International Journal of Pure and Applied Mathematics ————————————————————————– Volume 25 No. 2 2005, 173-179
UNIQUENESS, STABILITY AND CONTINUOUS DEPENDENCE OF SOLUTIONS OF PARABOLIC SYSTEM Haiqing Zhao1 § , Enmin Feng2 , Zhijun Li3 1,2 Department
of Applied Mathematics Dalian University of Technology Dalian, 116024, P.R. CHINA 1 e-mail:
[email protected] 3 State Key Lab of Coastal and Offshore Engineering Dalian University of Technology Dalian, 116024, P.R. CHINA Abstract: In this paper, we prove the uniqueness and stability of solutions and the continuous dependence on parameter variables for parabolic problem ∂u ∂ (α(u) ∂u ∂t = ∂x ∂x ) + f (x, t, u), (x, t) ∈ Ω × I, u(x, 0) = Φ(x), x ∈ Ω, u(0, t) = p(t), u(l, t) = q(t), t ∈ I. AMS Subject Classification: 35K40 Key Words: uniqueness, stability, continuous dependence, parabolic equation 1. Introduction In the study of heat diffusivity of the Arctic ice, spatially one-dimensional nonlinear parabolic equations need dealing with, i.e. the following problem: ∂u ∂ (α(u) ∂u ∂t = ∂x ∂x ) + f (x, t, u), (x, t) ∈ Ω × I, (1) u(x, 0) = Φ(x), x ∈ Ω, u(0, t) = p(t), u(l, t) = q(t) t ∈ I,
where α(u) = a − bu is the heat diffusivity, a, b ∈ R+ ; I = [0, T ] is time and Ω = [0, l] ⊂ R is the smooth domain; f, Φ, p, q are the smooth functions. Received:
July 10, 2005
§ Correspondence
author
c 2005, Academic Publications Ltd.
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Differential theory for parabolic equations is an object of longstanding interest. For the recent developments in parabolic theory we refer to (see [3], [5], [1], [2], [4], and so on). However, α(u) included in all the above papers are constant functions or not explicitly dependent on u, which certainly restricts the range of applications of solution theory. In this paper, based on the sea ice model, we employ the integration to deal with the parabolic equations with α(u) linear dependent on u. The paper is organized as follows. In Section 2, we introduce two lemmas whose results will be used in the proof of theorems. Next, we give the uniqueness, stability and continuous dependence by the method to integration for problem (1). We draw conclusions in Section 3.
2. Main Results Throughout this paper we assume u ∈ C 2 (Ω × I) and u(x, t) is bounded for every (x, t) ∈ Ω × I; k·k denotes the norm in R; f is Lipschitizian in u, namely, k(f (x, t, u1 ) − f (x, t, u2 ))k ≤ M ku1 − u2 k ,
M > 0.
Lemma 1. For t ∈ I, u ∈ C 2 (Ω × I), it is obvious to obtain the following results: R R (i) let J(t) = Ω u2 (x, t)dx, then J ′ (t) = 2 Ω u ∂u ∂t dx;
∂ 2 α(x,t) (ii) assume α(u) = a − bu, a, b ∈ R+ , then ∂x2 < L1 , L1 ∈ R+ . Theorem 2. For fixed a, b ∈ R+ , α(u) = a − bu, assume that (1) exists solutions, then the solution is unique.
Proof. Let u1 (x, t) and u2 (x, t) be the solutions of (1), then it is obvious that u=u1 − u2 satisfies the following equations: ∂u ∂ = ∂t ∂x
∂u2 ∂u1 − α(u2 ) α(u1 ) ∂x ∂x
+ f (x, t, u1 ) − f (x, t, u2 ), (x, t) ∈ Ω × I, (2)
with the initial value u(x, 0) = 0,
x ∈ Ω,
(3)
and the boundary conditions u(0, t) = 0,
u(l, t) = 0,
t ∈ I.
(4)
UNIQUENESS, STABILITY AND CONTINUOUS... Let J(t) =
R
Ω
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u2 (x, t)dx, according to Lemma 1, (2), (3) and (4), we have
∂ ∂u2 ∂u1 − α(u2 ) J (t) = 2 u α(u1 ) + f (x, t, u1 ) − f (x, t, u2 ) dx ∂x ∂x ∂x Ω Z Z ∂u1 ∂u2 ∂u = −2 α(u1 ) (f (x, t, u1 ) − f (x, t, u2 )) udx. − α(u2 ) dx+2 ∂x ∂x ∂x Ω Ω Z
′
According to kf (x, t, u1 ) − f (x, t, u2 )k ≤ M ku1 − u2 k = M kuk and α(u) = a − bu, we get Z Z ∂u1 ∂u2 ∂u ′ J (t) ≤ −2 α(u1 ) − α(u2 ) dx + 2 M kuk kuk dx ∂x ∂x ∂x Ω Ω Z 2 Z Z ∂u ∂u2 ∂u ∂u1 = −2a − u2 dx + 2M dx + 2b u1 u2 dx ∂x ∂x ∂x Ω ∂x Ω Ω Z Z ∂2u ∂2u u 2 dx − b u 2 (u1 + u2 ) dx + 2M J(t). = 2a Ω ∂x Ω ∂x Based on the integration by parts, J ′ (t) can be rewritten by ∂2u (α(u1 ) + α(u2 )) u 2 dx + 2M J(t) ∂x Ω 2 Z Z ∂ (α(u1 ) + α(u2 )) ∂u ∂u udx + 2M J(t). dx − = − (α(u1 ) + α(u2 )) ∂x ∂x ∂x Ω Ω ′
J (t) =
Z
Due to the positive heat diffusivity, we arrive at that 2 Z ∂u (α(u1 ) + α(u2 )) dx ≥ 0. ∂x Ω Therefore, ′
J (t) ≤ −
Z
Ω
∂ (α(u1 ) + α(u2 )) ∂u udx + 2M J(t) ∂x ∂x Z 2 ∂ (α(u1 ) + α(u2 )) 2 1 u dx + 2M J(t) = 2 Ω ∂x2
Z 2
∂ α(u1 ) ∂ 2 α(u2 ) 2 1
≤
∂x2 + ∂x2 u dx + 2M J(t). 2 Ω
In terms of Lemma 1, let L = L1 + 2M, then we have Z J ′ (t) ≤ L u2 dx = LJ(t). Ω
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Therefore, J ′ (t) ≤ LJ(t), i.e. J ′ (t) − LJ(t) ≤ 0. We multiply both sides of the inequality by e−Lt : J ′ (t)e−Lt − LJ(t)e−Lt =
d J(t)e−Lt ≤ 0. dt
Integrating the above inequality on [0, t], t ∈ I, then J(t) ≤ J(0)eLt = 0. Considering that J(t) ≥ 0, we have J(t) = 0. Hence u2 (x, t) = 0, i.e. u1 = u2 . That is to say that, the solution of (1) is unique for fixed a, b ∈ R+ . Theorem 3. For t ∈ I, α(u) = a − bu, a, b ∈ R+ , let u1 (x, t) and u2 (x, t) be the solutions of (1) corresponding to different initial values Φ1 (x) and Φ2 (x), respectively. The solution of (1) is stable, i.e. kΦ1 (x) − Φ2 (x)k < δ, then ku1 − u2 k < ε. Proof. Similar toR the proof of RTheorem 2, let L = L1 + 2M , we obtain J(t) ≤ J(0)eLt < eLt Ω δ2 dx ≤ eLT Ω δ2 dx, which yields ku1 − u2 k < ε.
Lemma 4. For t ∈ I, u ∈ C 2 (Ω × I), let α1 = a1 − b1 u, α2 = a2 − b2 u and u1 (x, t), u2 (x, t) be the solutions of (1) corresponding to α = α1 and α = α2 , respectively, u = u1 − u2 , then: R
∂ 2 u1 ∂ 2 u2 (i) Ω u( ∂x2 + ∂x2 ) dx < L2 ,
R
2 u2 +u2 (ii) Ω ∂∂xu2 1 2 2 dx < L3 ,
R 2
(iii) Ω u2 ∂∂xu2 dx < L4 , where a1 , b1 , a2 , b2 , L2 , L3 , L4 ∈ R+ . of (1), so u1 (x, t) and Proof. u1 (x, t) and u2 (x, t) are the solutions
t)
u22 (x,
∂ u1
∂ 2 u1 ∂ 2 u2 2 are bounded. Because u ∈ C (Ω × I) and u( ∂x2 + ∂x2 ) ≤ kuk · ( ∂x2 +
2
2 2
∂ u2
∂x2 ), there exists a constant Ll2 ∈ R+ such that u( ∂∂xu21 + ∂∂xu22 ) ≤ Ll2 . Therefore, we prove that (i) holds. A similar process is considered, one can obtain (ii) and (iii).
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Theorem 5. Let α1 = a1 − b1 u, α2 = a2 − b2 u and u1 (x, t), u2 (x, t) be the solutions of (1) corresponding to α = α1 and α = α2 , respectively. Assume ka1 − a2 k < δ, kb1 − b2 k < η, a1 , a2 , b1 , b2 ∈ R+ , then ku1 − u2 k < ε. Proof. u = u1 − u2 satisfies the above (3), (4) and the following equation: ∂u ∂ ∂u2 ∂u1 = − α2 α1 + f (x, t, u1 ) − f (x, t, u2 ), (x, t) ∈ Ω × I. (5) ∂t ∂x ∂x ∂x R Let J(t) = Ω u2 (x, t)dx, then, we get ′
J (t) = 2
Z
u Ω
∂ ∂x
∂u2 ∂u1 − α2 α1 + f (x, t, u1 ) − f (x, t, u2 ) dx ∂x ∂x Z ∂u2 ∂u ∂u1 − α2 dx = −2 α1 ∂x ∂x ∂x Ω Z +2 u (f (x, t, u1 ) − f (x, t, u2 )) dx . (6) Ω
In terms of k(f (x, t, u1 ) − f (x, t, u2 ))k ≤ M ku1 − u2 k, (6) can be rewritten by Z Z ∂u2 ∂u ∂u1 ′ − α2 dx + 2M J (t) ≤ −2 α1 u2 dx . ∂x ∂x ∂x Ω Ω According to α1 = a1 − b1 u and α2 = a2 − b2 u, then we get Z ∂u2 ∂u1 ∂u2 ∂u ∂u1 ′ − a2 − b1 u1 − b2 u2 dx + 2M J(t) . J (t) = −2 a1 ∂x ∂x ∂x ∂x ∂x Ω In addition,
∂u1 ∂u2 2 a1 − a2 ∂x ∂x ∂u1 ∂u1 ∂u2 ∂u1 ∂u2 ∂u2 = (a1 − a2 ) + a2 − + a1 − + (a1 − a2 ) , ∂x ∂x ∂x ∂x ∂x ∂x ∂u2 1 and 2 b1 u1 ∂u can be written in the same way. So − b u 2 2 ∂x ∂x 2 Z ∂ u1 ∂ 2 u2 ∂2u J (t) = (a1 + a2 ) u 2 dx + (a1 − a2 ) u + dx ∂x2 ∂x2 Ω ∂x Ω Z Z 2 2 (b1 + b2 ) ∂ u u1 + u22 ∂2u dx − + 2M J(t) − (b1 − b2 ) u 2 (u1 + u2 )dx 2 2 2 Ω ∂x Ω ∂x ′
Z
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H. Zhao, E. Feng, Z. Li
≤ δL2 + ηL3 +
Z
a1 + a2 − (b1 + b2 )
Ω
(u1 + u2 ) 2
u
∂2u dx + 2M J(t). ∂x2
Note that −(b1 + b2 )
(u1 + u2 ) (u1 − u2 ) = (b1 − b2 ) − b1 u1 − b2 u2 , 2 2
then Z b1 − b2 ∂2u ∂2u (α1 + α2 ) u 2 dx − J (t) = δL2 + ηL3 + u2 2 dx + 2M J(t) ∂x 2 ∂x Ω Ω Z 2 ∂u (α1 + α2 ) dx ≤ δL2 + ηL3 + ηL4 /2 − Ω ∂x Z ∂u ∂ − u (α1 + α2 ) dx + 2M J(t) Ω ∂x ∂x Z ∂2 1 u2 2 (α1 + α2 ) dx + 2M J(t) ≤ δL2 + ηL3 + ηL4 /2 + 2 Ω ∂x Z ≤ δL2 + ηL3 + ηL4 /2 + L1 u2 dx + 2M J(t). ′
Z
Ω
Let L1 + 2M = L, therefore we have J ′ (t) − LJ(t) ≤ δL2 + ηL3 + ηL4 /2.After integrating the above inequality, we get J(t) ≤ (δL2 + ηL3 + ηL4 /2) (eLt − 1)/L ≤ (δL2 + ηL3 + ηL4 /2) (eLT − 1)/L,
t ∈ [0, T ].
It follows that ku1 − u2 k < ε.
3. Conclusion In summary, we have shown that the solution of the nonlinear parabolic equation with α(u) dependent on u is unique, stable and continuous dependent on the parameter variables.
Acknowledgements This research was supported by the National Natural Science Foundation of China (40233032) and China Social Commonweal Project (2003DEB5J057).
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