R E S E A R C H
Open Access
Continuous dependence on data for a
solution of the quasilinear parabolic equation
with a periodic boundary condition
Fatma Kanca
1*and Irem Sakinc Baglan
2*Correspondence:
fatma.kanca@khas.edu.tr
1Department of Information
Technologies, Kadir Has University, Istanbul, 34083, Turkey
Full list of author information is available at the end of the article
Abstract
In this paper we consider a parabolic equation with a periodic boundary condition and we prove the stability of a solution on the data. We give a numerical example for the stability of the solution on the data.
1 Introduction
Consider the following mixed problem:
∂u ∂t – ∂u ∂x = f (x, t, u), (x, t)∈ D := { < t < T, < x < π}, () u(, t) = u(π , t), t∈ [, T], () ux(, t) = ux(π , t), t∈ [, T], () u(x, ) = ϕ(x), x∈ [, π] ()
for a quasilinear parabolic equation with the nonlinear source term f = f (x, t, u).
The functions ϕ(x) and f (x, t, u) are given functions on [, π ] and ¯D× (–∞, ∞)
respec-tively. Denote the solution of problem ()-() by u = u(x, t). The existence, uniqueness and convergence of the weak generalized solution of problem ()-() are considered in []. The numerical solution of problem ()-() is considered [].
In this study we prove the continuous dependence of the solution u = u(x, t) upon the data ϕ(x) and f (x, t, u). In [], a similar iteration method is used with this kind of a local boundary condition for a nonlinear inverse coefficient problem for a parabolic equation. Then we give a numerical example for the stability.
2 Continuous dependence upon the data
In this section, we will prove the continuous dependence of the solution u = u(x, t) using an iteration method. The continuous dependence upon the data for linear problems by different methods is shown in [, ].
Theorem Under the following assumptions, the solution u = u(x, t) depends continuously
upon the data.
©2013 Kanca and Baglan; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
(A) Let the function f (x, t, u) be continuous with respect to all arguments in ¯D× (–∞, ∞)
and satisfy the following condition: f(t, x, u) – f (t, x,˜u) ≤b(x, t)|u – ˜u|,
where b(x, t)∈ L(D), b(x, t)≥ ,
(A) f (x, t, )∈ C[, π ], t[, π ],
(A) ϕ(x)∈ C[, π ].
Proof Let φ ={ϕ, f } and φ = {ϕ, f } be two sets of data which satisfy the conditions (A
)-(A).
Let u = u(x, t) and v = v(x, t) be the solutions of problem ()-() corresponding to the data φ and φ respectively, and
f(t, x, ) – f (t, x, ) ≤ε for ε≥ .
The solutions of ()-(), u = u(x, t) and v = v(x, t), are presented in the following form, respectively: u(t) = ϕ+ π t π fξ, τ , Au(ξ , τ )dξ dτ, uck(t) = ϕcke–(k) t + π t π fξ, τ , Au(ξ , τ )e–(π k)(t–τ )coskξ dτ , () usk(t) = ϕske–(k) t + π t π fξ, τ , Au(ξ , τ )e–(π k)(t–τ )sinkξ dτ . Let Au(ξ , τ ) =u(τ ) + ∞ k=[uck(τ ) cos kξ + usk(τ ) sin kξ ]. v(t) = ϕ+ π t π fξ, τ , Av(ξ , τ )dξ dτ, vck(t) = ϕcke–(k) t + π t π fξ, τ , Av(ξ , τ )e–(π k)(t–τ )coskξ dτ , () vsk(t) = ϕske–(k) t + π t π fξ, τ , Av(ξ , τ )e–(π k)(t–τ )sinkξ dτ . Let Av(ξ , τ ) =v(τ ) + ∞ k=[vck(τ ) cos kξ + vsk(τ ) sin kξ ].
From the condition of the theorem, we have u()(t) and v()(t)∈ B. We will prove that
the other sequential approximations satisfy this condition.
u(N+) (t) = u() (t) + π t π fξ, τ , Au(N)(ξ , τ )dξ dτ, u(N+)ck (t) = u()ck(t) + π t π fξ, τ , Au(N)(ξ , τ )e–(k)(t–τ )coskξ dτ , () u(N+)sk (t) = u()sk(t) + π t π fξ, τ , Au(N)(ξ , τ )e–(k)(t–τ )sinkξ dτ , v(N+) (t) = v() (t) + π t π fξ, τ , Av(N)(ξ , τ )dξ dτ,
v(N+)ck (t) = v()ck(t) + π t π fξ, τ , Av(N)(ξ , τ )e–(k)(t–τ )coskξ dτ , () v(N+)sk (t) = v()sk(t) + π t π fξ, τ , Av(N)(ξ , τ )e–(k)(t–τ )sinkξ dτ , where u() (t) = ϕ, u()ck(t) = ϕcke–(k) t , u()sk(t) = ϕske–(k) t and v() (t) = ϕ, v()ck(t) = ϕcke–(k)t, v() sk(t) = ϕske–(k) t .
First of all, we write N = in ()-(). We consider u()(t) – v()(t)
u()(t) – v()(t) =u () (t) – v () (t) + ∞ k= u()ck(t) – v()ck(t)+u()sk(t) – v()sk(t) = (ϕ– ϕ) + π t π fξ, τ , Au()(ξ , τ )– fξ, τ , Av()(ξ , τ ) dξ dτ + (ϕck– ϕck)e–(k) t + π t π fξ, τ , Au()(ξ , τ )– fξ, τ , Av()(ξ , τ ) × e–(π k)(t–τ )cosπ kξ dξ dτ + (ϕ sk– ϕsk)e–(k) t + π t π fξ, τ , Au()(ξ , τ )– fξ, τ , Av()(ξ , τ ) × e–(π k)(t–τ )sinπ kξ dξ dτ . ()
Adding and subtracting t π f(ξ , τ , ) dξ dτ , t π e–(k)(t–τ )f(ξ , τ , ) cos π kξ dξ dτ , t π e–(k)(t–τ )f(ξ , τ , ) sin π kξ dξ dτ
to both sides and applying the Cauchy inequality, Hölder inequality, Lipschitz condition and Bessel inequality to the right-hand side of () respectively, we obtain
u()(t) – v()(t) ≤u() (t) – v() (t)+ ∞ k= u()ck(t) – v()ck(t)+u()sk(t) – v()sk(t) ≤ ϕ – ϕ + √ T + π √ π t π b(ξ , τ ) dξ dτ u()(t) + √ T + π √ π t π b(ξ , τ ) dξ dτ v()(t) + √ T + π √ π t π f(ξ , τ , ) – f(ξ , τ , ) dξ dτ ,
AT=ϕ – ϕ + √ T + π √ π b(x,t) ¯ u()(t)+ √ T + π √ π b(x,t) v()(t) + √ T + π √ π f – f , ϕ – ϕ = max|ϕ– ϕ| + ∞ k= max|ϕck– ϕck| + max |ϕsk– ϕsk|. For N = , u()(t) – v()(t) ≤ |u () (t) – v () (t)| + ∞ k= u() ck(t) – v () ck+u () sk(t) – v () sk(t) ≤ √ T + π √ π t π b(ξ , τ ) dξ dτ AT + √ T + π √ π t π b(ξ , τ ) dξ dτ AT. For N = , u()(t) – v()(t) ≤|u() (t) – v () (t)| + ∞ k= u()ck(t) – v()ck(t)+u()sk(t) – v()sk(t) ≤ √ T + π √ π t π b(ξ , τ )u()(t) – v()(t)dξ dτ + √ T + π √ π t π b(ξ , τ )u()(t) – v()(t)dξ dτ ≤ √ T + π √ π AT t b(ξ , τ ) τ π b(ξ, τ) dξdτ dξ dτ + √ T + π √ π AT t b(ξ , τ ) τ b(ξ, τ) dξdτ dξ dτ ≤ √ T + π √ π AT √ t b(ξ , τ ) dξ dτ + √ T + π √ π AT √ t b(ξ , τ ) dξ dτ .
In the same way, for a general value of N , we have u(N+)(t) – v(N+)(t) ≤ |u (N+) (t) – v (N+) (t)| + ∞ k= u(N+)ck (t) – v(N+)ck (t)+u(N+)sk (t) – v(N+)sk (t) ≤ AT· aN= aN ϕ – ϕ + C(t) + Mf – f , ()
where aN = √ T + π √ π N AT √ N! t π b(ξ , τ ) dξ dτ N + √ T + π √ π N AT √ N! t π b(ξ , τ ) dξ dτ N and M= √ T + π √ π N .
(The sequence aN is convergent, then we can write aN≤ M, ∀N.)
It follows from the estimation ([, pp.-]) that limN→∞u(N+)(t) = u(t).
Then let N→ ∞ for the last equation
u(t) – v(t) ≤Mϕ – ϕ + Mf – f ,
where M= M· M.
Iff – f ≤ ε and ϕ – ϕ ≤ ε, then |u(t) – v(t)| ≤ ε.
3 Numerical example
In this section we consider an example of numerical solution of ()-() to test the stability of this problem. The numerical procedure of ()-() is considered in [].
Figure 1 The exact and numerical solutions of u(x, 1). The exact and numerical solutions of u(x, 1), (–) for ε= 0, (–·) forε= 0.05, (..) forε= 0.01, the exact solution is shown with a dashed line.
Example Consider the problem ∂u ∂t – ∂u ∂x = u, () u(x, ) = sin x, x∈ [, π], () u(, t) = u(π , t), t∈ [, T], ux(, t) = ux(π , t), t∈ [, T]. ()
It is easy to see that the analytical solution of this problem is
u(x, t) = sin x exp(–t).
In this example, we take f (x, t, u) = f (x, t, u) + ε and ϕ(x) = ϕ(x) + ε for different ε values. The comparisons between the analytical solution and the numerical finite difference solution for ε = , , ε = , values when T = are shown in Figure .
The computational results presented are consistent with the theoretical results.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
FK conceived the study, participated in its design and coordination and prepared computing section. ISB participated in the sequence alignment and achieved the estimation.
Author details
1Department of Information Technologies, Kadir Has University, Istanbul, 34083, Turkey.2Department of Mathematics,
Kocaeli University, Kocaeli, 41380, Turkey.
Acknowledgements
Dedicated to Professor Hari M Srivastava.
Received: 7 January 2013 Accepted: 29 January 2013 Published: 14 February 2013
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doi:10.1186/1687-2770-2013-28
Cite this article as: Kanca and Baglan: Continuous dependence on data for a solution of the quasilinear parabolic
