Electronic Journal of Differential Equations, Vol. 2025 (2025), No. 39, pp. 1–14.

ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu

DOI: 10.58997/ejde.2025.39

BLOW-UP SOLUTIONS FOR DAMPED RAO-NAKRA BEAMS WITH

SOURCE TERMS

MIRELSON M. FREITAS, MAURO L. SANTOS, CARLOS A. RAPOSO,

ANDERSON A. RAMOS, JORGE FERREIRA

Abstract. This article concerns the blow-up of solutions for a damped Rao-Nakra beam equa-
tion with nonlinear source terms at arbitrary initial energy levels. We estimate the lower and

upper bounds of the lifespan of the blow-up solution and the blow-up rate by considering both

linear and nonlinear weak damping terms.

1. Introduction

In this article, we study the Rao-Nakra beam model with nonlinear source terms and nonlinear
damping

ρ1h1utt − E1h1uxx − k(−u+ v + αwx) + g1(ut) = f1(u, v, w), in (0, 1)× R+,

ρ3h3vtt − E3h3vxx + k(−u+ v + αwx) + g2(vt) = f2(u, v, w), in (0, 1)× R+,

ρhwtt + EIwxxxx − kα(−u+ v + αwx)x + g3(wt) = f3(u, v, w), in (0, 1)× R+,

(1.1)

with initial conditions

u(x, 0) = u0(x), ut(x, 0) = u1(x), in (0, 1),

v(x, 0) = v0(x), vt(x, 0) = v1(x), in (0, 1),

w(x, 0) = w0(x), wt(x, 0) = w1(x), in (0, 1),

(1.2)

and Dirichlet boundary conditions

u(0, t) = u(1, t) = 0, in R+,

v(0, t) = v(1, t) = 0, in R+,

w(0, t) = w(1, t) = 0, in R+.

(1.3)

Rao-Nakra sandwich beam was derived from the following general three-layer laminated beam
model developed in 1999 by Liu-Trogdon-Yong [16],

ρ1h1utt − E1h1uxx − τ = 0, (1.4)

ρ3h3vtt − E3h3vxx + τ = 0, (1.5)

ρhwtt + EIwxxxx −G1h1(wx + ϕ1)x −G3h3(wx + ϕ3)x − h2τx = 0, (1.6)

ρ1I1ϕ1,tt − E1I1ϕ1,xx − h1

2
τ +G1h1(wx + ϕ1) = 0, (1.7)

ρ3I3ϕ3,tt − E3I3ϕ3,xx − h3

2
τ +G3h3(wx + ϕ3) = 0. (1.8)

The parameters hi, ρi, Ei, Gi, Ii > 0 are the thickness, density, Young’s modulus, shear modulus,
and moments of inertia of the i-th layer for i = 1, 2, 3, from the bottom to the top, respectively.
In addition, ρh = ρ1h1 + ρ2h2 + ρ3h3 and EI = E1I1 + E3I3.

2020 Mathematics Subject Classification. 35B40, 35B41, 37B55, 37L30.
Key words and phrases. Rao-Nakra beam; nonlinear damping; source term; blow-up.
©2025. This work is licensed under a CC BY 4.0 license.
Submitted November 5, 2024. Published April 10, 2025.

1



2 M. M. FREITAS, M. L. SANTOS, C. A . RAPOSO, A. A. RAMOS, J. FERREIRA EJDE-2025/39

The Rao-Nakra system [22]

ρ1h1utt − E1h1uxx − k(−u+ v + αwx) = 0, in (0, L)× R+,

ρ3h3vtt − E3h3vxx + k(−u+ v + αwx) = 0, in (0, L)× R+,

ρhwtt + EIwxxxx − αk(−u+ v + αwx)x = 0, in (0, L)× R+,

is obtained from (1.4)-(1.8) when we consider the core material to be linearly elastic, i.e., τ = 2G2γ
with the shear strain

γ =
1

2h2
(−u+ v + αwx) and α = h2 +

1

2
(h1 + h3),

where k := G2

h2
, G2 = E2

2(1+ν) is the shear modulus, and −1 < ν < 1
2 is the Poisson ratio.

In [13], it was studied the Rao-Nakra system with internal damping,

ρ1h1utt − E1h1uxx − k(−u+ v + αwx) + a0ut = 0, in (0, 1)× R+, (1.9)

ρ3h3vtt − E3h3vxx + k(−u+ v + αwx) + a1vt = 0, in (0, 1)× R+, (1.10)

ρhwtt + EIwxxxx − αk(−u+ v + αwx)x + a3wt = 0, in (0, 1)× R+, (1.11)

and was proved that the polynomial stability occurs when there is only one viscous damping acting
either on the beam equation or one of the wave equations.

Now, we present a brief review of the literature. the Rao-Nakra with both internal damping and
Kelvin-Voigt damping was considered in [14], and the polynomial stability when two of the three
equations are directly damped was obtained. Méndez et al. [17] proved the lack of exponential
stability when the Kelvin-Voigt damping terms act on the first and third equations in the Rao-
Nakra sandwich beam model. Then, the system was proved to have polynomial decay. Exact
controllability results for the multilayer Rao-Nakra plate system with locally distributed control
in a neighborhood of a portion of the boundary were obtained in [7, 8]. Boundary controllability for
the Rao-Nakra beam equation has been studied in [9, 10, 19, 20, 21]. Rao-Nakra sandwich beam
equation with internal damping and time delay was analyzed in [23]. Exponential stabilization and
observability inequality for Rao-Nakra sandwich beam with time-varying weight and time-varying
delay was proved in [3]. By using semigroup theory, they obtained well-posedness, and exponential
stability. In [24], well-posedness and exponential stability were proved for the Rao-Nakra sandwich
beam with Cattaneo’s law for heat conduction. Exponential and general energy decay rates for
a Rao-Nakra sandwich beam equation with time-varying weights and frictional damping terms
acting complementarily in the domain were obtained in [1],.

Blow-up solutions have been investigated in several works. For wave equations with nonlinear
damping and source terms see [18]. For systems of nonlinear wave equations with damping and
source terms, see [2]. For a viscoelastic Kirchhoff-type equation with logarithmic nonlinearity
and strong damping, see [4]. For Kirchhoff type equation with variable-exponent nonlinearity, see
[15, 25]. For the Timoshenko beam with nonlinear damping and source terms, see [26] and its
references. By the way, blow-up results for the Rao-Nakra beam were not analyzed previously. In
this manuscript, we consider (1.9)-(1.11) in a general context, and we investigate the competition
between a nonlinear stabilization mechanism and a nonlinear source term. We estimate the lower
and upper bound of the lifespan of the blow-up solution and the blow-up rate by considering both
linear and nonlinear weak damping terms.

This manuscript is organized as follows. Section 2 introduces notation and preliminary results.
Section 3 presents the main results: the blow-up of solutions at high initial energy for both linear
and nonlinear weak damping. We establish some technical lemmas in Section 4 to prove the main
results. Finally, in Section 5, we prove the finite time blow-up of solutions by using the so-called
concavity method.

2. Preliminaries

The following notation will be used for the rest of this article:

∥u∥p = ∥u∥Lp(0,L), ⟨u, v⟩ = ⟨u, v⟩L2(0,1).



EJDE-2025/39 BLOW-UP SOLUTIONS FOR DAMPED RAO-NAKRA BEAMS 3

Similarly, for z = (u, v, w) and z̃ = (ũ, ṽ, w̃) we will use

∥z∥p :=
(
∥u∥pp + ∥v∥pp + ∥w∥pp

)1/p

, ⟨z, z̃⟩ := ⟨u, ũ⟩+ ⟨v, ṽ⟩+ ⟨w, w̃⟩.

Let us consider the Hilbert spaces

H = L2(0, 1)× L2(0, 1)×H1(0, 1), V = H1
0 (0, 1)×H1

0 (0, 1)×H2(0, 1) ∩H1
0 (0, 1).

with inner products

⟨z, z̃⟩V = E1h1⟨ux, ũx⟩+ E3h3⟨vx, ṽx⟩+ EI⟨wxx, w̃xx⟩+ κ⟨−u+ v + αwx,−ũ+ ṽ + αw̃x⟩, (2.1)

and

⟨U, Ũ⟩H = ⟨z, z̃⟩V + ⟨z1, z̃1⟩, (2.2)

for z = (u, v, w), z̃ = (ũ, ṽ, w̃), z1 = (u1, v1, w1), z̃1 = (ũ1, ṽ1, w̃1), and U = (z, z1), Ũ = (z̃, z̃1).
The corresponding norms are

∥z∥2V = E1h1∥ux∥22 + E3h3∥vx∥22 + EI∥wxx∥22 + κ∥ − u+ v + αwx∥22, (2.3)

and

∥U∥2H = ∥z∥2V + ∥z1∥22. (2.4)

Assumption 2.1.

(i) Damping: g1, g2, g3 : R → R are continuous, monotone increasing functions with g1(0) =
g2(0) = g3(0) = 0. In addition, the following growth conditions: there exist positive
constants α and β such that for all s ∈ R,

α|s|m+1 ⩽ g1(s)s ⩽ β|s|m+1, m ⩾ 1,

α|s|r+1 ⩽ g2(s)s ⩽ β|s|r+1, r ⩾ 1,

α|s|l+1 ⩽ g3(s)s ⩽ β|s|l+1, l ⩾ 1.

(2.5)

(ii) Sources: fj ∈ C2(R) and there is a positive constant C such that

|∇fj(z)| ⩽ C
(
|u|p−1

+ |v|p−1
+ |w|p−1

+ 1
)
, j = 1, 2, 3 and p ⩾ 1. (2.6)

There exists a positive function F ∈ C2(R2) such that

∇F = F = (f1, f2, f3). (2.7)

There exists α0 > 0 such that

F (z) ⩾ α0

(
|u|p+1

+ |v|p+1
+ |w|p+1

)
. (2.8)

Furthermore, F is homogeneous of order p+ 1, that is

F (λz) = λp+1F (z), ∀λ > 0, z ∈ R3. (2.9)

(iii) Coefficients: ρ1h1 = ρ2h2 = ρh = 1.

Remark 2.2. It is easy to see that f1, f2 and f3 are also homogeneous functions of degree p and
there exists a positive constant C such that

fj(z) ⩽ C(|u|p + |v|p + |w|p), j = 1, 2, 3. (2.10)

We also recall the definition of weak solution of problem (1.1)-(1.3). Let

W = (Lm+1((0, 1)× (0, T ))× Lr+1((0, 1)× (0, T ))× Ll+1((0, 1)× (0, T ))).

Definition 2.3. A vector-valued function z = (u, v, w) is called a weak solution of (1.1)-(1.3) on
[0, T ] if:

(i) z ∈ C([0, T ];V ), (z(0), z′(0)) = (z0, z1) ∈ H;

(ii) zt ∈ C([0, T ]; (L2(0, 1))
3
) ∩W ;



4 M. M. FREITAS, M. L. SANTOS, C. A . RAPOSO, A. A. RAMOS, J. FERREIRA EJDE-2025/39

(iii) z = (u, v, w) satisfies

⟨z′(t), θ(t)⟩ − ⟨z1, θ(0)⟩+
∫ t

0

(−⟨z′(τ), θt(τ)⟩+ ⟨z(τ), θ(τ)⟩)dτ +

∫ t

0

⟨G (z′(τ)), θ(τ)⟩dτ

=

∫ t

0

⟨F (z(τ)), θ(τ)⟩dτ ,

for all t ∈ [0, T ] and test functions θ in

Θ = {θ = (θ1, θ2, θ3) ∈ C([0, T ];V ), θt ∈ L1(0, T ; (L2(0, 1))
3
)},

where

G (z) = (g1(u), g2(v), g3(w)), F (z) = (f1(u, v, w), f2(u, v, w), f3(u, v, w)).

Moreover, we know that if z is a weak solution of problem (1.1)-(1.3) on [0, T∞) where T∞ is
maximal existence time, then we have the energy identity

E(t) = E(0)−
∫ t

0

⟨G (z′(τ)), z′(τ)⟩dτ , ∀t ∈ [0, T∞), (2.11)

where

E(t) = 1

2

(
∥z′(t)∥22 + ∥z(t)∥2V

)
−
∫ 1

0

F (z(x, t))dx. (2.12)

By using a standard continuation procedure for ODE’s to conclude that, if T∞ < ∞, then

lim
t→T∞

(
∥z′(t)∥22 + ∥z(t)∥2V

)
= ∞.

Combining this with (2.11) and (2.12), we obtain

lim
t→T∞

∫ 1

0

F (z(x, t))dx = ∞.

3. Main results

3.1. Blow-up at high initial energy with linear weak damping. In this subsection, we
consider problem (1.1)-(1.3) with g1(s) = g2(s) = g3(s) = λs where λ > 0.

Theorem 3.1. Suppose that Assumption 2.1 holds and that the initial data (z0, z1) ∈ H satisfies

∥z1∥22 − 2⟨z0, z1⟩+ α∗E(0) < 0, (3.1)

where

α∗ =
2(p+ 1)

(p− 1)S2
2

, Sp = inf
z∈V \{0}

∥z∥V
∥z∥p

.

Suppose further that E(0) > 0 and z0 ∈ N−. Then the weak solution of (1.1)-(1.3) blows up in
finite time. Furthermore, we have the following upper bound of the lifespan:

T∞ ⩽
4

p− 1

ζ +
√
ζ + β∗∥z0∥22
β∗

, (3.2)

where

ζ =
2λ

p− 1
∥z0∥22 − (z0, z1)2, β∗ =

(p− 1)S2
2

p+ 1

[
∥z0∥22 −

2(p+ 1)

(p− 1)S2
2

E(0)
]
.

Next, we give a lower bound for the lifespan and a blow-up rate.

Theorem 3.2. Under the assumptions in Theorem 3.1. We have the following lower bound

T∞ ⩾
∫ ∞

K(0)

dz

E(0) + z + 2p+1C2S−2p
p (E(0) + z)

p , (3.3)

where K(t) =
∫ 1

0
F (z(x, t))dx.

Next, we introduce another way for obtaining a lower bound of the lifespan.



EJDE-2025/39 BLOW-UP SOLUTIONS FOR DAMPED RAO-NAKRA BEAMS 5

Theorem 3.3. Under the assumptions in Theorem 3.1. We have the lower bound

T∞ ⩾
1

p− 1
ln(1 + 2−p−1C−2S2p

p E1−p(0)). (3.4)

For the blow-up rate, we have

∥z(t)∥p+1
V ≳ ∥z(t)∥p+1

p+1 ≳ K(t) ⩾ X−1(T∞ − t), ∀t ∈ [0, T∞), (3.5)

where X−1 is an inverse function of the function

X(s) =

∫ ∞

s

dz

E(0) + z + 2p+1C2S−2p
p (E(0) + z)

p , ∀s ∈ [0,∞).

3.2. Blow-up at high initial energy with nonlinear weak damping.

Theorem 3.4. Suppose that max{m, r, l} < p and (z0, z1) ∈ H satisfies

⟨z0, z1⟩ > ME(0) > 0,

then the weak solution blows up in finite time, where

M =
q

q + 1

(α
β

)q/q[ϵ0(p+ 1)2α0

β(1− θ)

]−1/q
,

where ϵ0 is a root of the equation

q

q + 1

(α
β

)− q+1

q
[ (p+ 1)

2
α0ϵ0

β(1− θ)

]−1/q
=

(p+ 1)(1− ϵ0)

α(ϵ0)
,

such that

ϵ1 =
(α
β

) q
q+1

[ϵ0(p+ 1)
2
α0

β(1− θ)

]
]

1
q+1 < 1

where

α(ϵ) = 2

√
[
(p+ 1)(1− ϵ)

2
+ 1][κ(ϵ)− (p+ 1)

2
θα0ϵ

2(1− θ)
],

κ(ϵ) = [
(p+ 1)(1− ϵ)

2
− 1]S2

2 ,

q = max{m, r, l}, q = min{m, r, l},

θ = max{θ1, θ2, θ3} =
p− q

p− 1
, θ = min{θ1, θ2, θ3} =

p− q

p− 1
.

4. Technical lemmas

to prove the theorems above, we need the following lemmas.

Lemma 4.1 ([6]). Let δ > 0, T > 0 and let h be a Lipschitzian function over [0, T ). Assume that
h(0) ⩾ 0 and h′(t) + δh(t) > 0 for a.e. t ∈ (0, T ). Then h(t) > 0 for all t ∈ (0, T ).

Lemma 4.2. Suppose that λ > 0. Let

z0 = (u0, v0, w0) ∈ N− = {z ∈ V : I(z) = ∥z∥2V − (p+ 1)

∫ 1

0

F (z(x))dx < 0}, (4.1)

and z1 = (u1, v1, w1) ∈ (L2(0, 1))3 such that

⟨z0, z1⟩ ⩾ 0. (4.2)

Then the map t 7→ ∥z(t)∥22 is strictly increasing as long as z(t) ∈ N−.



6 M. M. FREITAS, M. L. SANTOS, C. A . RAPOSO, A. A. RAMOS, J. FERREIRA EJDE-2025/39

Proof. Let Ψ(t) = ∥z(t)∥22 and G(t) = Ψ′(t) = 2⟨z′(t), z(t)⟩. By multiplying the first equation and
the second equation in (1.1) by u, v and w, respectively, and adding the two equations together,
we have

⟨z′′(t), z(t)⟩+ λ⟨z′(t), z(t)⟩ = (p+ 1)

∫ 1

0

F (z(x, t))dx− ∥z(t)∥2V = −I(z(t)). (4.3)

By using (4.3) and direct calculations, we obtain

G′(t) = 2∥z′(t)∥22 + 2⟨z′′(t), z(t)⟩

= 2∥z′(t)∥22 + 2[−∥z(t)∥2V + (p+ 1)

∫ 1

0

F (z(x, t))dx− λ

2
G(t)],

which yields (with z(t) ∈ N−) that

G′(t) + λG(t) = 2∥z′(t)∥22 + 2
[
(p+ 1)

∫ 1

0

F (z(x, t))dx− ∥z(t)∥2V
]
> 0.

Therefore, by Lemma 4.1, we have Ψ′(t) = G(t) > 0. Thus, Ψ(t) is strictly increasing. The proof
is complete. □

We now prove the invariance set of N− for E(0) > 0.

Lemma 4.3. Suppose that (3.1) holds. Then the solution z of problem (1.1)-(1.3) with E(0) > 0
belong to N−, provided z0 ∈ N−.

Proof. We proceed by contradiction, by the continuity of I(z(·)) in t, we suppose that there exists
a first time t0 ∈ (0, T∞) such that I(z(t0)) = 0 and I(z(t)) < 0 for t ∈ [0, t0). By the Cauchy-
Schwarz inequality and Lemma 4.2, we have

Ψ(t) = ∥z(t)∥22 > ∥z0∥22 ⩾ 2⟨z0, z1⟩ − ∥z1∥22 > α∗E(0), ∀t ∈ (0, t0).

From the continuity of z(t) with respect to t, we have

Ψ(t0) = ∥z(t0)∥22 > α∗E(0).

By the definition of total energy functional E and Lemma 4.2, we obtain

E(0) ⩾ 1

2
∥z′(t0)∥22 + (

1

2
− 1

p+ 1
)∥z(t0)∥2V +

I(z(t0))

p+ 1
⩾

(p− 1)S2
2

2(p+ 1)
∥z(t0)∥22

which yields that

∥z(t0)∥22 ⩽
2(p+ 1)

(p− 1)S2
2

E(0).

This implies that

α∗E(0) < Ψ(t0) = ∥z(t0)∥22 ⩽
2(p+ 1)

(p− 1)S2
2

E(0) = α∗E(0),

which contradicts with E(0) > 0. The proof is complete. □

5. Proofs

In this section, we prove the finite time blow-up of solutions by using the so-called concavity
method, which was first introduced by Levine [11, 12].

Proof of Theorem 3.1. Arguing by contradiction, we suppose that the solution z is a global so-
lution. By Lemmas 4.2 and 4.3, we know that z(t) ∈ N− and Ψ(t) = ∥z(t)∥22 > ∥z0∥22 ⩾
2⟨z0, z1⟩ − ∥z1∥22 > α∗E(0) for all t ∈ [0,∞). Next, for T0 > 0, β0 > 0, τ0 > 0 specified later, we
may consider the function η : [0, T0] −→ [0,∞) defined by

η(t) = ∥z(t)∥22 + λ

∫ t

0

∥z(s)∥22ds+ λ(T0 − t)∥z0∥22 + β0(t+ τ0)
2, ∀t ∈ [0, T0]. (5.1)



EJDE-2025/39 BLOW-UP SOLUTIONS FOR DAMPED RAO-NAKRA BEAMS 7

By direct calculation, we obtain

η′(t) = 2⟨z′(t), z(t)⟩+ λ∥z(t)∥22 − λ∥z0∥20 + 2β0(t+ τ0)

= 2⟨z′(t), z(t)⟩+ 2λ

∫ t

0

⟨z′(s), z(s)⟩ds+ 2β0(t+ τ0).
(5.2)

Moreover, by using (4.3), we can easily obtain

η′′(t) = 2∥z′(t)∥22 + 2⟨z′′(t), z(t)⟩+ 2λ⟨z′(t), z(t)⟩+ 2β0

= 2∥z′(t)∥22 + 2β0 − 2I(z(t)).
(5.3)

Notice that η(t) ⩾ β0τ
2
0 > 0 for all t ∈ [0, T0] and η′(0) = 2⟨z1, z0⟩+ 2β0τ0 > 0.

By using Cauchy-Schwarz inequality, we can easily obtain

(η′(t))
2

4
= (⟨z′(t), z(t)⟩+ λ

∫ t

0

⟨z′(s), z(s)⟩ds+ β0(t+ τ0))
2

⩽ [∥z(t)∥22 + λ

∫ t

0

∥z(s)∥22ds+ β0(t+ τ0)
2
](∥z′(t)∥22 + λ

∫ t

0

∥z′(s)∥22ds+ β0)

⩽ η(t)(∥z′(t)∥22 + λ

∫ t

0

∥z′(s)∥22ds+ β0).

(5.4)

From (5.1)-(5.3) and (5.4), we obtain the estimate

η′′(t)η(t)− (p+ 3)(η′(t))
2

4
⩾ η(t)ξ(t), ∀t ∈ [0, T0], (5.5)

where

ξ(t) = −(p+ 1)∥z′(t)∥22 − λ(p+ 3)

∫ t

0

∥z′(s)∥22ds− 2I(z(t))− (p+ 1)β0. (5.6)

On the other hand, from (2.11) and (2.12), we deduce that

E(0) = 1

2
∥z′(t)∥2 + p− 1

2(p+ 1)
∥z(t)∥2V +

I(z(t))

p+ 1
+ λ

∫ t

0

∥z′(s)∥22ds,

or equivalently

− (p+ 1)∥z′(t)∥2 − λ(p+ 3)

∫ t

0

∥z′(s)∥22ds− 2I(z(t))

= (p− 1)∥z(t)∥2V + λ(p− 1)

∫ t

0

∥z′(s)∥22ds− 2(p+ 1)E(0).

Therefore, from (5.6), we have

ξ(t) = (p− 1)∥z(t)∥2V − 2(p+ 1)E(0) + (p− 1)λ

∫ t

0

∥z′(s)∥22ds− (p+ 1)β0

⩾ (p− 1)S2
2∥z0∥22 − 2(p+ 1)E(0)− (p+ 1)β0.

(5.7)

Choose β0 ∈ (0, β∗] where

β∗ =
(p− 1)S2

2

p+ 1
∥z0∥22 − 2E(0) = (p− 1)S2

2

p+ 1
[∥z0∥22 −

2(p+ 1)

(p− 1)S2
2

E(0)] > 0,

then (5.7) leads to ξ(t) > 0 for all t ∈ [0, T0]. Therefore, (5.5) yields that

η(t) ⩾ η(0)
[
1− (p− 1)η′(0)t

4η(0)

]− 4
p−1 , ∀t ∈ [0, T0]. (5.8)

We choose τ0 ∈ (τ∗,∞) where

τ∗ = {

{
0 if ζ = 2λ

p−1∥z0∥
2
2 − (z0, z1)2 ⩽ 0,

ζ
β∗

if ζ > 0,



8 M. M. FREITAS, M. L. SANTOS, C. A . RAPOSO, A. A. RAMOS, J. FERREIRA EJDE-2025/39

and T0 ∈ [ 2
p−1

β0τ
2
0+∥z0∥2

2

β0τ0−ζ ,∞), then we have

T∗ =
4η(0)

(p− 1)η′(0)
=

2(∥z0∥22 + λT0∥z0∥22 + β0τ
2
0 )

(p− 1)((z0, z1)2 + β0τ0)
∈ [0, T0].

Therefore, (5.8) gives us limt→T∗ η(t) = ∞. This is a contradiction with the fact that the solution
is global and it shows that the solution blows up at finite time.

To derive the upper bound for T∞, we know that

T∞ ⩽
2

p− 1

β0τ
2
0 + ∥z0∥22
β0τ0 − ζ

=
2

p− 1
f(β0, τ0), ∀(β0, τ0) ∈ (0, β∗]× (τ∗,∞).

By direct calculation, we have

fτ0(β0, τ0) =
β0(β0τ

2
0 − 2ζτ0 − ∥z0∥22)
(β0τ0 − ζ)

2 = 0 ⇐⇒ τ±0 =
ζ ±

√
ζ2 + β0∥z0∥22
β0

.

Therefore, for any (β0, τ0) ∈ (0, β∗]× (τ∗,∞), we have

f(β0, τ0) ⩾ f(β0, τ
+
0 ) = 2τ+0 = 2

ζ +
√

ζ + β0∥z0∥22
β0

⩾ 2
ζ +

√
ζ + β∗∥z0∥22
β∗

.

This fact implies

T∞ ⩽
4

p− 1

ζ +
√
ζ + β∗∥z0∥22
β∗

.

The proof is complete. □

Proof of Theorem 3.2. First, we know that ∀t ∈ [0, T∞),

E(t) =
1

2
(∥z′(t)∥22 + ∥z(t)∥2V ) = E(t) +

∫ 1

0

F (z(x, t))dx ⩽ E(0) +
∫ 1

0

F (z(x, t))dx.

To obtain the lower bound of the blow-up time T∞, we define the auxiliary functional

K(t) =

∫ 1

0

F (z(x, t))dx, ∀t ∈ [0, T∞).

It is clear that limt→T∞ K(t) = ∞. By direct calculation and using Cauchy inequality, we find

K ′(t) =

∫ 1

0

F (z(x, t))z′(x, t)dx

⩽
∫ 1

0

|F (z(x, t))z′(x, t)|dx

⩽
1

2
∥z′(t)∥22 +

1

2

∫ 1

0

|F (z(x, t))|2dx

⩽
1

2
∥z′(t)∥22 + 2C2∥z(t)∥2pp

⩽ E(0) +K(t) + 2C2S−2p
p ∥z(t)∥2pV

⩽ E(0) +K(t) + 2p+1C2S−2p
p (E(0) +K(t))p.

This fact implies, for any t1, t2 ∈ [0, T∞) with t1 < t2, that

t2 − t1 ⩾
∫ K(t2)

K(t1)

dz

E(0) + z + 2p+1C2S−2p
p (E(0) + z)

p . (5.9)

In (5.9), let t2 → T∞ and t1 = 0, we obtain

T∞ ⩾
∫ ∞

K(0)

dz

E(0) + z + 2p+1C2S−2p
p (E(0) + z)

p .

On other hand, in (5.9), let t2 → T∞ and t1 = t ∈ [0, T∞), we obtain

T∞ − t ⩾
∫ ∞

K(t)

dz

E(0) + z + 2p+1C2S−2p
p (E(0) + z)

p = X(K(t)). (5.10)



EJDE-2025/39 BLOW-UP SOLUTIONS FOR DAMPED RAO-NAKRA BEAMS 9

We note that the function X is continuous and strictly decreasing on (0,∞). Therefore the inverse
function X−1 : X(0,∞) −→ (0,∞) is also continuous and strictly decreasing. Then (5.10) leads to

∥z(t)∥p+1
V ≳ ∥z(t)∥p+1

p+1 ≳ K(t) ⩾ X−1(T∞ − t), ∀t ∈ [0, T∞).

The proof is complete. □

Proof of Theorem 3.3. We put E(t) = 1
2 (∥z

′(t)∥22 + ∥z(t)∥2V ) for all t ∈ [0, T∞). It is clear that
E(t) > 0 for all t ∈ [0, T∞) and limt→T∞ E(t) = ∞. By direct calculation, we obtain

E′(t) = −λ∥z′(t)∥22 + ⟨F (z(t)), z′(t)⟩

⩽
1

2
∥z′(t)∥22 +

1

2
∥F (z(t))∥22

⩽
1

2
∥z′(t)∥22 + 2C2∥z(t)∥2pp

⩽
1

2
∥z′(t)∥22 + 2C2S−2p

p ∥z(t)∥2pV
⩽ E(t) + 2p+1C2S−2p

p Ep(t).

(5.11)

We put Σ(t) = −E1−p(t)
p−1 . By direct calculation, we have

Σ′(t) = E′(t)E−p(t) ⩽ (E(t) + 2p+1C2S−2p
p Ep(t))E−p(t)

= 2p+1C2S−2p
p + E1−p(t)

= 2p+1C2S−2p
p − (p− 1)Σ(t).

(5.12)

We deduce from (5.12) that

exp[(p− 1)t]Σ(t)− Σ(0) ⩾
2p+1C2S−2p

p

p− 1
{exp[(p− 1)t]− 1},

or equivalently

t ⩾
1

p− 1
ln(

2p+1C2S−2p
p + E1−p(0)

2p+1C2S−2p
p + E1−p(t)

). (5.13)

By letting t → T∞ in (5.13), we conclude that the estimate (3.4) holds. The proof is complete. □

Proof of Theorem 3.4. Assume that z is a global solution to (1.1)-(1.3). Without loss of generality,
we may assume that E(t) ⩾ 0 for all t ∈ [0,∞) (See [5, Theorem 2.8]). We put Γ(t) = ⟨z′(t), z(t)⟩
for all t ∈ [0,∞). By direct calculation, we have

Γ′(t) = ∥z′(t)∥22 + ⟨z′′(t), z(t)⟩

= ∥z′(t)∥22 − ∥z(t)∥2V + (p+ 1)

∫ 1

0

F (z(x, t))dx− ⟨G (z′(t)), z(t)⟩

= [
(p+ 1)(1− ϵ)

2
+ 1]∥z′(t)∥22 + [

(p+ 1)(1− ϵ)

2
− 1]∥z(t)∥2V

+ ϵ(p+ 1)

∫ 1

0

F (z(x, t))dx− ⟨G (z′(t)), z(t)⟩ − (p+ 1)(1− ϵ)E(t)

⩾ [
(p+ 1)(1− ϵ)

2
+ 1]∥z′(t)∥22 + [

(p+ 1)(1− ϵ)

2
− 1]∥z(t)∥2V

+ ϵ(p+ 1)α0∥z(t)∥p+1
p+1 − ⟨G (z′(t)), z(t)⟩ − (p+ 1)(1− ϵ)E(t).

(5.14)

For the fourth term on the right-hand side of (5.14), we have

⟨G (z′(t)), z(t)⟩ = ⟨g1(u′(t)), u(t)⟩+ ⟨g2(v′(t)), v(t)⟩+ ⟨g3(w′(t)), w(t)⟩.

From Assumption 2.1, for any ϵ1 ∈ (0, 1), by using Hölder’s and Young’s inequalities, we obtain

|⟨g1(u′(t)), u(t)⟩| ⩽
∫ 1

0

|g1(u′(x, t))u(x, t)|dx



10 M. M. FREITAS, M. L. SANTOS, C. A . RAPOSO, A. A. RAMOS, J. FERREIRA EJDE-2025/39

⩽ β

∫ 1

0

|u′(x, t)|m|u(x, t)|dx

⩽
βm+1α−mϵm+1

1

m+ 1
∥u(t)∥m+1

m+1 +
mαϵ

−m+1
m

1

m+ 1
∥u′(t)∥m+1

m+1.

From the convexity of the function y 7→ xy

y in y for x > 0 and y > 0, we obtain

1

m+ 1
∥u(t)∥m+1

m+1 ⩽
θ1
2
∥u(t)∥22 +

1− θ1
p+ 1

∥u(t)∥p+1
p+1,

where θ1 = p−m
p−1 > 0. Then, we obtain

|⟨g1(u′(t)), u(t)⟩| ⩽ βm+1α−mϵm+1
1 (

θ1
2
∥u(t)∥22 +

1− θ1
p+ 1

∥u(t)∥p+1
p+1) +

mαϵ
−m+1

m
1

m+ 1
∥u′(t)∥m+1

m+1.

Similarly,

|⟨g2(v′(t)), v(t)⟩|t ⩽ βr+1α−rϵr+1
1 (

θ2
2
∥v(t)∥22 +

1− θ2
p+ 1

∥v(t)∥p+1
p+1) +

rαϵ
− r+1

r
1

r + 1
∥v′(t)∥r+1

r+1,

where θ2 = p−r
p−1 > 0, and

|⟨g3(w′(t)), w(t)⟩| ⩽ βl+1α−lϵl+1
1 (

θ3
2
∥w(t)∥22 +

1− θ3
p+ 1

∥w(t)∥p+1
p+1) +

lαϵ
− l+1

l
1

l + 1
∥w′(t)∥l+1

l+1,

where θ3 = p−l
p−1 > 0. We put

q = max{m, r, l}, q = min{m, r, l},

θ = max{θ1, θ2, θ3} =
p− q

p− 1
, θ = min{θ1, θ2, θ3} =

p− q

p− 1
,

then

q

q + 1
⩾

m

m+ 1
,

q

q + 1
⩾

r

r + 1
,

q

q + 1
⩾

l

l + 1
,

ϵ
− q+1

q

1 ⩾ ϵ
−m+1

m
1 , ϵ

− q+1

q

1 ⩾ ϵ
− r+1

r
1 , ϵ

− q+1

q

1 ⩾ ϵ
− l+1

l
1 .

We denote

Λ(t) = Γ(t)− ϵ
− q+1

q

1

q

q + 1
E(t).

By using Assumption 2.1, we have

E ′(t) = −⟨G (z′(t)), z′(t)⟩ ⩽ −α(∥u′(t)∥m+1
m+1 + ∥v′(t)∥r+1

r+1 + ∥w′(t)∥l+1
l+1).

By direct calculation and using above estimates, we obtain

Λ′(t) = Γ′(t)− ϵ
− q+1

q

1

q

q + 1
E ′(t)

⩾ [
(p+ 1)(1− ϵ)

2
+ 1]∥z′(t)∥22

+ [
(p+ 1)(1− ϵ)

2
− 1]∥z(t)∥2V + ϵ(p+ 1)α0∥z(t)∥p+1

p+1

− βm+1α−mϵm+1
1 (

θ1
2
∥u(t)∥22 +

1− θ1
p+ 1

∥u(t)∥p+1
p+1)

− mαϵ
−m+1

m
1

m+ 1
∥u′(t)∥m+1

m+1

− βr+1α−rϵr+1
1 (

θ2
2
∥v(t)∥22 +

1− θ2
p+ 1

∥v(t)∥p+1
p+1)



EJDE-2025/39 BLOW-UP SOLUTIONS FOR DAMPED RAO-NAKRA BEAMS 11

− rαϵ
− r+1

r
1

r + 1
∥v′(t)∥r+1

r+1

− βl+1α−lϵl+1
1 (

θ3
2
∥w(t)∥22 +

1− θ3
p+ 1

∥w(t)∥p+1
p+1)

− lαϵ
− l+1

l
1

l + 1
∥w′(t)∥l+1

l+1 − (p+ 1)(1− ϵ)E(t)

+ ϵ
− q+1

q

1

q

q + 1
α(∥u′(t)∥m+1

m+1 + ∥v′(t)∥r+1
r+1 + ∥w′(t)∥l+1

l+1)

⩾ [
(p+ 1)(1− ϵ)

2
+ 1]∥z′(t)∥22

+ {[ (p+ 1)(1− ϵ)

2
− 1]S2

2 − βm+1α−mϵm+1
1 θ1

2
}∥u(t)∥22

+ {[ (p+ 1)(1− ϵ)

2
− 1]S2

2 − βr+1α−rϵr+1
1 θ2

2
}∥v(t)∥22

+ {[ (p+ 1)(1− ϵ)

2
− 1]S2

2 − βl+1α−lϵl+1
1 θ3

2
}∥w(t)∥22

+ [ϵ(p+ 1)α0 −
βm+1α−mϵm+1

1 (1− θ1)

p+ 1
]∥u(t)∥p+1

p+1

+ [ϵ(p+ 1)α0 −
βr+1α−rϵr+1

1 (1− θ2)

p+ 1
]∥v(t)∥p+1

p+1

+ [ϵ(p+ 1)α0 −
βl+1α−lϵl+1

1 (1− θ3)

p+ 1
]∥w(t)∥p+1

p+1

+ α(ϵ
− q+1

q

1

q

q + 1
− ϵ

−m+1
m

1

m

m+ 1
)∥u′(t)∥m+1

m+1

+ α(ϵ
− q+1

q

1

q

q + 1
− ϵ

− r+1
r

1

r

r + 1
)∥v′(t)∥r+1

r+1

+ α(ϵ
− q+1

q

1

q

q + 1
− ϵ

− l+1
l

1

l

l + 1
)∥w′(t)∥l+1

l+1 − (p+ 1)(1− ϵ)E(t)

⩾ [
(p+ 1)(1− ϵ)

2
+ 1]∥z′(t)∥22

+ {[ (p+ 1)(1− ϵ)

2
− 1]S2

2 − (
β

α
)
q βϵ

q+1

1 θ

2
}∥z(t)∥22

+ [ϵ(p+ 1)α0 −
βm+1α−mϵm+1

1 (1− θ1)

p+ 1
]∥u(t)∥p+1

p+1

+ [ϵ(p+ 1)α0 −
βr+1α−rϵr+1

1 (1− θ2)

p+ 1
]∥v(t)∥p+1

p+1

+ [ϵ(p+ 1)α0 −
βl+1α−lϵl+1

1 (1− θ3)

p+ 1
]∥w(t)∥p+1

p+1 − (p+ 1)(1− ϵ)E(t). (5.15)

We choose ϵ1 > 0 such that

ϵ(p+ 1)α0 − (
β

α
)q
βϵ

q+1

1 (1− θ)

p+ 1
= 0

equivalently

(α
β

)q ϵ(p+ 1)
2
α0

β(1− θ)
= ϵ

q+1

1



12 M. M. FREITAS, M. L. SANTOS, C. A . RAPOSO, A. A. RAMOS, J. FERREIRA EJDE-2025/39

equivalently

ϵ1 =
(α
β

) q
q+1

[ϵ(p+ 1)
2
α0

β(1− θ)

] 1
q+1 .

We observe that if

ϵ1 =
[(α

β

)q ϵ(p+ 1)
2
α0

β(1− θ)

] 1
q+1 < 1,

then

ϵ(p+ 1)α0 −
βm+1α−mϵm+1

1 (1− θ1)

p+ 1
⩾ ϵ(p+ 1)α0 − (

β

α
)q
βϵ

q+1

1 (1− θ)

p+ 1
= 0,

ϵ(p+ 1)α0 −
βr+1α−rϵr+1

1 (1− θ2)

p+ 1
⩾ ϵ(p+ 1)α0 − (

β

α
)q
βϵ

q+1

1 (1− θ)

p+ 1
= 0,

ϵ(p+ 1)α0 −
βl+1α−lϵl+1

1 (1− θ3)

p+ 1
⩾ ϵ(p+ 1)α0 − (

β

α
)q
βϵ

q+1

1 (1− θ)

p+ 1
= 0,

and

(p+ 1)
2
θα0ϵ

2(1− θ)
= (

β

α
)q
βϵ

q+1

1 θ

2
,

ϵ
− q+1

q

1

q

q + 1
=

q

q + 1

(α
β

)−q/q[ϵ(p+ 1)
2
α0

β(1− θ)

]−1/q
.

Therefore, (5.15) gives us

Λ′(t) = Γ′(t)− q

q + 1

(α
β

)−q/q[ϵ(p+ 1)
2
α0

β(1− θ)

]−1/qE ′(t)

⩾ [
(p+ 1)(1− ϵ)

2
+ 1]∥z′(t)∥22 + [κ(ϵ)− (p+ 1)

2
θα0ϵ

2(1− θ)
]∥z(t)∥22 − (p+ 1)(1− ϵ)E(t),

(5.16)

where

κ(ϵ) = [
(p+ 1)(1− ϵ)

2
− 1]S2

2 .

We note that κ(0) > 0. Then we can take ϵ > 0 small enough such that

κ(ϵ)− (p+ 1)
2
θα0ϵ

2(1− θ)
> 0.

Using the Cauchy inequality, we have

[
(p+ 1)(1− ϵ)

2
+ 1]∥z′(t)∥22 + [κ(ϵ)− (p+ 1)

2
θα0ϵ

2(1− θ)
]∥z(t)∥22 ⩾ α(ϵ)Γ(t),

where

α(ϵ) = 2

√
[
(p+ 1)(1− ϵ)

2
+ 1][κ(ϵ)− (p+ 1)

2
θα0ϵ

2(1− θ)
].

Therefore, (5.16) leads to

Λ′(t) = Γ′(t)− q

q + 1

(α
β

)−q/q[ϵ(p+ 1)
2
α0

β(1− θ)

]−1/qE ′(t)

⩾ α(ϵ)Γ(t)− (p+ 1)(1− ϵ)E(t)

⩾ α(ϵ)[Γ(t)− (p+ 1)(1− ϵ)

α(ϵ)
E(t)].

(5.17)

It is easy to see that

lim
ϵ→1

[κ(ϵ)− (p+ 1)
2
θα0ϵ

2(1− θ)
] < 0.



EJDE-2025/39 BLOW-UP SOLUTIONS FOR DAMPED RAO-NAKRA BEAMS 13

Hence, there exists ϵ∗ ∈ (0, 1) such that

κ(ϵ)− (p+ 1)
2
θα0ϵ

2(1− θ)
> 0, α(ϵ) > 0, ∀ϵ ∈ (0, ϵ∗), α(ϵ∗) = 0.

Furthermore,

lim
ϵ→0

q

q + 1

(α
β

)− q
q
[ϵ(p+ 1)

2
α0

β(1− θ)

]−1/q
= ∞,

lim
ϵ→ϵ∗

q

q + 1

(α
β

)− q
q
[ϵ(p+ 1)

2
α0

β(1− θ)

]−1/q
> 0,

and

lim
ϵ→0

(p+ 1)(1− ϵ)

α(ϵ)
> 0, lim

ϵ→ϵ∗

(p+ 1)(1− ϵ)

α(ϵ)
= ∞.

Then by continuity, there exists ϵ0 ∈ (0, ϵ∗) ⊂ (0, 1) such that

q

q + 1

(α
β

)−q/q[ϵ0(p+ 1)
2
α0

β(1− θ)

]−1/q
=

(p+ 1)(1− ϵ0)

α(ϵ0)
= γ∗ > 0.

Choose ϵ = ϵ0, (5.17) implies

Γ(t) ⩾ Λ(t) ⩾ exp(α(ϵ0)t), ∀t ∈ [0,∞).

So, we have the estimate

∥z(t)∥22 ≳
∫ t

0

Γ(s)ds ≳ exp(α(ϵ0)t), ∀t ∈ [0,∞). (5.18)

By using H’́older’s inequality, we have

∥z(t)∥2 ≲ ∥u(t)∥2 + ∥v(t)∥2 + ∥w(t)∥2

≲
∫ t

0

∥u′(s)∥2ds+
∫ t

0

∥v′(s)∥2ds+
∫ t

0

∥w′(s)∥2ds

≲
∫ t

0

∥u′(s)∥m+1ds+

∫ t

0

∥v′(s)∥r+1ds+

∫ t

0

∥w′(s)∥l+1ds

≲ t
m

m+1

(∫ t

0

∥u′(s)∥m+1
m+1ds

) 1
m+1

+ t
r

r+1

(∫ t

0

∥v′(s)∥r+1
r+1ds

) 1
r+1

+ t
l

l+1

(∫ t

0

∥w′(s)∥l+1
l+1ds

) 1
l+1

≲ t
m

m+1 + t
r

r+1 + t
l

l+1 , ∀t ∈ [0,∞),

which contradicts (5.18). Therefore, the weak solution blows up in finite time. The proof is
compete. □

References

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Mirelson M. Freitas

Department of Mathematics, University of Braśılia, Braśılia 0910-900, Brazil

Email address: m.m.freitas@mat.unb.br

Mauro L. Santos

Institute of Exact and Natural Sciences, Federal University of Pará, Belém - PA 66075-110, Brazil
Email address: ls@ufpa.br

Carlos A. Raposo (corresponding author)
Faculty of Mathematics, Federal University of Pará, Salinópolis - PA 68721-000, Brazil

Email address: carlosraposo@ufpa.br

Anderson A. Ramos

Faculty of Mathematics, Federal University of Pará, Salinópolis - PA 68721-000, Brazil

Email address: ramos@ufpa.br

Jorge Ferreira

Department of Exact Sciences, Federal Fluminense University, Volta Redonda - RJ 27255-126, Brazil
Email address: jorge ferreira@id.uff.br


	1. Introduction
	2. Preliminaries
	3. Main results
	3.1. Blow-up at high initial energy with linear weak damping
	3.2. Blow-up at high initial energy with nonlinear weak damping

	4. Technical lemmas
	5. Proofs
	References