Weakgeneralizedandnumericalsolutionforaquasilinearpseudo-parabolicequationwithnonlocalboundarycondition RESEARCHOpenAccess

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(1)Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. RESEARCH. Open Access. Weak generalized and numerical solution for a quasilinear pseudo-parabolic equation with nonlocal boundary condition Irem Baglan1 and Fatma Kanca2* * Correspondence: fatma.kanca@khas.edu.tr 2 Department of Management Information Systems, Kadir Has University, Istanbul, 34083, Turkey Full list of author information is available at the end of the article. Abstract This paper investigates the one dimensional mixed problem with nonlocal boundary conditions, for the quasilinear parabolic equation. Under some natural regularity and consistency conditions on the input data, the existence, uniqueness, convergence of the weak generalized solution, and also continuous dependence upon the data of the solution are shown by using the generalized Fourier method. We construct an iteration algorithm for the numerical solution of this problem.. 1 Introduction D denotes the domain D := { < x < ,  < t < T}. Consider the equation ∂u ∂  u ∂ u –  – ε  = f (x, t, u), ∂t ∂x ∂x ∂t. (). with the initial condition x ∈ [, ],. u(x, ) = ϕ(x),. (). the nonlocal boundary condition u(, t) = u(, t), ux (, t) = ,. t ∈ [, T],. t ∈ [, T],. () (). for a quasilinear pseudo-parabolic equation with the nonlinear source term f = f (x, t, u). ¯ × (–∞, ∞), respecThe functions ϕ(x) and f (x, t, u) are given functions on [, ] and D tively. ε ∈ [, ε ] is a small parameter, ε ≥ . Denote the solution of problem ()-() by u = u(x, t, ε). This problem was investigated with different boundary conditions by various researchers by using Fourier methods []. ©2014 Baglan and Kanca; 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..

(2) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 2 of 22. In this study, we consider the initial-boundary value problem ()-() with nonlocal boundary conditions ()-(). The periodic nature of ()-() type boundary conditions is demonstrated in []. In this study, we prove the existence, uniqueness convergence of the weak generalized solution, continuous dependent upon the data of the solution; and we construct an iteration algorithm for the numerical solution of this problem. We analyze computationally convergence of the iteration algorithm, as well as treating a test example. We will use the weak solution approach from [] for the considered problem ()-(). According to [, ] we assume the following definitions. ¯ is called test function if it satisfies the following Definition  The function v(x, t) ∈ C  (D) conditions:. v(x, T) = ,. v(, t) = v(, t),. vx (, t) = ,. ∀t ∈ [, T] and ∀x ∈ [, ].. ¯ satisfying the integral identity Definition  The function u(x, t, ε) ∈ C(D) . ∂ v ∂v ∂  v +  –ε  u + f (x, t, u)v dx dt ∂t ∂x ∂x ∂t    ∂  v(x, ) + ϕ(x) v(x, ) – ε dx = , ∂x  T.  . (). for arbitrary test function v = v(x, t), is called a generalized (weak) solution of problem ()-().. 2 Reducing the problem to countable system of integral equations Consider the following system of functions on the interval [, ]: X (x) = ,. Xk– (x) =  cos(πkx),. Xk (x) = ( – x) sin(πkx), Y (x) = x,. (). k = , , . . . ,. Yk– (x) = x cos(πkx),. Yk (x) = sin(πkx),. k = , , . . . .. (). The system of functions () and () arise in [] for the solution of a nonlocal boundary value problem in heat conduction. It is easy to verify that the systems of function () and () are biorthonormal on [, ]. They are also Riesz bases in L [, ] (see [, ]). We will use the Fourier series representation of the weak solution to transform the initial-boundary value problem to the infinite set of nonlinear integral equations. Any solution of ()-() can be represented as. u(x, t, ε) =. ∞ k=. uk (t, ε)Xk (x),. ().

(3) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 3 of 22. where the functions uk (t, ε), k = , , , . . . , satisfy the following system of equations: . t. u (t, ε) = ϕ +. f (τ ) dτ ,   – (π k) t. uk (t, ε) = ϕk e. +ε(π k). +.   + ε(πk). uk– (t, ε) = ϕk– – ϕk +. . t. fk (τ )e.  – (π k) (t–τ ) +ε(π k). dτ ,. . (π k) t (πk) – +ε(π k) ϕ k e  + ε(πk) t  (πk) + fk– (τ ) –  – (t – τ )fk (τ )  + ε(πk)   + ε(πk)  – (π k) (t–τ ). ·e. +ε(π k). dτ ,. where . . ϕk =. ϕ(x)Yk (x) dx, . . fk (x) =. . f (x, t, u)Yk (x) dx. . Definition  Denote the set. u(t, ε) = u (t, ε), uk (t, ε), uk– (t, ε), k = , , . . . ,. . of functions continuous on [, T] satisfying the condition ∞ . .

(4). max uk (t, ε) + max uk– (t, ε) < ∞, max u (t, ε) +. ≤t≤T. k=. ≤t≤T. ≤t≤T. by B. Let ∞ . .

(5). u(t, ε) = max u (t, ε) + max uk (t, ε) + max uk– (t, ε). ≤t≤T. k=. ≤t≤T. ≤t≤T. be the norm in B. It can be shown that B is a Banach space []. We denote the solution of the nonlinear system () by {u(t, ε)}. Theorem  () Assume the function f (x, t, u) is continuous with respect to all arguments in ¯ × (–∞, ∞) and satisfies the following condition: D. f (t, x, u) – f (t, x, u) ˜ ˜ ≤ b(t, x)|u – u|, where b(x, t) ∈ L (D), b(x, t) ≥ , () f (x, t, ) ∈ C  [, ], t ∈ [, ], () ϕ(x) ∈ C  [, ]. Then the system () has a unique solution in D.. ().

(6) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 4 of 22. Proof For N = , , . . . , let us define an iteration for the system () as follows:. u(N+) (t, ε) = u()   (t, ε) +. +. ∞ . t . . . . . f ξ , τ , u(N)  (τ , ε). u(N) k (τ , ε)( – ξ ) sin πkξ. + u(N) k– (τ , ε) cos πkξ. k=. u(N+) (t, ε) = u() k k (t, ε) +. +. ∞ .   + ε(πk). t. . e . –(π k) (t–τ ) +ε(π k). . u(N) k (τ , ε)( – ξ ) sin πkξ. . ξ dξ dτ ,. . f ξ , τ , u(N)  (τ , ε). + u(N) k– (τ , ε) cos πkξ. . k=. · sin πkξ dξ dτ , () u(N+) k– (t, ε) = uk– (t, ε) +. +. ∞ .   + ε(πk). t. . e . –(π k) (t–τ ) +ε(π k). . u(N) k (τ , ε)( – ξ ) sin πkξ. . . f ξ , τ , u(N)  (τ , ε). + u(N) k– (τ , ε) cos πkξ. k=.  · ξ cos πkξ dξ dτ +  + ε(πk) +. ∞ . t. . e . –(π k) (t–τ ) +ε(π k). . . () . f ξ , τ , u(N)  (τ ). + u(N) k– (τ ) cos πkξ. u(N) k (τ )( – ξ ) sin πkξ. . . k=. · (t – τ ) sin πkξ dξ dτ. (πk) e f ξ , τ , u(N)  (τ , ε)  + ε(πk)   ∞ (N) (N) + uk (τ , ε)( – ξ ) sin πkξ + uk– (τ , ε) cos πkξ.  –  + ε(πk). t. . –(π k) (t–τ ) +ε(π k). k=. · (t – τ ) sin πkξ dξ dτ ,. where, for simplicity, we let Au(N) (ξ , τ , ε) = u(N)  (τ , ε) + . ∞ (N) uk (τ , ε)( – ξ ) sin πkξ + u(N) k– (τ , ε) cos πkξ , k=. u(N+) (t, ε) = u()   (t, ε) + (t, ε) = u() u(N+) k k (t, ε) +. t . . f ξ , τ , Au(N) (ξ , τ , ε) ξ dξ dτ ,. .   + ε(πk). · sin πkξ dξ dτ ,. t. .  – (π k) (t–τ ). e . +ε(π k). f ξ , τ , Au(N) (ξ , τ , ε). . ().

(7) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. () u(N+) k– (t, ε) = uk– (t, ε). +. t.   + ε(πk). .  – (π k) (t–τ ). f ξ , τ , Au(N) (ξ , τ , ε) ξ cos πkξ dξ dτ.  – (π k) (t–τ ). f ξ , τ , Au(N) (ξ , τ , ε). e . +ε(π k). . t.  +  + ε(πk). Page 5 of 22. . e . +ε(π k). . · (t – τ ) sin πkξ dξ dτ t  (π k) (t–τ ) (πk)  – – e +ε(π k) f ξ , τ , Au(N) (ξ , τ , ε)    + ε(πk)    + ε(πk) · (t – τ ) sin πkξ dξ dτ ,  – (π k) t. () where u()  (t, ε) = ϕ , uk (t, ε) = ϕk e. +ε(π k). , u() k– (t, ε) = (ϕk– – ϕk +. (π k) ϕ )× +ε(π k) k.  – (π k) t. . From the condition of the theorem we have u() (t, ε) ∈ B. We will prove that the other sequential approximations satisfy this condition. Let us write N =  in (). e. +ε(π k). t. () u()  (t, ε) = u (t, ε) +. . Adding and subtracting () u()  (t, ε) = u (t, ε) +. t. . f ξ , τ , Au() (ξ , τ , ε) dξ dτ .. . t . f (ξ , τ , ) dξ dτ to both sides of the last equation, we obtain. . t . f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , ) dξ dτ.  . . f (ξ , τ , ) dξ dτ .. + . . Applying the Cauchy Inequality to the last equation, we have. (). u (t, ε) ≤ |ϕ | + . . . dτ . .  t . t. +.  t . t. . f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , ) dξ. . . . f (ξ , τ , ) dξ. dτ . .  dτ. .. . Applying the Lipschitzs Condition to the last equation, we have. () √. u (t, ε) ≤ |ϕ | + t  +. √. t . t . . b(ξ , τ ) Au() (ξ , τ , ε) dξ. . . . f (ξ , τ , ) dξ. t . . Let us use. (). Au (ξ , τ ) ≤ u() (τ , ε) ..  dτ. .. .  dτ. .  dτ.

(8) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 6 of 22. Taking the maximum of both sides of the last inequality yields the following: √ . max u()  (t, ε) ≤ |ϕ | + T b(x, t) L.  (D). ≤t≤T.  – (π k) t. u() k (t, ε) = ϕk e. +ε(π k).   + ε(πk). +. √ () u (t, ε) + T f (x, t, ). L (D). t. .  – (π k) (t–τ ). e . +ε(π k). ,. f ξ , τ , Au() (ξ , τ , ε). . · sin πkξ dξ dτ ..  Adding and subtracting +ε(π k) the last equation, we obtain. t.  – (π k) (t–τ ).  e. . +ε(π k). f (ξ , τ , ) sin πkξ dξ dτ to both sides of.  – (π k) t. u() k (t, ε) = ϕk e +. +ε(π k).   + ε(πk). t. .  – (π k) (t–τ ) . e . +ε(π k). f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , ). . · sin πkξ dξ dτ t  (π k) (t–τ )  – + e +ε(π k) f (ξ , τ , ) sin πkξ dξ dτ .  + ε(πk)   Applying the Cauchy Inequality to the last equation, we have. (). u (t, ε). k . t. ≤ |ϕk | +.  ) – (π k) (t–τ . e. +ε(π k).  dτ. . ·. t . +. t.   + ε(πk)  ) – (π k) (t–τ . e. +ε(π k). . . f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , ) sin πkξ dξ. .  dτ. .  t dτ. . .   + ε(πk). . . . f (ξ , τ , ) sin πkξ dξ.  .. dτ. . Taking the summation of both sides with respect to k and using the Hölder Inequality yield the following:  ∞ ∞ ∞ . . ()  . u (t, ε) ≤ |ϕk | + √ k  π k= k  (  + ε(πk) ) k= k= ·. ∞  t . . (). . . f ξ , τ , Au (ξ , τ , ε) – f (ξ , τ , ) sin πkξ dξ.  k=. .  ∞   + √  π k= k  (  + ε(πk) ) ·. ∞  t  k=. . .  f (ξ , τ , ) sin πkξ dξ. dτ. .. .  dτ.

(9) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 7 of 22. Using the Bessel Inequality in the last inequality, we obtain ∞ ∞ ∞ t  . . () . u (t, ε) ≤ |ϕk | + √ f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , ) k    k=  k= k= .  · sin πkξ dξ  + √  . . dτ. ∞  t.  k=. .  f (ξ , τ , ) sin πkξ dξ. dτ. .. . Applying the Lipschitzs Condition to the last equation and taking the maximum of both sides of the last inequality yield the following: ∞ k=. ∞. . ().  . max uk (t, ε) ≤ |ϕk | + √ b(x, t) L (D) u() (t, ε) B  ≤t≤T   k=.  + √ f (x, t, ) L (D) ,    (π k) t (πk) – () uk– (t, ε) = ϕk– – ϕk + ϕk e +ε(π k)  + ε(πk) t  (π k) (t–τ )  – + e +ε(π k) f ξ , τ , Au() (ξ , τ , ε) ξ cos πkξ dξ dτ  + ε(πk)   t  (π k) (t–τ )  – e +ε(π k) f ξ , τ , Au() (ξ , τ , ε) +  + ε(πk)   · (t – τ ) sin πkξ dξ dτ t  (π k) (t–τ )  (πk) – +ε(π k) – e f ξ , τ , Au() (ξ , τ , ε)  + ε(πk)    + ε(πk) · (t – τ ) sin πkξ dξ dτ . Adding and subtracting   + ε(πk)   + ε(πk)   + ε(πk). t. .  – (π k) (t–τ ). e . t. .  – (π k) (t–τ ). e . f (ξ , τ , )ξ cos πkξ dξ dτ ,. +ε(π k). f (ξ , τ , )(t – τ ) sin πkξ dξ dτ ,. . t. .  – (π k) (t–τ ). e . +ε(π k). . . +ε(π k). (πk) f (ξ , τ , )(t – τ ) sin πkξ dξ dτ  + ε(πk). to both sides of the last equation and applying the derivative to ϕk , we obtain (t, ε) = ϕk– + u() k–. (π k) t  (πk) – ϕ + ϕ e +ε(π k) π  k  k  + ε(πk) k t  –(π k) (t–τ )  + e +ε(π k) f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , )   + ε(πk)  .

(10) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 8 of 22. · ξ cos πkξ dξ dτ t  –(π k) (t–τ )  + e +ε(π k) f (ξ , τ , )ξ cos πkξ dξ dτ  + ε(πk)   t  –(π k) (t–τ )  e +ε(π k) f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , ) +  + ε(πk)   · (t – τ ) sin πkξ dξ dτ t  –(π k) (t–τ )  + e +ε(π k) f (ξ , τ , )(t – τ ) sin πkξ dξ dτ  + ε(πk)   t  –(π k) (t–τ ) (πk)  e +ε(π k) –   + ε(πk)    + ε(πk) · f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , ) (t – τ ) sin πkξ dξ dτ t  –(π k) (t–τ )  (πk) +ε(π k) e +  + ε(πk)    + ε(πk) · f (ξ , τ , )(t – τ ) sin πkξ dξ dτ . Applying the Cauchy Inequality to the last equation, we have the following:. (). u (t, ε). k–. . . . . ϕ. ϕk +    π k  + ε(πk) k t (π k) (t–τ )  – e +ε(π k) dτ +. ≤ |ϕk– | +. . t .   f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , )   + ε(πk)     · ξ cos πkξ dξ dτ ·. .  ) – (π k) (t–τ . t. +. e. +ε(π k).  t . f (ξ , τ , )ξ cos πkξ dξ. dτ. . .   ) – (π k) (t–τ . t. +. e. +ε(π k). ·. . .   + ε(πk). . . f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , ). .  · (t – τ ) sin πkξ dξ .  ) – (π k) (t–τ . t. +. e. +ε(π k). . dτ. . t . . .  dτ.  dτ. . ·. t . +. t.   + ε(πk)  ) – (π k) (t–τ . e . +ε(π k). . . . f (ξ , τ , )(t – τ ) sin πkξ dξ .  dτ.  dτ.  dτ.

(11) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. ·. t .   + ε(πk). .  . . · (t – τ ) sin πkξ dξ +. t.  ) – (π k) (t–τ . e. +ε(π k). Page 9 of 22. (πk) f ξ , τ , Au() (ξ , τ , ε) – f (ξ , τ , )   + ε(πk) . dτ  . dτ. . ·. t .   + ε(πk). . . (πk) f (ξ , τ , )(t – τ ) sin πkξ dξ  + ε(πk). .  dτ. .. Taking the summation of both sides with respect to k and using the Hölder, Bessel, and Lipschitzs Inequalities yields the following: ∞ ∞ ∞ . . () π . . . u (t, ε) ≤ + |ϕ | + ϕk √ √ k– k–  π  k= k= k=  + √    + √   |t| + √  . t . .  b (ξ , τ ) u() (τ , ε) dξ dτ . . . t . . f  (ξ , τ , ) dξ dτ . . . t . . .  b (ξ , τ ) u() (τ , ε) dξ dτ. . .   |t|  + √ f (ξ , τ , ) dξ dτ      t .  |t|π + √ b (ξ , τ ) u() (τ , ε) dξ dτ     t  |t|π  + √ f (ξ , τ , ) dξ dτ .     t . Taking the maximum of both sides of the last inequality yields the following: ∞ k=. max u() k– (t, ε). ≤t≤T. ≤. ∞ k=. |ϕk– | +. ∞  + π  . . ϕk √   k=.   + √ b(x, t) L (D) u() (t, ε) B + √ f (x, t, ) L (D)      . |T| |T| + √ b(x, t) L (D) u() (t, ε) B + √ f (x, t, ) L (D)      . |T|π |T|π + √ b(x, t) L (D) u() (t, ε) B + √ f (x, t, ) L (D) .      .

(12) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 10 of 22. Finally we have the following inequality: ∞ .

(13). () (). u (t, ε) =  max u() (t, ε) +  max u()  k (t, ε) + max uk– (t, ε) B ≤t≤T. k=. ≤t≤T. ≤t≤T. ∞ ∞  + π  . . ϕk |ϕk | + |ϕk– | + ≤ |ϕ | +  √  k= k= √ √ √ √. ()  ( +  π) |T| b(x, t) u (t, ε) +  T+ + L (D) B   √ √ √ √.  ( +  π) |T| f (x, t, ) +  T+ + . L (D)   Hence u() (t, ε) ∈ B. In the same way, for a general value of N we have ∞ (N).

(14). (N). u (t, ε) =  max u(N) (t, ε) + . u (t, ε) + max u(N) (t, ε). max  k k– B ≤t≤T. k=. ≤t≤T. ≤t≤T. ∞ ∞  + π  . . ϕk |ϕk | + |ϕk– | + √  k= k= √ √ √ √. (N–)  ( +  π) |T| b(x, t) u +  T+ + (t, ε) B L (D)    √ √ √ √.  ( +  π) |T| f (x, t, ) +  T+ + , L (D)  . ≤ |ϕ | + . considering the induction hypothesis that u(N–) (t, ε) ∈ B, we deduce that u(N) (t, ε) ∈ B, and by the principle of mathematical induction we obtain. u(t, ε) = u (t, ε), uk (t, ε), uk– (t, ε), k = , , . . . ∈ B.. . Now we prove that the iterations u(N+) (t, ε) converge in B, as N → ∞. We have u() (t, ε) – u() (t, ε) ∞ () () () () =  u() (t, ε) – u (t, ε) +  uk (t, ε) – u()   k (t, ε) + uk– (t, ε) – uk– (t, ε) k=. t. . =. e . –(π k) (t–τ ) +ε(π k). .  +  + ε(πk) + –.   + ε(πk)   + ε(πk). f ξ , τ , Au() (ξ , τ , ε) ξ dξ dτ. t. . e . . e. –(π k) (t–τ ) +ε(π k). f ξ , τ , Au() (ξ , τ , ε) (t – τ ) sin πkξ dξ dτ. –(π k) (t–τ ) +ε(π k). (πk) f ξ , τ , Au() (ξ , τ , ε)   + ε(πk). . t. . e . f ξ , τ , Au() (ξ , τ , ε) ξ cos πkξ dξ dτ. . t . –(π k) (t–τ ) +ε(π k). . · (t – τ ) sin πkξ dξ dτ ..

(15) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to the right side of (), respectively, we obtain. (). u (t, ε) – u() (t, ε). ∞ . () (). u (t, ε) – u() (t, ε) + u() (t, ε) – u() (t, ε). ≤  u()  (t, ε) – u (t, ε) +  k k k– k– k=. √. √ t  .  ( +  π) |T| + b (ξ , τ ) dξ dτ u() (t, ε).     √ √ √ t   √  ( +  π) |T|  +  T+ + f (ξ , τ , ) dξ dτ .    . . √ ≤  T+. √. Let √ √ √ t   √.  ( +  π) |T|  + b (ξ , τ ) dξ dτ u() (t, ε). AT =  T +     √ √ √ t   √  ( +  π) |T| + +  T+ f  (ξ , τ , ) dξ dτ ,     () u() (t, ε) – u() (t, ε) =  u()  (t, ε) – u (t, ε) +. ∞ () () () uk (t, ε) – u() k (t, ε) + uk– (t, ε) – uk– (t, ε) . k=. Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to the right hand side of (), respectively, we obtain. (). u (t, ε) – u() (t, ε). (). ≤  u()  (t, ε) – u (t, ε) +. ∞ . (). u (t, ε) – u() (t, ε) + u() (t, ε) – u() (t, ε). k k k– k– k=. √ √ t    ( +  π) |T| + b (ξ , τ ) dξ dτ AT ,     () () () () u (t, ε) – u (t, ε) =  u (t, ε) – u (t, ε) . √ ≤  T+. √. +. ∞ () () () uk (t, ε) – u() k (t, ε) + uk– (t, ε) – uk– (t, ε) . k=. Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to the right hand side of (), respectively, we obtain. (). u (t, ε) – u() (t, ε). ∞ . () (). +. u (t, ε) – u() (t, ε) + u() (t, ε) – u() (t, ε). ≤  u() (t, ε) – u (t, ε)   k k k– k– k=. Page 11 of 22.

(16) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 12 of 22. √ √ t  . ().   ( +  π)   (). + b (ξ , τ ) u (t, ε) – u (t, ε) dξ dτ     √ √ √ √  ( +  π)   + ≤  T+ AT   t  τ   · b (ξ , τ ) b (ξ , τ ) dξ dτ dξ dτ . √ ≤  T+. . . . √ ≤  T+. √. . √. . √ t     ( +  π)     + AT √ b (ξ , τ ) dξ dτ .     . √. In the same way, for a general value of N we have. (N+). u (t, ε) – u(N) (t, ε). (t, ε) – u(N) ≤  u(N+)   (t, ε) +. ∞ . (N+). (N+) (N). u (t, ε) – u(N) k k (t, ε) + uk– (t, ε) – uk– (t, ε) k=. √ √  N t   ( +  π) |T| N AT + b (ξ , τ ) dξ dτ √   N!   √ √ √ N √ N  ( +  π) |T|  + ≤  T+ AT √ b(x, t) L (D) .    N! . √ ≤  T+. √. Then the last inequality shows that the u(N+) (t, ε) converge in B. Now let us show limN→∞ u(N+) (t, ε) = u(t, ε). Noting that. t . (N). . f ξ , τ , Au(ξ , τ , ε) – f ξ , τ , Au (ξ , τ , ε) ξ dξ dτ.  . ∞ t  –(π k) (t–τ ). . + . e +ε(π k).  + ε(πk)   k=. . · f ξ , τ , Au(ξ , τ , ε) – f ξ , τ , Au(N) (ξ , τ , ε) sin πkξ dξ dτ. ∞ t  –(π k) (t–τ ). . + . e +ε(π k).  + ε(πk)   k=. . (N) · f ξ , τ , Au(ξ , τ , ε) – f ξ , τ , Au (ξ , τ , ε) ξ cos πkξ dξ dτ. ∞ t  –(π k) (t–τ ). . + . e +ε(π k).  + ε(πk)   k=. . (N) · (t – τ ) f ξ , τ , Au(ξ , τ , ε) – f ξ , τ , Au (ξ , τ , ε) sin πkξ dξ dτ. ∞ t  –(π k) (t–τ ).  (πk). + . e +ε(π k) .  + ε(πk)    + ε(πk) k=. ().

(17) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 13 of 22. . (N) · (t – τ ) f ξ , τ , Au(ξ , τ , ε) – f ξ , τ , Au (ξ , τ , ε) sin πkξ dξ dτ. √ √ √ √.  ( +  π) |T| b(x, t) u(τ , ε) – u(N) (τ , ε) , ≤  T+ + L (D) B    it follows that if we prove limN→∞ u(τ , ε) – u(N) (τ , ε) B = , then we may deduce that u(t, ε) satisfies (). To this aim we estimate the difference u(t, ε) – u(N+) (t, ε) B ; after some transformation we obtain. u(t, ε) – u(N+) (t, ε). =  u (t, ε) – u(N+) (t, ε).  +. ∞ . . uk (t, ε) – u(N+) (t, ε) + uk– (t, ε) – u(N+) (t, ε). k. k–. k=. t . ≤ . f ξ , τ , Au(ξ , τ , ε) – f ξ , τ , Au(N) (ξ , τ , ε) ξ dξ dτ. . . ∞ t  –(π k) (t–τ ). . + . e +ε(π k).  + ε(πk)   k=. . · f ξ , τ , Au(ξ , τ , ε) – f (ξ , τ , Au(N) (ξ , τ , ε)] sin πkξ dξ dτ. ∞ t  –(π k) (t–τ ). . + . e +ε(π k).  + ε(πk)   k=. . (N) · f ξ , τ , Au(ξ , τ , ε) – f ξ , τ , Au (ξ , τ , ε) ξ cos πkξ dξ dτ. ∞ t  –(π k) (t–τ ). . + . e +ε(π k).  + ε(πk)   k=. . (N) · (t – τ ) f ξ , τ , Au(ξ , τ , ε) – f ξ , τ , Au (ξ , τ , ε) sin πkξ dξ dτ. ∞ t  –(π k) (t–τ ).  (πk). + . e +ε(π k) .  + ε(πk)    + ε(πk) k=. . · (t – τ ) f ξ , τ , Au(ξ , τ , ε) – f ξ , τ , Au(N) (ξ , τ , ε) sin πkξ dξ dτ .. Adding and subtracting f (ξ , τ , Au(N+) (ξ , τ , ε)) under the appropriate integrals to the right hand side of the inequality we obtain √ √ √ √. u(t, ε) – u(N+) (t, ε) ≤  T +  + ( +  π) |T|   t  .  · b (ξ , τ ) u(τ , ε) – u(N+) (τ , ε) dξ dτ . .

(18) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 14 of 22. √ √ √  ( +  π) |T| + +  T+   t  . · b (ξ , τ ) dξ dτ u(N+) (t, ε) – u(N) (t, ε) B . . √. . . Applying the Gronwall Inequality to the last inequality and using inequality (), we have. u(t, ε) – u(N+) (t, ε) ≤. B. √ √ √ √.   ( +  π) |T| (N+) b(x, t) (N+) AT  T + + L (D) N!   √ √ √ √.  ( +  π) |T|  b(x, t))  . + · exp  T + L (D)   . (). For the uniqueness, we assume that problem ()-() has two solutions u, v. Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to the right hand side of |u(t, ε) – v(t, ε)|, respectively, we obtain. u(t, ε) – v(t, ε)  . √ ≤  T+. √ √ . .  ( +  π) |T|  t   + b (ξ , τ ) u(τ , ε) – v(τ , ε) dξ dτ ,    . √. applying the Gronwall Inequality to the last inequality we have u(t, ε) = v(t, ε). . The theorem is proved.. 3 Solution of problem (1)-(4) Using the solution of the system () we compose the series u (t, ε) + . ∞ . uk (t, ε)( – x) sin πkx + uk– (t, ε) cos πkx .. k=. It is evident that the series converges uniformly on D. Therefore the sum. u(ξ , τ , ε) = u (τ , ε) + . ∞ . uk (τ , ε)( – ξ ) sin πkξ + uk– (τ , ε) cos πkξ ,. k=. is continuous on D. We have ul (ξ , τ , ε) = u (τ , ε) + . ł . uk (τ , ε)( – ξ ) sin πkξ + uk– (τ , ε) cos πkξ .. k=. From the conditions of Theorem  and from lim ul (ξ , τ , ε) = u(ξ , τ , ε),. l→∞. ().

(19) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 15 of 22. it follows that lim f ξ , τ , ul (τ , ξ , ε) = f ξ , τ , u(ξ , τ , ε) .. l→∞. Using ul (ξ , τ , ε) and ϕl (x) = ϕ + . l . ϕk ( – x) sin πkx + ϕk– cos πkx ,. x ∈ [, ],. k=. on the left hand side of () we denote the obtained expression by Jl : ∂v ∂  v ∂  v + ε  u(l) (x, t, ε) + f x, t, u(l) (x, t, ε) v(x, t) dx dt ∂t ∂x ∂x ∂t    ∂  v(x, ) + ϕ(l) (x) v(x, ) – ε dx. ∂x . Jl =. T.  . (). Applying the integration by parts the formula on the right hand side, the last equation, and using the conditions of Theorem , we can show that lim Jl = . l→∞. This shows that the function u(x, t, ε) is a generalized (weak) solution of problem ()-(). The following theorem shows the existence and uniqueness results for the generalized solutions of problem ()-(). Theorem  Under the assumptions of Theorem , problem ()-() possesses a unique gen¯ in the following form: eralized solution u = u(x, t) ∈ C(D) u(x, t, ε) = u (t, ε) + . ∞ uk (t, ε)( – x) sin πkx + uk– (t, ε) cos πkx . k=. 4 Continuous dependence upon the data In this section, we shall prove the continuous dependence of the solution u = u(x, t, ε) using an iteration method. Theorem  Under the conditions of Theorem , the solution u = u(x, t, ε) depends continuously upon the data. Proof Let φ = {ϕ, f } and φ = {ϕ, f } be two sets of data which satisfy the conditions of Theorem . Let u = u(x, t, ε) and v = v(x, t, ε) be the solutions of problem ()-() corresponding to the data φ and φ, respectively, and. f (t, x, ) – f (t, x, ) ≤ ε ,. for ε ≥ ..

(20) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 16 of 22. The solution v = v(x, t, ε) is defined by the following forms, respectively: . t. v (t, ε) = ϕ  + . f  (τ ) dτ ,.  – (π k) t. vk (t, ε) = ϕ k e. +ε(π k). .  +  + ε(πk). . t. f k (τ )e.  – (π k) (t–τ ) +ε(π k). dτ ,. (π k) t t –(π k) (t–τ ) (πk)  –  +ε(π k) ϕ e + e +ε(π k) vk– (t, ε) = ϕ k– – ϕ k +  + ε(πk) k  + ε(πk)  (πk) · f k– (τ ) –  – (τ ) dτ , (t – τ )f k  + ε(πk) where . . . ϕk =. ϕ(x)Yk (x) dx,. f k (x) =. . . f (x, t, u)Yk (x) dx. . For simplicity, let us write ∞ . Av(N) (ξ , τ ) = v(N)  (τ , ε) + . (N) v(N) k (τ , ε)( – ξ ) sin πkξ + vk– (τ ) cos πkξ ,. k=. v(N+) (t, ε) = v()   (t, ε) +. t . . f ξ , τ , Av(N) (ξ , τ , ε) ξ dξ dτ ,. . t  –(π k) (t–τ )  e +ε(π k)  + ε(πk)   · f ξ , τ , Av(N) (ξ , τ , ε) sin πkξ dξ dτ , t  –(π k) (t–τ )  (N+) () e +ε(π k) vk– (t, ε) = vk– (t, ε) +  + ε(πk)   · f ξ , τ , Av(N) (ξ , τ , ε) ξ cos πkξ dξ dτ t  –(π k) (t–τ ) πk e +ε(π k) f ξ , τ , Av(N) (ξ , τ , ε) –  + ε(πk)   (t, ε) = v() v(N+) k k (t, ε) +. · (t – τ ) sin πkξ dξ dτ ,  – (π k) t.  – (π k) t. () +ε(π k) , v() (t) = (ϕ +ε(π k) . From the where v() k– – πkϕ k )e  (t) = ϕ  , vk (t) = ϕ k e k– condition of the theorem we have v() (t, ε) ∈ B. We will prove that the other sequential approximations satisfy this condition. First of all, we consider u() (t, ε) – v() (t, ε):. u() (t, ε) – v() (t, ε) =. . () u()  (t, ε) – v (t, ε) +  . = (ϕ – ϕ  ) + . f ξ , τ , Au() (ξ , τ , ε) – f ξ , τ , Av() (ξ , τ , ε) ξ dξ dτ .  –(π k) t. + (ϕk – ϕ k )e. () () () u() k (t, ε) – vk (t, ε) + uk– (t, ε) – vk– (t, ε). k=. t . ∞ .

(21) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. t. . + . –(π k) (t–τ ) f ξ , τ , Au() (ξ , τ , ε) – f ξ , τ , Av() (ξ , τ , ε) e +ε(π k) sin πkξ dξ dτ. . – (π k) t + (ϕk– – ϕ k– ) – πk(ϕk – ϕ k ) e +ε(π k) t  + f ξ , τ , Au() (ξ , τ , ε) – f ξ , τ , Av() (ξ , τ , ε) . ·e. . –(π k) (t–τ ) +ε(π k). ξ cos πkξ dξ dτ t  (t – τ ) f ξ , τ , Au() (ξ , τ , ε) – f ξ , τ , Av() (ξ , τ , ε) – πk . ·e. –(π k) (t–τ ) +ε(π k). . sin πkξ dξ dτ .. Adding and subtracting t. t. . . f (ξ , τ , )ξ dξ dτ , . t.  . f (ξ , τ , )e . t. . f (ξ , τ , )e . t. . f (ξ , τ , )e. –(π k) (t–τ ) +ε(π k). sin πkξ dξ dτ ,. –(π k) (t–τ ) +ε(π k). ξ cos πkξ dξ dτ ,.  . f (ξ , τ , )e.  . sin πkξ dξ dτ ,. .  t. –(π k) (t–τ ) +ε(π k). . . . . f (ξ , τ , )ξ dξ dτ ,. . . . . –(π k) (t–τ ) +ε(π k). ξ cos πkξ dξ dτ ,. . t. . f (ξ , τ , )e. –πk . sin πkξ dξ dτ ,. . t. . f (ξ , τ , )e. –πk . –(π k) (t–τ ) +ε(π k). –(π k) (t–τ ) +ε(π k). sin πkξ dξ dτ. . to both sides of the last equation. Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to the right side of (), respectively, we obtain. (). u (t, ε) – v() (t, ε). ∞ . () (). +. u (t, ε) – v() (t, ε) + u() (t, ε) – v() (t, ε). ≤  u() (t, ε) – v (t, ε)   k k k– k– k=. ≤  max |ϕ – ϕ  | + . ∞ . max |ϕk – ϕ k | + max |ϕk– – ϕ k– |. k=. √ ∞. .  |T| max ϕk – ϕ k. +  k=. √ √ √ t   √.  ( +  π) |T| + ·  T+ b (ξ , τ ) dξ dτ u() (t, ε).     . Page 17 of 22.

(22) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 18 of 22. √ √ √ t  .  ( +  π) |T|  + +  T+ b (ξ , τ ) dξ dτ v() (t, ε).     √ √ √ t   √  ( +  π) |T|  +  T+ + f  (ξ , τ , ) – f (ξ , τ , ) dξ dτ ,     . √. ϕ – ϕ =  max |ϕ – ϕ  | + . ∞ max |ϕk – ϕ k | + max |ϕk– – ϕ k– | k=. +.  + π √ . ∞ . . max ϕk – ϕ k .. k=. Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to u() (t, ε) – v() (t, ε), respectively, we obtain. (). u (t, ε) – v() (t, ε). ∞ . (). () (). u (t, ε) – v() (t, ε) + u() (t, ε) – v() (t, ε). ≤  u (t, ε) – v (t, ε) +  k k k– k– k=. √. √ t    ( +  π) |T|  + b (ξ , τ ) dξ dτ AT     √ √ √ t   √  ( +  π) |T|  + +  T+ b (ξ , τ ) dξ dτ AT .    . . √ ≤  T+. √. Applying the Cauchy Inequality, the Hölder Inequality, the Lipschitzs Condition, and the Bessel Inequality to u() (t, ε) – v() (t, ε), respectively, we obtain. (). u (t, ε) – v() (t, ε). ∞ . (). () (). u (t, ε) – v() (t, ε) + u() (t, ε) – v() (t, ε). ≤  u (t, ε) – v (t, ε) +  k k k– k–. . √ ≤  T+ ·. t . k=. √. √  ( +  π) |T| +  . √. .  b (ξ , τ ) u() (t, ε) – v() (t, ε) dξ dτ . . . . √ √ √  ( +  π) |T| + +  T+   t  . ().   (). · b (ξ , τ ) u (t, ε) – v (t, ε) dξ dτ . √. . . √ √  ( +  π) |T|  + AT   t  τ   · b (ξ , τ ) b (ξ , τ ) dξ dτ dξ dτ. . √ ≤  T+. . . √. . . √ √ √ √  ( +  π) |T|  + +  T+ AT   .

(23) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. ·. t . . . τ. . . . b (ξ , τ ) . . Page 19 of 22. . . . b (ξ , τ ) dξ dτ dξ dτ . . √ √ √ t    √  ( +  π) |T|   + ≤  T+ AT √ b (ξ , τ ) dξ dτ      √ √ √ t     √  ( +  π) |T|   + +  T+ AT √ b (ξ , τ ) dξ dτ .      In the same way, for a general value of N we have. (N+). u (t, ε) – v(N+) (t, ε). ≤  u(N+) (t, ε) – v(N+) (t, ε).   +. ∞ . (N+). u (t, ε) – v(N+) (t, ε) + u(N+) (t, ε) – v(N+) (t, ε). k. k. k–. k–. k=. √ √  N t   ( +  π) |T| N AT + b (ξ , τ ) dξ dτ √   N!   √ √ √ N  N t  √  ( +  π) |T| AT  + +  T+ b (ξ , τ ) dξ dτ √   N!   √ √ √ √ N   ( +  π) |T| N + ≤  T+ AT √ b(x, t) L (D)    N! √ √ √ √ N  ( +  π) |T| N  + +  T+ AT √ b(x, t) L (D)    N! ≤ AT · aN = aN ϕ – ϕ + C(t) + M f – f. . √ ≤  T+. √. where √ √  N t   ( +  π) |T| N   + b (ξ , τ ) dξ dτ √   N!   √ √ √  N t  √  ( +  π) |T| N   + +  T+ b (ξ , τ ) dξ dτ √   N!  . . √ aN =  T +. √. and √ √ √  ( +  π) |T| N + . M =  T +   . √. (The sequence aN is convergent; then we can write aN ≤ M, ∀N .) It follows from the estimation [] that limN→∞ u(N+) (t) = u(t); then let N → ∞ for the last equation,. u(t) – v(t) ≤ M ϕ – ϕ + M f – f , where M = M · M . If f – f ≤ ε and ϕ – ϕ ≤ ε, then |u(t, ε) – v(t, ε)| ≤ ε.. .

(24) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 20 of 22. 5 Numerical procedure for the nonlinear problem (1)-(4) We construct an iteration algorithm for the linearization of problem ()-() for ε = : ∂u(n) ∂  u(n) – = f x, t, u(n–) ,  ∂t ∂x u(n) (, t) = u(n) (, t),. (x, t) ∈ D,. t ∈ [, T],. t ∈ [, T],. u(n) x (, t) = ,. () (). x ∈ [, ].. u(n) (x, ) = ϕ(x),. (). (). f (x, t). Then problem ()-() can be written as Let u(n) (x, t) = v(x, t) and f (x, t, u(n–) ) = a linear problem: ∂v ∂  v – = f (x, t), ∂t ∂x v(, t) = v(, t),. (x, t) ∈ D,. t ∈ [, T],. (). t ∈ [, T],. vx (, t) = , v(x, ) = ϕ(x),. (). (). x ∈ [, ].. (). We use the finite difference method to solve ()-(). We subdivide the intervals [, ] and [, T] into M and N subintervals of equal lengths, h = M and τ = NT , respectively. Then we add the line x = (M + )h to generate the fictitious point needed for the second boundary condition. We choose the implicit scheme, which is absolutely stable and has a second order accuracy in h and a first order accuracy in τ . The implicit monotone difference scheme for ()-() is as follows: vi,j+ – vi,j a =  (vi–,j+ – vi,j+ + vi+,j+ ) + fi,j+ , τ h vi, = ϕi ,. v,j = vM,j ,. vx,Mj = ,. where  ≤ i ≤ M and  ≤ j ≤ N are the indices for the spatial and time steps, respectively, vi,j is the approximation to v(xi , tj ), fi,j = f (xi , tj ), ϕi = ϕ(xi ), xi = ih, tj = jτ []. At the t =  level, adjustment should be made according to the initial condition and the compatibility requirements.. 6 Numerical examples In this section, we will consider an example of numerical solution of the nonlinear problem ()-(). These problems were solved by applying the iteration scheme and the finite difference scheme which were explained in Section . The condition. – u(n) error(i, j) := u(n+) i,j i,j ∞ with error(i, j) := – was used as a stopping criterion for the iteration process..

(25) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. Page 21 of 22. Figure 1 The exact and numerical solutions of u(x, 1). The exact and numerical solutions of u(x, 1) for ε = 0, the exact solution is shown with a dashed line.. Figure 2 The exact and numerical solutions of u(x, 1). The exact and numerical solutions of u(x, 1) for ε = 0.05, the exact solution is shown with a dashed line.. Figure 3 The exact and numerical solutions of u(x, 1). The exact and numerical solutions of u(x, 1), (––) for ε = 0, (··) for ε = 0.05, (..) for ε = 0.1, the exact solution is shown with a dashed line.. Example  Consider the problem ∂ u ∂u ∂  u –  – ε  =  + (π) ( + ε) u + ( + ε) π cos(πx) exp(t) , ∂t ∂x ∂x ∂t u(x, ) = ( – x) sin πx, u(, t) = u(, t),. x ∈ [, π],. t ∈ [, T],. ux (, t) = ,. t ∈ [, T].. It is easy to check that the analytical solution of this problem is u(x, t) = et ( – x) sin πx..

(26) Baglan and Kanca Advances in Difference Equations 2014, 2014:277 http://www.advancesindifferenceequations.com/content/2014/1/277. The comparisons between the analytical solution and the numerical finite difference solution for different ε values when T =  are shown in Figures  and . We show in Figure  the analytical solution for ε =  and the numerical solution for ε = , ε = , , ε = , .. Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript. Author details 1 Department of Mathematics, Kocaeli University, Kocaeli, 41380, Turkey. 2 Department of Management Information Systems, Kadir Has University, Istanbul, 34083, Turkey. Received: 18 September 2013 Accepted: 29 April 2014 Published: 30 Oct 2014 References 1. Halilov, H, Çiftçi, I: On Fourier method for a quasilinear pseudo-parabolic equation with periodic boundary condition. Int. J. Pure Appl. Math. 52, 717-727 (2009) 2. Nakhushev, AM: Equations of Mathematical Biology. Vysshaya Shkola, Moscow (1995) 3. Conzalez-Velasco, EA: Fourier Analysis and Boundary Value Problems. Academic Press, New York (1995) 4. Chandirov, GI: On mixed problem for a class of quasilinear equation. Ph.D. thesis, Tibilisi (1970) 5. Il’in, VA: Solvability of mixed problem for hyperbolic and parabolic equation. Usp. Mat. Nauk 15, 97-154 (1960) 6. Ionkin, N: Solution of a boundary-value problem in heat conduction with a nonclassical boundary condition. Differ. Equ. 13, 204-211 (1977) 7. Ismailov, M, Kanca, F: An inverse coefficient problem for a parabolic equation in the case of nonlocal boundary and overdetermination conditions. Math. Methods Appl. Sci. 34, 692-702 (2011) 8. Kanca, F, Ismailov, M: Inverse problem of finding the time-dependent coefficient of heat equation from integral overdetermination condition data. Inverse Probl. Sci. Eng. 20, 463-476 (2012) 9. Ladyzhenskaya, DA: Boundary Value Problems of Mathematical Physics. Springer, New York (1985) 10. Samarskii, AA: The Theory of Difference Schemes. Dekker, New York (2001). 10.1186/1687-1847-2014-277 Cite this article as: Baglan and Kanca: Weak generalized and numerical solution for a quasilinear pseudo-parabolic equation with nonlocal boundary condition. Advances in Difference Equations 2014, 2014:277. Page 22 of 22.

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