The dual approach to stationary and evolution quasilinear PDEs · in lower regular spaces, a...

Nonlinear Differ. Equ. Appl. (2016) 23:4c© 2016 Springer International Publishing1021-9722/16/020001-22published online February 12, 2016DOI 10.1007/s00030-016-0367-0

Nonlinear Differential Equationsand Applications NoDEA

The dual approach to stationaryand evolution quasilinear PDEs

Alessandro Selvitella

Abstract. In this paper, we study several quasilinear PDEs with a particu-lar algebraic structure. In the case of stationary solutions for a quasilinearSchrodinger equation, Colin and Jeanjean (Nonlinear Anal 56:213–226,2004) implemented a dual approach to prove existence and qualitativeproperties of the solutions. The method takes advantage of the particularunderlying structure of that quasilinear PDE, which is essentially semilin-ear. The main goal of this manuscript is to show that the dual approach issuccessful in a broader set of problems, especially in the stationary casesinvolving more general quasilinear terms and the p-Laplacian. We provealso that this approach works for some quasilinear heat equations, butfails for the complete evolution of quasilinear Schrodinger equations. Thereason of the failure seems related to the extra structure of the complexplane.

Mathematics Subject Classification. 35Q55, 35J60, 35K55.

1. Introduction

We consider the following quasilinear Schrodinger equation:

i∂tφ(t, x) + Δφ(t, x) + λφ(t, x)Δ|φ(t, x)|2 + |φ(t, x)|p−1φ(t, x) = 0, (1)

where i is the imaginary unit, N ≥ 1 is the space dimension, p > 1 is theexponent of the semilinear term, the domain is (t, x) ∈ (0,∞) × RN andφ : RN → C is the wave function. This equation appears in different physicalmodels, such as the superfluid film equation in plasma physics. We refer to [6]and [7], for a more detailed bibliography on the physical background.

The mathematical theory for this equation is more involved than the semi-linear case, especially for what concerns the short time dynamics. In the studyof the Cauchy problem, the presence of the quasilinear term causes the phe-nomenon called loss of derivatives. To overcome this problem, mathematiciansassume high regularity on the initial datum and prove local well-posednessin Sobolev spaces of high order. Because of this lack of local wellposedness

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4 Page 2 of 22 A. Selvitella NoDEA

in lower regular spaces, a Gagliardo and Nirenberg type inequality [6] cannotguarantee global wellposedness in the energy space. We refer to Colin [4], Colinet al. [6], Kenig et al. [7], Lange et al. [8] and Poppenberg [14,15], Selvitellaand Wang [20] for the main results, concerning well-posedness of this problemand the long time behavior of solutions. To our knowledge, the latest resultsin this area are due to Marzuola et al. in [12] and [13].

We have a better understanding for what concerns the existence of par-ticular types of stationary solutions, called standing waves, namely solutionswhich model particles at rest in the form φ(t, x) = u(x)eiνt. Here, u : RN → Csolves the quasilinear elliptic equation:

− Δu − uΔ|u|2 + νu − |u|p−1u = 0, (2)

Many authors have studied solutions to this problem. See, for example, Liuet al. [9], Liu et al. [10], Liu and Wang [11] and Colin and Jeanjean [5]. Themost complete result concerning ground states of (2) is due to Colin et al.[6]. In [6], the authors prove that ground states exist, are real and positive(up to phase shifts) classical solutions, they decay exponentially at infinitywith their first and second derivatives (actually with all their derivatives, seeSelvitella [18]) and, moreover, they are radial with respect to some point andnondegenerate (see Selvitella [19]).

The method employed by Colin and Jeanjean in [5] and Colin et al. in [6],relies on a magic transformation which sends the original problem to a semi-linear one. The method takes advantage of the particular algebraic structureof this quasilinear PDE. The transformation is defined through the followingODE: {

F ′(s) = 1√1+2F 2(s)

F (0) = 0.

The main goal of this manuscript is to show that this approach is successfulin a broader set of problems. We start by considering the equation

− Δu − ul′(|u|2)Δl(|u|2) = ωu − f(|u|2)u =: g(u). (3)

For this type of problems, we can prove the following theorem.

Theorem 1.1. Suppose that• (H0) g(s) is locally Holder continuous in [0,+∞);• (H1) for N ≥ 3

−∞ < lim infs→0

g(s)s

≤ lim sups→0

g(s)s

= −ν < 0,

while for N = 1, 2,

lims→0

g(s)s

= −ν < 0;

• (H2) for N ≥ 3

lims→+∞

g(s)√1 + 2l′(F 2(s))F 2(s)s

N+2N−2

= 0,

NoDEA The dual approach to stationary Page 3 of 22 4

while for N = 2, for any α > 0, there exists Cα such that

|g(s)| ≤ Cαeαs2,

for any s ≥ 0.• (H3) Define k(s) := g(s)√

1+2l′(F 2(s))F 2(s), then, for N ≥ 2, there exists s0

such that∫ s0

0k(r)dr = 0 and k(s0) > 0, while for N = 1, there exists s0

such that∫ s

0k(r)dr < 0 for all s ∈ (0, s0),

∫ s0

0k(r)dr = 0 and k(s0) > 0;

• (H4) the function l has the following regularity: l ∈ C2(R+,R).

Then, Eq. (3) admits a solution Q(x) ∈ H1(RN ) satisfying the following prop-erties:

• Q(x) > 0 for any x ∈ RN ;• Q(x) is spherically symmetric;• Q(x) is smooth;• Q(x) and its derivatives up to order two are exponentially decaying.

Remark 1.2. An important thing to notice here is that the critical exponentp∗ varies accordingly to the growth of the function l. For example, in the casel(v) = v, the critical exponent is p∗ = 3 + 4

N−2 (see for example [6]).

Remark 1.3. We note that this type of result is very similar to the resultsthat have been proven by Adachi and Watanabe [1]. We refer to [1] and thereferences therein for more details.

Another application of this dual approach involves the p-Laplacian. Anexample of results that we can prove with transformation theory is the follow-ing.

Theorem 1.4. Consider the following equation

Δpu + uΔp|u|2 + |x|−βg(u) = 0, in Ω (4)

where β < p, g : R+0 → R is continuous and Ω = Rn if β ≤ 0, while

Ω = RN\{0} if β ∈ (0, p). Consider T , the solution of{T ′(s) = 1

(1+2p−1T p(s))1/p

T (0) = 0.

Assume the following conditions on the nonlinear term g:• (F0) f(u) := g(T (u))

(1+2p−1T (u)p)1/p is continuous in R+0 ;

• (F1) there exists a > 0 and q > 1 such that limu→0+ u1−qf(u) = −a;• (F2) there exists u∗ > 0 such that f(u∗) ≥ 0 and

∫ u∗

0f(t)dt > 0;

• (F3) limu→∞ u1−p∗β f(u) = 0, d∗

β := pn−βn−p .

Then, Eq. (4), with β < p, admits a radial ground state u ∈ D1,prad(R

N )∩Lq(Rn)bounded by u∗. Moreover,

• u ∈ C1,θloc (RN\{0}) for some θ ∈ (0, 1);

• |Du|p−2Du ∈ C1(RN\{0}) and u solves (4) pointwise in RN\{0};• |Du(x)| → 0 as |x| → +∞ and |D(T−1u)(x)| = O(|x|−(N−1)/(p−1)) as

|x| → 0;


• u is continuous at x = 0, (x,Du(x))RN ≤ 0 in RN\{0}, and ‖u‖∞ =u(0) ∈ (u0, u

∗], where u0 := inf{v > 0 :∫ v

0f(t)dt > 0};

• if 1 < p ≤ 2, then T−1(u) ∈ H2,ploc (RN\{0}); if, furthermore, β < N

p′ , thenT−1(u) ∈ H2,p

loc (RN ).

If 1 < q < p, then u s compactly supported in RN and is a fast decay solutionof (4) such that T−1(u) ∈ H1,p(RN ). Furthermore, u has the regularity in RN

as described in the following table

1 < p ≤ 2 p > 2β < 1 C2(RN ) C1(RN )1 ≤ β < p C

0,(p−β)/(p−1)loc (RN ) C

0,(p−β)/(p−1)loc (RN )

While, if q ≥ p, then u is positive in RN , u ∈ C2(RN\{0}) and u hasthe following regularity in RN as described in the following table

1 < p ≤ 2 p > 2β < 2 − p C2(RN ) C2(RN )β = 2 − p C2(RN ) C1,1

loc (RN )2 − p < β < 1 C2(RN ) C

1,(1−β)/(p−1)loc (RN )

1 ≤ β < p C0,(p−β)/(p−1)loc (RN ) C

0,(p−β)/(p−1)loc (RN )

Moreover, u is a fast decay solution of (4) and, in particular, |x|(n−p)/(p−1)

T−1(u) is decreasing in [R,∞) for R sufficiently large and approaches a limitl ≥ 0 as r → +∞. While l > 0, then T−1(u) ∈ H1,p(RN ) if and only ifN > p2; while, if l = 0 and N > p2, then T−1(u) ∈ H1,p(RN ).

Remark 1.5. The novelty of this result lies in the contemporary presence ofboth the terms Δpu and uΔpu

2. In the case in which uΔu2 is absent, wereduce our problem to Theorem 2.5 below proven by [16]. Instead, if the Δpuis absent, we can call u2 =: v and apply again Theorem 2.5. When both theterms Δpu and uΔpu

2 are present, Theorem 2.5 cannot be applied and oneneeds our approach.

Remark 1.6. We note that the transformation trick does not work if we aimto transform the p-Laplacian into the ordinary Laplacian. In fact, in that case,the transformation reduces to the identity and the PDE remains unchanged.This transformation trick seems therefore useful every time we want to removea lower order coefficient in front of the main derivative term, but not when wewant to modify the derivative term.

Since the dual approach seemed pretty successful in the stationary case,we may wonder what happens in the case of solutions with non trivial dynam-ics. We start by considering a quasilinear reaction–diffusion equation.

Theorem 1.7. Consider the following quasilinear heat equation

ut = Δu + uΔ|u|2 + g(u). (5)


with initial condition u(0) = u0 ≥ 0 and with g(u) ≤ 0 for any u ≥ 0 and0 otherwise. We assume that x ∈ M := TN , the N -dimensional torus. Ifu0 = F (v0), with F such that{

F ′(s) = 1√1+2F 2(s)

F (0) = 0.

and with v0 = F−1(u0) ∈ Hs(M), s > N2 + 1, then, there exists a unique

solution such that

v = F−1(u) ∈ C([0, T ),Hs(M)) ∩ C∞((0, T ) × M),

which persists as long as ‖u(t)‖Cr is bounded, given r > 0.

Remark 1.8. This result is a simple application of well known theorems, afterthe change of variables performed through F . However, we decided to discussthis theorem, because important to show the wide applicability of the dualapproach. We refer to [21] for more possible applications of this method in thesetting of quasilinear reaction–diffusion systems.

More interesting is to understand what happens to the complete evolutionof the quasilinear Schrodinger equation.

Theorem 1.9. Suppose that u solves the equation

iut = −Δu − uΔ|u|2 − g(|u|2)uwith hypotheses on the nonlinearity g which ensure local well-posedness. Sup-pose that there exists a holomorphic transformation F that eliminates the gra-dient terms in this equation. Then, u is real up to a phase shift constant inspace and time. Moreover, u is stationary.

Remark 1.10. Essentially, this means that the method fails in the non-stationary case for equation of evolution type, which mix real and imaginaryparts.

One might still wonder if the reason is that the requirement of holo-morphicity for F in the previous theorem is too strict. The answer is in thefollowing theorem.

Theorem 1.11. Suppose that u solves the equation

iut = −Δu − uΔ|u|2 − g(|u|2)uwith hypotheses on the nonlinearity g which ensure local well-posedness. Sup-pose that there exists a transformation F , not necessarily holomorphic, thateliminates the gradient terms in this equation. Then, u is real up to a phaseshift constant in space and time. Moreover, u is stationary.

Remark 1.12. Note that the only difference between these two theorems is theregularity requirement on the transformation.

This implies that in general, if one studies the complete evolution, it isnot possible, through this strategy, to simplify the original quasilinear problemto a semilinear one.


Remark 1.13. We think that the dual approach can be potentially adaptedto more general combinations of derivative terms and nonlinear terms. Forexample, it would be interesting to understand what happens for non-localequations and for more general evolution problems.

The remaining part of the paper is organized as follows. In Sect. 2, wecollect some results concerning the inverse of partitioned matrices, the Cauchy–Riemann equations, the results of Berestycki and Lions [2] and Berestycki et al.[3] on the semilinear problem, a result of the survey of Pucci and Servadei [16]on the p-Laplacian and a result concerning strong parabolic equations presentin [21]. In Sect. 3, we prove Theorem 1.1. In Sect. 4, we prove Theorem 1.4.In Sect. 5, we prove Theorem 1.7. In Sect. 6, we prove Theorem 1.9. In Sect.7, we prove Theorem 1.11.

2. Preliminaries

We collect some results which we will need later in our discussion and whichhelp us to frame our problem in the literature.

2.1. The inverse of a partitioned matrix

In the proof of Theorem 1.11, we need an expression for the inverse of apartitioned matrix. In particular, we need the following lemma.

Lemma 2.1. Consider an N × N matrix B partitioned in the following way:

B :=(

B11 B12

B21 B22

).

Then, if the inverse exists, it takes the following form:

B−1 :=

((B11−B12B

−122 B21

)−1 (−B−111 B12(B22−B21B

−111 B12

)−1(−B−122 B21(B11−B12B

−122 B21

)−1 (B22−B21B

−111 B12

)−1

).

2.2. The Cauchy–Riemann equations

The Cauchy–Riemann equations are a necessary condition that a complexfunction must satisfy to be differentiable. Here is the precise statement:

Theorem 2.2. Suppose that f ′(z0) exists at a point z0 ∈ O ⊂ C. Then thefollowing equation must be satisfied:

∂

∂zf = 0.

This condition is called Cauchy–Riemann equations.

Proof. The proof is a simple calculation. We refer to [17] for a proof of thiswell known fact. �

Remark 2.3. The Cauchy–Riemann equations can be rewritten in several dif-ferent forms, which all can be deducted by the previous condition. We refer to[17] for more details.


At least at the level of continuity, we can identify C and R2, through the

following transformation

u �→i (u1, u2)T �→F (v1, v2)T �→i−1v.

We remind that if F = (F1, F2) and F1, F2 are real differentiable, then i−1 ◦Fis real differentiable, but not necessarily complex differentiable. In fact (i−1 ◦F ◦ i)u is complex differentiable if and only if Cauchy–Riemann equations aresatisfied.

2.3. The semilinear problem

The following is a classical result due to Berestycki and Lions [2] and Berestyckiet al. [3].

Theorem 2.4. Consider the following equation:

− Δv = k(v), v ∈ H, (6)

which is the Euler–Lagrange equation of

J(v) :=12

∫Rn

dx|∇v|2 −∫Rn

dxK(v), (7)

where K(t) =∫ t

0k(s)ds is differentiable. Assume the following conditions:

• (A) k(s) ∈ C(R+,R) and k(s) = 0 for s ∈ R−;• (B) for N ≥ 3

−∞ < lim infs→0

k(s)s

≤ lim sups→0

k(s)s

= −ν < 0,

while for N = 1, 2,

lims→0

k(s)s

= −ν < 0;

• (C) for N ≥ 3, lims→+∞k(s)

sN+2N−2

= 0, while for N = 2, for any α > 0,

there exists Cα such that

|k(s)| ≤ Cαeαs2,

for any s ≥ 0;• (D) for N ≥ 2, there exists s0 such that K(s0) = 0 and k(s0) > 0,

while for N = 1, there exists s0 such that K(s) < 0 for all s ∈ (0, s0),K(s0) = 0 and k(s0) > 0.

Define

m := inf{J(v)|v ∈ H\{0},−Δv = k(v)}.

Then m > 0 and there exists a least energy solution Q(x) of Eq. (6) such that• Q(x) > 0 for any x ∈ RN;• Q(x) is spherically symmetric;• Q(x) is smooth;• Q(x) and its derivatives up to order two are exponentially decaying.


2.4. The p-Laplacian with weights

There is a wide literature concerning the p-Laplace operator. Here, we report aresult exposed in the survey [16]. We refer to [16] for a more detailed literaturereview and for different results.

Theorem 2.5. Consider the following equation

Δpu + |x|−βf(u) = 0, in Ω (8)

where β < p, f : R+0 → R is continuous and Ω = Rn if β ≤ 0, while Ω =

RN\{0} if β ∈ (0, p). Assume the following conditions on the nonlinear termf :

• (F0) f is continuous in R+0 ;

• (F1) there exists a > 0 and q > 1 such that limu→0+ u1−qf(u) = −a;• (F2) there exists u∗ > 0 such that f(u∗) ≥ 0 and

∫ u∗

0f(t)dt > 0;

• (F3) limu→∞ u1−p∗β f(u) = 0, d∗

β := pn−βn−p .

Then, Eq. (8), with β < p, admits a radial ground state u ∈ D1,prad(R

N ) ∩Lq(RN ) bounded by u∗. Moreover,

• u ∈ C1,θloc (RN\{0}) for some θ ∈ (0, 1);

• |Du|p−2Du ∈ C1(RN\{0}) and u solves (8) pointwise in RN\{0};• |Du(x)| → 0 as |x| → +∞ and |Du(x)| = O(|x|−(N−1)/(p−1)) as |x| → 0;• u is continuous at x = 0, (x,Du(x))RN ≤ 0 in RN\{0}, and ‖u‖∞ =

u(0) ∈ (u0, u∗], where u0 := inf{v > 0 :

∫ v

0f(t)dt > 0};

• if 1 < p ≤ 2, then u ∈ H2,ploc (RN\{0}); if, furthermore, β < n

p′ , thenu ∈ H2,p

loc (RN ).

If 1 < q < p, then u s compactly supported in RN and it is a fast decay solutionof (8) of class H1,p(RN ). Furthermore, u has the regularity in RN as describedin the following table

1 < p ≤ 2 p > 2β < 1 C2(RN ) C1(RN )1 ≤ β < p C

0,(p−β)/(p−1)loc (RN ) C

0,(p−β)/(p−1)loc (RN )

While, if q ≥ p, then u is positive in RN , u ∈ C2(RN\{0}) and u hasthe following regularity in RN as described in the following table

1 < p ≤ 2 p > 2β < 2 − p C2(RN ) C2(RN )β = 2 − p C2(RN ) C1,1

loc (RN )2 − p < β < 1 C2(RN ) C

1,(1−β)/(p−1)loc (RN )

1 ≤ β < p C0,(p−β)/(p−1)loc (RN ) C

0,(p−β)/(p−1)loc (RN )

Moreover, u is a fast decay solution of (8) and, in particular, |x|(n−p)/(p−1)

u is decreasing in [R,∞) for R sufficiently large and approaches a limit l ≥ 0as r → +∞. While l > 0, then u ∈ H1,p(RN ) if and only if N > p2; while, ifl = 0 and N > p2, then u ∈ H1,p(RN ).


2.5. Strong parabolic quasilinear equations

In this subsection, we collect some results about strong parabolic equations.We start with the definition of strong parabolicity.

Definition 2.6. Consider the following system

∂u

∂t=

N∑j,k=1

Ajk(t, x,D1xu)∂j∂ku + B(t, x,D1

xu).

We say that this system is strongly parabolic if it satisfies the following con-dition

N∑j,k=1

Ajk(t, x,D1x(u))ξjξk ≥ C0|ξ|2IdK×K ,

where we say that a pair of symmetric K × K matrices A1 and A2 satisfyA1 ≥ A2, if and only if A1 − A2 is a positive semidefinite matrix.

Concerning strong parabolic equations, we have the following theoremfrom [21].

Proposition 2.7. Assume that the system

∂u

∂t=

N∑j,k=1

Ajk(t, x, u)∂j∂ku + B(t, x, u)

is strongly parabolic and that u(0) = u0. Here, u takes values in RK and Ajk

is a symmetric K × K matrix. We assume that Ajk and B are smooth intheir arguments. We assume that x ∈ M = TN , the N -dimensional torus. Ifu0 ∈ Hs(M), s > N

2 + 1, then there exists a unique solution such that

u ∈ C([0, T ),Hs(M)) ∩ C∞((0, T ) × M),

which persists as long as ‖u(t)‖Cr is bounded, given r > 0.

3. Proof of Theorem 1.1

In this section, we prove Theorem 1.1. Therefore, consider the equation

Δu − ul′(|u|2)Δl(|u|2) + ωu + f(|u|2)u = 0. (9)

Use the transformation u = F (v) with F : R+ → R. Plugging this inside eachterm, we get the following. First:

∂

∂xjl(|u|2) = 2F (v)

∂

∂xjF (v)l′(|u|2) = 2F (v)F ′(v)l′(|u|2) ∂

∂xjv.

Taking another derivative, we get

∂

∂xj

∂

∂xjl(|u|2) =

∂

∂xj

(2F (v)F ′(v)l′(|u|2) ∂

∂xjv

)

= 2(F ′)2l′(F 2)∂

∂xjv

∂

∂xjv + (2FF ′)2l′′(F 2)

∂

∂xjv

∂

∂xjv


+ 2Fl′(F 2)F ′′ ∂

∂xjv

∂

∂xjv + 2FF ′l′(F 2)

∂2

∂x2j

v.

Therefore

Δl(|u|2) = |∇v|2(2(F ′)2l′(F 2) + (2FF ′)2l′′(F 2)

+2Fl′(F 2)F ′′)+ 2FF ′l′(F 2)Δv.

Plugging this inside Eq. (9), we get

−(F ′′|∇v|2 + F ′Δv) − Fl′(F 2)(2(F ′)2l′(F 2)

+(2FF ′)2l′′(F 2) + 2Fl′(F 2)F ′′)|∇v|2 +

2Fl′(F 2)F ′′ + 2FF ′l′(F 2)Δv + ωF + f(|F (v)|2)F (v) = 0,

which properly reorganized becomes:

−Δv(F ′ + 2FF ′l′(F 2)

)+ ωF + f(|F (v)|2)F (v)

−|∇v|2 (F ′′ + 2(F ′)2l′(F 2) + (2FF ′)2l′′(F 2) + 2Fl′(F 2)F ′′) = 0.

Our goal is to eliminate the second line of the previous equation. To do that,we need to find F such that:

F ′′(1 + 2F 2(l′)2(F 2)) + 2(l′2(F 2))F (F ′)2 + 4F 3l′′(F 2)l′(F 2)(F ′)2.

We can rewrite this equation as

−F ′′

F ′ =2(l′)2FF ′ + 4F 3l′′l′(F ′)2

1 + 2F 2(l′2),

which can be rewritten as

−F ′′

F ′ =12

ddv

(1 + 2F 2(l′2)

)1 + 2F 2(l′2)

.

Integrating this side by side, we get

− ln |F ′| =12

ln(1 + 2F 2(l′)2

),

using the condition F ′(0) = 1. Therefore, we have{F ′ = 1√

1+2F 2(l′)2

F (0) = 0.(10)

If we can get a F with good properties (see Lemma 3.1 below), we reducedEq. (9) to the following semilinear equation

−(

1√1 + 2F 2(l′)2

+2F 2(l′)2√

1 + 2F 2(l′)2

)Δv = f(F 2(v))F (v) − ωF (v),

which can be simplified to

− Δv =f(F 2(v))F (v) − ωF (v)√

1 + 2F 2(l′)2. (11)

Now, we have to analyze the solutions of the ODE (10).


Lemma 3.1. The solution F of the equation{F ′ = 1√

1+2(l′)2F 2

F (0) = 0.

satisfies the following properties:• F is uniquely defined, smooth and invertible;• |F ′(v)| ≤ 1, for all v ∈ R;• F (v)

v → 1 as v → 0.

Proof. The proof is straightforward and corresponding to Lemma 2.1 of [5]. �

As an application of Lemma 3.1, we have that

k(v) :=f(F 2(v))F (v) − ωF (v)√

1 + 2F 2(l′)2

satisfies the hypotheses of Theorem 2.4. Therefore, by Theorem 2.4, we obtainthe conclusion of Theorem 1.1.

Remark 3.2. See also [1] for completeness.


In this section, we prove Theorem 1.4. Therefore, consider the equation

Δpu + uΔpu2 + |x|−βg(u) = 0. (12)

Again, we try to find a transformation of the form u = T (v) with T : R+ → Rwhich exploits the semilinear structure of our problem. We get

∇u = ∇(T (v)) = T ′∇v.

We study the transformed equation term by term. First, we analyze the p-Laplacian term:

Δpu = div(|∇u|p−2∇u) = div(|T ′|p−2T ′|∇v|p−2∇v)

=∂

∂xj

(|T ′|p−2T ′|∇v|p−2 ∂v

∂xj

)

= T ′|p−2T ′|∇v|p−2Δv + |T ′|p−2T ′′|∇v|p−2 ∂v

∂xj

∂v

∂xj

+ (p − 2)|T ′|p−3T ′′|∇v|p−2T ′ T ′

|T ′|∂v

∂xj

∂v

∂xj

+ |T ′|p−2T ′ ∂

∂xj

(| ∂

∂xlv|p−2

) p−22 ∂v

∂xj

= T ′|p−2T ′|∇v|p−2Δv + |T ′|p−2T ′′|∇v|p−2|∇v|2+ |T ′|p−2T ′|∇v|p−2|∇v|2 + |T ′|p−2T ′∇|∇v|p−2 · ∇v

= T ′|p−2T ′|∇v|p−2Δv + |T ′|p−2T ′′|∇v|p


+ |T ′|p−2T ′′|∇v|p + |T ′|p−2T ′div∣∣∣∇v|p−2 · ∇v) − |T ′|p−2T ′Δv

= |T ′|p−2T ′′|∇v|p + |T ′|p−2T ′′|∇v|p + |T ′|p−2T ′div(|∇v|p−2 · ∇v)

Therefore:

Δpu = |T ′|p−2T ′Δpv + (p − 1)|∇v|p|T ′|p−2T ′′. (13)

Second, we consider Δpu2 and get

Δpu2 = div

(|2TT ′∇v|p−22TT ′∇v

)= 2p−1|TT ′|p−2TT ′Δpv

+ 2p−1(p − 1)|∇v|p|TT ′|p−2(TT ′

)′.

Putting these two computations together, we get

Δpu + uΔpu2 + |x|−βg(u) = |T ′|p−2T ′Δpv + (p − 1)|∇v|p|T ′|p−2T ′′

+T(2p−1|TT ′|p−2TT ′Δpv +2p−1(p − 1)|∇v|p|TT ′|p−2

(TT ′

)′)+|x|−βg(u)

= Δpv(|T ′|p−2T ′ + 2p−1|TT ′|p−2TT ′

)+ |∇v|p

((p − 1)|T ′|p−2T ′′ + 2p−1(p − 1)|TT ′|p−2

(TT ′

)′)+ |x|−βg(u).

Remark 4.1. Note that we have used that TT ′ is positive.

Now, we choose the following condition for the transformation T :

|T ′|p−2T ′′ + 2p−1|TT ′|p−2(TT ′

)′= 0,

in order to eliminate the coefficient of |∇v|p. We get

T ′′(1 + 2p−1|T |p) + 2p−1|T |p−2TT ′′ = 0.

Dividing by T ′ and by 1 + 2p−1|T |p, we obtain

−T ′′

T ′ =2p−1|T |p−2T

1 + 2p−1|T |pand so

−pd

dvlog|T ′| =

d

dvlog(1 + 2p−1|T |p).

The equation for T becomes{T ′(s) = 1

(1+2p−1T p(s))1/p

T (0) = 0.

With this condition, we get

Δpu + Δpu2 + |x|−βg(u) =

1T ′ Δpv + |x|−βg(T (v))

and so our original problem is reduced to

Δpv + |x|−β g(T (v))(1 + 2p−1T p(v))1/p

= 0.


The solution T has properties similar to the ones of the function F describedin Lemma 3.1 (see also [5]). Therefore, this equation falls in the hypothesesof Theorem 2.5 and so, by Theorem 2.5, we get the thesis. This concludes theproof of the theorem.

Remark 4.2. An interesting fact here is that, in absence of the extra termuΔ|u|2, the transformation T reduces to the identity. In fact:

Δpu = |T ′|p−2T ′Δpv + (p − 1)|∇v|p|T ′|p−2T ′′

and so to get rid of the term with |∇v|p, one needs |T ′| = 0 or T ′′ = 0 and sowith the use of the initial conditions one gets T (u) = u. Therefore, with thismethod, there is no way to reduce the p-Laplace equation to the 2-Laplaceequation.


In this section, we prove Theorem 1.7. Therefore, we consider the followingquasilinear reaction–diffusion equation

ut = Δ + uΔ|u|2 + g(u).

Here, g(u) satisfies the hypotheses of Theorem 1.7. We apply again the trans-formation u = F (v) with F : R+ → R. Similarly as in the previous sections,we get

ut = F ′(v)vt,

Δu = F ′′(v)|∇v|2 + F ′Δv.

and

Δ|u|2 = |∇v|2 (2(F ′)2 + 2FF ′′)+ 2FF ′Δv.

Plugging this inside our equation, we get:

F ′(v)vt =F ′′(v)|∇v|2+F ′Δv +F |∇v|2 (2(F ′)2+ 2FF ′′)+ 2FF ′Δv + g(F (v)).

Therefore, as before, if we take F such that{F ′(s) = 1√

1+2(l′)2F 2(s)

F (0) = 0,

we reduce our equation to

F ′vt = F ′(1 + 2F 2)Δv + g(F (v))

and hence to

vt = (1 + 2F 2)Δv +√

1 + 2(l′)2F 2g(F (v)).

This equation satisfies the hypotheses of Proposition 2.7, with

Ajj(t, x, v) = 1 + 2F 2(v)

and

Ajk(t, x, v) = 0


for j �= k, k = 1, . . . , N and B(t, x, v) :=√

1 + 2F 2g(F (v)). Therefore, we getlocal existence in time for our problem.

Remark 5.1. To get global estimates in time at least in L2(M), we notice that,multiplying

vt = (1 + 2F 2)Δv +√

1 + 2F 2g(F (v))

by v side by side, integrating by parts and using the periodicity of M = TN ,we get

12∂t‖v‖2L2(M) = −

∫M

(|∇v|2 (1 + 2F 2)3/2 + 4vF

(1 + 2F 2)1/2+√

1 + 2F 2g(F (v)))

.

Now, if g(F (v)) ≤ 0, we get

‖v(t)‖2L2(M) ≤ ‖v(0)‖2L2(M),

which gives a uniform in time global bound in L2(M).


In this section, we prove Theorem 1.9. Therefore, we consider the equation:

iut + Δu + uΔ(|u|2) + g(|u|2)u = 0. (14)

Rewriting it in an expanded way, it becomes

iut = −Δu − 2u|∇u|2 − 2u�(uΔu) − g(|u|2)u. (15)

Suppose that F is a holomorphic complex function on all C and consideru = F (v). Then we have

iut = iF ′(v)vt,

alsoΔu =

∑j ∂xj

∂xj(F (v))

=∑

j∂

∂xj

(F ′(v) · ∂v

∂xj

)=∑

j

(F ′′(v) ∂v

∂xj+ F ′ ∂2v

∂x2j

)= F ′′(v)|∇v|2 + F ′Δv.

Hence, the Eq. (15) for u can be transformed to an equation for v, which is infact

−iF ′(v)vt = −F ′′(v) · |∇v|2 − F ′(v)Δv − 2F (v)|F ′(v)|2 ·|∇v|2 − 2F (v)� (

FF ′′(v)|∇v|2 + FF ′Δv), (16)

and so

−iF ′(v)vt = |∇v|2[−F ′′(v) − 2F (v)|F ′(v)|2 − 2F�(F (v)F ′′(v))

]−F ′(v)Δv − 2F (v)�(F (v)F ′(v)Δv).

(17)

We want to eliminate the term with a gradient in the nonlinearity, thereforewe need that the following complex ODE is satisfied:

F ′′ + 2F |F ′|2 + 2F�(FF ′′) = 0. (18)


Now, we study the Eq. (18). Multiplying the equation by F , then

F ′′F + 2|F |2|F ′|2 + 2|F |2�(FF ′′) = 0.

The second and third term are both real valued and this makes the first termreal valued too, i.e., F ′′F is real. Let us exploit this condition. Since F isholomorphic, then ∂

∂z F = 0 and so F = F (z) = a(z) + ib(z). The condition�(F ′′F ) = 0 becomes

0 = �(F ′′F ) = b′′(z)a(z) − a′′(z)b(z).

We add and sum on both the sides the quantity a′b′ and get therefore

(b′a)′ = (a′b)′,

which, using the condition 0 = F (0) = a(0) + ib(0), becomes

b′a = a′b.

Integrating by parts again, we get

ln |a| = ln |b| + C

with C ∈ R. This implies that

|a(z)| = γ|b(z)|for some γ > 0 and so that F (z) = b(z)(1 ± iγ) or F (z) = b(z)eiθ with{

b′ = 1√1+2b2

b(0) = 0.

and with θ ∈ [0, 2π). This implies that u must be real up to a constant.Plugging this condition in Eq. (14), we get that ut = 0. Therefore u mustbe stationary and real up to a phase shift. This completes the proof of thetheorem.

Remark 6.1. This result is, in some sense, to be expected, because the requestof being holomorphic to a complex function can be thought as a reductionof dimensionality of that complex function. This does not affect the ellipticproblem, where the solution is essentially real and so lower dimensional, butdoes affect the complete evolution, which is fully complex.


In this section, we give a complete proof of Theorem 1.11. Consider the equa-tion

iut = −Δu − uΔ|u|2 + λ|u|p−1u.

We decompose the solution u and the equation into real and imaginary part.Therefore, the system satisfied by (u1, u2) with u = u1 + iu2, is{−u1t = −Δu1 − 2u1 (|∇u1|2 + |∇u2|2) − 2u1 (u1Δu1 + u2Δu2) + λ|u2

1 + u22|p−1u1

u2t = −Δu2 − 2u2 (|∇u1|2 + |∇u2|2) − 2u2 (u1Δu1 + u2Δu2) + λ|u21 + u2

2|p−1u2


Now, we use the transformation:{u1 = F (v, w)u2 = G(v, w)

We concentrate on one term at the time. First, we compute the time derivative:

u1t = Fvvt + Fwwt

and similarly

u2t = Gvvt + Gwwt.

Second, the Laplacian:

Δu1 =n∑

i=1

∂

∂xi

(∂

∂xiF (v, w)

)

=n∑

i=1

∂

∂xi

(∂

∂xiF (v, w)

)=

n∑i=1

∂

∂xi(Fvvxi

+ Fwwxi)

=n∑

i=1

∂

∂xi

(Fvv|vxi

|2 + Fvwvxiwxi

+ Fwvwxivxi

+ Fww|wxi|2)

= Fvv|∇v|2 + 2Fvw∇v∇w + Fww|∇w|2 + FvΔv + FwΔw.

Similarly:

Δu2 = Gvv|∇v|2 + 2Gvw∇v∇w + Gww|∇w|2 + GvΔv + GwΔw.

Then, the gradient terms:

|∇u1|2 = F 2v |∇v|2 + 2FvFw∇v∇w + F 2

w|∇w|2

and also

|∇u2|2 = G2v|∇v|2 + 2GvGw∇v∇w + G2

w|∇w|2.We can plug everything inside the equation and get the system:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

−Fvvt − Fwwt = −(1 + 2F 2)(Fvv|∇v|2

+2Fvw∇v∇w + Fww|∇w|2 + FvΔv + FwΔw)

−2F((F 2

v + G2v)|∇v|2 + 2(FvFw + GvGw)∇v∇w + (F 2

w + G2w)|∇w|2)

−2FG(Gvv|∇v|2 + 2Gvw∇v∇w + Gww|∇w|2

+GvΔv + GwΔw)

+ λ|F 2 + G2| p−12 F ;

Gvvt + Gwwt = −(1 + 2G2)(Gvv|∇v|2 + 2Gvw∇v∇w

+Gww|∇w|2 + GvΔv + GwΔw)

−2G((F 2

v + G2v)|∇v|2 + 2(FvFw + GvGw)∇v∇w + (F 2

w + G2w)|∇w|2)

−2FG(Fvv|∇v|2 + 2Fvw∇v∇w + Fww|∇w|2

+FvΔv + FwΔw)

+ λ|F 2 + G2| p−12 G.


Therefore, to eliminate the gradient terms we need to solve the solving quasi-linear system of coupled PDEs:⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

−Fvv(1 + 2F 2) − 2F (F 2v + G2

v) − 2FGGvv = 0,−2Fvw(1 + 2F 2) − 4F (FvFw + GvGw) − 4FGGvw = 0,

−Fww(1 + 2F 2) − 2F (F 2w + G2

w) − 2FGGww = 0,−Gvv(1 + 2G2) − 2G(F 2

v + G2v) − 2FGFvv = 0,

−2Gvw(1 + 2G2) − 4F (FvFw + GvGw) − 4FGFvw = 0,−Gww(1 + 2G2) − 2G(F 2

w + G2w) − 2FGFww = 0.

We can rewrite this system in the following matrix form⎡⎢⎢⎢⎢⎢⎢⎣

−(1 + 2F 2) 0 0 −2FG 0 00 −2(1 + 2F 2) 0 0 −4FG 00 0 −(1 + 2F 2) 0 0 −2FG

−2FG 0 0 −(1 + 2G2) 0 00 −4FG 0 0 −2(1 + 2G2) 00 0 −2FG 0 0 −(1 + 2G2)

⎤⎥⎥⎥⎥⎥⎥⎦

⎡⎢⎢⎢⎢⎢⎢⎣

Fvv

Fvw

Fww

Gvv

Gvw

Gww

⎤⎥⎥⎥⎥⎥⎥⎦

=

⎡⎢⎢⎢⎢⎢⎢⎣

2F (F 2v + G2

v)4F (FvFw + GvGw)

2F (F 2w + G2

w)2G(F 2

v + G2v)

4G(FvFw + GvGw)2G(F 2

w + G2w)

⎤⎥⎥⎥⎥⎥⎥⎦

.

We can rewrite this in a more compact form. Define

B :=

[B11 B12

B21 B22

]

=

⎡⎢⎢⎢⎢⎢⎢⎣

−(1 + 2F 2) 0 0 −2FG 0 00 −2(1 + 2F 2) 0 0 −4FG 00 0 −(1 + 2F 2) 0 0 −2FG

−2FG 0 0 −(1 + 2G2) 0 00 −4FG 0 0 −2(1 + 2G2) 00 0 −2FG 0 0 −(1 + 2G2)

⎤⎥⎥⎥⎥⎥⎥⎦

,

F ′′ :=

⎡⎢⎢⎢⎢⎢⎢⎣

Fvv

Fvw

Fww

Gvv

Gvw

Gww

⎤⎥⎥⎥⎥⎥⎥⎦

and

h :=

⎡⎢⎢⎢⎢⎢⎢⎣

2F (F 2v + G2

v)4F (FvFw + GvGw)

2F (F 2w + G2

w)2G(F 2

v + G2v)

4G(FvFw + GvGw)2G(F 2

w + G2w)

⎤⎥⎥⎥⎥⎥⎥⎦

.

Our system takes the form:

BF ′′ = h.


Here B11 = −(−1+2F 2)M , B12 = B21 = −2FGM and B21 = −(−1+2G2)M ,where

M :=

⎡⎣1 0 0

0 2 00 0 1

⎤⎦ .

This particular form turns out to be very helpful and the solution of this systemis not unfeasible as it could seem at first sight. This matrix B is partitionedinto four squared diagonal blocks. We use Lemma 2.1, to compute the inverseof the matrix B:

B−1 :=

((B11 − B12B

−122 B21

)−1 (−B−111 B12(B22 − B21B

−111 B12

)−1(−B−122 B21(B11 − B12B

−122 B21

)−1 (B22 − B21B

−111 B12

)−1

).

We compute entry by entry:(B11 − B12B

−122 B21

)−1=(

−(1 + 2F 2)M − 4F 2G2

1 + 2G2MM−1M

)

= −(

1 + 2(F 2 + G2)1 + 2G2

)−1

M−1 = − 1 + 2G2

1 + 2(F 2 + G2)

⎡⎣1 0 0

0 0.5 00 0 1

⎤⎦ .

Now, (−B−111 B12(B22 − B21B

−111 B12

)−1

=1

1 + 2F 2M−1(−2FG)M

(−(1 + 2G2)M − 4F 2G2

1 + 2F 2MM−1M

)

= − 2FG

1 + 2F 2

1 + 2F 2

1 + 2[F 2 + G2]M−1

=2FG

1 + 2(F 2 + G2)M−1 =

2FG

1 + 2(F 2 + G2)

⎡⎣1 0 0

0 0.5 00 0 1

⎤⎦ .

Therefore, also(−B−122 B21(B11 − B12B

−122 B21

)−1=

2FG

1 + 2(F 2 + G2)M−1

=2FG

1 + 2(F 2 + G2)

⎡⎣1 0 0

0 0.5 00 0 1

⎤⎦ .

And finally(B22 − B21B

−111 B12

)−1=(

−(1 + 2G2)M − 4F 2G2

1 + 2F 2MM−1M

)

= −(

1 + 2(F 2 + G2)1 + 2F 2

)−1

M−1 = − 1 + 2F 2

1 + 2(F 2 + G2)

⎡⎣1 0 0

0 0.5 00 0 1

⎤⎦ .

We can rewrite our system as follows

F ′′ = B−1h,


with B−1 taking the following form:

B−1 :=

[− 1+2G2

1+2(F 2+G2)M−1 − 2FG

1+2(F 2+G2)M−1

− 2FG1+2(F 2+G2)M

−1 − 1+2F 2

1+2(F 2+G2)M−1

].

Now, we apply the matrix B−1 to the vector h to obtain the vector F ′′. Thisgives the following system:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

Fvv = −2FF 2

v +G2v

1+2(F 2+G2) ,

Fvw = −2F FvFw+GvGw

1+2(F 2+G2) ,

Fww = −2FF 2

w+G2w

1+2(F 2+G2) ,

Gvv = −2GF 2

v +G2v

1+2(F 2+G2) ,

Gvw = −2GFvFw+GvGw

1+2(F 2+G2) ,

Gww = −2GF 2

w+G2w

1+2(F 2+G2) .

At first look, it seems very complicated. Assume that both F and G are notidentically zero. We deduce that

Fvv

F=

Gvv

G

Fvw

F=

Gvw

G

Fww

F=

Gww

G,

by taking ratios side by side. Let us consider the first equality. We get:

FvvG = GvvF.

Adding FvGv to both sides, we get:

FvvG + FvGv = GvvF + FvGv,

which becomes

(FvG)v = (GvF )v.

Integrating by parts and using the condition G(v, 0) = F (v, 0) = 0 (samecondition as in [5] and [6]), we obtain

FvG = GvF

and so (since both F and G must be not identically zero) one gets

Fv

F=

Gv

G

which implies, integrating side by side, that

G = F.

This condition decouples the system and reduces to three the number of inde-pendent equations of our original problem of six coupled equations. Therefore,we have to solve: ⎧⎪⎨

⎪⎩Fvv = − 4FF 2

v

1+4F 2 ,

Fvw = − 4FFvFw

1+4F 2 ,

Fww = − 4FF 2w

1+4F 2 .


Let us consider the first equation:

Fvv = − 4FF 2v

1 + 4F 2.

This is the same equation (there it was in one single variable) present in [5]and [6] in the stationary case. It can be rewritten as:

Fv =1√

1 + 4F 2

with F (0, w) = 0 and similarly

Fw =1√

1 + 4F 2

with F (v, 0) = 0. But this implies that

Fv = Fw,

which is a transport equation, whose solution is

F (v, w) = F (v + w) = F0(v + w) = 0,

because of our initial conditions. Therefore, the only solution to our systemwith initial conditions G(v, 0) = F (v, 0) = G(0, w) = F (0, w) = 0 is F = G =0, which is absurd, because we assumed that both F �= 0 and G �= 0. Thisimplies that at least one between F and G must be zero and so that u mustbe purely real or purely imaginary (and so real up to a constant phase shift).This implies again that u must be also stationary. This concludes the proof ofthe theorem.

Acknowledgements

I thank my family and Victoria Ban for their constant support. I thank Prof.Yun Wang for several discussions about Section 6. I thank my supervisor Prof.Narayanaswamy Balakrishnan for his constant help and inspiring guidance. Ithank the referees for their useful comments which lead to an improvement ofthe manuscript.

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Alessandro SelvitellaDepartment of Mathematics and StatisticsMcMaster University1280 Main Street WestHamiltonON L8S-4L8Canadae-mail: [email protected]

Received: 6 May 2015.

Accepted: 2 January 2016.

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The dual approach to stationary and evolution quasilinear PDEs · in lower regular spaces, a...

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