Nonexistence of solutions to fractional parabolic problem with general nonlinearities

Rendiconti del Circolo Matematico di Palermo Series 2 https://doi.org/10.1007/s12215-023-00932-1 Nonexistence of solutions to fractional parabolic problem with general nonlinearities Lihong Zhang1·Yuchuan Liu1·Juan J. Nieto2·Guotao Wang1 Received: 26 April 2023 / Accepted: 27 June 2023 © The Author(s) 2023 Abstract In this content, we investigate a class of fractional parabolic equation with general nonlinearities ∂z(x,t) ∂t−( +λ)β 2z(x,t)=a(x1)f(z), where aand fare nondecreasing functions. We first prove that the monotone increasing property of the positive solutions in x1direction. Based on this, nonexistence of the solutions are obtained by using a contradiction argument. We believe these new ideas we introduced will be applied to solve more fractional parabolic problems. Keywords Fractional parabolic equation ·General nonlinearity ·Tempered fractional Laplacian ·Monotonicity Mathematics Subject Classification 35R11 ·35K91 All authors contributed equally to this work. BJuan J. Nieto [email protected] Lihong Zhang [email protected] Yuchuan Liu [email protected] Guotao Wang [email protected] 1School of Mathematics and Computer Science, Shanxi Normal University, Taiyuan, Shanxi 030031, China 2CITMAga, Departamento de Estatistica, Análise Matemática e Optimización, Universidade de Santiago de Compostela, 15782 Santiago de Compostela, Spain 123 L. Zhang et al. 1 Introduction As we all know, the nonexistence of solutions to indefinite elliptic and parabolic problems have been studied extensively. In [1–3], for the following elliptic problem with nonlinearities and the regular Laplacian −u(x)=a(x1)up(x), x∈Rn,1<p<∞. the Liouville theorems were studied. In [4], Chen and Zhu applied the extension method [5] to transform the problem to a local one. They derived the nonexistence of positive solutions to the following equation: (−)su(x)=x1up(x), where 1 2<s<1and1<p<∞.In[6], Poláˇcik and Quittner introduced indefinite parabolic problem with the regular Laplacian as follows, ∂z(x,t) ∂t−z(x,t)=a(x1)zp(x,t), (x,t)∈Rn×R, where x=(x1,x2,···,xN)∈Rn,t∈R,ais nondecreasing continuous function. They received the nonexistence of bounded positive solutions of the above equation. In [7], Chen, Wu and Wang studied the indefinite fractional parabolic equation ∂u(x,t) ∂t+(−)su(x,t)=x1up(x,t), (x,t)∈Rn×R, where 0 <s<1and1 <p<∞and they obtained the monotone increasing and nonexistence of positive bounded solutions. More details can be seen in [8–11]. In 2023, with the aid of the direct method of moving planes, we [12] studied a tempered fractional Laplacian parabolic equation with logarithmic nonlinearity, asymptotic symmetry and monotonicity of radial solution of the parabolic equation were obtained. As a supplement and continuation of our above research results, in this work, we will study the nonexistence of solutions for a class of tempered fractional Laplacian parabolic problem with general nonlinearity, which will further enrich the theory of tempered fractional Laplacian parabolic problem. To our knowledge, the nonexistence of solutions to parabolic equation with general nonlinearity is rarely studied. Here, we mainly focus on the following equation: ∂z(x,t) ∂t−( +λ)β 2z(x,t)=a(x1)f(z), (1.1) where aand fare nondecreasing functions and the tempered fractional Laplacian operator is defined as ( +λ)β 2z(x,t)=−Cn,β,λ P.V.Rn z(x,t)−z(y,t) eλ|x−y||x−y|n+βdy, where β∈(0,2),λis a sufficient small positive constant and Cn,β,λ =( n 2) 2πn 2|(−β)|.P.V. presents the cauchy principle value and (t)=∞ 0st−1e−sds is the Gamma function. Obviously, when z∈C1,1 loc Lβ,( +λ) β 2z(x,t)is well defined, where Lβ={z(·,t)∈ L1 loc(Rn)|Rn |z(x,t)| 1+|x|n+βdx <+∞}. The fractional Laplacian  2 βis the generator of the β-stable L´evy process, in which the second and all higher order moments diverge. It sometimes is referred to as a shortcoming 123 Nonexistence of solutions to fractional... when applied to physical processes. So a parameter λis introduced to temper the L´evy process. Tempered L´evy process is the scaling limit of the tempered L´evy flight, which makes the L´evy flight a more suitable physical model. Moreover, the tempered fractional Laplacian equation governs the probability distribution function of the position of the particles and some works on the tempered fractional Laplacian have been done by scholars. For example, in [13], Zhang, Deng and Fan developed the finite difference schemes for the tempered fractional Laplacian equation with the generalized Dirichlet type boundary condition. In [14], Zhang, Deng and Karniadakis established numerical methods in the Riesz basis Galerkin framework with respect to the tempered fractional Laplacian. In [15], Zhang, Hou, Ahmad and Wang studied the Choquard equation involving a generalized nonlinear tempered fractional p- Laplacian operator. In addition, more results on tempered fractional Laplacian operator can be found in [16–19]. The nonlocal property of the fractional Laplacian operator creates some difficulties to study it. To overcome this difficulty, an extension method was introduced by Caffarelli and Slivestre [20], which converts the nonlocal problem into a high dimensional local one. In addition, the method of moving planes in integral forms also has been widely used to study the nonlocal problems, please see [21,22]. However, some nonlocal operators cannot be solved by the above method. In [23], Chen, Li and Li put forward a novel approach: a direct method of moving planes method, which is a new idea to solve the fractional Laplacian problems. By using direct method of moving planes, in [24], Wang and Ren devoted to a nonlinear Schr¨odinger equation with the fractional Laplacian and Hardy potential and in [25], Zhang and Nie studied two nonlinear equations concerning Logarithmic Laplacian. Recently, in [26], in view of nonlocal parabolic problems, Chen, Wang and Niu developed the asymptotic method of moving planes and applied it on bounded or unbounded domains. Numerous results can be seen in [27–29]. In this article, we study parabolic equation involving the general nonlinearity by the direct method of moving planes. A mass of elliptic equations involving general nonlinearity have been studied by many authors. Here, we make a new attempt to study parabolic equation with the general nonlinearity to obtain monotonicity and nonexistence of its solution. 2 Preliminaries in order for the lemma to work, we introduce the following notations. We define Tα={x=(x1,x2,···,xn)∈Rn|x1=α, for ∈R} being the moving planes and α={x∈Rn|x1<α} being the region to the left of Tα. Also, xα=(2α−x1,x2,···,xn) is the reflection of xabout Tα. Meanwhile, we denote zα(x,t)=z(xα,t), Zα(x,t)=zα(x,t)−z(x,t). In order to continue the proof, we show the following lemma. 123 L. Zhang et al. Lemma 2.1 [9] Given any N(t)>0, there exists a positive constant k0such that if N(t)≤ −N(0),then C |x1(t)−α|β>k0>0,(2.1) where x(t)=(x1(t), ···,xn(t)) is a minimum point of ¯ Zα(x,t)in αfor each fixed t. 3 Main results For this part, our main results are given. The monotonicity of solutions in x1direction and the nonexistence of positive solutions are established by Theorem.3.1 and Theorem.3.2 respectively. The main content of Theorems are as follows. Theorem 3.1 Let z(x,t)∈(C1,1 loc (Rn)∩Lβ)×C1(R)be a positive bounded classical solution of (1.1), assume that (1.1) satisfies the following conditions: (H1)a(p)≤0,forp≤0; (H2)a(p)>0somewhere for p >0; (H2)f is positive and locally Lipschitz continuous. Then z(x,t)is monotone increasing in x1direction. Proof From the equation (1.1), we deduce that ∂Zα(x,t) ∂t−( +λ)β 2Zα(x,t) =a(xα 1)f(zα)−a(x1)f(z) =(a(xα 1)−a(x1)) f(zα)+a(x1)( f(zα)−f(z)) ≥a(x1)( f(zα)−f(z)) =a(x1)N(α, x)Zα(x,t) where N(α, x)=f(zα)−f(z) zα−z. Meanwhile, we impose the condition that N(α, x)is nonnegative. Next, we consider the following problem ∂Zα(x,t) ∂t−( +λ)β 2Zα(x,t)≥a(x1)N(α, x)Zα(x,t), (x,t)∈α×R, Zα(x,t)=−Zα(xα,t), (x,t)∈α×R, (3.1) Step 1.As usual, we want to show that Zα(x,t)≥0,(x,t)∈α×R,for is sufficiently negative.(3.2) The assumption that zis bounded, which cannot guarantee the minimum of Zαcan be obtained. To overcome this difficulty, we introduce an auxiliary function ¯ Zα(x,t)=Zα(x,t) h(x), where h(x)=|x−(α +1)e1|owith e1=(1,0,···,0),ois a small positive constant. Based on above, we know that the sign of ¯ Zα(x,t)is same as Zα(x,t). 123 Nonexistence of solutions to fractional... Letting |x|→+∞,wehave lim |x|→+∞ ¯ Zα(x,t)→0.(3.3) In the following processes, we will consider ¯ Zα(x,t). Accordingto(3.3), we deduce that there is x(t),then ¯ Zα(x(t), t)=inf x∈α ¯ Zα(x,t), for arbitrary fixed t∈R. Next, we infer that if ¯ Zα(x(t), t)<0,∂¯ Zα ∂t(x(t), t)≥−C |x1(t)−α|β¯ Zα(x(t), t). (3.4) In reality, on the basis of a similar calculation as (22) in [9], we have if Zα(x(t), t)<0,−( +λ)β 2Zα(x(t), t)≤C |x1(t)−α|βZα(x(t), t). Combined with (3.1), then ∂Zα ∂t(x(t), t)≥−C |x1(t)−α|βZα(x(t), t). According to the definition of ¯ Zα(x,t), we deduce (3.4). For arbitrary fixed t∈R,let N(t):= ¯ Zα(x(t), t)=inf x∈α ¯ Zα(x,t). Proving (3.2) is equivalent to prove the following (3.5) N(t)≥0,∀t∈R.(3.5) Now we proceed with the proof of (3.5). Suppose that (3.5) is invalid, there is a t∈R,then −N(0):= N(t)=¯ Zα(x(t), t)<0.(3.6) For arbitrary ¯ t<t, we set up a subsolution m(t)=−¯ Re−k0(t−¯ t), here k0is defined in (2.1)and −¯ R=inf α×R ¯ Zα(x,t). We show that ¯ Zα(x,t)≥m(t), (x,t)∈α×[ ¯ t,t].(3.7) Think about the function V(x,t)=¯ Zα(x,t)−m(t), (x,t)∈α×[ ¯ t,t]. From the construction of m(t),wehave V(x,t)=¯ Zα(x,t)−m(t)=¯ Zα(x,t)−(−¯ R)≥0,(x,t)∈α×{ ¯ t}; 123 L. Zhang et al. and V(x,t)=¯ Zα(x,t)−m(t)=−m(t)≥0,(x,t)∈Tα×[ ¯ t,t]. Assume that (3.7) is not true, there is (x(ˆ t), ˆ t)∈α×(¯ t,t],then V(x(ˆ t), ˆ t)=inf α×(¯ t,t] V(x,t)<0,(3.8) ∂V ∂t(x(ˆ t), ˆ t)≤0.(3.9) On one hand, in view of the definition of V(x,t),then ¯ Zα(x(ˆ t), ˆ t)=inf α ¯ Zα(x,ˆ t)<m(ˆ t)<0. Hence, from (3.4), one has ∂¯ Zα ∂t(x(ˆ t), ˆ t)≥−C |x1(ˆ t)−α|β¯ Zα(x(ˆ t), ˆ t). (3.10) But on the other, by (3.9), we derive that V(x(ˆ t), ˆ t)≤V(x(t), t), that is ¯ Zα(x(ˆ t), ˆ t)−¯ Zα(x(t), t)≤m(ˆ t)−m(t)≤0 due to monotone increasing property of m(t). As a result, N(ˆ t)=¯ Zα(x(ˆ t), ˆ t)≤¯ Zα(x(t), t)=N(t)=−N(0). (3.11) Taking account of Lemma 2.1,by(3.11), then C |x1(ˆ t)−α|β>k0>0. Combined with (3.10), we derive that ∂¯ Zα ∂t(x(ˆ t), ˆ t)≥−k0¯ Zα(x(ˆ t), ˆ t). (3.12) Then from (3.9), one has −k0m(ˆ t)=∂m ∂t(ˆ t)≥∂¯ Zα ∂t(x(ˆ t), ˆ t)≥−k0¯ Zα(x(ˆ t), ˆ t), which concludes that V(x(ˆ t), ˆ t)=¯ Zα(x(ˆ t), ˆ t)−m(ˆ t)≥0, which yields a contradiction to V(x(ˆ t), ˆ t)<0. Therefore, we derive that (3.7) holds. It means that ¯ Zα(x,t)≥m(t), (x,t)∈α×[ ¯ t,t]. For any ¯ t, the above formula is true. Letting ¯ t→−∞,wehavem(t)→0. Consequently, ¯ Zα(x,t)≥0,(x,t)∈α×(−∞,t], which contradicts to (3.6). Therefore, (3.5)istrue and so does (3.2). 123 Nonexistence of solutions to fractional... Remark 3.1 Using a proof similar to Step 1, we deduce that for arbitrary α>0, ¯ Zα(x,t)is asolutionto(3.1)and ¯ Zα(x(t), t)=inf x∈α ¯ Zα(x,t)<0, it follows that x1(t)>0. The above will be employed in Step 2. Step 2.On account of (3.2), we move the Tαas long as the inequality holds. Let α0=sup{α|Zν(x,t)≥0,∀(x,t)∈ν×R,ν ≤α}. Now we verify that α0=+∞.(3.13) We use a contradiction argument. Assume that 0 <α 0<+∞, then in view of the definition of α0, there is a sequence αkα0such that inf αk×RZαk(x,t)<0. Let ¯ Zαk(x,t)=Zαk(x,t) h(x), where h(x)is mentioned earlier. Then obviously, −Nk:= inf αk×R ¯ Zαk(x,t)<0.(3.14) Since t∈R, the minimum point of ¯ Zαkmight not be obtained for finite value t. For more information about ∂¯ Zαk(x,t) ∂t, we pick a sequence tk,andx(tk)and τk0, then ¯ Zαk(x(tk), tk)=inf αk ¯ Zαk(·,t)=−Nk+τkNk.(3.15) We construct an auxiliary function  Zαk(x,t)=¯ Zαk(x,t)−τkNkϑk(t), here ϑk(t)=ϑ(t−tk),ϑ(t)∈C∞ 0(R),|ϑ(t)|≤1and ϑ(t)=1|t|≤1 2, 0|t|≥2. Next we study the value of  Zαk(x,t)in αk×(tk−2,tk+2). In view of the definition of  Zαk(x,t),wehave  Zαk(x(tk), tk)=−Nk. Otherwise, when |t−tk|≥2,  Zαk(x,t)=¯ Zαk(x,t)≥−Nk. To sum up, the minimum point of  Zαk(x,t)is obtained in αk×(tk−2,tk+2). We denote it as (x(ˆ tk), ˆ tk).Thatis  Zαk(x(ˆ tk), ˆ tk)=inf αk×R(x,t)<0. 123 L. Zhang et al. Hence, ∂ Zαk ∂t(x(ˆ tk), ˆ tk)=0, which means that |∂¯ Zαk ∂t(x(ˆ tk), ˆ tk)|=|τkNk ∂ϑk ∂t|≤τkNk.(3.16) Combined the definition of Nkin (3.14)and  Zαk(x(ˆ tk), ˆ tk)≤ Zαk(x(tk), tk), we have −Nk≤¯ Zαk(x(ˆ tk), ˆ tk)≤¯ Zαk(x(tk), tk)=−Nk+τkNk.(3.17) From the definition of  Zαk(x,t),wehave ¯ Zαk(x(ˆ tk), ˆ tk)=inf αk (x,ˆ tk)<0. According to Remark 3.1, we know that x1(ˆ tk)>0. Therefore, we could assume 0 < x1(ˆ tk)<α 0+1. Then by a similar process as (22) in [9], we deduce −( +λ)β 2Zαk(x(ˆ tk), ˆ tk)≤C |x1(ˆ tk)−αk|βZαk(x(ˆ tk), ˆ tk). (3.18) Obverse that there is a positive number m1,then a(x1(ˆ tk))N(α, x)≤m1. Together with (3.1), (3.18)andZαk(x(ˆ tk), ˆ tk)<0, we arrive at ∂Zαk ∂t(x(ˆ tk), ˆ tk)+C |x1(ˆ tk)−αk|βZαk(x(ˆ tk), ˆ tk) ≥a(x1(ˆ tk))N(α, x)Zαk(x(ˆ tk), ˆ tk) ≥m1Zαk(x(ˆ tk), ˆ tk). (3.19) For above inequality, we divide h(x(ˆ tk), ˆ tk)on both sides. Then we obtain ∂¯ Zαk ∂t(x(ˆ tk), ˆ tk)+C |x1(ˆ tk)−αk|β¯ Zαk(x(ˆ tk), ˆ tk)≥m1¯ Zαk(x(ˆ tk), ˆ tk). (3.20) Together with (3.16),(3.17)and(3.20), we divide −Nkon both sides, one can arrive at C |x1(ˆ tk)−αk|β≤m1 2,for økis small,(3.21) which concludes that |x1(ˆ tk)−αk|≥m2>0 and |x1(ˆ tk)−α0|≥m2 2>0.(3.22) 123 Nonexistence of solutions to fractional... When kis sufficiently large, one has ∂Zαk ∂t(x(ˆ tk), ˆ tk)+C |x1(ˆ tk)−αk|βZαk(x(ˆ tk), ˆ tk) ≥(a(xα 1(ˆ tk)) −x1(ˆ tk)) f(Zαk)N(α, x)Zαk(x(ˆ tk), ˆ tk) ≥m3>0, (3.23) where we use the fact that fis locally Lipschitz continuous and Zαk(x,t)⇒Zα0(x,t)≥0. On account of Zαk(x(ˆ tk), ˆ tk), from (3.23), we derive that ∂Zαk ∂t(x(ˆ tk), ˆ tk)≥m3>0.(3.24) Next we let ˆ Zαk(x,t)=Zαk(x+x(ˆ tk), t+ˆ tk), from (3.24), we arrive at ∂ˆ Zαk ∂t(0,0)≥m3>0.(3.25) In view of [30], we have ˆ Zαk1+ε,β(1+ε) t,x≤m4,∀(x,t)∈×(−T,T)⊂⊂ Rn×R, it concludes that there is a subsequence of (x(ˆ tk), ˆ tk)and when k→+∞, ˆ Zαk(x,t)→ˆ Zα0(x,t), ∂ˆ Zαk ∂t(x,t)→∂ˆ Zα0 ∂t(x,t). On account of 0<x1(ˆ tk)≤αk, and αk→α0,k→+∞. Hence, there is a subsequence of x1(ˆ tk)and 0 ≤x0 1≤α0,then x1(ˆ tk)→x0 1. Now we think about ˆ Zα0(x,t). Obviously, one has ˆ Zα0(x,t)≥0,(x,t)∈α0−x0 1×R. Taking account of Zαk(x(ˆ tk), ˆ tk)<0, we arrive at ˆ Zα0(0,0)=0=inf α0−x0 1 ×R ˆ Zα0(x,t). 123