A nonexistence result for the semi-linear Moore-Gibson-Thompson equation with nonlinear memory on the Heisenberg group

1.    Introduction and Preliminaries

The main goal of this paper is concerned with the nonexistence of global weak solutions for the following semi-linear Moore–Gibson–Thompson equation with nonlinear mixed damping term

u ttt + u tt + ^{( —} A g ) m u -

( - A h ) ² u t

t

j^(t - srw ds

(1.1)

subject to the following initial conditions

u(n, 0) = uo(n),   ut(n, 0) = ui(n),   utt(n, 0) = U2(n),   П € R2n+1,          (1.2)

(0 2022 Georgiev, S. G. and Hakem, A.

where 0 <  y < 1 , a € (0,2] , m ^ 1 , p >  1 and A h is the Kohn-Laplace operator on the (2n + 1) -dimensional Heisenberg group. The operator ( - A h ) 2 accounts for anomalous diffusion. In the paper by W. Chen and A. Palmieri [1], it is investigated the blow-up of the solutions for the following semi linear Cauchy problem for MGT equation in the conservative case with nonlinearity of derivative type

( euttt + utt - Au - eAut = lutlp,   x E Rn, t > 0,

((u, ut, utt)(x, 0) = e(uq,ui,u2)(x), x E Rn, where p > 1 and e is a positive parameter describing the size of the initial data. More precisely, they proved that there exists a positive constant eq such that for any e E (0, eq] the solution u blows up in finite time. Furthermore, the upper bound estimate for the lifespan

1    n — i\

Ce ' p - i ² ) е—,

1

holds, where C >  0 is an independent of e constant and P Gi _a (n) = n+y is the so called Glassey exponent. The MGT was previously analyzed by several authors from a different point of view. For instance, see the papers [1–4] and references therein for a variety of problems related to this equations. Recently, T. Dao and A. Z. Fino in [5] have proved blow-up results to determine the critical exponents for the following Cauchy problem for semi-linear structurally damped wave model with nonlinear memory

σ ^t

utt — Au + ц(—A) 2 ut = J(t — s)-Y|u(s)|p ds, x E Rn, t > 0 ,

u(x, 0) = uq(x),                                     ut(x, 0) = u1(x), where ц > 0, a E (0, 2) for some 7 E (0,1) and p > 1. Using a modified test function method, if p ^ pc = 1 +---2 + (1 /)(2  ^   ,    [ fui(x) + (—A)2)uo(x) dx > 0, max[n — 2 + y(2 — a), 0] J \              /

R 2n+1

it was shown that there is no global (in time) weak solution. The problem of nonexistence of global solutions in the Heisenberg group has received specific attention in recent years. For instance, see the papers [6–9] and references therein. For more details on Heisenberg groups and partial differential equations in Heisenberg groups, we refer the reader to [7–11] and the references therein.

Motivated by above papers, we investigate the problem (1.1), (1.2) for nonexistence of global weak solutions by using the method of the test function. Our main result is as follows.

Theorem 1.1. Let 0 < a C 2, a = min(a, 1) and n ^ 1. We assume that the initial data (uo,u1,u2) E H°(R2n+1) x H2(R2n+1) x L2(R2n+1) satisfy the following relation j (ui(n)+ u2(n) + (—Ah) 2 uo(n)) dn > 0.

(1.3)

(1.4)

_R 2 n +i

If or

q + 2 — a

p C pc =     4,

(2 — a)Y + Q — 2m p < 1, γ

then there exists no global nontrivial weak solution to (1.1)-(1.2).

The paper is organized as follows. In the next section, we give some auxiliary results. In Section 3 we prove our main result.

2.    Auxiliary Results

For the sake of the reader, in this section we give some known facts about the Heisenberg group H and the operator A h .

The Heisenberg group H whose points will be denoted by n = (x,У,т) , is the Lie group ( R ²ⁿ⁺¹ ; о ) with the non-commutative group operation о defined by

n о n = (x + x,y + У,т + т' + 2 (x • y‘ - x‘ • y))

for all n = (x, y, т), n‘ = (x‘, y‘, т‘) € Rn x Rn x R, where • denotes the standard inner product in Rn. This group operation endows H with the structure of a Lie group. The Laplacian Ah over H is obtained from the vector fields Xi = -X + 2yiyT, Y = yy — 2xidT and we have

n

A h = £ (X ² + Y .

Observe that the vector field T = yT. does not appear in the equality above. This fact makes us presume a “loss of derivative” in the variable τ . The compensation comes from the relation

X,Y j ] = - 4T, i,j € 1, 2, 3, ...,n.

Then the Heisenberg group H is a nilpotent Lie group of order 2. Explicit computation gives the expression

.          / d ²      d ² d ²            d ²         , _{2     2}x д ² A

^H = Ц (dx ^{2 +} дy ^{2 + 4} y i дх^дт - x i дy i дт ⁺ ( x i ⁺ y i ^ дт ² ) ^.

A natural group of dilatations on H is given by

6a(n) = (Ax, Ay, А2т), A> 0, whose Jacobian determinant is AQ, where Q = 2n + 2 is the homogeneous dimension of H. The operator Ah is a degenerate elliptic operator. It is invariant with respect to the left translation of H and homogeneous with respect to the dilations δλ . More precisely, we have

A h (u(n о n ’ )) = (A H u)(n о n ‘ ), A h (u о 5 a ) = A ² (A h u ) о 6 a , n,n' € H .

The natural distance from η to the origin is introduced by Folland and Stein, see [10],

n

2\ 4

| n | H = ^{т 2 +} 52 ^{( x} 2 ⁺ y ² )

Now, we will collect some preliminary knowledge that will be used hereafter.

Definition 2.1 [12] . A function f : [a, b] ^ R is said to be absolutely continuous if there exists a Lebesgue summable function ^ € L^ a, b) such that

t

f (t) = f (a) + / ^(s) ds, t € [a, b].

a

We denote by AC[0, T] the space of all functions which are absolutely continuous on [0, T] with 0 < T <  to .

Definition 2.2 [12]. Let f € L ¹ (0-T) with T >  0 . The Riemann-Liouville left- and right-sided fractional integrals of order a € (0,1) are defined by

t

I ' f (t) ⁼ r1 _a )/(t — «'"^f (s) ds-  t> 0,                 ( ² '1)

and

T

I 't f (t) = Г0) /(t - s) ^{- (1 - a)} f (s) ^ds- t <  T-                   (2.2)

t respectively, where Г is the Euler gamma function.

Definition 2.3 [12]. Let f € AC [0,T] with T >  0 . The Riemann-Liouville left- and right-sided fractional derivatives of order € a € (0,1) are defined by

t

Df(t) = ^-f(t) = f(i^)S / ^{( t} - ^“f ⁽ s) ds, t >  ^{0       (2} - ³⁾

T dtf№ = -^f (t) = -Г(Г^dt /(t- s)-af

(2.4)

respectively.

Proposition 2.1 [12] . Let T >  0 and a € (0,1) . The fractional integration formula

by parts

TT

I ^{f ( t ) D} a | t g ^{( t ) dt} = I g ^{( t ) D} t \ T ^{f ( t ) dt}, 0                         0

(2.5)

is valid for every f € It^ ( L ^p (0,T)) and g € I 0 _t ( L ^q (0,T)) such that p + q ^ 1 + a with p,q >  1 , where

I't(Lp(0,T)) = {f = I'th, h € Lp(0,T)} and

I 0 t ( L (0,T )) = {f = l o at h, h € L q (0,T )} .

Proposition 2.2 [13] . Let T >  0 and a € (0,1) . Then, we have the following identities

D o \ _t I o \ _t f (t) = f(t), a-e- t € (0,T) for all f € L ^r (0,T) with 1 <  r <  to,     (2.6)

( _ 1) ^m D ^m D« _T f = D mt + ^a f for all f € AC ^m+1 [0,T],             (2.7)

where

ACm+1[0; T] = f : [0,T] -^ R, such that Dm f € AC[0,T]} and Dm = ddtm is the usual m times derivative-

Lemma 2.1 [13] . Let T >  0 , 0 < a < 1 and m ^ 0 - For all t € [0, T ] , we have

D m + “ (1   t ^^в- ^Г( в ⁺¹ )    T— ( m + a ) (1   t V"'^        (2 81

^D t \ T V - T) =Г(в + 1 - m - a) T    Ч¹ - T)      .        ^{( )}

Lemma 2.2 [5]. Let T > 0, 0 < a < 1, m > 0 and p > 1. Then, we have

Tp j ^(t)-P-1 |Dm+a^(t)|p-T dt = CT1-(m+a) p-т. 0

(2.9)

Before to proceed with the proof of our main result, we will give a definition for weak solutions for (1.1)–(1.2).

Definition 2.4. A function u E L p ( (0,T), L p ( R ²ⁿ⁺¹ ) ) П L ¹ ( (0,T), L ² ( R ²ⁿ⁺¹ ) ) is called a local weak solution of (1.1)-(1.2) subject to the initial data (u o ,u i ,U 2 ) E H ° ( R ²ⁿ⁺¹ ) x H ² ( R ²ⁿ⁺¹ ) x L ² ( R ²ⁿ⁺¹ ) if the following equality

т

Г(а) j j

0 R 2n+1

I 0 |t ^{( | u (} n^{,t ) | p} Mn,^ dndt +

/

(U 1 (n) + U 2 (n) + ( - A h ) ² uo) v(n, 0) dn

R ^{2 n +1}

• j Mn)

R 2n+1

+ U1(n)>t(n, 0) dn +

/

R 2n+^T

Uo(n)^tt (n, 0) dn

т

= - / /

0 R 2n+^T

T u(n,t)^ttt (n,t) dndt +

0 R 2n+^T

u(n,t)^tt(n,t) dndt

т

- / /

0 R 2n+^T

T

u(n, t)(-AH)m^(n, t) dn dt +

0 R 2n+^T

u(n, t)( - A h ) ² ^ t (n,t) dn dt,

holds for any regular function

^ E C1 ((0, T]; H° (R2n+1)) П C ((0,T]; H2 (R2n+1)) П C2 ((0, T]; L2 (R2n+1)) , such that ^(n, T) = 0, ^t(n,T) = 0 and ^tt(n,T) = 0 for all n E R2n+1. The solution u is called global if T = +w.

3.    Proof of the Main Result

Throughout this section, with C we will denote a positive constant whose value may change from line to line. The proof of our main result is based on a contradiction. Suppose that u is a global weak solution to (1.1)–(1.2). Then u satisfies the following equation

т

^{Г( а )} j j I 0 |t ^{( | u (} n^{,t ) | p )} ^ ⁽ n^{,t )} dndt ⁺ j ( u 1 (n)+ ^U 2 ⁽ n ^{) + ( - A} H ) ² u 0 ^ v ⁽ n, ⁰⁾ dn 0 R 2n⁺¹                                 R 2n⁺¹

• - j (U0(n) + U1(n))^t(n, 0) dn + j U0(n)vtt(n, 0) dn

т                        T

(3.1)

= —j j u(n,t)^ttt (n,t) dndt + j j u(n,t)^tt (n,t) dndt

т

- / /

0 R 2n+^T

T

u(n, t)(-AH)m^(n, t) dn dt +

0 R 2n+^T

u(n, t)(-AH)2 ^t(n, t) dn dt,

where a = 1 — y € (0,1), for all test function ^ such that ^(n,T) = ^t(n,T) = ^tt(n,T) = 0 and for all T >> 1. We define the following auxiliary functions фR(n) = ф (R) ,  ^(n,t)= фR(n)^(t),  ^(t) = (1 - T) , where φ is a non-negative smooth function such that

ф(х) = ф( | х | ), ф(0) = 1,   0 < ф(г) <  1, for r ^ 0.

(3.2)

Moreover, ф is decreasing and ф(г) ^ 0 , as r ^ to sufficiently fast. Then, we define the test function as follows

^ ⁽ n,f) = D aT Ф ⁽ п^,ф ) = Mn^)D t“T Mt))-

From (3.1), using (2.7) and (2.8), we have

T

Г(а) j j I o 4t ( | u(n,tr) D 0|T 95(n,t) dndt — CT 0 _R 2n+1

1 - α

^? j ^{( u} 0 ⁽ n) ⁺ uiM^ R ⁽ n) dn

R ²n+1

+ CT

^a j ( u i (n) + U 2 (n) + ( — ^ H ) ² u o ) Ф R (n) dn + CT ^{2 а}

_R 2n+1

T

= / 1 u(n, t)ФR(n)D2\+a(^(t)) dn dt — o R2n+1

¹ j u o (n)Ф R (n) dn R 2n+1

T j j u(n,t)ФR(n)Dt3+a(ф(t)) dndt    (3.3)

0 _R 2n+1

T

— / 1 u(n,t)D “\_T (^ф(t))( — A h ) ^m ф R (n) dndt

0 _R 2n+1

T

⁺ / 1 ^{u (} n^{,t ) D} “\T ⁽ ^ ^{( t ))( — A} h ) ^{2 ( ф} R ⁽ n)) dndt.

0 _R 2n+1

Using (2.5) and then (2.6), we arrive at

I 1 + CT

^а j ( u i (n) + U 2 (n) + ( - ^ H ) 2 u o ) Ф R (n) dn

_R 2n+1

- CT

1 - α

¹ j (u o (n) + uW n (n) dn + CT

_R 2n+1

2 - α

^? j u o (n)Ф R (n) dn _R 2n+1

T

T

= C / S u(n,t# R (n^D^ (^(t)) dndt - cj j u(n,t)Ф R (n)D ³ +^a (W)) dndt

0 R 2n+1

0 R 2n+1

(3.4)

— C J I u(n,t)D ^a i T №(t))( — A h ) m ф R ^n) dndt

^{0 {|} n | H > ^{R }}

T

+ cj I u(n,t)D a_T MtM — A h ) ² (Ф R (n)) dndt = A (^ф) + В(^ф) + C (^ф) + D (^ф). 0 _R 2n+1

Here

T

I 1 - //

| u(n, t) | ^p 0(n, t) dn dt.

0 _R 2n+1

On the other hand, using Holder’s inequality with p + p- — 1 , where p ‘ is the conjugate of p , we can proceed with the estimate for A (0) as follows

T

| A (0) 1 0 cj J | u(n,t) I Ф R (n) | D 2 | + ^a (0(t)) | dndt

0 _R 2n+1

T

• — cj J Hn^t)^ ^p (n,t)0 - ^p (n,t)Ф R (n) | D 2+ ^a (0(t)) | dndt

0 _R 2n+1

T 1

0 CI-

p ^′

D^a(0(t))l

dη dt .

0 _R 2n+1

At this stage, we pass to the scaled variables t— t and n —(x, y,T), such that

τxy

^{T —} r 2 ,x ^— R’ y ^— R.

Using Lemma 2.2, one has ¹Q1

| A (0) | 0 CI - R p ' T p ' ²   .

(3.5)

Similarly, we obtain

T

¹ Г Г             _ _        - p_ ^{1              |} p '

| C ^{( 0 ) | 0 ci} 2 ^p[ у у    ^{( ф} R ⁽ n)) p ^{( 0 ( t ))} p | D aT ^{( 0 ( t ))} |

0 {|n|H>R} x |(-Aн)mфR(n)|P' dndt) P 0 CI21 RQ 2mTp'

- α

(3.6)

where

T

I 2 — / /

| u(n, t) | ^p

t) dn dt.

0 {| n l H > R }

Now, we can proceed with the estimate for D (0) and B (0) in the following manner

T

¹                         _- p ^′        _- p ^′                  p ^′

| D ^{( 0 ) | 0} CIH у J ^{( ф} R ⁽ n ^{)) p ( 0 ( t )) p} ^1| Т ^{( 0 ( t ))} |

0 _R 2 n +1

x

( - A h ) 2 ф R (n)

| p dndt ) p

¹ Q 1

0 CI - R p ' ° T p '    ¹

(3.7)

and

T

| B (^) | <  Cj j l u(n,t)^ R (n) | D _t 3+ ^a (^(t)) | dndt 0 R 2n+1

^ C/ p

T

= Cj j Hn,t№ p (n>t)<£ ^— p ⁽ n,t ^{) ф} R ⁽ n) | ^D 3+ ^a C0 ^{( t ))} | dndt 0 R 2n+1

T                    _— p‘    .          ^| p '       'V ¹ Q X

J J ФR(n)(ф(t)') p |D3+a (W))| dndt)   ^ Clip Rp' Tp'

0 R 2n+1                                             ²

(3.8)

3 — a

.

Combining the estimates (3.5)–(3.8) into (3.4), one deduces that

Ii + CT■ j (ui(n)

_R 2n+1

+ U2(n) + (-Ah) 2 uo) ФR(n) dn

- CT-1-a j (uo(n)

_R 2n+1

+ U1(n))ФR(n) dn + CT 2 a

/

_R 2n+1

/ л t rp ( ^Q^p- ¹y — a — 1 oQn^ ¹7 — 2 — a oQn^ ¹7 — 3 — a\ <  CI { R p' T p'      + R p' T p'      + R p' T p'      I

Hence, we get

Ii + CT■ j (ui(n)

_R 2 n +1

Uo(n)ФR(n) dn

Q..

+ CI ₂ ^p R p ' T p ' ^a .

+ U 2 (n) + ( - A h ) ² u o (n)) Ф R (n) dn

^{1 Q                                  1} Q—O

^ Cl i p R p ' T p ' ^a T — ^{— 2} + T - R "  + T ^{— 3}) + CI ₂ ^p R p ' ^2m T p ' ^a

+ CT ^{i — a} j (u o (n)+ U 1 (n))Ф R (n) dn - CT ^{2 — a} j _R 2 n +1                                          _R 2 n +1

Uo(n)ФR(n) dn-

(3-9)

(3.10)

Because (1.3) holds and ф R (n) ^ 1 , as R ^ to , there exists a sufficiently large constant R o > 0 such that we have

j (ui (n) + U2 (n) + (-Ah)2 Uo (n)) ФR (n) dn > 0,

_R 2 n +1

(3.11)

for all R > R o . It is clear that the inequality (1.4) is equivalent to 1 - ap + ^ Qg: - ^I pAm С 0 . So, we have to consider the following two cases.

Case 1. If 1 - ap + | ^Q g - ^mm <  0 , then we take R = T ^2—7 . Hence, using (3.10) and (3.11), we have that

¹„1-ct^{Q 2 m}

I _i <  Cl i p T p ' ^a+ (2 - aV ² - s

+ C (T—i—a - T-2—a) j uo(n)ФR(n) dn + CT—i—a j U1(n)ФR(n) dn-R2n+1                              R2n+1

Thanks to the Young inequality for

1                   Q2m a = Ii , b = Tp' a+ (2-s)p' 2-5 ,

we conclude that

1         О 1      / । Q 2p‘m

AI1 ^ CT1-aP‘ + 2Q2 - — p′      p′

+ c ( T ^{- 1 - a} - T ^{-2- a}) I U)W r (n) dn + CT ^{- 1 - a} У uiW R (n) dn- (3.12)

R ^{2 n +1}                               R ^{2 n +1}

If 1 — ap + ^ QC — ^2 — m m < 0 , then by letting T ^ + ^ , we deduce that u = 0 . By invoking (3.10), we obtain

j (ui(n) + U2(n) + (—Ah) 2 uo(n)^ ФR(n) dn

R 2n+1

^ CT

1 j U1(nWR(n) dn + C (T 1 — T 2) j uo(n)ФR(n) dn-R2n+1                                      R2n+1

Hence, passing to the limit in the above inequality as T ^ + to , one obtains a contradiction with (1.3).

Case 2. If 1 — ap + 2 Q^ — T pm = 0 , then using (3.12), we obtain

pI1 ^ CC + c (T-1-a — T-2-a) j u0(n)фR(n) dn + CT-1-a j U1(n#R(n) dn-R2n+1                              R2n+1

Hence, it follows that I 1 ^ C , as T ^ + to . By the dominated convergence theorem, one has

+ ^

1J

0 R 2n+1

T

| u(n, t) | ^p dndt = lim t ^ + ^

0 _R 2n+1

|u(n, t)|p

which yields u E L p ((0, + to ) x R ^2n+r) . On the other hand, repeating the same calculations 1     _ 1

as above with R = T ^{2 - 2} L ^{2 - 2} , where 1 ^ L < R is large enough such that when R ^ + ro we do not have L ^ + ro at the same time, we arrive at the following inequality

‘ ⁺ 2 - S ^

¹ /     2(1 - 2)

I 1 <  CI 1 p   T ^_ "- 2^

_ Q          4-32  _ ^Q          (T-2 _ ^Q

L (2 - 2)p ‘ + T ^- "2-S- L (2 - 2)p ‘ + T ^- 2-S L (2 - 2)p

¹ Q 1 2m + I ₂ p L (2 - 2)p ‘ ^{+ 2} - ²

+ C(T

1 — a

—

T

2 - a

U0(n)ФR(n) dn + CT 1 a

_R 2n+1

/

_R 2n+1

U1(n)ФR(n) dn-

Applying the Young inequality for

a = I 1 p , 1

a = I 1 , 1

.a = I 1 ,

2(1-2)  _ Q b = T 2-2 L (2-2)p‘ ,

2 - 2 __ Q __I ²

b = T_ 2-S L (2-2)p‘ + 2-2 , one concludes that

A ii p ^′

c

C p ^′

( 2(1 - S)p ‘       Q

T 2 - 5 L 2 - 5

(4 — 35)p ‘        Q

+ T —   2 - 5 L (2 - 5)

⁽ ст ~ ⁵⁾p '     ^Q-V\ ^{1 Q} _L mn        , i -

+ T - 2-5 L - 2-5    + I P L (² — ⁵W ^{+ 2 - 5} + C(T ^{- 1 - a} - T ^{-2- a}^

cl

R 2n+1

U o (n)Ф R (n) dn

+ CT —1—a I U1(n)ФR(n) dn. (3.13)

_R 2 n +1

We have to distinguish the following two cases:

• If a E (0,1] , then a = a . Consequently, using the fact that u E L p ((0, + to )

one has

x R ²ⁿ⁺¹) ,

lim I2 = lim

T → + ∞     T → + ∞

/

|u(n, t) |p

{| n | H T L ^{2 — 5} }

Taking into account the inequality (3.13), by letting T ^ + ro , it follows that

+ ∞