# Nonlinear Singular BVP of Limit Circle Type and the Presence of Reverse-Ordered Upper and Lower Solutions

Authors: Amit K. Verma; Lajja Verma
Journal: International Journal of Differential Equations
 (2011)

---

## Abstract

We consider the following class of nonlinear singular differential equation-(p(x)y′(x))′+q(x)f(x,y(x),p(x)y′(x))=0,  0<x<1 subject to the Neumann boundary condition y′(0)=y′(1)=0. Conditions on p(x) and q(x) ensure that x=0 is a singular point of limit circle type. A simple approximation scheme which is iterative in nature is considered. The initial iterates are upper and lower solutions which can be ordered in one way (v0≤u0) or the other (u0≤v0).


---

## Body

## 1. Introduction

The upper and lower solution technique is the most promising technique as far as singular boundary value problems, are concerned [1]. Recently, lot of activities are there regarding upper and lower solutions technique (see [2, 3] and the references therein). To see the application of the similar kind of problems, one should see the references of [3]. In most of the results, upper and lower solutions are well ordered, that is, u0≥v0. As far as reverse-ordered upper and lower solutions are considered, that is, u0≤v0, the literature is not that rich. Though references are there for nonsingular boundary value problem, but singular boundary value problems require further exploration. The details of the work done for the nonsingular problem when upper and lower solutions are in reverse order can be seen in [4, 5]. To fill this gap in the present paper, we consider the following singular BVP:(1.1)-(p(x)y′(x))′+q(x)f(x,y(x),p(x)y′(x))=0,0<x<1,y′(0)=0,y′(1)=0.(A1)
Letp(x) satisfy the following conditions.
(i)
p(0)=0 and p>0 in (0,1).(ii)
p∈C[0,1]∩C1(0,1).(A2)
Letq(x) satisfy the following conditions.
(i)
q(x)>0 in (0,1) and q(x)∈C(0,1].(ii)
∫01q(x)dx<∞.(iii)
limx→0(q(x))/(p′(x))=0.(iv)
∫01(1/p(x))(∫0xq(s)ds)1/2dx<∞.In this paper, we consider a computationally simple iterative scheme defined by(1.2)-(pyn′)′+λqyn=-qf(x,yn-1,pyn-1′)+λqyn-1,0<x<1,yn′(0)=0,yn′(1)=0.
Starting with upper and lower solutions, we generate monotone sequences. To generate these monotonic sequences, we need the existence of some differential inequalities. To prove these differential inequalities, we analyze the corresponding singular IVP and extract properties of the solutions and their derivative.We have arranged the paper in four sections. In Section2, we discuss some elementary results, for example, maximum principles and existence of two differential inequalities. Then using these elementary results, we establish existence results for well-ordered upper and lower solutions in Section 3 and for reverse-ordered upper and lower solutions in Section 4. In Section 5, we conclude this paper with some remarks.


## 2. Preliminaries

Leth(x)∈C[0,1], and let λ∈ℝ0(ℝ0=ℝ∖{0}), let A∈ℝ and let B∈ℝ. Now, consider the following class of linear singular problems:(2.1)-(p(x)y′(x))′+λq(x)y(x)=q(x)h(x),0<x<1,(2.2)y′(0)=A,y′(1)=B.
The corresponding homogeneous system (eigenvalue problem) is given by(2.3)-(p(x)y′(x))′+λq(x)y(x)=0,0<x<1,(2.4)y′(0)=0,y′(1)=0.
The solution of the nonhomogeneous problem (2.1)-(2.2) can be written as follows:(2.5)w(x)=z1(x)[∫0xq(t)h(t)z0(t)Wp(z1,z0)dt+Az1′(0)]+z0(x)[∫x1q(t)h(t)z1(t)Wp(z1,z0)dt+Bz0′(1)],
where z0(x,λ) is the solution of(2.6)-(p(x)z0′(x))′+λq(x)z0(x)=0,0<x<1,z0(0)=1,z0′(0)=0,z1(x,λ) is the solution of(2.7)-(p(x)z1′(x))′+λq(x)z1(x)=0,0<x<1,z1(1)=1,z1′(1)=0,
and Wp(z1,z0)=p(t)(z1z0′-z1′z0). By replacing x with 1-x in (2.6), it is easy to verify that (2.8)z1(x)=z0(1-x),
for both positive and negative values of λ.Remark 2.1.
Existence ofz0(x) and z1(x) satisfying the IVP (2.6) and IVP (2.7), respectively, is an immediate consequence of the result due to O'Regan [6, Theorem  2.1, page 432].Remark 2.2.
LetLq2(0,1) be a Hilbert space with inner product defined by
(2.9)〈f,g〉=∫01q(x)f(x)g(x)¯dx.
From (A2) (iv), it can easily be verified that x=0 is a singular point of limit circle type (see [6, Remark (i) page 434]) in Lq2(0,1). Thus, we have pure point spectrum [7, page 125]. It is easy to show that the eigenvalues are real, simple, and negative.Remark 2.3.
Sincez0 and z1 are two linearly independent solutions of (2.3), the eigenvalues of the eigenvalue problem (2.3)-(2.4) will be the zeros of z0′(1,λ). Since z0′(1,λ) is an analytic function of λ so its zeros will be isolated and they all will be negative. Let them be -λ0,-λ1,-λ2,…, where λi>0 for i=0,1,2,…. Now, we have -λ0 as the first negative zero of z0′(1,λ) or in other words first negative eigenvalue of (2.3)-(2.4).
Sincez0(x,λ) does not change its sign for -λ0<λ<0 and z0(0,λ)=1; therefore, z0(x,λ)>0 for all x∈[0,1] and for all -λ0<λ<0.Remark 2.4.
Using (2.6), z1(x)=z0(1-x), it is easy to prove that if λ>0 then for all x∈(0,1], z0(x)>1, and z0′(x)>0 and for all x∈[0,1), we have z1(x)>1 and z1′(x)<0.Remark 2.5.
Using Remark2.3, z1(x)=z0(1-x), and the differential equation (2.6), it is easy to prove that if -λ0<λ<0 for all x∈[0,1), z0(x)>0 and z1′(x)>0 and for all x∈(0,1], we have z0′(x)<0 and z1(x)>0.Remark 2.6.
Letλ>0 and let h∈C[0,1]. If h≥0(orh≤0), then
(2.10)∫0xq(t)h(t)z0(t)Wp(z1,z0)dt,∫x1q(t)h(t)z1(t)Wp(z1,z0)dt
are nonnegative (or nonpositive).Remark 2.7.
Let-λ0<λ<0 and let h∈C[0,1]. If h≥0(orh≤0), then
(2.11)∫0xq(t)h(t)z0(t)Wp(z1,z0)dt,∫x1q(t)h(t)z1(t)Wp(z1,z0)dt
are nonpositive (or nonnegative).Proposition 2.8 (Maximum Principle).
Letλ>0. If A≤0, B≥0 (or A≥0,B≤0) and h∈C[0,1] is such that h≥0 (or h≤0), then w(x)≥0 (or w(x)≤0), where w(x) is the solution of (2.1)-(2.2).Proposition 2.9 (Antimaximum Principle).
Let-λ0<λ<0. If A≤0, B≥0 (orA≥0,B≤0) and h∈C[0,1] is such that h≥0(or h≤0), then w(x)≤0(or w(x)≥0), where w(x) is the solution of (2.1)-(2.2).Now, we derive conditions onλ which will help us to prove the monotonicity of the solutions generated by the iterative scheme (1.2).Lemma 2.10.
LetM and N∈ℝ+. If λ>0 is such that
(2.12)λ≥M(1-N∫01q(x)dx)-1,
then for all x∈[0,1],
(2.13)(M-λ)z0(x)+Np(x)z0′(x)≤0.Proof.
Integrating (2.6) from 0 to x and using the fact that z0′(x)>0 in (0,1], we get
(2.14)p(x)z0′(x)≤λz0(x)∫01q(x)dx.
Therefore, we get (M-λ)z0(x)+Np(x)z0′(x)≤(M-λ)z0+Nλz0(x)∫01q(x)dx. Hence, (2.13) will hold if (M-λ)+Nλ∫01q(x)dx≤0. Hence the result.Lemma 2.11.
LetM and N∈ℝ+. If -λ0<λ<0 is such that -(∫01(1/p(x))∫0xq(t)dtdx)-1<λ≤-M and
(2.15)(M+λ)(1+λ∫011p(x)∫0xq(t)dtdx)-Nλ∫01q(x)dx≤0,
then for all x∈[0,1],
(2.16)(M+λ)z0(x)-Np(x)z0′(x)≤0.Proof.
Using (2.6) and Remark 2.5, it can be deduced that z0(x) and p(x)z0′(x) are decreasing functions of x for -λ0<λ<0, thus
(2.17)(M+λ)z0(x)-Np(x)z0′(x)≤(M+λ)z0(1)-Np(1)z0′(1).
Now using (2.6), we get -p(1)z0′(1)≤(-λ)∫01q(x)dx and z0(1)>1+λ∫01(1/p(x))∫0xq(t)dtdx. This completes the proof.Note 2.
In Lemma2.11, we arrive at the integral∫01(1/p(x))∫0xq(t)dtdxwhich is an improper integral, and it should be convergent. Using the assumption (A2) (iv) and Remark(i)and(ii)at [6, page 434] its convergence can be established.


## 3. Well-Ordered Upper and Lower Solutions

Let us define upper and lower solutions.Definition 3.1.
A functionu0∈C[0,1]∩C2(0,1] is an upper solution of (1.1) if
(3.1)-(pu0′)′+qf(x,u0,pu0′)≥0,0<x<1,u0′(0)≤0≤u0′(1).Definition 3.2.
A functionv0∈C[0,1]∩C2(0,1] is a lower solution of (1.1) if
(3.2)-(pv0′)′+qf(x,v0,pv0′)≤0,0<x<1,v0′(0)≥0≥v0′(1).
Now, for every n, the problem (1.2) has a unique solution yn+1 given by (2.5) with h(x)=-f(x,yn,pyn′)+λyn, A=0, and B=0.In this section, we show that for the proposed scheme (1.2) a good choice of λ is possible so that the solutions generated by the approximation scheme converge monotonically to solutions of (1.1). We require a number of results.Lemma 3.3.
Letλ>0. If un is an upper solution of (1.1) and un+1 is defined by (1.2), then un+1≤un.Proof.
Letwn=un+1-un, then
(3.3)-(pwn′)′+λqwn=(pun′)′-qf(x,un,pun′)≤0,wn′(0)≥0,wn′(1)≤0,
and using Proposition 2.8, we have un+1≤un.Proposition 3.4.
Assume that(H1)
there exist upper solution(u0) and lower solution (v0) in C[0,1]∩C2(0,1] such that v0≤u0 for all x∈[0,1],(H2)
the functionf:D→ℝ is continuous on(3.4)D:={(x,y,py′)∈[0,1]×R×R:v0≤y≤u0},(H3)
there existsM≥0 such that for all (x,τ,pv′),(x,σ,pv′)∈D,(3.5)f(x,τ,pv′)-f(x,σ,pv′)≥M(τ-σ),(τ≤σ),(H4)
there existN≥0 such that for all (x,u,pv1′)(x,u,pv2′)∈D,(3.6)|f(x,u,pv1′)-f(x,u,pv2′)|≤N|pv2′-pv1′|.
Let λ>0 be such that λ≥M(1-N∫01q(x)dx)-1. Then the functions un+1 defined recursively by (1.2) are such that, for all n∈ℕ,  (i)
un is an upper solution of (1.1).(ii)
un+1≤un.Proof.
We prove the claims by the principle of mathematical induction. Sinceu0 is an upper solution and by Lemma 3.3  u0≥u1; therefore, both the claims are true for n=0.
Further, let the claims be true forn-1, that is, un-1 is an upper solution and un-1≥un. Now, we are required to prove that un is an upper solution and un+1≤un. To prove this, let w=un-un-1, then we have
(3.7)-(pun′)′+qf(x,un,pun′)≥p[(M-λ)w-N(sign⁡w′)pw′].
Thus, to prove that un is an upper solution, we are required to prove that
(3.8)(M-λ)w-N(sign⁡w′)pw′≥0.
Now, since w satisfies
(3.9)-(pw′)′+λqw=(pun-1′)′-qf(x,un-1,pun-1′)≤0,w′(0)≥0,w′(1)≤0,
from Proposition 2.8, we have w≤0 for λ>0. Now, putting the value of w from (2.5) in (3.8), and in view of h=(pun-1′)′-qf(x,un-1,pun-1′)≤0, we deduce that to prove (3.8) it is sufficient to prove that
(3.10)(M-λ)z0-N(sign⁡w′)pz0′≤0,(M-λ)z1-N(sign⁡w′)pz1′≤0,
for all x∈[0,1]. Since z1=z0(1-x), using Remark 2.6, the above inequalities will be true if for all x∈[0,1] we have
(3.11)(M-λ)z0(x)+Np(x)z0′(x)≤0.
Which is true (Lemma 2.10). Therefore, (3.8) holds, and hence un is an upper solution.
Now applying Lemma3.3, we deduce that un+1≤un. This completes the proof.Similarly, we can prove the following two results (Lemma3.5, Proposition 3.6) for lower solutions.Lemma 3.5.
Letλ>0. If vn is a lower solution of (1.1) and vn+1 is defined by (1.2), then vn≤vn+1.Proposition 3.6.
Assume that(H1), (H2), (H3), and (H4) hold, and let λ>0 be such that λ≥M(1-N∫01q(x)dx)-1. Then the functions vn+1 defined recursively by (1.2) are such that for all n∈ℕ,  (i)
vn is a lower solution of (1.1).(ii)
vn≤vn+1.In the next result, we prove that upper solutionun is larger than lower solution vn for all n.Proposition 3.7.
Assume that(H1), (H2), (H3), and (H4) hold, and let λ>0 such that λ≥M(1-N∫01q(x)dx)-1and for all x∈[0,1](3.12)f(x,v0,pv0′)-f(x,u0,pu0′)+λ(u0-v0)≥0.
Then for all n∈ℕ, the functions un and vn defined recursively by (1.2) satisfy vn≤un.Proof.
We define a function(3.13)hi(x)=f(x,vipvi′)-f(x,ui,pui′)+λ(ui-vi),i∈N.
It is easy to see that for all i∈ℕ, wi=ui-vi satisfies the following differential equation:
(3.14)-(pwi′)′+λqwi=q{f(x,vi-1,pvi-1′)-f(x,ui-1,pui-1′)+λ(ui-1-vi-1)}=qhi-1.
Now to prove this proposition again, we use the principle of mathematical induction. For i=1, we have h0≥0, and w1 is the solution of (2.1)-(2.2) with A=0 and B=0. Using Proposition 2.8, we deduce that w1≥0, that is, u1≥v1.
Now, letn≥2, let hn-2≥0, and let un-1≥vn-1, then we are required to prove that hn-1≥0 and un≥vn. First, we show that for all x∈[0,1] the function hn-1 is nonnegative. Indeed, we have
(3.15)hn-1=f(x,vn-1,pvn-1′)-f(x,un-1,pun-1′)+λ(un-1-vn-1)≥-[(M-λ)wn-1+N(sign⁡wn-1′)pwn-1′].
Here wn-1 is a solution of (2.1) with h(x)=hn-2≥0, A=0, and B=0. Arguments similar to Proposition 3.4 can be used to prove that hn-1≥0. Now, we have hn-1≥0, wn′(0)=0, and wn′(1)=0, thus from Proposition 2.8, we deduce that wn≥0, that is, un≥vn.Lemma 3.8.
Iff(x,u,pu′) satisfies (H1), (H2), and  (H5)
for all(x,u,pu′)∈D, |f(x,u,pu′)|≤φ(|pu′|), where φ:[0,∞)→(0,∞) is continuous and satisfies.(3.16)∫0∞dsφ(s)>∫01q(x)dx,
then there exists R0>0 such that any solution of
(3.17)-(pu′)′+qf(x,u,pu′)≥0,0<x<1,u′(0)=0=u′(1)
with u∈[v0,u0], for all x∈[0,1], satisfies ∥pu′∥∞<R0.Proof.
Consider an interval[x,x0]⊂[0,1] such that
(3.18)∀s∈[x,x0),u′(s)<0,u′(x0)=0.
Now using (H5), we have
(3.19)(pu′)′≤qφ(|pu′|),
and after integrating it from x to x0 and using (H5), we have
(3.20)-pu′≤R0.
Similarly for the interval [x0,x], we have
(3.21)pu′≤R0.
Thus
(3.22)‖pu′‖∞≤R0.
In the same way, we can prove the following result for lower solutions.Lemma 3.9.
Iff(x,v,pv′) satisfies (H1), (H2), and (H5), then there exists R0>0 such that any solution of
(3.23)-(pv′)′+qf(x,v,pv′)≤0,0<x<1,v′(0)=0=v′(1)
with v∈[v0,u0], for all x∈[0,1], satisfies ∥pv′∥∞<R0.Now we are in a situation to prove our final result for the case when upper and lower solutions are well ordered.Theorem 3.10.
Assume(H1), (H2), (H3), (H4), and (H5) are true. Let λ>0 be such that
(3.24)λ≥M(1-N∫01q(x)dx)-1,
and for all x∈[0,1],
(3.25)f(x,v0,pv0′)-f(x,u0,pu0′)+λ(u0-v0)≥0.
Then the sequences {un} and {vn} defined by (1.2) converge monotonically to solutions ũ(x) and ṽ(x) of (1.1). Any solution z(x) of (1.1) in D satisfies
(3.26)ṽ(x)≤z(x)≤ũ(x).Proof.
Using Lemma3.3 to Lemma 3.9 and Proposition 3.4 to Proposition 3.7, we deduce that the sequences {un} and {vn} are monotonic (u0≥u1≥u2⋯≥un≥vn⋯≥v2≥v1≥v0) and are bounded by v0 and u0 in C[0,1], and by Dini's theorem, they converge uniformly to ũ and ṽ (say). We can also deduce that the sequences {pun′} and {pvn′} are uniformly bounded and equicontinuous in C[0,1], and by Arzela-Ascoli theorem, there exists uniformly convergent subsequences {punk′} and {pvnk′} in C[0,1]. It is easy to observe that un→ũ and vn→ṽ imply pun′→pũ′ and pṽn′→pṽ′.
Solution of (1.2) is given by (2.5) where h(x)=-f(x,yn-1,pyn-1′)+λyn-1. Since the sequences are uniformly convergent taking limit as n→∞, we get ũ and ṽ as the solutions of the nonlinear boundary value problem (1.1). Any solution z(x) in D plays the role of u0. Hence, z(x)≥ṽ(x). Similarly z(x)≤ũ(x).Remark 3.11.
When the source function is derivative independent, that is,N=0, in this case we can choose λ=M.


## 4. Upper and Lower Solutions in Reverse Order

In this section, we consider the case when the upper and lower solutions are in reverse order, that is,(4.1)u0(x)≤v0(x).
For this, we require opposite one-sided Lipschitz condition, and we assume that  (F1)
there exists upper solution (u0) and lower solution (v0) in C[0,1]∩C2(0,1] such that u0≤v0 for all x∈[0,1],(F2)
the functionf:D0→ℝ is continuous on(4.2)D0:={(x,y,py′)∈[0,1]×R×R:u0≤y≤v0},(F3)
there existsM≥0 such that for all (x,τ̃,pv′),(x,σ̃,pv′)∈D0,(4.3)f(x,σ̃,pv′)-f(x,τ̃,pv′)≥-M(σ̃-τ̃),(τ̃≤σ̃),(F4)
there existN≥0 such that for all (x,u,pv1′)(x,u,pv2′)∈D0,(4.4)|f(x,u,pv1′)-f(x,u,pv2′)|≤N|pv2′-pv1′|.
Here again we define the approximation scheme by (1.2) and use the Antimaximum principle. We make a good choice of λ so that the sequences thus generated converge to the solution of the nonlinear problem. Similar to Section 3, we require the following  lemmas and propositions.Lemma 4.1.
Let-λ0<λ<0. If un is an upper solution of (1.1) and un+1 is defined by (1.2), then un+1≥un.Proof.
Letwn=un+1-un, then
(4.5)-(pwn′)′+λqwn=(pun′)′-qf(x,un,pun′)≤0,wn′(0)≥0,wn′(1)≤0,
and using Proposition 2.8, we have un+1≥un.Proposition 4.2.
Assume that(F1), (F2), (F3), and (F4) hold. Let -λ0<λ<0 be such that -(∫01(1/p(x))∫0xq(t)dtdx)-1<λ≤-M and (M+λ)(1+λ∫01(1/p(x))∫0xq(t)dtdx)-Nλ∫01q(x)dx≤0. Then the functions un+1 defined recursively by (1.2) are such that, for all n∈ℕ,  (i)
un is an upper solution of (1.1);(ii)
un+1≥un.Proof.
Using Remarks2.5 and 2.7, Lemmas 2.11 and 4.1, and on the lines of the proof of Proposition 3.4, this proposition can be deduced easily.
In the same way, we can prove the following results for the lower solutions.Lemma 4.3.
Let-λ0<λ<0. If vn is a lower solution of (1.1) and vn+1 is defined by (1.2), then vn≥vn+1.Proposition 4.4.
Assume that(F1), (F2), (F3), and (F4) hold. Let -λ0<λ<0 be such that -(∫01(1/p(x))∫0xq(t)dtdx)-1<λ≤-M and (M+λ)(1+λ∫01(1/p(x))∫0xq(t)dtdx)-Nλ∫01q(x)dx≤0. Then the functions vn+1 defined recursively by (1.2) are such that, for all n∈ℕ,  (i)
vn is a lower solution of (1.1);(ii)
vn≥vn+1.In the next result, we prove that lower solutionvn is larger than upper solution un for all n.Proposition 4.5.
Assume that(F1), (F2), (F3), and (F4) hold. Let -λ0<λ<0 be such that -(∫01(1/p(x))∫0xq(t)dtdx)-1<λ≤-M and
(4.6)(M+λ)(1+λ∫011p(x)∫0xq(t)dtdx)-Nλ∫01q(x)dx≤0,
and for all x∈[0,1],
(4.7)f(x,v0,pv0′)-f(x,u0,pu0′)+λ(u0-v0)≥0.
Then, for all n∈ℕ, the functions un and vn defined recursively by (1.2) satisfy vn≥un.Now similar to Lemmas3.8 and 3.9, we state the following two results. These results establish a bound on p(x)u′(x) and p(x)v′(x).Lemma 4.6.
Iff(x,u,pu′) satisfies (F1), (F2), and  (F5)
for all(x,u,pu′)∈D0, |f(x,u,pu′)|≤φ(|pu′|), where φ:[0,∞)→(0,∞) is continuous and satisfies(4.8)∫0∞dsφ(s)>∫01q(x)dx,
then there exists R0>0 such that any solution of
(4.9)-(pu′)′+qf(x,u,pu′)≥0,0<x<1,u′(0)=0=u′(1)
with u∈[u0,v0], for all x∈[0,1], satisfies ∥pu′∥∞<R0.Lemma 4.7.
Iff(x,v,pv′) satisfies (F1), (F2), and (F5), then there exists R0>0 such that any solution of
(4.10)-(pv′)′+qf(x,v,pv′)≤0,0<x<1,v′(0)=0=v′(1)
with v∈[u0,v0], for all x∈[0,1], satisfies ∥pv′∥∞<R0.Finally we arrive at the theorem similar to Theorem3.10.Theorem 4.8.
Assume(F1), (F2), (F3), (F4), and (F5) are true. Let -λ0<λ<0 be such that -(∫01(1/p(x))∫0xq(t)dtdx)-1<λ≤-M and (M+λ)(1+λ∫01(1/p(x))∫0xq(t)dtdx)-Nλ∫01q(x)dx≤0, and for all x∈[0,1],
(4.11)f(x,v0,pv0′)-f(x,u0,pu0′)+λ(u0-v0)≥0.
Then the sequences{un} and {vn} defined by (1.2) converge monotonically to solutions ũ(x) and ṽ(x) of (1.1). Any solution z(x) of (1.1) in D0 satisfies
(4.12)ũ(x)≤z(x)≤ṽ(x).Proof.
Using Lemma4.1 to Lemma 4.7 and Proposition 4.2 to Proposition 4.5, we deduce that
(4.13)u0≤u1≤u2≤⋯≤un≤vn⋯≤v1≤v0.
Now similar to the proof of Theorem 3.10, the result of this theorem can be deduced.Remark 4.9.
When the source function is derivative independent, that is,N=0, in this case we can choose λ=-M.


## 5. Conclusion

We establish some existence results under quite general conditions onp(x), q(x), and f(x,y,py′). We prove some fundamental differential inequalities which enables us to prove the monotonicity of the sequences {un} and {vn}. For this we have analyzed the singular differential equation -(py′)′+λqy=0 and derived properties of the solutions and their derivatives. This work generalizes our previous work [3]. Lot of exploration is still left. For example, one can consider different type of boundary conditions, and one can also try to remove the Lipschitz condition.



---

*Source: 986948-2011-09-13.xml*