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Finite time blow-up for a wave equation with dynamic boundary condition at critical and high energy levels in control systems

  • We study the initial boundary value problem of linear homogeneous wave equation with dynamic boundary condition. We aim to prove the finite time blow-up of the solution at critical energy level or high energy level with the nonlinear damping term on boundary in control systems.

    Citation: Xiaoqiang Dai, Chao Yang, Shaobin Huang, Tao Yu, Yuanran Zhu. Finite time blow-up for a wave equation with dynamic boundary condition at critical and high energy levels in control systems[J]. Electronic Research Archive, 2020, 28(1): 91-102. doi: 10.3934/era.2020006

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  • We study the initial boundary value problem of linear homogeneous wave equation with dynamic boundary condition. We aim to prove the finite time blow-up of the solution at critical energy level or high energy level with the nonlinear damping term on boundary in control systems.



    In this paper, we mainly discuss the initial boundary value problem of linear homogeneous wave equation with dynamic boundary condition

    uttΔu=0    in(0,)×Ω, (1.1)
    u(x,t)=0    on[0,)×Γ0, (1.2)
    uν=Q(ut)+f(u)    on[0,)×Γ1, (1.3)
    u(x,0)=u0(x),ut(x,0)=u1(x)    onΩ, (1.4)

    where ΩRn, n1 is a regular, bounded and connected domain with boundary Ω=Γ0Γ1, Γ0Γ1=, where Γ0 and Γ1 are measurable over Ω endowed with the (n1) dimensional Lebesgue measures λn1(Γ0) and λn1(Γ1); Δ is the Laplacian operator with respect to the x; Q(ut)=|ut|m2ut, f(u)=|u|p2u, m2, p2. These properties of Ω, Γ0 and Γ1 will be assumed throughout the paper. The initial data are u0H1(Ω) and u1L2(Ω), with the compatibility condition u0=0 on Γ0. We always assume that λn1(Γ0)>0 and λn1(Γ1)>0 throughout the paper.

    For the wave equation with nonlinear dynamic boundary condition like problem (1.1)-(1.4), arising in the physical models and control systems, there have been many papers dealing with the existence and blow-up of the solution. In [4]-[7], [12]-[18] and [33], the global existence and decay properties of the solution of the problem (1.1)-(1.4) were proved for arbitrarily large initial data when f0 or f(x,u)u0 if Q and f were under some special assumptions. When f(x,u)u0, which means f is a source term, the situation is quite different. If f(x,u)=|u|p2u, p>2 and Q0, when the (n1)-dimensional Lebesgue measure λn1(Γ0) and λn1(Γ1) are assumed to be positive, the authors of [18] obtained the finite time blow-up when the initial energy is negative for problem (1.1)-(1.4). For the same problem above, Levine and Smith [11] proved the global existence of the solution when initial data u0 and u1 are very small. Zhang et al. [30] considered the Kirchhoff equation with dynamic boundary condition. They obtained the energy decay and blow-up of a solution with negative and small positive initial energy. Recently, some authors have studied the viscoelastic wave equation with boundary damping and source terms. In [9], Lee et al. proved the global existence and exponential growth solution. In [16], they proved the blow-up result of solutions under suitable conditions of the initial data, and in [17] they studied the existence and decay of solutions for a viscoelastic wave equation with acoustic boundary conditions. In [25], the author studied problem (1.1)-(1.4) when Q(ut)=|ut|m2ut and f(u)=|u|p2u, and showed that the solution of problem (1.1)-(1.4) globally exists in time for arbitrary initial data when 2pm, in opposition with the finite time blow-up occurring when m=2<p. In [2] they proved more general existence and stability results by using a natural tool for the problem–monotone operator theory that contrast with the Schauder fixed point arguments used in [25]. For the same problem with p>m, Zhang and Hu [29] obtained the nonexistence of the solution under the energy level E(0)<d when the initial data are in the unstable set. As well known, in the frame of potential well theory, the variational arguments are usually taken by considering different levels of the initial data, see its applications to differential kinds of model equations in[21], [27] and [28], which are very different from the other aspects of the studies on the wave equations [3], [10], [11], [22], [23], [26] and [31]. However, no results were obtained about the finite time blow-up of the solution for problem (1.1)-(1.4) when Q(ut)=|ut|m2ut and f(u)=|u|p2u, p>m at critical energy level E(0)=d or high energy level E(0)>d. The main purpose of this paper is to get the finite time blow-up result of the solution of the problem (1.1)-(1.4) when Q(ut)=|ut|m2ut and f(u)=|u|p2u, p>m for the critical initial data and arbitrarily large initial data. We mainly adapt the method introduced by Vitillaro in [24] to study the solution of problem (1.1)-(1.4) when Q(ut)=|ut|m2ut and f(u)=|u|p2u, p>m at critical energy level and the convexity method introduced by Gazzola and Squassina in [8] to get the finite time blow-up of the solution at high energy level.

    In Section 2, we introduce some basic setup, notations and some known results of the solution to problem (1.1)-(1.4). In section 3, we prove the blow-up result of the solution when E(0)=d. Finally in Section 4, we obtain the blow-up result when E(0)>d.

    First we denote

    =L2(Ω), q=Lq(Ω), q,Γ1=Lq(Γ1), 1q,

    and

    H1Γ0(Ω)={uH1(Ω)u|Γ0=0}, (2.1)

    (u|Γ0 is in the sense of trace). We can endow H1Γ0(Ω) the equivalent norm upp,Γ1=u because of the Poincarè inequality (see [17]) and the fact that λn1(Γ0)>0. We also define some useful functionals

    J(u)=12u21pupp,Γ1, (2.2)
    I(u)=u2upp,Γ1, (2.3)
    E(t)=12ut2+12u21pupp,Γ1. (2.4)

    All these functionals are defined on H1Γ1(Ω). According to the definition of E(t), we have

    E(0)=12u12+12u021pu0pp,Γ1.

    We also use the trace-Sobolev embedding H1Γ1(Ω)Lp(Γ1) for 2p<r introduced in [1], where

    r={2(n1)n2, if n3;+, if n=1,2. (2.5)

    We also have the embedding inequality

    upp,Γ1Cu, (2.6)

    where C is the embedding constant. Then we introduce the unstable set V defined by

    V={(u0,u1)H1Γ1(Ω)×L2(Ω)I(0)<0,0<E(0)=d}, (2.7)

    where d is the mountain pass level, characterized as

    d=infuH1Γ0(Ω),u|Γ10(supλ>0J(λu)). (2.8)

    We define

    λ1:=Cpp2. (2.9)

    It has been proved in [25] that (121p)λ21 is the potential well depth, that is

    d=(121p)λ21. (2.10)

    In [25], the author have proved the local and global existence of the solution for problem (1.1)-(1.4). Now we introduce some results in [25] as follows:

    Theorem 2.1. (Local existence of the solution) Suppose that m>1,2p<r and m>rr+1p. Then, given initial data u0H1Γ0(Ω) and u1L2(Ω), there is a T>0 and a weak solution u of the problem (1.1)-(1.4) on (0,T)×Ω such that uC([0,T];H1Γ0(Ω)C1([0,T];L2(Ω))), utLm((0,T)×Γ1),

    E(t)+tsuτ(τ)mm,Γ1dτ=E(s), (2.11)

    holds for 0stT.

    In this section, we mainly show the finite time blow-up of the solution when initial data are at critical level. In order to prove the finite time blow-up of the solution, we first prove some basic lemmas.

    Lemma 3.1. (Invariant manifolds) We suppose that m>1,2p<r and m>rr+1p. Let

    V={(u0,u1)H1Γ0(Ω)×L2(Ω)u0>λ1,0<E(0)=d},

    then we have V=V.

    Proof. First we show that VV. Let (u0,u1)V, then we have I(0)<0 (i.e u02<u0pp,Γ1). By using the embedding inequality (2.6), we can obtain that

    u02<u0pp,Γ1Cpu0p.

    Hence we have u0>Cpp2=λ1. So we get that VV. Then we show that VV. Let (u0,u1)V, then we have u0>λ1 and E(0)=d. Supposing by contradiction that I(0)0, we have

    u02u0pp,Γ1.

    Combining

    E(0)=12u122+12u0221pu0pp,Γ1,

    we can get

    12u0221pu0pp,Γ1d.

    Then we obtain that

    d(121p)u02.

    Since u0>λ1, it follows that

    d>(121p)λ21=d,

    which leads to a contradiction. This completes the proof.

    Lemma 3.2. (Invariant manifolds and boundness) Suppose that m>1,2p<r and m>rr+1p. Let (u0,u1)V and u(x,t) be the weak solution of the problem (1.1)-(1.4) on [0,Tmax). Then (u(t,),ut(t,)) remains inside V for any [0,Tmax). Furthermore, we have

    u(t)2<u(t)pp,Γ1,  t[0,tmax),u(t)p,Γ1>Cλ1,  t[0,tmax),u(t)>λ1,  t[0,tmax).

    Proof. From I(0)<0 and the continuity of I(u) respecting to t, it follows that there exists a sufficiently small t1>0, such that I(u)<0 for 0<t<t1. Combining (2.11), we set that d1=E(t1), then we have

    0<d1=dt10uτ(τ)mm,Γ1dτ<d.

    We choose t=t1 as the initial time, by the same proceeding in Theorem 2.3 [12], we can get that (u(t,),ut(t,)) remains inside V for any t[0,Tmax) (Here we have already known the invariance when E(0)<d, so by selecting t1 as the initial time, we can have E(t1)<d again, then we can use the invariance conclusion for E(t1)<d). Furthermore, noting I(u)<0 for all t[0,Tmax), according to the definition of I(u), we have

    u(t)2<u(t)pp,Γ1, t[0,Tmax). (3.1)

    Moreover, according to Lemma 3.1, we can obtain that

    u(t)>λ1, t[0,Tmax). (3.2)

    Then, by using (2.9) and (3.1), we have

    u(t)p,Γ1>Cλ1, t[0,Tmax). (3.3)

    This completes the proof.

    By the similar method in [24], we can prove that in the manifold V={(u0,u1)H1Γ0(Ω)×L2(Ω)u0>λ1,0<E(0)=d}, there is a constant λ2 between u(t) and λ1, i.e., there is a λ2>λ1 such that u(t)λ2>λ1. This will be given by the following lemma. And this lemma will be used to prove the finite time blow-up for the critical case E(0)=d.

    Lemma 3.3. Suppose that m>1,2p<r and m>rr+1p. Let (u0,u1)V and u(x,t) be the weak solution of the problem (1.1)-(1.4) on [0,Tmax). There is a λ2>λ1 such that u(t)λ2>λ1.

    Proof. According to Lemma 3.2, we have I(u)<0 for all t[0,Tmax). By (2.6), we have

    E(t)=12ut2+12u21pupp,Γ112u21pupp,Γ112u21pCpup:=g(u), (3.4)

    where g(λ)=12λ21pCpλp for λ0. It is easy to see that g takes its maximum at λ=λ1, with g(λ1)=d, being strictly decreasing for λλ1, and g(λ) as λ. Combining the fact that E(t) is decreasing when t[0,Tmax) and E(0)=d, we can continue to argue as follows. By the continuity of u(), there are only two possibilities:

    (a) there is a t00 such that E(t0)<d and u(t0)>λ1;

    (b) there is an ε0>0 such that E(t)=d on [0,ε0).

    In the first case, we choose t0 as the initial time. Due to the fact that E(t) and g(λ) are both decreasing and continuous, there exists a λ2>λ1 such that E(t0)=g(λ2). We now claim that

    u(t0)λ2.

    Suppose for contradiction that u(t0)<λ2 for some t[0,Tmax). By using (3.4) and the fact that g(λ) is a decreasing function, we have

    E(t0)g(u(t0))>g(λ2)=E(t0),

    which leads to a contradiction. We can also obtain that

    12u212λ22>12λ12.

    We then choose λ0=12λ12 and E1=(1p2)λ0. By doing the same process in Theorem 2.3 in [24], we can conclude the proof for the first case.

    In the second case, for t[0,ε0), we have E(t)=d. By using (2.11), we have

    t0uτ(τ)mm,Γ1dτ=0  t[0,ε0).

    Due to the fact that ut(t)mm,Γ10, we have ut=0 and u(t)=u0 on [0,ε0). Suppose for contradiction that u0<λ2 for t[0,ε0). We can obtain that

    d=E(0)=12u021pu0pp,Γ1g(u0)>g(λ2)=E(0)=d,

    which leads to a contradiction. This completes the proof.

    Theorem 3.4. (Finite time blow-up of solutions for E(0)=d) Assume that 1<m<p, m>rr+1p, 2p<r. If (u0,u1)V, then the solution of problem (1.1)-(1.4) blows up in finite time.

    Proof. Arguing by contradition, we assume that there exists a global weak solution of problem (1.1)-(1.4). We set H(t)=dE(t). By Theorem 2.1, E(t) is decreasing about t. So H(t) is an increasing function, then we have

    H(t)H(0)=dE(0)=0, t0. (3.5)

    Next, by using the definition of E(t), we have

    H(t)d12u(t)22+1pu(t)pp,Γ1, t0. (3.6)

    By using (3.2) and (2.10), we can obtain that

    d12u(t)22d12λ21=1pλ21<0.

    Combining (3.6), we have

    H(t)1pu(t)pp,Γ1d12u(t)22<0.

    Then (3.5) tells

    H(0)H(t)<1pu(t)pp,Γ1, t0. (3.7)

    Next, using the definition of E(t) and (3.6), we have

    ddt(u,ut)=ut2u(t)22+u(t)pp,Γ1Γ1|ut|m2utudσ=2ut2+(12p)u(t)pp,Γ12E(t)Γ1|ut|m2utudσ=2ut2+(12p)u(t)pp,Γ12d+2H(t)Γ1|ut|m2utudσ.

    Then, According to Lemma 3.3, we choose a λ2, such that u(t)λ2>λ1, then by using (3.3), we have

    ddt(u,ut)2ut2+(12p2d(Cλ2)p)u(t)pp,Γ1   +2H(t)Γ1|ut|m2utudσ=2ut2+C1u(t)pp,Γ1+2H(t)Γ1|ut|m2utudσ, (3.8)

    where C1=12p2d(Cλ2)p>0. To obtain ddt(u,ut), we first estimate the last term in (3.8). By Hölder's inequality, we obtain

    |Γ1|ut|m2utudσ|utm1m,Γ1up,Γ1=utm1m,Γ1u1pmp,Γ1upmp,Γ1, (3.9)

    in which 1p+1m=1, and then, by (3.7), Hölder's inequality, Young inequality and the fact that H(t)=utmm,Γ1, we obtain that

    |Γ1|ut|m2utudσ|u1pmp,Γ1upmp,Γ1utm1m,Γ1C2H1p1m(t)upmp,Γ1utm1m,Γ1C3(εmupp,Γ1+εmutmm,Γ1)Hˉα(t)C3(εmupp,Γ1+εmH(t))Hˉα(t), (3.10)

    for any ε>0, where ˉα=1m1p>0 and 1m+1m=1, and we denote C1,C2, ..., as suitable positive constants. Let 0<α<ˉα, by (3.5) and (3.10), we have

    |Γ1|ut|m2utudσ|C3(εmHˉα(0)upp,Γ1+εmH(t)Hα(t)Hαˉα(0)). (3.11)

    Now we introduce an auxiliary function

    Z(t)=H1α(t)+δΩuutdx,

    where δ is a small positive constant which will be decided later. By (3.8) and (3.11), we have

    Z(t)(1α)Hα(t)H(t)   +δ(2ut2+C1upp,Γ1+2H(t)Γ1|ut|m2utudσ)(1αδC3εmHαˉα(0))Hα(t)H(t)   +δ(C1C3εmHˉα(0))upp,Γ1+2δut2+2δH(t). (3.12)

    Let δ<(1α)C13εmHˉαα(0), so the first term on the right hand side of (3.12) is positive. Moreover, if we choose ε sufficiently small, we can obtain that

    C1C3εmHˉα(0)12C1,

    further,

    Z(t)12C0δupp,Γ1+2δut2+2δH(t)C4δ(upp,Γ1+ut2+H(t)). (3.13)

    Letting δ sufficiently small, we have Z(0)>0. And noting that H(t) is an increasing function, we have Z(t)Z(0) for t0. Now we set r=11α, since α<ˉα<1, it is evident that 1<r<ˉr:=11ˉα. According to the following inequality

    |a+b|r2r1(|a|r+|b|r) forr1,

    Young inequality and Cauchy-Schwarz inequality, we have

    Zr(t)2r1(H(t)+δrutrur)C4(H(t)+ut2+u112α). (3.14)

    Now by choosing α sufficiently small, we have

    u112α1+u2. (3.15)

    Using Poincaré inequality and combining (3.1), (3.14) and (3.15), we have

    Zr(t)C5(H(t)+ut2+u2)C6(upp,Γ1+ut2+H(t)). (3.16)

    In turning by (3.13) and (3.16), and the fact that r>1, we obtain that

    Z(t)C7Zr(t). (3.17)

    Solving (3.17), we can obtain that there exist positive constants C8 and C9, such that

    Zr1(t)1C8t+C9.

    Then we have

    limtC9C8Zr1(t)=.

    So Z(t) is not global. This completes the proof.

    In this section, we mainly discuss the problem (1.1)-(1.4) without damping term, i.e., Q(ut)=0. We mainly adapt the convexity method introduced in [8].

    Lemma 4.1. Let γ>0, T>0 and h(t) be a Lipschitzian function over [0,T). Assume that h(0)0 and h(t)+γh(t)>0 for all t(0,T). Then h(t)>0 for all t(0,T).

    Theorem 4.2. (Finite time blow-up of solutions for E(0)>d.) Assume that Q(ut)=0, 1<m<p, m>rr+1p, 2p<r. If

    I(0)<0,E(0)>d,Ω u0u1dx0,u02>2pp2E(0),

    then the solution of problem (1.1)-(1.4) blows up in finite time.

    Proof. We will prove the result by two steps.

    Step I. We first show that

    I(u(t))<0andu2>2pp2E(0),    t0. (4.1)

    Arguing by contradiction, we suppose by the continuity of I(t) that there exists a first time t0>0 such that I(u(t0))=0. Then we consider the L(t) function defined by

    L(t):=u2.

    We have

    L(t)=2Ω uutdx.

    From the definition of I(u), we obtain that

    L(t)=2ut22I(u).

    Noticing that

    I(u)0,    t(0,t0].

    As L(t)0 and L(0)=Ω u0u1dx0 holds for all t(0,t0], by Lemma 4.1, we can obtain that L(t)>0 of all t(0,t0). So we can know that L(t) is strictly increasing on (0,t0]. Thus,

    L(t)u02>2pp2E(0),    t(0,t0].

    As a consequence, we have

    L(t0)>2pp2E(0).

    On the other hand, combining the fact that E(t) is a decreasing function, we have

    E(t0)E(t)<E(0),    t(0,t0].

    According to the assumption, when I(t0)=0, we have

    u(t0)22pp2E(0).

    From uH1Γ0(Ω)=u for uH1Γ0(Ω), we have

    L(t0)=u(t0)2u(t0)2H1Γ0(Ω)=u(t0)22pp2E(0)

    which leads to a contradiction. Thus we have proved that

    I(u)<0,    t[0,Tmax).

    By the discussion above, we see that L(t) is strictly increasing on [0,Tmax) provided I(u)<0, which implies

    u2L(t)>2pp2E(0),    t[0,Tmax). (4.2)

    Step II. Now we prove the finite time blow-up of the solution. As discussed above, we assume that u is global first. Then for any t>0, we use Cauchy-Schwarz inequality

    uutΩuutdx

    to get

    L(t)L(t)p+24(L(t))2=2u2(ut2I(u))p+24(2Ωuutdx)22u2(ut2I(u))(p+2)u2ut2u2ξ(t),

    where

    ξ(t)=2pE(0)+(p2)u2.

    According to (4.2), we have

    u2>2pp2E(0).

    So we can obtain that

    2pE(0)<(p2)u2.

    Then we have ξ(t)>0. Further we have

    L(t)L(t)p+24(L(t))20.

    Thus

    (L(t)α)=αLα+2(t)(L(t)L(t)(1+α)(L(t))2)<0,

    where α=p24>0. Hence, it proves that Lα(t) reaches 0 in finite time, then there exists a T[0,) such that

    limtTL(t)=.

    Finally we prove that L(t) is not global. This completes the proof.

    The author thanks the support from the Natural Science Foundation of Jiangsu Province (BK20160564) and Jiangsu key R & D plan(BE2018007).



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