1)/(R2R1).

2.2. Case of the boundary conditions (1.14)

Theorem 2.2.

If vC1[R1,R2] is such that (1.14) are satisfied, rN1ϕ(v) is differentiable, L(v)(r)[0,b) and L(v) is increasing on [R1,R2], then v is nonnegative, decreasing and concave. In addition, for any c(R1,R2), one has

(2.2) v(c)R2cR2R1v(R1).
Proof.

If we let h=L(v), then by integration we obtain

v(r)=ϕ1(r1NR1rτN1h(τ)𝑑τ)

and

S(h)(r)=v(r)=rR2ϕ1(s1NR1sτN1h(τ)𝑑τ)𝑑s.

Since h is nonnegative, these formulas immediately imply that v is nonnegative and decreasing.

To show that v is concave we need to prove that v is decreasing, equivalently, that the function η(r)=r1NR1rτN1h(τ)𝑑τ is increasing. Indeed, using the monotonicity of h, one has

η(r) = h(r)N1rNR1rτN1h(τ)𝑑τ
h(r)N1rNh(r)R1rτN1𝑑τ
= h(r)N1Nh(r)(1(R1r)N)
0.

Finally, since the graph of the concave function v is over the line joining the points (R1,v(R1)) and (R2,0), if c is any point in (R1,R2), we have (2.2). ∎

Note that under the assumptions of Theorem 2.2, in (2.2), one has v(c)=minr[R1,c]v(r) and v(R1)=maxr[R1,R2]v(r)=v. Therefore,

minr[R1,c]v(r)k2v,

where k2=(R2c)/(R2R1).

2.3. Case of the boundary conditions (1.10)

Theorem 2.3.

For each function hL1(R1,R2) not identically zero satisfying h(r)0 a.e. on(R1,R2) and hL1<b(R1/R2)N1, the equation L(v)=h endowed with the boundary conditions (1.10) has a unique nonzero nonnegative solution v which is concave and such that for any c[0,(R2R1)/2), one has:

(2.3) minr[c1,c2]v(r)1R2R1(R2R12c)v,

where c1=Rmc, c2=Rm+c and Rm=(R1+R2)/2.

Proof.

Let v be a nonnegative solution. Since h is not identically zero, v is nonzero and since it vanishes at R1 and R2, any maximum point R is interior and so v(R)=0. Integrating from R to r then gives

(2.4) v(r)=ϕ1(r1NRrτN1h(τ)𝑑τ).

This shows that v is decreasing on [R1,R2]. Hence v is concave on [R1,R2]. Let R be such that v(R)=v=maxr[R1,R2]v(r).First assume that RRm. The concavity of v implies that the graph of v restricted to [R,R2] is over the line joining the points (R,v(R)) and (R2,0) which at its turn is over the line joining the points (R1,v(R)),(R2,0), of equation y=v(R)R2R1(R2x). Thus, since c2[R,R2], we have

v(c2)v(R)R2R1(R2c2)=1R2R1(R2R12c)v(R).

In addition the graph of v on [R1,Rm] is over the line joining the points (R1,0),(R2,v(R)). Then

v(c1)v(R)R2R1(c1R1)=1R2R1(R2R12c)v(R).

As a result

minr[c1,c2]v(r)=min{v(c1),v(c2)}1R2R1(R2R12c)v.

The proof of the case RRm is similar.

Next, integration in (2.4) gives the representation formulas (for the solution operator)

(2.5) v(r)=R1rϕ1(s1NRsτN1h(τ)𝑑τ)𝑑s(r[R1,R2]),
v(r)=rR2ϕ1(s1NRsτN1h(τ)𝑑τ)𝑑s(r[R1,R2]).

To prove the existence of a solution, in virtue of (2.5) and (2.4) it is enough to prove the existence of a number R(R1,R2) such that

R1R2ϕ1(s1NRsτN1h(τ)𝑑τ)𝑑s=0.

This immediately follows since the continuous function

tR1R2ϕ1(s1NtsτN1h(τ)𝑑τ)𝑑s(t[R1,R2])

takes values of opposite sign at the ends R1 and R2.

To prove the uniqueness of the solution, assume that v1 and v2 are two nonnegative solutions and let R,R′′ be two of their maximum points, respectively. Using the representation formula (2.4) it is easy to see that (v2v1) preserves its sign on the whole interval (R1,R2), positive or negative depending on the ordering between R and R′′. Thus v2v1 is monotone and being zero at the ends of the interval it must be identically zero. Hence v1=v2.

3. Existence and localization

As mentioned above, the key ingredient together with Harnack inequalities to obtain positive solutions in this paper will be the fixed point index in cones. In particular, we recall the well–known homotopy version of Krasnosel’skiĭ fixed point theorem in cones.

Theorem 3.1 (Krasnosel’skiĭ).

Let X be a Banach space, K a cone of X and Ω1 and Ω2 two relatively open and bounded subsets of K with 0Ω1Ω¯1Ω2. Let N:KK be a completely continuous operator satisfying one of the following two conditions:

(i):

λuN(u) for all uKΩ1 and all λ1; and there exists hK{0} such that uN(u)+λh for all uKΩ2 and all λ0.

(ii):

λuN(u) for all uKΩ2 and all λ1; and there exists hK{0} such that uN(u)+λh for all uKΩ1 and all λ0.

Then N has a fixed point uK such that uΩ2Ω¯1.

In the sequel, consider the Banach space of continuous functions C[R1,R2] endowed with the usual maximum norm v=maxr[R1,R2]|v(r)| and denote by P the cone of nonnegative functions in C[R1,R2].

3.1. Case of the boundary conditions (1.12)

By a solution of (1.8)–(1.12) we mean a function vC1[R1,R2] with v(R1)=0=v(R2) such that v(a,a), rn1ϕ(v)W1,1[R1,R2] and satisfies (1.8). We will look for nonnegative nontrivial solutions on [R1,R2].

It is clear that v is a nonnegative solution of (1.8)–(1.12) if and only if v is a fixed point of the operator

T1(v)(r)=R1rϕ1(s1NsR2τN1g(τ)f(v(τ))𝑑τ)𝑑s.

If f is such that f(s)R1R2g(τ)𝑑τ<(R1R2)N1b for all s+, then T1 is defined on the whole cone P and T1(P)P. Moreover, T1 is completely continuous as follows from the Arzelà–Ascoli theorem.

Here, for a fixed c(R1,R2), we will look for fixed points of the operator T1 in a subcone of P, namely,

K1={vP:minr[c,R2]v(r)k1v},

where k1=(cR1)/(R2R1). By the Harnack inequality given by Theorem 2.1, it is easy to check that the operator T1 maps the cone K1 into itself.

Now, for any numbers α,β>0, consider the open (in K1) sets

(3.1) Vα:={vK1:v<α}

and

(3.2) Wβ:={vK1:minr[c,R2]v(r)<β}.

Note that VβWβVβ/k1, so Wβ is bounded.

We are in the position to apply Theorem 3.1 in order to obtain existence and localization results for problem (1.8)–(1.12). In this way, we localize a solution in the set WβV¯α if β>α and in the set VαW¯β if α>β/k1.

We will use the following notations:

A:=R1R2g(τ)𝑑τandB:=cR2g(τ)𝑑τ.

Also, for any α,β>0, we denote

Mα:=max{f(s):s[0,α]}andmβ:=min{f(s):s[β,β/k1]}.
Theorem 3.2.

Assume that

(3.3) f(s)<bA(R1R2)N1for all s+.

In addition assume that there exist α,β>0 such that

(3.4) (R2R1)ϕ1((R2/R1)N1AMα) <α,
(3.5) (cR1)ϕ1(Bmβ) >β.
(10):

If α<β, then problem (1.8)–(1.12) has a positive solution v such that α<v<β/k1.

(20):

If α>β/k1, then problem (1.8)–(1.12) has a positive solution v such that β<v<α.

Proof.

We shall apply Theorem 3.1. First, let us see that

T1(v)<αfor all vK1 with v=α,

which clearly implies that λvT1(v) for all vK1Vα and all λ1. Indeed, for vK1 with v=α, we have that f(v(s))Mα and so from (3.4) it follows that

T1(v) =R1R2ϕ1(s1NsR2τN1g(τ)f(v(τ))𝑑τ)𝑑s
R1R2ϕ1(Mαs1NR1R2τN1g(τ)𝑑τ)𝑑s
(R2R1)ϕ1((R2/R1)N1AMα)<α,

as wished.

On the other hand, let us prove that vT1(v)+λh for all vK1Wβ and all λ0 with h1. Notice that for vK1 with minr[c,R2]v(r)=β, we have that βv(r)β/k1 for all r[c,R2], and thus mβf(v(r)) for all r[c,R2]. Hence, for any r[c,R2],

T1(v)(r) R1cϕ1(s1NsR2τN1g(τ)f(v(τ))𝑑τ)𝑑s
R1cϕ1(s1NcR2τN1g(τ)f(v(τ))𝑑τ)𝑑s
(cR1)ϕ1(Bmβ).

Now, (3.5) implies that T1(v)(r)>β=minr[c,R2]v(r) for all r[c,R2], which clearly ensures that vT1(v)+λ for all vK1Wβ and all λ0.

Now, if α<β, then V¯αWβ, so Theorem 3.1 guarantees that the operator T1 has at least a fixed point in WβV¯αVβ/k1V¯α. But if one has α>β/k1, then WβVβ/k1V¯β/k1Vα and thus Theorem 3.1 implies that the operator T1 has at least a fixed point located in VαW¯βVαV¯β.

Note that condition (3.3) trivially holds if b=+. Obviously, if ϕ is a classical or a bounded homeomorphism, i.e., if a=+, then conditions (3.4) and (3.5) can be rewritten as

Mαϕ(C1α)<C2andmβϕ(C3β)>C4,

with suitable positive constants C1, C2, C3 and C4 as come from (3.4)-(3.5).

Hence, if we are only interested on the existence and not on the localization of the solutions, we can establish sufficient conditions for the existence of the numbers α and β satisfying the inequalities above. They are given by asymptotic conditions on the ratio f/ϕ at 0 and at infinity.

Theorem 3.3.

Assume that the following conditions are satisfied:a=+,

(3.6) lim supx0ϕ(τx)ϕ(x)<+,lim supx+ϕ(x)ϕ(τx)<+for all τ>0

and

f0:=limx0+f(x)ϕ(x)=+,f:=limx+f(x)ϕ(x)=0.

Then problem (1.8)–(1.12) has at least one positive solution.

Proof.

First we show that there exists β>0 such that

(3.7) mβ>C4ϕ(C3β).

By (3.6), with τ=C3, there exist L>0 and ρ>0 such that

LC4ϕ(x)>ϕ(C3x)for all x(0,ρ).

Now, since f0=+, there exists τ>0 (we may assume τ<ρ) such that

f(x)Lϕ(x)for all x(0,τ].

Hence, the fact that ϕ is increasing implies that

minx[τk1,τ]f(x)Lϕ(τk1).

Then, taking β=τk1, one has

mβ:=minx[β,β/k1]f(x)Lϕ(β)>C4ϕ(C3β),

and so (3.7) holds.

Secondly, we prove that there exists α>β/k1 such that

(3.8) Mα<C2ϕ(C1α).

By (3.6), with τ=C1, there exist L~>0 and ρ~>0 such that

C2L~ϕ(C1x)>ϕ(x)for all x(ρ~,+).

Since f=0, there exists σ>0 such that

f(x)σ+12L~ϕ(x)for all x0.

Now, it follows from the fact that ϕ: is an increasing unbounded homeomorphism that there exists α>0 such that 2L~σϕ(α). Thus,

f(x)1L~ϕ(α)for all x[0,α],

and so

Mα:=maxx[0,α]f(x)1L~ϕ(α)<C2ϕ(C1α),

that is, (3.8) holds.

Finally, the conclusion follows from Theorem 3.2. ∎

Similarly, an existence result can be obtained if f is sublinear at 0 and superlinear at infinity with respect to ϕ.

Theorem 3.4.

Assume that ϕ is a classical homeomorphism such that

(3.9) lim supx0ϕ(x)ϕ(τx)<+,lim supx+ϕ(τx)ϕ(x)<+for all τ>0

and f satisfies

f0=0,f=+.

Then problem (1.8)–(1.12) has at least one positive solution.

Remark 3.5.

Note that if ϕ is bounded, then condition f=+ is not possible, since limx+ϕ(x)=b and f must be bounded.

Note that if ϕ is singular (i.e., a<+, b=+), then condition (3.4) is trivially satisfied for α large enough and so the existence of a positive solution for problem (1.8)–(1.12) is ensured provided that there exists a positive number β satisfying (3.5). This holds if f is superlinear at 0 with respect to ϕ, i.e., f0=+. Thus we have

Theorem 3.6.

Assume that ϕ is a singular homeomorphism such that

(3.10) lim supx0ϕ(τx)ϕ(x)<+for all τ>0

and f satisfies

f0=+.

Then problem (1.8)–(1.12) has at least one positive solution.

Obviously, the localization of solutions given by Theorem 3.2 allows us to derive multiplicity results provided that there exist several couples of positive numbers (α,β) satisfying assumptions (3.4)–(3.5). Some conclusions in this line are collected in the following

Theorem 3.7.

Assume that condition (3.3) holds.

  1. (1)

    Let (αi)1ik, (βi)1ik (k) be sets of positive numbers with αi<βik1αi+1 for each i. If the assumptions of Theorem 3.2 hold for each couple (αi,βi), then problem (1.8)–(1.12) has k different solutions vi such that αi<vi<βi/k1.

  2. (2)

    Let (αi)1ik, (βi)1ik (k) be sets of positive numbers with αi<βi<k1αi+1 for each i. If the assumptions of Theorem 3.2 hold for each couple (αi,βi), then problem (1.8)–(1.12) has 2k1 different solutions vi, wj (i=1,,k,j=1,,k1) such that

    αi<vi,minr[c,R2]vi(r)<βi and minr[c,R2]wj(r)>βj,wj<αj+1.
  3. (3)

    Let (αi)i, (βi)i be two sequences of positive numbers with αi<βik1αi+1 for each i. If the assumptions of Theorem 3.2 hold for each couple (αi,βi), then problem (1.8)–(1.12) has infinitely many different solutions vi such that αi<vi<βi/k1.

Proof.

Let us prove cases (1) and (2).

  1. (1)

    For each i, since αi<βi, Theorem 3.2 ensures that problem (1.8)–(1.12) has a positive solution vi such that αi<vi<βi/k1. Now, it suffices to remark that βi/k1αi+1 implies that vi<vi+1, so there exist at least k different such solutions.

  2. (2)

    For each i, since αi<βi, we can derive from the proof of Theorem 3.2 a better localization result: the solution vi belongs to the set WβiV¯αi, that is,

    αi<vi,minr[c,R2]vi(r)<βi.

    On the other hand, for each j{1,,k1}, since βj<k1αj+1, Theorem 3.2 also implies that problem (1.8)–(1.12) has a positive solution wj located in the set Vαj+1W¯βj. Thus,

    minr[c,R2]wj(r)>βj,wj<αj+1.

    Since αj+1<βj+1, one has that wj<wj+1. Finally, the estimations

    minr[c,R2]vn(r)<βj<minr[c,R2]wj(r) and wj<αj+1<vm,

    for n{1,,j} and m{j+1,,k}, show that wj is also distinct from any vi and so problem (1.8)–(1.12) has at least 2k1 different solutions.

The proof of case (3) is analogous and thus we omit it. ∎

3.2. Case of the boundary conditions (1.14)

By a solution of (1.8)–(1.14) we mean a function vC1[R1,R2] with v(R1)=0=v(R2) such that v(a,a), rn1ϕ(v)W1,1[R1,R2] and satisfies (1.8). We will look for nonnegative nontrivial solutions on [R1,R2].

It is clear that v is a nonnegative solution of (1.8)–(1.14) if and only if v is a fixed point of the operator T2:PP defined as

T2(v)(r)=rR2ϕ1(s1NR1sτN1g(τ)f(v(τ))𝑑τ)𝑑s,

which is a completely continuous operator.

Let us assume that the functions f and g satisfy the following monotonicity assumptions:

  1. (Hf)

    f is decreasing on + and 0<f(0)<bR1R2g(τ)𝑑τ;

  2. (Hg)

    rN1g(r) is increasing on [R1,R2].

For a fixed c(R1,R2), we consider the following subcone of P:

K2={vP:v is decreasing and minr[R1,c]v(r)k2v},

where k2=(R2c)/(R2R1).

Note that the operator T2 maps the cone K2 into itself. Indeed, take vK2 and let us show that w:=T2(v) belongs to K2. Since f and g are nonnegative, then w is nonnegative and decreasing. Moreover, the monotonicity assumptions on f and g given by (Hf) and (Hg) together with the fact that v is decreasing imply that the function rrN1g(r)f(v(r)) is increasing. Thus,

rL(w)(r)=rN1g(r)f(v(r))

is increasing on [R1,R2]. Then Theorem 2.2 ensures that w satisfies that

minr[R1,c]w(r)k2w,

so wK2.

For any numbers α,β>0, define the sets Vα and Wβ as in (3.1) and (3.2), with K2 instead of K1. Then the following existence and localization result for problem (1.8)–(1.14) can be proved as an application of Theorem 3.1, which guarantees the existence of a fixed point of T2 in WβV¯α or in VαW¯β.

We will use the following notation:

A:=R1R2g(τ)𝑑τandB:=R1cg(τ)𝑑τ.

Moreover, for any α,β>0, denote

Mα:=max{f(s):s[0,α]}=f(0)andmβ:=min{f(s):s[β,β/k2]}=f(β/k2).

It is obvious that the following result can be proved in a similar way to Theorem 3.2, so we omit the proof here.

Theorem 3.8.

Assume that conditions (Hf) and (Hg) hold and that there exist α,β>0 such that

(3.11) (R2R1)ϕ1(AMα) <α,
(3.12) (R2c)ϕ1((R1/R2)N1Bmβ) >β.
(10):

If α<β, then problem (1.8)–(1.14) has a positive solution v such that α<v<β/k2.

(20):

If α>β/k2, then problem (1.8)–(1.14) has a positive solution v such that β<v<α.

Remark 3.9.

If we take into account that f is decreasing, then conditions (3.11) and (3.12) can be rewritten as

(3.13) (R2R1)ϕ1(Af(0)) <α,
(3.14) (R2c)ϕ1((R1/R2)N1Bf(β/k2)) >β.

Note that condition (3.13) is always satisfied for α sufficiently large since the left-hand side in the inequality is independent of α. Furthermore, from the fact that f is continuous with f(0)>0, it follows that condition (3.14) holds for any β close enough to 0.

In view of Theorem 3.8 and Remark 3.9, it is clear that problem (1.8)–(1.14) is always solvable under assumptions (Hf) and (Hg). Thus we have

Corollary 3.10.

If conditions (Hf) and (Hg) hold, then problem (1.8)–(1.14) has at least one positive solution.

Remark 3.11.

Observe that multiplicity results cannot be derived from Theorem 3.8. Indeed, since AB and f is decreasing, one has

Af(0)(R1/R2)N1Bf(β/k2),

and so

(R2R1)ϕ1(Af(0))>(R2c)ϕ1((R1/R2)N1Bf(β/k2)).

Therefore, any α satisfying (3.13) must be bigger than any β for which (3.14) holds.

Remark 3.12.

Observe that the results contained in Section 3.2 remain valid for R1=0, i.e., in the ball.

Note that problem (1.8)–(1.14) with R1=0 and R2=1, that is, in the unit ball, was considered in [16], but the results are not comparable since there f was assumed to be nondecreasing.

3.3. Case of the boundary conditions (1.10)

By a solution of (1.8)–(1.10) we mean a function vC1[R1,R2] with v(R1)=0=v(R2) such that v(a,a), rn1ϕ(v)W1,1[R1,R2] and satisfies (1.8). We will look for nonnegative nontrivial solutions on [R1,R2].

To construct the fixed point operator, we need the following technical result, similar to Lemma 1 in [4].

Denote

Db={hP:hL1<b(R1R2)N1}.
Lemma 3.13.

For each function hDb, there exists R(R1,R2) such that

γ=R1RτN1h(τ)𝑑τ

is the unique number γ satisfying

s1N[R1sτN1h(τ)𝑑τγ](b,b)

and

R1R2ϕ1(s1N[R1sτN1h(τ)𝑑τγ])𝑑s=0.

Moreover, the function Qϕ:Db,Qϕ(h)=γ is continuous.

Proof.

The existence of R with the desired property follows from the proof of Theorem 2.3. Note that for any hDb, one has

s1N[R1sτN1h(τ)𝑑τγ](b,b)for all s[R1,R2].

Indeed

|s1N[R1sτN1h(τ)𝑑τγ]|=|s1NRsτN1h(τ)𝑑τ|(R2R1)N1hL1<b.

For uniqueness, assume that there exist γi (i=1,2) such that

R1R2ϕ1(s1N[R1sτN1h(τ)𝑑τγi])𝑑s=0.

Now, by the mean value theorem for integration, there exists s0[R1,R2] such that

ϕ1(s01N[R1s0τN1h(τ)𝑑τγ1])ds=ϕ1(s01N[R1s0τN1h(τ)𝑑τγ2])ds.

This clearly implies that γ1=γ2.

Finally, for the continuity of Qϕ, let {hn}nDb such that hnh0Db in C[R1,R2]. We may assume that Qϕ(hn)γ0. Passing to limit we find that

R1R2ϕ1(s1N[R1sτN1h0(τ)𝑑τγ0])𝑑s=0,

and so γ0=Qϕ(h0), as wished. ∎

In addition, the solution operator

S:DbC[R1,R2],S(h)(r)=R1rϕ1(s1N[R1sτN1h(τ)𝑑τQϕ(h)])𝑑s,

is monotone as shows the next lemma. The proof follows similar ideas to those in [10].

Lemma 3.14.

Let h1,h2Db, h1h2 a.e. on [R1,R2], and let v1,v2C1[R1,R2] be such that for i=1,2, one has vi(R1)=0=vi(R2) and

L(vi)(r)=hi(r)for r(R1,R2).

Then v1v2 on [R1,R2].

Proof.

Assume to the contrary that v1v2. Then there exists an interval [t1,t2], with R1t1<t2R2, such that v1<v2 on (t1,t2) and v1(ti)=v2(ti), i=1,2. Hence, by the mean value theorem, there exists R(t1,t2) such that (v1v2)(R)=0. Then

ϕ(v2)(r)ϕ(v1)(r)=1rN1RrsN1[h1(s)h2(s)]𝑑s.

Since h1h2, we deduce that ϕ(v2)(r)ϕ(v1)(r)0 on (R,t2). Thus, v2(r)v1(r) on (R,t2) which joint with v2(t2)=v1(t2) imply v1v2 on (R,t2), a contradiction. ∎

If f satisfies condition (3.3), then for each vP, the function h:=gf(v)Db and since h0, one has S(h)S(0)=0. Hence the operator

T3:PP,T3(v)=S(gf(v))

is well-defined. In addition, thanks to the continuity of Qϕ and the Arzelà-Ascoli theorem, it is completely continuous.

Notice that v is a nonnegative solution of (1.8)–(1.10) if and only if v is a fixed point of the operator T3. Here, for a fixed c(0,(R2R1)/2), we shall look for fixed points of the operator T3 in a subcone of P, namely,

K3={vP:minrIcv(r)k3v},

where k3=((R2R1)/2c)/(R2R1) and Ic=[Rmc,Rm+c]. By the Harnack inequality given by Theorem 2.3, it follows that the operator T3 maps the cone K3 into itself.

Now, for any numbers α,β>0, consider the relatively open sets

Vα:={vK3:v<α}andWβ:={vK3:minrIcv(r)<β}.

We will use the following notation:

A:=R1R2g(τ)dτandB:=min{RmcRmg(τ)dτ,RmRm+cg(τ)dτ.}.

Moreover, for any α,β>0, denote

Mα:=max{f(s):s[0,α]}andmβ:=min{f(s):s[β,β/k3]}.
Theorem 3.15.

Assume that f satisfies (3.3) and there exist α,β>0 such that

(3.15) (R2R1)ϕ1((R2R1)N1AMα) <α,
(3.16) k3(RmR1c)ϕ1(Bmβ) >β,
(3.17) k3(R2Rmc)ϕ1((Rm/R2)N1Bmβ) >β.
(10):

If α<β, then problem (1.8)–(1.10) has a positive solution v such that α<v<β/k3.

(20):

If α>β/k3, then problem (1.8)–(1.10) has a positive solution v such that β<v<α.

Proof.

We shall apply Theorem 3.1. First, let us show that

T3(v)<αfor all vK3 with v=α,

which clearly implies that λvT3(v) for all vK3Vα and all λ1. Indeed, for vK3 with v=α, we have that there exists R(R1,R2) such that T3(v)=T3(v)(R) and (T3(v))(R)=0. Thus,

T3(v)(r) = R1rϕ1(s1NRsτN1g(τ)f(v(τ))𝑑τ)𝑑s
= rR2ϕ1(s1NRsτN1g(τ)f(v(τ))𝑑τ)𝑑s(r[R1,R2]).

Since f(v(s))Mα for every s[R1,R2] and S is monotone, we have

T3(v)=T3(v)(R) =S(gf(v))(R)S(gMα)(R)=R1Rϕ1(s1NMαRsτN1g(τ)𝑑τ)𝑑s
=R1Rϕ1(s1NMαsRτN1g(τ)𝑑τ)𝑑s
(R2R1)ϕ1((R2R1)N1AMα)<α,

as wished.

On the other hand, let us prove that vT3(v)+λh for all vK3Wβ and all λ0 with h1. Notice that for vK3 with minrIcv(r)=β, we have that βv(r)β/k3 for all rIc, and thus mβf(v(r)) for all rIc. Hence, f(v(r))mβχIc(r) for all r[R1,R2] (where χIc denotes the characteristic function of Ic). Then Lemma 3.14 implies that

T3(v)(r)S(mβgχIc)(r),(r[R1,R2]).

Note that there is R(R1,R2) such that

S(mβgχIc)(r)=R1rϕ1(s1NmβRsτN1g(τ)χIc(τ)𝑑τ)𝑑s.

Now, suppose that RRm. Then

T3(v)(Rmc) S(mβgχIc)(Rmc)
=R1Rmcϕ1(s1NmβRmcRτN1g(τ)χIc(τ)𝑑τ)𝑑s
R1Rmcϕ1(s1NmβRmcRmτN1g(τ)χIc(τ)𝑑τ)𝑑s
(RmR1c)ϕ1(Bmβ)
>β/k3,

that is, T3(v)(Rmc)>β/k3v(r) for all rIc. In particular, T3(v)(Rmc)>v(Rmc).

Analogously, if RRm, then

T3(v)(Rm+c) S(mβgχIc)(Rm+c)
= Rm+cR2ϕ1(s1NmβRsτN1g(τ)χIc(τ)𝑑τ)𝑑s
= Rm+cR2ϕ1(s1NmβRmRm+cτN1g(τ)χIc(τ)𝑑τ)𝑑s
(R2Rmc)ϕ1((Rm/R2)N1Bmβ)
> β/k3.

we may prove that T3(v)(Rm+c)>β/k3v(r) for all rIc.

Therefore, vT3(v)+λ for all vK3Wβ and all λ0. The conclusion follows from Theorem 3.1. ∎

Remark 3.16.

If ϕ is odd then the two conditions (3.16)and (3.17) on β reduce to the unique inequality

(3.18) k3(R2Rmc)ϕ1((Rm/R2)N1Bmβ)>β.

We emphasize that if ϕ is a classical or bounded odd homeomorphism, then conditions (3.15) and (3.18) can be rewritten as

Mαϕ(C1α)<C2andmβϕ(C3β)>C4,

for certain positive constants C1, C2, C3 and C4. Therefore, existence results for sublinear and superlinear nonlinearities can be proven exactly as in Section 3.1.

Theorem 3.17.

Assume that ϕ is odd and that one of the following conditions holds:

(i):

f0=+, f=0 and ϕ is a classical or bounded homeomorphism satisfying (3.6).

(ii):

f0=0, f=+ and ϕ is a classical homeomorphism satisfying (3.9).

(iii):

f0=+ and ϕ is a singular homeomorphism satisfying (3.10).

Then problem (1.8)–(1.10) has at least one positive solution.

Remark 3.18.

Theorem 3.15 allows us to deduce the existence of multiple positive solutions for problem (1.8)–(1.10) provided that there are several pairs of positive numbers (α,β) satisfying conditions (3.15)–(3.17).

4. Numerical examples

From numerical point of view we will consider three distinct boundary value problems. In order to solve them we make use of the new and powerful MATLAB package Chebfun which is a product of the numerical analysis group at Oxford University led by Professor Trefethen (see for instance [17] and [18] to quote but a few).

The philosophy behind this package is non-standard in numerical analysis and can be summed up in the words of its initiator as “Feel symbolic but run at the speed of numerics”.

In short, the method implemented by Chebfun is a Chebyshev type collocation one. Chebfun tries to solve a BVP by using successively to approximate the solution Chebyshev polynomials on grids of size 17, 33, 65 until the spectral convergence is reached. The relative accuracy of each computation carried out by a Chebfun algorithm is usually about 16 digits, and in principle the user need have no knowledge of the underlying algorithms. However, when solving a nonlinear BVP, Chebfun provides useful information on the convergence of the Newtonian method used to solve nonlinear algebraic systems obtained by discretization. In addition, the behavior of the solution coefficients can be visualized (the way in which they decrease to the machine accuracy). We will display these two outputs for each of the three issues considered. In fact, we must emphasize that we have used Chebfun with excellent results in our previous works [7] and [14].

4.1. First example: a Dirichlet-Neumann problem

Consider the Dirichlet-Neumann problem for an equation involving a singular homeomorphism

(4.1) {(rv1v2)=rg(r)f(v),r(1,2)v(1)=v(2)=0,

where

g(r)=r+12r2+1,f(v)=v2+1.
Refer to caption
Figure 1. Graph of the numerical solution of problem (4.1). The initial guess for the initialization of the Newton procedure is v0:=1.

The residual Chebfun satisfies the operator is of order 1010 and the boundary conditions are satisfied exactly.

Refer to caption
Figure 2. Newton iterations (left panel) and the coefficients of Chebyshev solution when Chebfun solves problem (4.1).

From the left panel of Fig. 2 it is very clear that Newton method converges with an order of at most 2 and from the right panel of the same figure one can observe that a Chebyshev polynomial of order 16, with highly and smoothly decreasing coefficients is the solution of this problem.

4.2. Second example: a Neumann-Dirichlet problem

We now solve numerically the following problem

(4.2) {(rv1v2)=rg(r)f(v),r(0,1)v(0)=v(1)=0,

where

g(r)=er+12r,f(v)=1v2+1.
Refer to caption
Figure 3. Graph of the numerical solution of problem (4.2). The initial guess for the initialization of the Newton procedure is v0:=1.

The residual Chebfun satisfies the operator is of order 1011 and the boundary conditions are satisfied exactly.

Refer to caption
Figure 4. Newton iterations (left panel) and the coefficients of Chebyshev solution when Chebfun solves problem (4.2).

From the left panel of Fig. 4 it is very clear that Newton method converges with an order of at least 2 and from the right panel of the same figure one can observe that a Chebyshev polynomial of order 17, with highly decreasing coefficients is the solution of the problem.

4.3. Third example: a Dirichlet problem

The last example is giving by the Dirichlet problem

(4.3) {(rv1v2)=rg(r)f(v),r(1,2)v(1)=v(2)=0,

where

g(r)=1,f(v)=v+1v2+1.
Refer to caption
Figure 5. Graph of the numerical solution of problem (4.3). The initial guess for the initialization of the Newton procedure is v0:=1.

The residual Chebfun satisfies the operator is of order 1010 and the boundary conditions are satisfied exactly.

Refer to caption
Figure 6. Newton iterations (left panel) and the coefficients of Chebyshev solution when Chebfun solves problem (4.3).

From the left panel of Fig. 6 it is very clear that Newton method converges with an order of at least 2 and from the right panel of the same figure one can observe that a Chebyshev polynomial of order 24, with highly decreasing coefficients is the solution of the problem.

We must make an important remark at the end of these three examples. Spectral convergence occurs for solutions represented by Chebyshev polynomials of relatively small order (of the order of a few tens). Moreover and more important, the convergence is so fast that no rounding off plateau appears (see the right panels of the figures 2, 4 and 6).

Acknowledgements

Jorge Rodríguez-López was partially supported by Institute of Advanced Studies in Science and Technology of Babeş-Bolyai University of Cluj-Napoca (Romania) under a Postdoctoral Advanced Fellowship, project CNFIS-FDI-2021-0061; by Xunta de Galicia (Spain), project ED431C 2019/02 and AIE, Spain and FEDER, grant PID2020-113275GB-I00.

References

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