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Advances in Nonlinear Analysis

Editor-in-Chief: Radulescu, Vicentiu / Squassina, Marco


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Carleman estimates and null controllability of a class of singular parabolic equations

Runmei Du
  • School of Mathematics, Jilin University, Changchun 130012; and School of Basic Science, Changchun University of Technology, Changchun 130012, P. R. China
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/ Jürgen Eichhorn / Qiang Liu / Chunpeng Wang
Published Online: 2018-06-06 | DOI: https://doi.org/10.1515/anona-2016-0266

Abstract

In this paper, we consider control systems governed by a class of semilinear parabolic equations, which are singular at the boundary and possess singular convection and reaction terms. The systems are shown to be null controllable by establishing Carleman estimates, observability inequalities and energy estimates for solutions to linearized equations.

Keywords: Carleman estimate; null controllability; singular equation

MSC 2010: 93B05; 93C20; 35K67

1 Introduction

This paper concerns the null controllability of the system governed by the following first initial-boundary value problem:

xαut-uxx+c(x,t)u=h(x,t)χω,(x,t)QT=(0,1)×(0,T),(1.1)u(0,t)=u(1,t)=0,t(0,T),(1.2)u(x,0)=u0(x),x(0,1),(1.3)

where α>0, T>0, cL(QT), ω=(x0,x1) (0<x0<x1<1) is the control region and χω is the characteristic function of ω, hL2(QT) is the control function, and u0α with

α={ς is a measurable function in (0,1):xα/2ςL2(0,1)}.

System (1.1)–(1.3) is said to be null controllable if for each u0α there exists hL2(QT) such that its solution u satisfies u(,T)|(0,1)=0. It is noted that (1.1) is singular at the boundary x=0. Equations with such a boundary singularity arise in some physical problems such as the propagation of a thermal wave in an inhomogeneous medium [32, 36] and the Ockendon model for the flow in a channel of a fluid whose viscosity is temperature-dependent [21, 35]. By the coordinate transformation

y=xα+1,0<x<1,

problem (1.1)–(1.3) can be transformed into

u^t-(α+1)2(yα/(α+1)u^y)y+y-α/(α+1)c^(y,t)u^=y-α/(α+1)h^(y,t)χω^,(y,t)QT,(1.4)u^(0,t)=u^(1,t)=0,t(0,T),(1.5)u^(y,0)=u0(y1/(α+1)),y(0,1),(1.6)

where ω^=(x0α+1,x1α+1) and

u^(y,t)=u(y1/(α+1),t),c^(y,t)=c(y1/(α+1),t),h^(y,t)=h(y1/(α+1),t),(y,t)QT.

Here, (1.4) is degenerate at the boundary y=0, and its reaction term is singular.

Controllability theory, containing approximate controllability and exact controllability, has been widely investigated for semilinear uniformly parabolic equations over the last forty years and there have been a great number of results (see for instance [3, 20, 19, 18, 31] and the references therein for a detailed account). The study of the controllability for semilinear degenerate or singular parabolic equations just began about ten years ago and it is far from being completely solved although many results have been known. Parabolic equations may be not exact controllable for a general terminal datum since there is a smoothing effect for their solutions. As usual, one considers null controllability. That is to say, the zero function is chosen as the terminal datum.

There are some studies on the null controllability of systems governed by equations with boundary degeneracy, whose simple example is

ut-a0(xλux)x+c(x,t)u=h(x,t)χω,(x,t)QT,(1.7)

with a0>0, λ>0 and cL(QT). The degeneracy of (1.7) is classified as a weak one (0<λ<1) and a strong one (λ1), and different boundary conditions are prescribed for the two cases (see [2, 12, 13, 34, 39, 43]). Precisely, for each λ>0 and each u0L2(0,1), the problem of (1.7) subject to the following boundary and initial conditions is well-posed

u(0,t)=0 if 0<λ<1,(xλux)(0,t)=0 if λ1,t(0,T),(1.8)u(1,t)=0,t(0,T),(1.9)u(x,0)=u0(x),x(0,1).(1.10)

System (1.7)–(1.10) was proved to be null controllable if 0<λ<2 (see [2, 12, 13, 34]), while not if λ2 (cf. [11]). Besides, Wang [39] and Cannarsa et al. [7, 11, 10] proved that it is approximately controllable in L2(0,1) and regional null controllable, respectively, for each λ>0. Here, the approximate controllability in L2(0,1) means that for each u0,udL2(0,1) and ε>0 there exists hL2(QT) such that the solution u satisfies u(,T)-ud()L2(0,1)<ε, while the regional null controllability means that for each u0L2(0,1) and δ(0,1-x0) there exists hL2(QT) such that the solution u satisfies u(,T)|(x0+δ,1)=0. In [37, 24, 23], it was proved that the system of

ut-a0(xλux)x+c0xβu=h(x,t)χω,(x,t)QT,(1.11)

subject to (1.8)–(1.10) is null controllable, where a0>0, λ[0,2), 0<β<2-λ and c0, or a0=1, λ[0,1)(1,2), β=2-λ and c0-(1-λ)2/4. Since systems (1.1)–(1.3) and (1.4)–(1.6) are equivalent, choosing a0=(α+1)2 and λ=α/(α+1) in (1.7) and (1.11) shows that (1.1)–(1.3) is null controllable if x-αcL(QT) or x-βc is a constant in QT with some β(-α,2). However, the general case that cL(QT) is unknown yet. There are also other studies on the controllability for semilinear degenerate or singular parabolic equations such as [16, 22, 40, 41] for equations with first-order terms, [8, 9, 25, 26] for equations in nondivergence form, [26, 27, 28, 29] for equations with interior degeneracy, [4, 17, 14, 33, 38] for multi-dimensional equations and [1, 15] for coupled systems. Moreover, [5, 6] studied null controllability for heat equations with singular potentials.

The null controllability of system (1.1)–(1.3) is based on the Carleman estimate for solutions to the conjugate problem

xαvt+vxx-c(x,t)v=0,(x,t)QT,(1.12)v(0,t)=v(1,t)=0,t(0,T),(1.13)v(x,T)=vT(x),x(0,1),(1.14)

where vTα. A Carleman estimate is a weighted inequality that relates a global (weighted) energy with a weighted local norm of the solution. Since (1.12) is singular, the reaction term cannot be controlled by the diffusion term generally in establishing the Carleman estimate for solutions to problem (1.12)–(1.14). Therefore, there should be some further restrictions on ct or cxx. In the present paper, we prescribe the restriction on ct as x2ctL(QT). More generally, cL(QT) can be relaxed by xβcL(QT) with some β[0,2). Moreover, the Carleman estimate still holds if there are convection and reaction terms with suitable weights in (1.12). Precisely, we can establish the Carleman estimate of solutions to the problem of

xαvt+vxx+xα/2b(x,t)vx-c(x,t)v-xα/2-1γ(x,t)v=0,(x,t)QT,(1.15)

subject to (1.13) and (1.14), where b,xβc,x2ct,γL(QT) with some β[0,2). In establishing the Carleman estimate, the convection term and the second reaction term can be controlled by the diffusion term thanks to their weights. Since (1.15) is singular, solutions to problem (1.15), (1.13), (1.14) are weak and it is not convenient to estimate them. Thus we consider the regularized problem

(x+η)αvt+vxx+(x+η)α/2b(x,t)vx-c(x,t)v-(x+η)α/2-1γ(x,t)v=0,(x,t)QT,(1.16)v(0,t)=v(1,t)=0,t(0,T),(1.17)v(x,T)=vT(x),x(0,1),(1.18)

and estimate its solutions uniformly with respect to η, where 0<η<1, b,c,ct,γL(QT) and vTL2(0,1). The key for the Carleman estimate is the local one near the singular point x=0. To show the idea on the choice of the weights, we use the method of undetermined functions to determine the suitable weights. By complicated and detailed estimates, we establish the local Carleman estimate, uniformly with respect to η(0,1), near the singular point x=0. Combining this local Carleman estimate and the classical one for uniformly parabolic equations, we get the uniform Carleman estimate and thus the uniform observability inequality for solutions to problem (1.16)–(1.18), which imply the ones for solutions to problem (1.15), (1.13), (1.14) by a limit process as η0+. Owing to the uniform observability inequality, we can prove that the linear system

(x+η)αut-uxx+((x+η)α/2b(x,t)u)x+c(x,t)u+(x+η)α/2-1γ(x,t)u=h(x,t)χω,(x,t)QT,(1.19)u(0,t)=u(1,t)=0,t(0,T),(1.20)u(x,0)=u0(x),x(0,1),(1.21)

is null controllable and the control function is uniformly bounded by considering a family of functional minimum problems, where b,c,ct,γL(QT) and u0L2(0,1). It is noted that there is no other restriction on b and γ except for b,γL(QT). By a fixed point argument, we can show that the semilinear system of

(x+η)αut-uxx+(p~(x,t,u))x+q~(x,t,u)=h(x,t)χω,(x,t)QT,(1.22)

subject to (1.20) and (1.21) is null controllable and the control function is uniformly bounded, where p~ and q~ are two measurable functions in QT× such that for (x,t)QT and u,v,

p~(x,t,0)=0,|p~(x,t,u)-p~(x,t,v)|K(x+η)α/2|u-v|,(1.23)q~(x,t,0)=0,|q~(x,t,u)-q~(x,t,v)-c(x,t)(u-v)|K(x+η)α/2-1|u-v|(1.24)

with some K>0. Then, by the uniform estimates on the control functions and solutions of system (1.22), (1.20), (1.21) and a limit process as η0+, we prove that the semilinear system of the singular parabolic equation

xαut-uxx+(p(x,t,u))x+q(x,t,u)=h(x,t)χω,(x,t)QT=(0,1)×(0,T),(1.25)

subject to (1.2) and (1.3) is null controllable, where p and q are two measurable functions in QT× such that for (x,t)QT and u,v,

p(x,t,0)=0,|p(x,t,u)-p(x,t,v)|Kxα/2|u-v|,(1.26)q(x,t,0)=0,|q(x,t,u)-q(x,t,v)-c(x,t)(u-v)|Kxα/2-1|u-v|.(1.27)

In particular, in the case 0<α2, the system of

xαut-uxx+g(x,t,u)=h(x,t)χω,(x,t)QT,(1.28)

subject to (1.2) and (1.3) is null controllable, where g is a measurable function in QT× satisfying

g(x,t,0)=0,|g(x,t,u)-g(x,t,v)|K|u-v|,(x,t)QT,u,v.(1.29)

The paper is organized as follows: In Section 2, we prove the well-posedness of the singular problems (1.25), (1.2), (1.3) and (1.15), (1.13), (1.14) by doing energy estimates for solutions to the regularized problems (1.19)–(1.21) and (1.16)–(1.18), respectively. Carleman estimates and observability inequalities are proved in Section 3, where we first establish the uniform ones for solutions to the regularized problem (1.16)–(1.18) by complicated and detailed estimates and then get the ones for solutions to the singular problem (1.15), (1.13), (1.14) by a limit process. Subsequently, the null controllability is studied in Section 4. Owing to the uniform observability inequality, we show that the regularized system (1.19)–(1.21) is null controllable and the control function is uniformly bounded by considering a family of functional minimum problems. Then it follows from a fixed point argument that the semilinear system (1.22), (1.20), (1.21) is null controllable and the control function is uniformly bounded, which yields the null controllability of the semilinear singular system (1.25), (1.20), (1.21) by the uniform estimates and a limit process.

2 Energy estimates and well-posedness

In this section, we prove the well-posedness of the singular problems (1.25), (1.2), (1.3) and (1.15), (1.13), (1.14), which is based on uniform energy estimates for solutions to the regularized problems (1.19)–(1.21) and (1.16)–(1.18), respectively. More generally, instead of (1.15), we consider its semilinear case

xαvt+vxx+g^(x,t,v,vx)=0,(x,t)QT,(2.1)

where g^ is a measurable function in QT×2 such that for (x,t)QT and u,v,z,η,

g^(x,t,0,0)=0,|g^(x,t,u,z)-g^(x,t,v,η)-c(x,t)(u-v)|K(xα/2-1|u-v|+xα/2|z-η|).(2.2)

Definition 2.1.

A function uL2(0,T;H01(0,1)) with xα/2uL(0,T;L2(0,1)) is said to be a solution to problem (1.25), (1.2), (1.3) if

QT(-xαuςt+uxςx-p(x,t,u)ςx+q(x,t,u)ς)𝑑x𝑑t=QThχως𝑑x𝑑t+01xαu0(x)ς(x,0)𝑑x

for each ςC1(Q¯T) with ς(,T)|(0,1)=0 and ς(0,)|(0,T)=ς(1,)|(0,T)=0.

Definition 2.2.

A function vL2(0,T;H01(0,1))Hloc1(QT) with xα/2vL(0,T;L2(0,1)) is said to be a solution to problem (2.1), (1.13), (1.14) if

QT(xαvςt+vxςx-g^(x,t,v,vx)ς)𝑑x𝑑t=01xαvT(x)ς(x,T)𝑑x

for each ςC1(Q¯T) with ς(,0)|(0,1)=0 and ς(0,)|(0,T)=ς(1,)|(0,T)=0.

Remark 2.3.

Like the heat equation where α=0, both xαu and xαv are continuous from [0,T] to L2(0,1) with respect to the weak topology of L2(0,1), where u and v are solutions to problems (1.25), (1.2), (1.3) and (1.15), (1.13), (1.14), respectively. That is to say, u and v have some continuity with respect to the time variable so that they make sense at a time.

2.1 Energy estimates for solutions to linear problems

Since (1.25) and (1.15) are singular at the boundary x=0, the existence of solutions to problems (1.25), (1.2), (1.3) and (1.15), (1.13), (1.14) are based on energy estimates for solutions to the regularized problems (1.19)–(1.21) and (1.16)–(1.18), respectively. To do so, we first prove the following lemma similar to the Poincaré inequality.

Lemma 2.4.

Assume that 0<η<1 and vH1(0,1) with v(0)=0. Then for each δ(0,1] and λ[0,2) it holds that

0δ(x+η)-2v2(x)𝑑x40δv2(x)𝑑x,0δ(x+η)-λv2(x)𝑑x12-λδ2-λ0δv2(x)𝑑x.

Proof.

It follows from the Hölder inequality and the Fubini theorem that

0δ(x+η)-2v2(x)𝑑x=0δ(x+η)-2(0xv(y)𝑑y)2𝑑x0δ(x+η)-2(0x(y+η)-1/2𝑑y)(0x(y+η)1/2v2(y)𝑑y)𝑑x20δ(x+η)-3/2(0x(y+η)1/2v2(y)𝑑y)𝑑x=20δ(y+η)1/2v2(y)(yδ(x+η)-3/2𝑑x)𝑑y40δv2(y)𝑑y.

The Hölder inequality gives

0δ(x+η)-λv2(x)𝑑x=0δ(x+η)-λ(0xv(y)𝑑y)2𝑑x0δ(x+η)-λx(0xv2(y)𝑑y)𝑑x0δx1-λ𝑑x0δv2(x)𝑑x=12-λδ2-λ0δv2(y)𝑑y.

Consider problem (1.19)–(1.21).

Proposition 2.5.

Assume that α>0, 0β<2 and bL(QT),(x+η)βcL(QT),γL(QT)N for some N>0. Then for each hL2(QT) and u0L2(0,1) problem (1.19)–(1.21) admits a unique solution

uL2(0,T;H01(0,1))L(0,T;L2(0,1)).

Furthermore, the solution satisfies

(x+η)α/2uL(0,T;L2(0,1))+uL2(QT)+uxL2(QT)M((x+η)α/2u0L2(0,1)+hL2(QT))(2.3)

and for each ε(0,T),

0T-ε01(u(x,τ+ε)-u(x,τ))2𝑑x𝑑τMε1/(2+α)((x+η)α/2u0L2(0,1)2+hL2(QT)2),(2.4)

where M>0 depends only on N, T, α and β.

Proof.

The classical theory for parabolic equations (cf. [30, 42]) shows that problem (1.19)–(1.21) admits a unique solution uL2(0,T;H01(0,1))L(0,T;L2(0,1)). It suffices to prove (2.3) and (2.4). Without loss of generality, we can assume that u is a classical solution. Otherwise, one can consider the approximating problem with mollified b, c, γ, hχω and u0 (cf. [42]). For convenience, we use Ci (1i7) to denote generic constants depending only on N, T, α and β in the proof.

For each τ(0,T], multiplying (1.19) by u and integrating over Qτ=(0,1)×(0,τ) by parts with (1.20), we get

1201(x+η)αu2(x,τ)𝑑x+Qτux2𝑑x𝑑t=1201(x+η)αu02(x)𝑑x+Qτhχωu𝑑x𝑑t+Qτ(x+η)α/2buux𝑑x𝑑t-Qτcu2𝑑x𝑑t   -Qτ(x+η)α/2-1γu2𝑑x𝑑t1201(x+η)αu02(x)𝑑x+12(1+2αbL(QT)2)Qτu2𝑑x𝑑t+12Qτh2𝑑x𝑑t+12Qτux2𝑑x𝑑t   +(x+η)βcL(QT)Qτ(x+η)-βu2𝑑x𝑑t+2α/2γL(QT)Qτ(x+η)-1u2𝑑x𝑑t,

which, together with Lemma 2.4, yields

01(x+η)αu2(x,τ)𝑑x+Qτux2𝑑x𝑑t01(x+η)αu02(x)𝑑x+Qτh2𝑑x𝑑t+C10τ0δ(1+(x+η)-β+(x+η)-1)u2𝑑x𝑑t   +C10τδ1(1+(x+η)-β+(x+η)-1)u2𝑑x𝑑t01(x+η)αu02(x)𝑑x+Qτh2𝑑x𝑑t+C1(12δ2+12-βδ2-β+δ)0τ0δux2𝑑x𝑑t   +C1(δ-α+δ-α-β+δ-α-1)0τδ1(x+η)αu2𝑑x𝑑t

for each δ(0,1]. Choosing δ(0,1) such that

12δ2+12-βδ2-β+δ=min{12,12C1},

we obtain

01(x+η)αu2(x,τ)𝑑x+12Qτux2𝑑x𝑑t01(x+η)αu02(x)𝑑x+Qτh2𝑑x𝑑t+C2Qτ(x+η)αu2𝑑x𝑑t.(2.5)

Using the Gronwall inequality in (2.5), one gets that for each τ(0,T],

01(x+η)αu2(x,τ)𝑑xC3(01(x+η)αu02(x)𝑑x+Qτh2𝑑x𝑑t).(2.6)

Choosing τ=T in (2.5), together with (2.6), leads to

QTux2𝑑x𝑑tC4(01(x+η)αu02(x)𝑑x+QTh2𝑑x𝑑t).(2.7)

Then (2.3) follows from (2.6), (2.7) and Lemma 2.4.

Now we turn to proving (2.4). For each ε(0,T) and (x,τ)(0,1)×(0,T-ε), multiplying (1.19) by u(x,τ+ε)-u(x,τ) and then integrating over (τ,τ+ε) with respect to t, we get

(x+η)α(u(x,τ+ε)-u(x,τ))2=ττ+ε(uxx(x,t)-((x+η)α/2b(x,t)u(x,t))x)(u(x,τ+ε)-u(x,τ))𝑑t+ττ+ε(h(x,t)χω-(x+η)α/2-1γ(x,t)u(x,t))(u(x,τ+ε)-u(x,τ))𝑑t-ττ+εc(x,t)u(x,t)(u(x,τ+ε)-u(x,τ))𝑑t.(2.8)

Integrating (2.8) over (0,1)×(0,T-ε) by parts, together with (1.20), Lemma 2.4 and (2.3), leads to

0T-ε01(x+η)α(u(x,τ+ε)-u(x,τ))2𝑑x𝑑τ=-0T-ε01ττ+ε(ux(x,t)-(x+η)α/2b(x,t)u(x,t))(ux(x,τ+ε)-ux(x,τ))𝑑t𝑑x𝑑τ   +0T-ε01ττ+ε(h(x,t)χω-(x+η)α/2-1γ(x,t)u(x,t))(u(x,τ+ε)-u(x,τ))𝑑t𝑑x𝑑τ   -0T-ε01ττ+εc(x,t)u(x,t)(u(x,τ+ε)-u(x,τ))𝑑t𝑑x𝑑τ((T-ε)ττ+ε01(ux-(x+η)α/2bu)2𝑑x𝑑t)1/2(ε0T-ε01(ux(x,τ+ε)-ux(x,τ))2𝑑x𝑑τ)1/2+((T-ε)ττ+ε01(hχω-(x+η)α/2-1γu)2𝑑x𝑑t)1/2(ε0T-ε01(u(x,τ+ε)-u(x,τ))2𝑑x𝑑τ)1/2+(x+η)βcL(QT)((T-ε)ττ+ε01(x+η)-βu2𝑑x𝑑t)1/2(ε0T-ε01(x+η)-β(u(x,τ+ε)-u(x,τ))2𝑑x𝑑τ)1/22T1/2ε1/2uxL2(QT)(2uxL2(QT)2+2α+1bL(QT)2uL2(QT)2)1/2+2T1/2ε1/2uL2(QT)(2hL2(QT)2+2α+1γL(QT)2(x+η)-1uL2(QT)2)1/2+2T1/2ε1/2(x+η)βcL(QT)(x+η)-β/2uL2(QT)2C5ε1/2(uxL2(QT)2+uL2(QT)2+hL2(QT)2+(x+η)-β/2uL2(QT)2)C6ε1/2(01(x+η)αu02(x)𝑑x+QTh2𝑑x𝑑t).(2.9)

It follows from Lemma 2.4, (2.3) and (2.9) that

0T-ε01(u(x,τ+ε)-u(x,τ))2𝑑x𝑑τ=0T-ε0δ~(u(x,τ+ε)-u(x,τ))2𝑑x𝑑τ+0T-εδ~1(u(x,τ+ε)-u(x,τ))2𝑑x𝑑τ20T0δ~u2𝑑x𝑑t+δ~-α0T-εδ~1(x+η)α(u(x,τ+ε)-u(x,τ))2𝑑x𝑑τδ~20T0δ~ux2𝑑x𝑑t+C6δ~-αε1/2(01(x+η)αu02(x)𝑑x+QTh2𝑑x𝑑t)C7ε1/(2+α)(01(x+η)αu02(x)𝑑x+QTh2𝑑x𝑑t),

where δ~=min{1,ε1/(4+2α)}. ∎

Turn to problem (1.16)–(1.18).

Proposition 2.6.

Assume that α>0, 0β<2 and b,c,ct,γL(QT). Then for each vTL2(0,1), problem (1.16)–(1.18) admits a unique solution vL2(0,T;H01(0,1))L(0,T;L2(0,1))Hloc1(QT). Furthermore, the solution satisfies

(x+η)α/2vL(0,T;L2(0,1))+vL2(QT)+vxL2(QT)M1(x+η)α/2vTL2(0,1)(2.10)

and for each ε(0,T),

vL(0,T-ε;H1(0,1))+(x+η)α/2vtL2((0,1)×(0,T-ε))M1ε-1/2(x+η)α/2vTL2(0,1).(2.11)

If vTH01(0,1), then vL(0,T;H01(0,1))H1(QT)L2(0,T;H2(0,1)) and

vL(0,T;H1(0,1))+(x+η)α/2vtL2(QT)M2((x+η)α/2vTL2(0,1)+vTL2(0,1)).(2.12)

Here M1,M2>0 depend only on bL(QT), (x+η)βcL(QT), (x+η)2ctL(QT), γL(QT), T, α and β.

Proof.

The classical theory for parabolic equations (cf. [30, 42]) gives the well-posedness. It suffices to prove (2.10)–(2.12). Without loss of generality, we can assume that v is a classical solution. Otherwise, one can consider the approximating problem with mollified b, c, γ and vT (cf. [42]). For convenience, we use Ci (1i6) to denote generic constants depending only on bL(QT), (x+η)βcL(QT), (x+η)2ctL(QT), γL(QT), T, α and β in the proof.

Similarly to the proof of (2.3), one can prove (2.10). Below we prove (2.12). For each τ[0,T), multiplying (1.16) by vt and integrating over Q~τ=(0,1)×(τ,T) by parts, we get

Q~τ(x+η)αvt2𝑑x𝑑t+1201vx2(x,τ)𝑑x=1201(vT(x))2𝑑x-Q~τ(x+η)α/2bvxvt𝑑x𝑑t+1201c(x,T)vT2(x)𝑑x   -1201c(x,τ)v2(x,τ)𝑑x-12Q~τctv2𝑑x𝑑t+Q~τ(x+η)α/2-1γvvt𝑑x𝑑t1201(vT(x))2𝑑x+14Q~τ(x+η)αvt2𝑑x𝑑t+bL(QT)2Q~τvx2𝑑x𝑑t   +12(x+η)βcL(QT)01(x+η)-βvT2(x)𝑑x+12(x+η)βcL(QT)01(x+η)-βv2(x,τ)𝑑x   +(12(x+η)2ctL(QT)+γL(QT)2)Q~τ(x+η)-2v2𝑑x𝑑t+14Q~τ(x+η)αvt2𝑑x𝑑t,

which, together with Lemma 2.4, yields

Q~τ(x+η)αvt2𝑑x𝑑t+01vx2(x,τ)𝑑x01(vT(x))2𝑑x+C101(x+η)-βvT2(x)𝑑x+C1Q~τvx2𝑑x𝑑t+C101(x+η)-βv2(x,τ)𝑑x   +C1Q~τ(x+η)-2v2𝑑x𝑑t(1+C12-β)01(vT(x))2𝑑x+5C1Q~τvx2𝑑x𝑑t+C10δ(x+η)-βv2(x,τ)𝑑x+C1δ1(x+η)-βv2(x,τ)𝑑x(1+C12-β)01(vT(x))2𝑑x+5C1Q~τvx2𝑑x𝑑t+C12-βδ2-β0δvx2(x,τ)𝑑x   +C1δ-α-βδ1(x+η)αv2(x,τ)𝑑x

for each δ(0,1]. Choosing δ(0,1) such that C1δ2-β=min{1-β2,C22} and using (2.10), we obtain

Q~τ(x+η)αvt2𝑑x𝑑t+1201vx2(x,τ)𝑑x(1+C12-β)01(vT(x))2𝑑x+5C1Q~τvx2𝑑x𝑑t+C1δ-α-βδ1(x+η)αv2(x,τ)𝑑x(1+C12-β)01(vT(x))2𝑑x+5C1Q~τvx2𝑑x𝑑t+C201(x+η)αvT2(x)𝑑x.(2.13)

Using the Gronwall inequality in (2.13), we get that for each τ[0,T),

01vx2(x,τ)𝑑xC301(vT(x))2𝑑x+C401(x+η)αvT2(x)𝑑x.(2.14)

Choosing τ=0 in (2.13), together with (2.14), leads to

QT(x+η)αvt2𝑑x𝑑tC501(vT(x))2𝑑x+C601(x+η)αvT2(x)𝑑x.(2.15)

Then (2.12) follows from (2.14), (2.15) and Lemma 2.4.

Finally, let us prove (2.11). For each ε(0,T), inequality (2.10) implies that there exists ε~(0,ε) such that v(,T-ε~)H01(0,1) and

vx(,T-ε~)L2(0,1)(1ε0ε01vx2𝑑x𝑑t)1/2C6ε-1/2(x+η)α/2vTL2(0,1).(2.16)

Regarding v as a solution to problem (1.16)–(1.18) in (0,1)×(0,T-ε~), we get (2.11) from (2.12), (2.10) and (2.16). ∎

Remark 2.7.

Propositions 2.5 and 2.6 still hold if (x+η)2cL(QT)c0<14 (M,M1,M2 depend also on c0 in this case). The restriction c0<14 is owing to the first estimate in Lemma 2.4.

2.2 Well-posedness for semilinear singular problems

Theorem 2.8.

Assume that α>0, 0β<2, xβc,x2ctL(QT) and p~,q~ satisfy (1.23), (1.24). Then for each hL2(QT) and u0Iα, there exists uniquely a solution uL2(0,T;H01(0,1)) with xα/2uL(0,T;L2(0,1)) to problem (1.25), (1.2), (1.3).

Proof.

Assume that b,γL(QT) and bL(QT),γL(QT)K. Using the uniform estimates in Propositions 2.5 and 2.6, we can show the well-posedness of the problem of the linear equation

xαut-uxx+(xα/2b(x,t)u)x+c(x,t)u+xα/2-1γ(x,t)u=h(x,t)χω,(x,t)QT,

subject to (1.2) and (1.3). Furthermore, u satisfies

xα/2uL(0,T;L2(0,1))+uL2(QT)+uxL2(QT)M(xα/2u0L2(0,1)+hL2(QT))(2.17)

and for each ε(0,T),

0T-ε01(u(x,τ+ε)-u(x,τ))2𝑑x𝑑τMε1/(2+α)(xα/2u0L2(0,1)2+hL2(QT)2),(2.18)

where M>0 depends only on K, xβcL(QT), T, α and β. Using the uniform estimates (2.17) and (2.18), we can get the existence of solutions to problem (1.25), (1.2), (1.3) by a fixed point argument. And the uniqueness can be proved by the Holmgren method. The proof is so standard (see, for example, [39, Theorem 3.1]) that we omit the details here. ∎

Similarly, one can prove the well-posedness of problem (2.1), (1.13), (1.14).

Theorem 2.9.

Assume that α>0, 0β<2, xβc,x2ctL(QT) and g^ satisfies (2.2). Then for each vTIα there exists uniquely a solution vL2(0,T;H01(0,1))Hloc1(QT) with xα/2vL(0,T;L2(0,1)) to problem (2.1), (1.13), (1.14).

Remark 2.10.

Due to Remark 2.7, Theorems 2.8 and 2.9 still hold if xβcL(QT) (β[0,2)) is relaxed by x2cL(QT) with x2cL(QT)c0<14.

3 Carleman estimates and observability inequalities

In this section, we establish Carleman estimates and observability inequalities of solutions to the regularized problem (1.16)–(1.18) and the singular problem (1.15), (1.13), (1.14).

3.1 Estimates near the singular point

In this subsection, we consider the regularized problem (1.16)–(1.18) and we always assume that α>0, 0β<2, 0<x0<x1<1, 0<η<1, vTH01(0,1), and b,c,ct,γL(QT) satisfying

(x+η)βcL(QT)N,(x+η)2ctL(QT)N,bL(QT)K,γL(QT)K(3.1)

with some N,K>0. As shown in Proposition 2.6, problem (1.16)–(1.18) admits a unique solution

vL(0,T;H01(0,1))H1(QT)L2(0,T;H2(0,1)).

We first cut off v in the following way: Set

ω~=(2x0+x13,x0+2x13).(3.2)

Define

w(x,t)=ξ(x)v(x,t),(x,t)QT,(3.3)

where ξC([0,1]) satisfies

ξ(x){=1,x[0,2x0+x13],[0,1],xω~,=0,x[x0+2x13,1].(3.4)

It follows from (1.16) and (3.3) that w satisfies

(x+η)αwt+wxx-c(x,t)w=f(x,t),(x,t)QT,(3.5)

where

f(x,t)=-(x+η)α/2b(x,t)ξ(x)vx(x,t)+2ξ(x)vx(x,t)+ξ′′(x)v(x,t)+(x+η)α/2-1γ(x,t)ξ(x)v(x,t),(x,t)QT.(3.6)

Remark 3.1.

The convection term and the second reaction term of (1.16) are regarded as a known function in (3.5) since they can be controlled by the diffusion term for the Carleman estimate. However, the first reaction term of (1.16) has to be treated as a reaction term in (3.5).

Reformulate (3.5) in a similar way as for the classical Carleman estimate. For s>0, set

z(x,t)=esφ(x,t)w(x,t),φ(x,t)=θ(t)ψ(x),(x,t)QT.(3.7)

Here, θ takes the form of

θ(t)=1(t(T-t))4,t(0,T),(3.8)

as usual, which satisfies

|θ(t)|C0θ2(t),|θ′′(t)|C0θ2(t),t(0,T),(3.9)

with C0>0 depending only on T, while ψC2([0,1]) is a negative function and will be determined below (see (3.15)). It follows from (3.5) and (3.7) that z satisfies

Lsz=esφ((x+η)α(e-sφz)t+(e-sφz)xx-c(e-sφz))=fesφ,(x,t)QT.(3.10)

Decompose Lsz into

Lsz=Ls+z+Ls-z,(x,t)QT,

where

Ls+z=zxx-s(x+η)αφt(x,t)z+s2φx2(x,t)z-c(x,t)z,(x,t)QT,Ls-z=(x+η)αzt-2sφx(x,t)zx-sφxx(x,t)z,(x,t)QT.

It follows from (3.10) that

QT(x+η)-αLs+zLs-z𝑑x𝑑tQT(x+η)-αf2e2sφ𝑑x𝑑t.(3.11)

The following lemma gives the formula of the left-hand side of (3.11).

Lemma 3.2.

For each s>0,

QT(x+η)-αLs+zLs-z𝑑x𝑑t=sQT(2(x+η)-αφxx-α(x+η)-α-1φx)zx2𝑑x𝑑t+s3QT(2(x+η)-αφxx-α(x+η)-α-1φx)φx2z2𝑑x𝑑t-2s2QTφxφxtz2𝑑x𝑑t+s2QT((x+η)αφtt-((x+η)-αφxx)xx+2(x+η)-αcφxx)z2𝑑x𝑑t+12QTctz2dxdt+2sQT(x+η)-αcφxzzxdxdt+s0T(x+η)-αφxzx2dt|x=0.(3.12)

Proof.

Note that

QT(x+η)-αLs+zLs-z𝑑x𝑑t=QT(x+η)-αzxxLs-z𝑑x𝑑t-sQTφtzLs-z𝑑x𝑑t+s2QT(x+η)-αφx2zLs-z𝑑x𝑑t-QT(x+η)-αczLs-z𝑑x𝑑t.(3.13)

We compute the four integrals on the right-hand side of (3.13), respectively. Integrating by parts and using

z(x,0)=z(x,T)=0 for 0<x<1andz(0,t)=z(1,t)=zx(1,t)=0 for 0<t<T,

we get

QT(x+η)-αzxxLs-zdxdt=0Tzxztdt|x=0x=1-1201zx2dx|t=0t=T-s0T(x+η)-αφxzx2dt|x=0x=1+sQT((x+η)-αφx)xzx2dxdt-s0T(x+η)-αφxxzzxdt|x=0x=1+s20T((x+η)-αφxx)xz2dt|x=0x=1-s2QT((x+η)-αφxx)xxz2dxdt+sQT(x+η)-αφxxzx2𝑑x𝑑t=s0T(x+η)-αφxzx2dt|x=0+sQT(2(x+η)-αφxx-α(x+η)-α-1φx)zx2dxdt-s2QT((x+η)-αφxx)xxz2𝑑x𝑑t,-sQTφtzLs-zdxdt=-s201(x+η)αφtz2dx|t=0t=T+s2QT(x+η)αφttz2dxdt+s20Tφtφxz2dt|x=0x=1-s2QTφtxφxz2𝑑x𝑑t=s2QT(x+η)αφttz2𝑑x𝑑t-s2QTφtxφxz2𝑑x𝑑t,s2QT(x+η)-αφx2zLs-zdxdt=s2201φx2z2dx|t=0t=T-s2QTφxφxtz2dxdt-s30T(x+η)-αφx3z2dt|x=0x=1+2s3QT(x+η)-αφx2φxxz2𝑑x𝑑t-αs3QT(x+η)-α-1φx3z2𝑑x𝑑t=-s2QTφxφxtz2𝑑x𝑑t+s3QT(2(x+η)-αφxx-α(x+η)-α-1φx)φx2z2𝑑x𝑑t,-QT(x+η)-αczLs-zdxdt=-1201cz2dx|t=0t=T+12QTctz2dxdt+2sQT(x+η)-αcφxzzxdxdt+sQT(x+η)-αcφxxz2𝑑x𝑑t=12QTctz2𝑑x𝑑t+2sQT(x+η)-αcφxzzx𝑑x𝑑t+sQT(x+η)-αcφxxz2𝑑x𝑑t.

Then (3.12) follows by substituting the above four identities into (3.13). ∎

Remark 3.3.

The penultimate term in (3.12) is 2sQT(x+η)-αcφxzzx𝑑x𝑑t, which, integrated by parts, is equal to -sQT((x+η)-αcφx)xz2𝑑x𝑑t if cxL(QT). Note that the assumption on c is only c,ctL(QT). So this term is treated as an integral not of z2 but of zzx.

Owing to (3.12), in order to get a global weighted energy, one should choose ψ such that

2(x+η)-αψ′′(x)-α(x+η)-α-1ψ(x)>0,ψ(x)0,0<x<1.(3.14)

A choice of ψC2([0,1]) being a negative function and satisfying (3.14) is

ψ(x)=12+α(x+η)2+α-22+α,x(0,1),(3.15)

which satisfies

2(x+η)-αψ′′(x)-α(x+η)-α-1ψ(x)=2+α,ψ(x)=(x+η)1+α>0,0<x<1.(3.16)

For such ψ, one gets the following lemma.

Lemma 3.4.

There exist two positive constants s0 and M0, depending only on N, T, α and β, such that for each ss0,

QT(x+η)-αLs+zLs-z𝑑x𝑑tM0QT(sθzx2+s3(x+η)2+2αθ3z2)𝑑x𝑑t.(3.17)

Proof.

Substituting the definition of φ into (3.12) and using (3.16), we get after a direct calculation that

QT(x+η)-αLs+zLs-z𝑑x𝑑t(2+α)sQTθzx2𝑑x𝑑t+(2+α)s3QT(x+η)2+2αθ3z2𝑑x𝑑t-2s2QT(x+η)2+2αθθz2𝑑x𝑑t   +s2QT(x+η)αψθ′′z2𝑑x𝑑t+(1+α)sQTcθz2𝑑x𝑑t   +12QTctz2𝑑x𝑑t+2sQT(x+η)cθzzx𝑑x𝑑t.(3.18)

We estimate the last five terms on the right-hand side of (3.18). On one hand, one obtains from (3.9), the Hölder inequality and Lemma 2.4 that

|-2s2QT(x+η)2+2αθθz2𝑑x𝑑t+s2QT(x+η)αψθ′′z2𝑑x𝑑t|2C0s2QT(x+η)2+2αθ3z2𝑑x𝑑t+C02sQT(x+η)αθ2z2𝑑x𝑑t(2C0s2+C02s2)QT(x+η)2+2αθ3z2𝑑x𝑑t+116QT(x+η)-2θz2𝑑x𝑑t(2C0+C02)s2QT(x+η)2+2αθ3z2𝑑x𝑑t+14QTθzx2𝑑x𝑑t.(3.19)

On the other hand, a direct calculation and Lemma 2.4 show

|(1+α)sQTcθz2𝑑x𝑑t+12QTctz2𝑑x𝑑t+2sQTc(x+η)θzzx𝑑x𝑑t|(1+α)(x+η)βcL(QT)sQT(x+η)-βθz2𝑑x𝑑t+(T2)8(x+η)2ctL(QT)QT(x+η)-2θz2𝑑x𝑑t   +2(x+η)2+βc2L(QT)2sQT(x+η)-βθz2𝑑x𝑑t+12sQTθzx2𝑑x𝑑t12sQTθzx2𝑑x𝑑t+C1s0T0δ(x+η)-βθz2𝑑x𝑑t+C1s0Tδ1(x+η)-βθ3z2𝑑x𝑑t   +C1QT(x+η)-2θz2𝑑x𝑑t(3.20)12sQTθzx2𝑑x𝑑t+12-βδ2-βC1s0T0δθzx2𝑑x𝑑t+C1δ-β-2-2αs0Tδ1(x+η)2+2αθ3z2𝑑x𝑑t+4C1QTθzx2𝑑x𝑑t(12+12-βδ2-βC1)sQTθzx2𝑑x𝑑t+C1δ-β-2-2αsQT(x+η)2+2αθ3z2𝑑x𝑑t+4C1QTθzx2𝑑x𝑑t(3.21)

for each δ(0,1], where C1>0 depends only on N, T, α and β. Choose δ(0,1) so small that

δ2-βC1=min{1-β2,C12}.

Then it follows from (3.21) that

|(1+α)sQTcθz2𝑑x𝑑t+12QTctz2𝑑x𝑑t+2sQT(x+η)cθzzx𝑑x𝑑t|(s+4C1)QTθzx2𝑑x𝑑t+C2sQT(x+η)2+2αθ3z2𝑑x𝑑t(3.22)

with C2>0 depending only on N, T, α and β. By substituting (3.19) and (3.22) into (3.18) and choosing s0=2C0+C02+4C1+C2+14, one gets (3.17). ∎

Remark 3.5.

In Lemma 3.4, M0 depends on ct, which is caused because we distribute cz into Ls+z. If cz is distributed into Ls-z, then M0 will depend on cx and cxx.

Remark 3.6.

Lemma 3.4 still holds if (x+η)βcL(QT)N (β[0,2)) is relaxed by

(x+η)2cL(QT)c0<14.

But s0 and M0 depend also on c0.

Below we prove the following Caccioppoli inequality.

Lemma 3.7.

There exists C>0 depending only on N, K, x0, x1, T, α and β such that for each s>0,

0Tω~vx2e2sφ𝑑x𝑑tC0Tωv2𝑑x𝑑t,(3.23)

where ω~ is defined in (3.2).

Proof.

Let ζC0(0,1) satisfy

ζ(x){=1,xω~,[0,1],xωω~,=0,x[0,1]ω.

For s>0, the definition of φ implies

0=0Tddt01(x+η)αζ2v2e2sφ𝑑x𝑑t=2sQT(x+η)αφtζ2v2e2sφ𝑑x𝑑t+2QT(x+η)αζ2vvte2sφ𝑑x𝑑t.(3.24)

Substituting (1.16) into (3.24) leads to

0=2sQT(x+η)αφtζ2v2e2sφ𝑑x𝑑t-2QTζ2vvxxe2sφ𝑑x𝑑t-2QT(x+η)α/2bζ2vvxe2sφ𝑑x𝑑t+2QTcζ2v2e2sφ𝑑x𝑑t+2QT(x+η)α/2-1γζ2v2e2sφ𝑑x𝑑t,s>0.(3.25)

Integrating by parts gives

-2QTζ2vvxxe2sφ𝑑x𝑑t=2QTζ2vx2e2sφ𝑑x𝑑t+4QTζζxvvxe2sφ𝑑x𝑑t+4sQTφxζ2vvxe2sφ𝑑x𝑑t,s>0.(3.26)

Owing to (3.25), (3.26) and the choice of ζ, one gets from the Hölder inequality that

QTζ2vx2e2sφ𝑑x𝑑t=-sQT(x+η)αζ2φtv2e2sφ𝑑x𝑑t-2QTζζxvvxe2sφ𝑑x𝑑t-2sQTφxζ2vvxe2sφ𝑑x𝑑t+QT(x+η)α/2bζ2vvxe2sφ𝑑x𝑑t-QTcζ2v2e2sφ𝑑x𝑑t-QT(x+η)α/2-1γζ2v2e2sφ𝑑x𝑑t12QTζ2vx2e2sφ𝑑x𝑑t+C~(s2+1)0Tωθ2v2e2sφ𝑑x𝑑t,s>0,

with C~>0 depending only on N, K, x0, x1, T, α and β. Therefore,

0Tω~vx2e2sφ𝑑x𝑑tQTζ2vx2e2sφ𝑑x𝑑t2C~(s2+1)0Tωθ2v2e2sφ𝑑x𝑑t,s>0.(3.27)

Then (3.23) follows from (3.27) with C=2C~sup{(s2+1)θ2e2sφ:s>0,(x,t)QT}. ∎

3.2 Carleman estimates

We are ready to establish the local Carleman estimate for solutions to problem (1.16)–(1.18), uniformly with respect to η(0,1), near the singular point x=0.

Proposition 3.8.

Assume that α>0, 0β<2, 0<x0<x1<1, 0<η<1 and b,c,ct,γL(QT) satisfying (3.1). Then there exist two positive constants s1>0 and M1>0, depending only on N, K, x0, x1, T, α and β, such that for each vTL2(0,1) and ss1 the solution v to problem (1.16)–(1.18) satisfies

0T0(2x0+x1)/3(sθvx2+s3(x+η)2+2αθ3v2)e2sφ𝑑x𝑑tM10Tωv2𝑑x𝑑t.(3.28)

Proof.

Without loss of generality, we can assume vTH01(0,1). Otherwise, (3.28) can be proved by a limit process. Due to (3.11) and (3.17), one gets that for each s>s0,

QT(sθzx2+s3(x+η)2+2αθ3z2)𝑑x𝑑t1M0QT(x+η)-αf2e2sφ𝑑x𝑑t,(3.29)

where z is given in (3.7). For ss0, substituting (3.7), (3.3) and (3.6) into (3.29) yields

0T0(2x0+x1)/3(sθvx2+s3(x+η)2+2αθ3v2)e2sφ𝑑x𝑑tsQTθwx2e2sφ𝑑x𝑑t+s3QT(x+η)2+2αθ3w2e2sφ𝑑x𝑑t2sQTθzx2𝑑x𝑑t+3s3QT(x+η)2+2αθ3z2𝑑x𝑑t3M0QT(x+η)-α(-(x+η)α/2bξvx+2ξvx+ξ′′v+(x+η)α/2-1γξv)2e2sφ𝑑x𝑑tM0Tω~(v2+vx2)e2sφ𝑑x𝑑t+M0T0(2x0+x1)/3θ(vx2+(x+η)-2v2)e2sφ𝑑x𝑑t,(3.30)

where M>0 depends only on N, K, x0, x1, T, α and β. Lemma 2.4 shows

0T0(2x0+x1)/3(x+η)-2θv2e2sφ𝑑x𝑑t40T0(2x0+x1)/3θ(vx+2sφxv)2e2sφ𝑑x𝑑t80T0(2x0+x1)/3θvx2e2sφ𝑑x𝑑t+32s20T0(2x0+x1)/3(x+η)2+2αθ3v2e2sφ𝑑x𝑑t.(3.31)

Choose s1=s0+64M. Then it follows from (3.30) and (3.31) that

0T0(2x0+x1)/3(sθvx2+s3(x+η)2+2αθ3v2)e2sφ𝑑x𝑑t2M0Tω~(v2+vx2)e2sφ𝑑x𝑑t,ss1,

which, together with (3.23), yields (3.28). ∎

Remark 3.9.

The assumption on ct, the factors (x+η)α/2 and (x+η)α/2-1 in the convection term and the second reaction term of (1.16) are necessary when one establishes the Carleman estimate in such a way.

Define

Φ(x,t)=ξ(x)φ(x,t)+(1-ξ(x))ϕ(x,t),(x,t)QT,

where ξ, φ are given by (3.4), (3.7), (3.8) and (3.15), while

ϕ(x,t)=(er(1-x)-e2r)θ(t),(x,t)QT,

with r being a positive constant. Then one can prove the following Carleman estimate.

Theorem 3.10 (Uniform Carleman estimate).

Assume that α>0, 0β<2, 0<x0<x1<1, 0<η<1 and b,c,ct,γL(QT) satisfying (3.1). Then there exist r>0, s2>0 and M2>0, depending only on N, K, x0, x1, T, α and β, such that for each vTL2(0,1) and ss2 the solution v to problem (1.16)–(1.18) satisfies

QT(sθvx2+s3(x+η)2+2αθ3v2)e2sΦ𝑑x𝑑tM20Tωv2𝑑x𝑑t.

Proof.

For convenience, we use Ci (1i4) to denote generic constants depending only on N, K, x0, x1, T, α and β. Let

v~(x,t)=(1-ξ(x))v(x,t)=v(x,t)-w(x,t),(x,t)QT.

Then v~ solves the following problem:

{(x+η)αv~t+v~xx-c(x,t)v~=f~(x,t),(x,t)(x0,1)×(0,T),v~(x0,t)=v~(1,t)=0,t(0,T),v~(x,T)=(1-ξ(x))vT(x),x(x0,1),

where

f~(x,t)=-(x+η)α/2b(x,t)(1-ξ(x))vx(x,t)-2ξ(x)vx(x,t)-ξ′′(x)v(x,t)+(x+η)α/2-1γ(x,t)(1-ξ(x))v(x,t),(x,t)(x0,1)×(0,T).

Due to the classical Carleman estimate (see [3]), there exist r>0, s~1>0 and M~1>0, depending only on N, K, x0, x1, T, α and β, such that for each ss~1,

0Tx01(sθv~x2+s3θ3v~2)e2sϕ𝑑x𝑑tM~1(0Tx01f~2e2sϕ𝑑x𝑑t+0Tω~v~2𝑑x𝑑t).

Therefore, for each ss~1,

0T(x0+2x1)/31(sθvx2+s3(x+η)2+2αθ3v2)e2sϕ𝑑x𝑑tC1(0Tω~vx2e2sϕ𝑑x𝑑t+0Tωv2𝑑x𝑑t).(3.32)

Combine (3.28) and (3.32) to get that for each smax{s~1,s1},

0T01(sθvx2+s3(x+η)2+2αθ3v2)e2sΦ𝑑x𝑑t=0T0(2x0+x1)/3(sθvx2+s3(x+η)2+2αθ3v2)e2sφ𝑑x𝑑t+0Tω~(sθvx2+s3(x+η)2+2αθ3v2)e2sΦ𝑑x𝑑t   +0T(x0+2x1)/31(sθvx2+s3(x+η)2+2αθ3v2)e2sϕ𝑑x𝑑t(M1+C1)0Tωv2𝑑x𝑑t+0Tω~(sθvx2+s3(x+η)2+2αθ3v2)e2sΦ𝑑x𝑑t+C10Tω~vx2e2sϕ𝑑x𝑑t.(3.33)

Similar to the proof of (3.23), one can prove the following Caccioppoli inequalities:

s0Tω~θvx2e2sΦ𝑑x𝑑tC20Tωv2𝑑x𝑑t,0Tω~vx2e2sϕ𝑑x𝑑tC30Tωv2𝑑x𝑑t,s>0.(3.34)

It is clear that

s30Tω~(x+η)2+2αθ3v2e2sΦ𝑑x𝑑tC40Tωv2𝑑x𝑑t,s>0.(3.35)

Then the theorem is proved by combining (3.33)–(3.35). ∎

3.3 Observability inequalities

Theorem 3.11 (Uniform observability inequality).

Assume that α>0, 0β<2, 0<x0<x1<1, 0<η<1 and b,c,ct,γL(QT) satisfying (3.1). Then there exists M3>0, depending only on N, K, x0, x1, T, α and β, such that for each vTL2(0,1) the solution v to problem (1.16)–(1.18) satisfies

01(x+η)αv2(x,0)𝑑xM30Tωv2𝑑x𝑑t.

Proof.

Without loss of generality, we can assume vTH01(0,1). Otherwise, the theorem can be proved by a limit process. For convenience, we use Ci (1i3) to denote generic constants depending only on N, K, x0, x1, T, α and β. Set

m(s)=min0<x<1,T/4t3T/4{sθ(t)e2sΦ(x,t),s3θ3(t)e2sΦ(x,t)},s>0.

Then for each s>0,

QT(sθvx2e2sΦ+s3(x+η)2+2αθ3v2e2sΦ)𝑑x𝑑tm(s)T/43T/401((x+η)2+2αv2+vx2)𝑑x𝑑t,

which, together with Theorem 3.10, leads to

T/43T/401((x+η)2+2αv2+vx2)𝑑x𝑑tC10Tωv2𝑑x𝑑t.(3.36)

For each τ(0,T), multiplying (1.16) by eMtv with M>0 to be determined, and then integrating over (0,1)×(0,τ) by parts, we get that

1201(x+η)αeMτv2(x,τ)𝑑x-1201(x+η)αv2(x,0)𝑑x=M2Qτ(x+η)αeMtv2𝑑x𝑑t+QτeMtvx2𝑑x𝑑t-Qτ(x+η)α/2beMtvvx𝑑x𝑑t   +QτceMtv2𝑑x𝑑t+Qτ(x+η)α/2-1γeMtv2𝑑x𝑑tM2Qτ(x+η)αeMtv2𝑑x𝑑t+12QτeMtvx2𝑑x𝑑t-2α-1bL(QT)2QτeMtv2𝑑x𝑑t   -(x+η)βcL(QT)Qτ(x+η)-βeMtv2𝑑x𝑑t-2α/2γL(QT)Qτ(x+η)-1eMtv2𝑑x𝑑tM2Qτ(x+η)αeMtv2𝑑x𝑑t+12QτeMtvx2𝑑x𝑑t-C2QτeMtv2𝑑x𝑑t   -C2Qτ(x+η)-βeMtv2𝑑x𝑑t-C2Qτ(x+η)-1eMtv2𝑑x𝑑t.(3.37)

Lemma 2.4 gives

QτeMtv2𝑑x𝑑t12δ20τ0δeMtvx2𝑑x𝑑t+0τδ1eMtv2𝑑x𝑑t,(3.38)Qτ(x+η)-βeMtv2𝑑x𝑑t12-βδ2-β0τ0δeMtvx2𝑑x𝑑t+0τδ1(x+η)-βeMtv2𝑑x𝑑t,(3.39)Qτ(x+η)-1eMtv2𝑑x𝑑tδ0τ0δeMtvx2𝑑x𝑑t+0τδ1(x+η)-1eMtv2𝑑x𝑑t(3.40)

for each δ(0,1]. Choose δ(0,1) such that

12δ2+12-βδ2-β+δ=min{12C2,1}.

It follows from (3.37)–(3.40) that

01(x+η)αeMτv2(x,τ)𝑑x-01(x+η)αv2(x,0)𝑑xMQτ(x+η)αeMtv2𝑑x𝑑t-C3Qτ(x+η)αeMtv2𝑑x𝑑t.(3.41)

Choose M=C3 in (3.41) to obtain

01(x+η)αv2(x,0)𝑑xeC3T01(x+η)αv2(x,t)𝑑x,0<t<T.(3.42)

Integrating (3.42) over (T4,3T4) and using Lemma 2.4, we get

01(x+η)αv2(x,0)𝑑x2TeC3TT/43T/401(x+η)αv2𝑑x𝑑t21+αTeC3TT/43T/401v2𝑑x𝑑t2αTeC3TT/43T/401vx2𝑑x𝑑t.(3.43)

Then the theorem follows from (3.36) and (3.43). ∎

Remark 3.12.

Proposition 3.8 and Theorems 3.10 and 3.11 still hold if (x+η)βcL(QT)N (β[0,2)) is relaxed by (x+η)2cL(QT)c0<14. But s1,s2,M1,M2,M3 depend also on c0.

As a corollary of Theorems 3.10 and 3.11, one can get the following Carleman estimate and observability inequality for the singular problem (1.15), (1.13), (1.14).

Theorem 3.13 (Carleman estimate, observability inequality).

Assume that α>0, 0β<2, 0<x0<x1<1 and b,xβc,x2ct,γL(QT). Then for each vTIα the solution v to problem (1.15), (1.13), (1.14) satisfies

QT(sθvx2+s3x2+2αθ3v2)e2sΦ𝑑x𝑑tM20Tωv2𝑑x𝑑t,ss2,

and

01xαv2(x,0)𝑑xM30Tωv2𝑑x𝑑t,

where M2, s2 and M3 are given in Theorems 3.10 and 3.11, which depend only on bL(QT), xβcL(QT), x2ctL(QT), γL(QT), x0, x1, T, α and β.

Proof.

Choose {vTη}0<η<1L2(0,1) and cηC1(Q¯T) such that

(x+η)α/2vTηL2(0,1)+(x+η)βcηL(QT)+(x+η)2ctηL(QT)xα/2vTL2(0,1)+xβcL(QT)+x2ctL(QT),0<η<1,(3.44)

and

limη001((x+η)α/2vTη(x)-xα/2vT(x))2𝑑x=0,limη0QT((x+η)βcη(x)-xβc(x))2𝑑x𝑑t=0.(3.45)

Denote by

vηL2(0,T;H01(0,1))L(0,T;L2(0,1))Hloc1(QT)

the solution to problem (1.16)–(1.18) with vT=vTη and c=cη. That is to say, for each ςC1(Q¯T) with ς(,0)|(0,1)=0 and ς(0,)|(0,T)=ς(1,)|(0,T)=0 it holds that

QT((x+η)αvηςt+vxηςx-(x+η)α/2bvxης+cηvης+(x+η)α/2-1γvης)𝑑x𝑑t=01(x+η)αvη(x,T)ς(x,T)𝑑x.(3.46)

Proposition 2.6 shows that

(x+η)α/2vηL(0,T;L2(0,1))+vηL2(QT)+vxηL2(QT)M(x+η)α/2vTηL2(0,1),(3.47)vηL(0,T-ε;H1(0,1))+(x+η)α/2vtηL2((0,1)×(0,T-ε))Mε-1/2(x+η)α/2vTηL2(0,1)(3.48)

for each ε(0,T), where M>0 depends on bL(QT), xβcL(QT), x2ctL(QT), γL(QT), T, α and β. Owing to Theorems 3.10 and 3.11, vη satisfies

QT(sθ(vxη)2+s3(x+η)2+2αθ3(vη)2)e2sΦ𝑑x𝑑tM20Tω(vη)2𝑑x𝑑t,ss2,(3.49)01(x+η)α(vη)2(x,0)𝑑xM30Tω(vη)2𝑑x𝑑t,(3.50)

where M2, s2 and M3 are given in Theorems 3.10 and 3.11, which depend only on bL(QT), xβcηL(QT), x2ctηL(QT), γL(QT), x0, x1, T, α and β. Due to (3.44), (3.47) and (3.48), there exist {ηn}n=1(0,1) with limnηn=0, and vL2(0,T;H01(0,1)) with xα/2vL(0,T;L2(0,1)), such that

vηnv and vxηnvxweakly in L2(QT),(3.51)vηnvin L2(ω×(0,T)) as n,(x+ηn)α/2vηn(x,0)xα/2v(x,0)weakly in L2(0,1) as n.(3.52)

For each ςC1(Q¯T) with ς(0,)|(0,T)=ς(1,)|(0,T)=0, choosing η=ηn in (3.46) and then letting n, together with (3.45) and (3.51), we get that v is the solution to problem (1.15), (1.13), (1.14). Thanks to (3.51) and (3.52), the proof of the theorem is complete by letting n in (3.49) and (3.50). ∎

Remark 3.14.

Owing to Remark 3.12, Theorem 3.13 still holds if xβcL(QT) (β[0,2)) is relaxed by x2cL(QT) with x2cL(QT)c0<1/4.

4 Null controllability

In this section, we prove the null controllability of system (1.25), (1.2), (1.3). It can be proved by using Theorem 3.13. However, (1.25) is singular and its solutions are weak. It is more convenient that one first shows the null controllability of the regularized system (1.22), (1.20), (1.21) by using the uniform observability inequality (Theorem 3.11) and then taking a limit by means of uniform energy estimates (Propositions 2.5 and 2.6).

4.1 Linear case

Lemma 4.1.

Assume α>0, 0β<2, 0<x0<x1<1, 0<η<1, (x+η)βcL(QT),(x+η)2ctL(QT)N and bL(QT),γL(QT)K with some N,K>0. Then for each u0L2(0,1) there exists hL2(QT) such that the solution u to problem (1.19)–(1.21) satisfies

u(x,T)=0,x(0,1).(4.1)

Furthermore, there exists M>0 depending only on N, K, x0, x1, T, α and β such that

hL2(QT)M(x+η)α/2u0L2(0,1).(4.2)

Proof.

For each ε(0,1), choose bεC1(Q¯T) to satisfy

bεL(QT)bL(QT)K,bε-bL2(QT)<ε.(4.3)

Consider the following minimum problem:

minhL2(QT)(QTh2𝑑x𝑑t+1ε01u2(x,T)𝑑x),(4.4)

where u is the solution to problem (1.19)–(1.21) with b=bε. By a standard argument (cf. [3]), one can prove that the minimum problem (4.4) admits a unique solution (uε,hε) with

hε=χωvεin QT,

where

uεL2(0,T;H01(0,1))L(0,T;L2(0,1))L(τ,T;H1(0,1))H1((0,1)×(τ,T))

(for each τ(0,T)) solves the problem

{(x+η)αutε-uxxε+((x+η)α/2bε(x,t)uε)x+c(x,t)uε+(x+η)α/2-1γ(x,t)uε=vε(x,t)χω,(x,t)QT,uε(0,t)=uε(1,t)=0,t(0,T),uε(x,0)=u0(x),x(0,1),

while vεL(0,T;H01(0,1))H1(QT) solves the conjugate problem

{(x+η)αvtε+vxxε+(x+η)α/2bε(x,t)vxε-c(x,t)vε-(x+η)α/2-1γ(x,t)vε=0,(x,t)QT,vε(0,t)=vε(1,t)=0,t(0,T),vε(x,T)=-1εuε(x,T),x(0,1).

That is to say, for each ςL2(0,T;H01(0,1))L(0,T;L2(0,1))H1(QT) and ϱL2(0,T;H01(0,1)) it holds that

QT(-(x+η)αuεςt+uxεςx-(x+η)α/2bεuεςx+cuες+(x+η)α/2-1γuες)𝑑x𝑑t=QTvεχως𝑑x𝑑t+01(x+η)αu0(x)ς(x,0)𝑑x-01(x+η)αuε(x,T)ς(x,T)𝑑x(4.5)

and

QT((x+η)αvtεϱ-vxεϱx+(x+η)α/2bεvxεϱ-cvεϱ-(x+η)α/2-1γvεϱ)𝑑x𝑑t=0.(4.6)

Choosing ς=vε in (4.5) and ϱ=uε in (4.6) yields

0Tω(vε)2𝑑x𝑑t+1ε01(x+η)α(uε)2(x,T)𝑑x=-01(x+η)αvε(x,0)u0(x)𝑑x.(4.7)

It follows from the Hölder inequality and Theorem 3.11 together with the first inequality in (4.3) that

|01(x+η)αvε(x,0)u0(x)𝑑x|(01(x+η)α(vε)2(x,0)𝑑x)1/2(01(x+η)αu02(x)𝑑x)1/2M~(0Tω(vε)2𝑑x𝑑t)1/2(01(x+η)αu02(x)𝑑x)1/2120Tω(vε)2𝑑x𝑑t+12M~201(x+η)αu02(x)𝑑x,

where M~>0 depends only on N, K, x0, x1, T, α and β. Combining this estimate with (4.7) leads to

12QT(hε)2𝑑x𝑑t+1ε01(x+η)α(uε)2(x,T)𝑑x12M~201(x+η)αu02(x)𝑑x.(4.8)

Due to the first inequality in (4.3), (4.8) and Proposition 2.5, there exist a positive sequence {εn}n=1(0,1) with limnεn=0, hL2(QT) and uL2(0,T;H01(0,1))L(0,T;L2(0,1)) such that

hεnh,uεnu,uxεnux  weakly in L2(QT) as n,(4.9)limn01(x+η)α(uεn)2(x,T)𝑑x=0.(4.10)

For each ςC1(Q¯T) with ς(0,)|(0,T)=ς(1,)|(0,T)=0, choosing ε=εn in (4.5) and then letting n, together with the second inequality in (4.3), (4.9) and (4.10), we get that

QT(-(x+η)αuςt+uxςx-(x+η)α/2buςx+cuς+(x+η)α/2-1γuς)𝑑x𝑑t=QThχως𝑑x𝑑t+01(x+η)αu0(x)ς(x,0)𝑑x.

Therefore, u is the solution to problem (1.19)–(1.21) and satisfies (4.1). Finally, (4.2) follows from (4.8) and (4.9). ∎

4.2 Semilinear case

Owing to Lemma 4.1 and Proposition 2.5, one can prove by a fixed point argument (see, for example, [7, 10]) that the semilinear system (1.22), (1.20), (1.21) is null controllable. The proof is standard and we omit it here.

Lemma 4.2.

Assume that α>0, 0β<2, 0<x0<x1<1, xβcL(QT),x2ctL(QT)N for some N>0, and p~,q~ satisfy (1.23), (1.24). Then for each u0L2(0,1) there exists hL2(QT) such that the solution u to problem (1.22), (1.20), (1.21) satisfies

u(x,T)=0,x(0,1).

Furthermore, there exists M>0 depending only on N, K, x0, x1, T, α and β such that

hL2(QT)M(x+η)α/2u0L2(0,1).

Remark 4.3.

Lemmas 4.1 and 4.2 still hold if (x+η)βcL(QT)N (β[0,2)) is relaxed by

(x+η)2cL(QT)c0<14.

But M depends also on c0.

Turn to the null controllability of system (1.25), (1.2), (1.3).

Theorem 4.4.

Assume α>0, 0β<2, 0<x0<x1<1, xβc,x2ctL(QT) and p,q satisfy (1.26), (1.27). Then system (1.25), (1.2), (1.3) is null controllable. More precisely, for each u0Iα there exists hL2(QT) such that the solution u to problem (1.25), (1.2), (1.3) satisfies

u(x,T)=0,x(0,1).(4.11)

Furthermore, there exists M>0, depending only on xβcL(QT), x2ctL(QT), K, x0, x1, T, α and β, such that

hL2(QT)Mxα/2u0L2(0,1).(4.12)

Proof.

Set

p^(x,t,u)=x-α/2p(x,t,u),q^(x,t,u)=x1-α/2(q(x,t,u)-c(x,t)u),(x,t,u)QT×.

Then (1.25) is equivalent to

xαut-uxx+(xα/2p^(x,t,u))x+c(x,t)u+xα/2-1q^(x,t,u)=h(x,t)χω,(x,t)QT.

Choose {u0η}0<η<1L2(0,1) and cηC1(Q¯T) such that

(x+η)α/2u0ηL2(0,1)+(x+η)βcηL(QT)+(x+η)2ctηL(QT)xα/2u0L2(0,1)+xβcL(QT)+x2ctL(QT),0<η<1,(4.13)

and

limη001((x+η)α/2u0η(x)-xα/2u0(x))2𝑑x=0,limη0QT((x+η)βcη(x)-xβc(x))2𝑑x𝑑t=0.(4.14)

Owing to Lemma 4.2, for each η(0,1) there exists hηL2(QT) with

hηL2(QT)M(x+η)α/2u0ηL2(0,1)(4.15)

such that the solution uηL2(0,T;H01(0,1))L(0,T;L2(0,1)) to the problem

{(x+η)αutη-uxxη+((x+η)α/2p^(x,t,uη))x+cη(x,t)uη+(x+η)α/2-1q^(x,t,uη)=hη(x,t)χω,(x,t)QT,uη(0,t)=uη(1,t)=0,t(0,T),uη(x,0)=u0η(x),x(0,1)

satisfies

uη(x,T)=0,x(0,1),

where M>0 depends only on xβcηL(QT), x2ctηL(QT), K, x0, x1, T, α and β. That is to say,

QT(-(x+η)αuηςt+uxηςx-(x+η)α/2p^(x,t,uη)ςx+cηuης+(x+η)α/2-1q^(x,t,uη)ς)𝑑x𝑑t=QThηχως𝑑x𝑑t+01(x+η)αu0η(x)ς(x,0)𝑑x(4.16)

for each ςC1(Q¯T) with ς(0,)|(0,T)=ς(1,)|(0,T)=0. Rewrite the equation of uη as the following linear equation:

(x+η)αutη-uxxη+((x+η)α/2bη(x,t)uη)x+cη(x,t)uη+(x+η)α/2-1γη(x,t)uη=hη(x,t)χω,(x,t)QT,

where

bη(x,t)={p^(x,t,uη(x,t))uη(x,t),uη(x,t)0,0,uη(x,t)=0,(x,t)QT,γη(x,t)={q^(x,t,uη(x,t))uη(x,t),uη(x,t)0,0,uη(x,t)=0,(x,t)QT.

Then Proposition 2.5 shows

(x+η)α/2uηL(0,T;L2(0,1))+uηL2(QT)+uxηL2(QT)M~((x+η)α/2u0ηL2(0,1)+hηL2(QT)2)(4.17)

and

0T-ε01(uη(x,τ+ε)-uη(x,τ))2𝑑x𝑑τM~ε1/(2+α)((x+η)α/2u0ηL2(0,1)2+hηL2(QT)2)(4.18)

for each ε(0,T), where M~>0 depends only on xβcηL(QT), K, T, α and β. Owing to (4.13), (4.15), (4.17) and (4.18), there exist ηn(0,1) (n1) with limnηn=0, EmQT (m1) with limmmeas(Em)=0, hL2(QT) and uL2(0,T;H01(0,1)) with xα/2uL(0,T;L2(0,1)), such that

hηnh,uηnu,uxηnux  weakly in L2(QT) as n,(4.19)uηnuuniformly in QTEm as n for each positive integer m.(4.20)

For each ζL2(QT), it holds that

|QT((x+ηn)α/2p^(x,t,uηn)-xα/2p^(x,t,u))ζ𝑑x𝑑t||Aδ((x+ηn)α/2p^(x,t,uηn)-xα/2p^(x,t,u))ζ𝑑x𝑑t|   +|BδEm((x+ηn)α/2p^(x,t,uηn)-xα/2p^(x,t,u))ζ𝑑x𝑑t|   +|BδEm((x+ηn)α/2p^(x,t,uηn)-xα/2p^(x,t,u))ζ𝑑x𝑑t|(x+ηn)α/2p^(x,t,uηn)-xα/2p^(x,t,u)L2(QT)(ζL2(Aδ)+ζL2(BδEm))   +|BδEm((x+ηn)α/2p^(x,t,uηn)-xα/2p^(x,t,u))ζ𝑑x𝑑t|,

where Aδ=(0,δ)×(0,T) and Bδ=(δ,1)×(0,T) with δ(0,1). Letting n, m and then δ0 in turn, we get from (4.17) and (4.20) that

limnQT((x+ηn)α/2p^(x,t,uηn)-xα/2p^(x,t,u))ζ𝑑x𝑑t=0,

which shows

(x+ηn)α/2p^(x,t,uηn)xα/2p^(x,t,u)weakly in L2(QT) as n.(4.21)

A similar discussion gives

(x+ηn)α/2-1q^(x,t,uηn)xα/2-1q^(x,t,u)weakly in L2(QT) as n.(4.22)

For each ςC1(Q¯T) with ς(0,)|(0,T)=ς(1,)|(0,T)=0, letting n in (4.16), together with (4.19), (4.21), (4.22) and (4.14), leads to

QT(-xαuςt+uxςx-xα/2p^(x,t,u)ςx+cuς+xα/2-1q^(x,t,u)ς)𝑑x𝑑t=QThχως𝑑x𝑑t+01xαu0(x)ς(x,0)𝑑x,

or equivalently to

QT(-xαuςt+uxςx-p(x,t,u)ςx+q(x,t,u)ς)𝑑x𝑑t=QThχως𝑑x𝑑t+01xαu0(x)ς(x,0)𝑑x.

Hence, u is the solution to problem (1.25), (1.2), (1.3) and satisfies (4.11). Finally, (4.12) follows from (4.15) and (4.19). ∎

Remark 4.5.

Owing to Remarks 4.3 and 2.7, Theorem 4.4 still holds if xβcL(QT) (β[0,2)) is relaxed by x2cL(QT) with x2cL(QT)c0<14.

Remark 4.6.

Theorem 4.4 still holds if ω(0,1) is a nonempty open set.

Corollary 4.7.

Assume that 0<α2, 0<x0<x1<1 and g satisfies (1.29). Then system (1.28), (1.2), (1.3) is null controllable.

Acknowledgements

The authors would like to express their sincerely thanks to the referees and the editor for their helpful comments on the original version of the paper.

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About the article

Received: 2016-12-12

Revised: 2018-01-18

Accepted: 2018-02-02

Published Online: 2018-06-06


Funding Source: National Natural Science Foundation of China

Award identifier / Grant number: 11222106

Award identifier / Grant number: 11571137

Supported by the German–Chinese research project on PDEs and the National Natural Science Foundation of China (nos. 11222106 and 11571137).


Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 1057–1082, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2016-0266.

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