Volume 32 No. 2 2019, 289-294
ISSN: 1311-1728 (printed version); ISSN: 1314-8060 (on-line version) doi:http://dx.doi.org/10.12732/ijam.v32i2.10
STABILITY INEQUALITIES FOR THE DELAY PSEUDO−PARABOLIC EQUATIONS Ilhame Amirali1 §, Seda Cati2, Gabil M. Amiraliyev3
1Department of Mathematics
Faculty of Arts & Sciences
Duzce University, 81620, Duzce, TURKEY
2Department of Mathematics
Faculty of Arts & Sciences
Duzce University, 81620, Duzce, TURKEY
3Department of Mathematics
Faculty of Arts & Sciences
Erzincan University, 24000, Erzincan, TURKEY
Abstract: This paper deals with the initial-boundary value problem for linear pseudo-parabolic equation. Using the method of energy estimates the stabil-ity bounds obtained for the considered problem. Illustrative example is also presented.
AMS Subject Classification: 65M12, 65M15, 65M22, 34K28
Key Words: Sobolev equations, pseudo-parabolic equations, stability bounds
*
In the domain Q = Ω × [0, T ]; Ω = [0, l], Q = Ω × (0, T ], Ω = (0, l) we consider the following initial-boundary value problem for a pseudo-parabolic equation with delay
Received: January 9, 2019 2019 Academic Publicationsc
∂u(x, t) ∂t − a(t) ∂3u(x, t) ∂t∂x2 = b(t) ∂2u(x, t) ∂x2 + c(t) ∂2u(x, t − r) ∂x2 + d(t)u(x, t) + f (x, t), (x, t) ∈ Q, (1) u(x, t) = φ(x, t), x∈ Ω, −r ≤ t ≤ 0, (2) u(0, t) = u(l, t) = 0, t∈ [0, T ] , a(t) ≤ α ≤ 0, (3) where r > 0 represents the delay parameter, a ≥ α > 0, b, c, d, f and φ are given sufficiently smooth functions satisfying certain regularity conditions to be specified. The above equations are usually called Sobolev type or pseudo-parabolic equations, which appear in engineering fields, such as, for instance, flows of fluids through fissured rock, heat condition involving a thermodynamic temperature and a conductive temperature, and quasistationary processes in semiconductors (see, e.g. [1]-[6]). This existence and uniqueness result for pseudo-parabolic equations without delay can be found, e.g., in [7]-[11]. In the present study, using the method of energy estimates we have obtained the stability bounds for the problem (1)−(3). Illustrative example is also given.
Lemma 1. Let δ(t) ≥ 0 be the continuous function such that δ(t) ≤ δ∗+
Z t
0 {c0
δ(s) + c1δ(s − r)}ds, t >0,
δ(t) = ϕ(t), −r ≤ t ≤ 0,
with nonnegative constants δ∗, c0, c1 and ϕ∈ C [−r, 0]. Then
δ(t) ≤ δ∗exp c0+ c1 Z 0 −r ϕ(s)ds .
Proof. After replacing s − r = η we observe that Z t 0 δ(s − r)ds = Z t−r −r δ(η)dη = ( R0 −rϕ(t)dt, 0 ≤ t ≤ r, R0 −rϕ(t)dt + Rt−r 0 δ(η)dη, t≥ r. Therefore we have δ(t) ≤ δ∗+ c0+ c1 Z 0 −r ϕ(t)dt Z t 0 δ(s)ds, which by using the Gronwall inequality completes the proof.
Theorem 2. For a, b, c, d ∈ C [0, T ], f ∈ C Q and ∂k
φ
∂φk ∈ C Ω × [−r, 0],
k= 0, 1 the solution of the delay boundary-value problem (1) − (3) satisfies the following stability bound:
αkuk2+ ∂u ∂x 2 ≤ " A+ c1 Z 0 −r αkφk2+ ∂φ ∂x 2 dt !# , 0 ≤ t ≤ T, (4) where (g, h) = Z l 0 g(x)h(x)dx, kgk2 = Z l 0 g2(x)dx, A= 2α kφk2+ ∂φ ∂x 2 + 4αT Z T 0 kf k 2ds, c0 = T max 4c2,2b2, c1 = 4α−2d 2 T, g= max [0,T ]|g(t)|.
Proof. Consider the identity ∂u ∂t, ∂u ∂t − a(t) ∂3u ∂t∂x2, ∂u ∂t = b(t) ∂ 2u ∂x2, ∂u ∂t + c(t) ∂ 2u(·, t − r) ∂x2 , ∂u ∂t + d(t) u,∂u ∂t + f(t),∂u ∂t . (5)
Next we will use the following relations ∂u ∂t, ∂u ∂t = ∂u ∂t 2 , ∂3u ∂t∂x2, ∂u ∂t = ∂ 2u ∂t∂x, ∂2u ∂t∂x = ∂2u ∂t∂x 2 , b(t) ∂ 2u ∂x2, ∂u ∂t = b(t) ∂u ∂x, ∂2u ∂t∂x ≤ µ1 ∂2u ∂t∂x 2 +¯b 2(t) 4µ1 ∂u ∂x 2 , c(t) ∂ 2u(·, t − r) ∂x2 , ∂u ∂t = c(t) ∂u(·, t − r) ∂x , ∂2u ∂t∂x ≤ µ2 ∂2u ∂t∂x 2 +c¯ 2(t) 4µ2 ∂u(·, t − r) ∂x 2 , d(t) u,∂u ∂t ≤ µ3 ∂u ∂t 2 +d¯ 2(t) 4µ3 kuk 2,
f(t),∂u ∂t ≤ µ4 ∂u ∂t 2 + 1 4µ4kf k 2.
Then from (5) we have
(1 − µ3− µ4) ∂u ∂t 2 + (a(t) − µ1− µ2) ∂2u ∂t∂x 2 ≤ ¯b 2 4µ1 ∂u ∂x 2 + c¯ 2 4µ2 ∂u(·, t − r) ∂x 2 + d¯ 2 4µ3 kuk 2+ 1 4µ4kf k 2. Choosing µ1= µ2= α 4, µ3 = µ4 = 1 4, we get ∂u ∂t 2 + α ∂2u ∂t∂x 2 ≤ ¯b 2 α ∂u ∂x 2 + 2 ¯d2kuk2+2¯c 2 α ∂u(·, t − r) ∂x 2 +2 kf k2. (6)
After integrating (6) on (0, t) and using the inequality g2(t) ≤ 2g2(0) + 2T Z t 0 |g ′ (s)|2ds, we obtain Z t 0 ∂u ∂t 2 ds≥ 1 2T kuk 2 −T1 kφk2, Z t 0 ∂2u ∂t∂x 2 ds≥ 1 2T ∂u ∂x 2 −T1 ∂φ ∂x 2 . Therefore the inequality (6) reduces to
αkuk2+ ∂u ∂x 2 ≤ 4αT ¯d2 Z t 0 kuk 2ds+ 2T¯b2Z t 0 ∂u ∂x 2 ds + 4T ¯c2 Z t 0 ∂u(·, t − r) ∂x 2 ds+ A. Denote δ(t) = α kuk2+ ∂u ∂x 2 ,
then δ(t) ≤ A + c0 Z t 0 δ(s)ds + c1 Z t 0 δ(s − r)ds. From here by fLemma1 we have
δ(t) ≤ " A+ c1 Z 0 −r αkφk2+ ∂φ ∂x 2! dt # exp (c0T + c1exp(c0T)t)
which immediately leads to (4).
Example ∂u ∂t − (1 + t) 2 ∂3u ∂t∂x2 = e −t∂2u ∂x2 + p 2 + t2∂2u(x, t − 1) ∂x2 + tu + t sin πx, 0 < t ≤ 1, 0 < x < 1, u(x, t) = te−t, 0 < x < 1, −1 ≤ t ≤ 0. Using the inequality (4) with
α = 1, ¯b = 1, ¯c =√3, d¯= 1, Z 0 −rkφk 2dt=Z 0 −r t 2 1 − e −2x dt = 1 − e−2x 6 , Z 0 −r ∂φ ∂x 2 dt= 1 − e −2x 6 ,
gives us the following stability estimate for the solution of our particular prob-lem: v(t) ≥ 0, kuk2+ ∂u ∂x 2 ≤ 2 + 12 1 − e −2x 6 exp(4T + 12 exp(4T )t).
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