Boundary layers in a two-point boundary value problem with fractional derivatives

Transcription

1 Boundary layers in a two-point boundary value problem with fractional derivatives J.L. Gracia and M. Stynes Institute of Mathematics and Applications (IUMA) and Department of Applied Mathematics, University of Zaragoza (Spain) National University of Ireland (Cork) and Beijing Computational Science Research Center (China) Third Fractional Calculus, Probability and Non-Local Operators: Applications and Recent Developments. Bilbao Nov Research partially supported by IUMA, project MTM P and Diputación General de Aragón

2 Talk overview 1 The Caputo fractional derivative problem: Preliminaries Boundary layers in the constant coefficient problem: Laplace transform Boundary layers in the variable-coefficient problem: Maximum principle/comparison principle An example with an interior layer 2 Some comments and results for the Riemann-Liouville fractional derivative problem 3 Conclusions

3 It models, for example, anomalous diffusion in some porous media flows. Preliminaries We consider the following two-point boundary value problem D δ u(x) + b(x)u (x) + c(x)u(x) = f (x) for x (0, 1), u(0) α 0 u (0) = γ 0, u(1) + α 1 u (1) = γ 1, where 1 < δ < 2 and D δ is the Caputo fractional derivative D δ g(x) := 1 Γ(2 δ) x t=0 (x t) 1 δ g (t) dt for 0 < x 1. The constants α 0, α 1, γ 0, γ 1 and functions b, c and f are given. Assume that α 1 0 and α 0 1 δ 1.

4 Existence, and bounds on derivatives of the solution [M. Stynes & JLG, IMA J. Numer. Anal. 2015] Let b, c, f be sufficiently smooth. Assume that c 0, α 1 0 and α 0 1/(δ 1). Then the boundary value problem has a unique solution u with u C 1 [0, 1] C q+1 (0, 1] (q depends on b, c, f ), and for all x (0, 1] there exists a constant C, which is independent of x but depends on the problem data including δ, such that { C if i = 0, 1, u (i) (x) Cx δ i if i = 2, 3,..., q + 1.

5 An example Set r(x) = x δ with 1 < δ < 2. It is solution of But note that D δ u(x) = Γ(δ + 1) for x (0, 1), u(0) 1 1 δ u (0) = 0, u(1) = 1. r (x) = δx δ 1, r (x) = δ(δ 1)x δ 2 and the second derivative blows up at x = 0.

6 Existence, and bounds on derivatives of the solution [M. Stynes & JLG, IMA J. Numer. Anal. 2015] Let b, c, f be sufficiently smooth. Assume that c 0, α 1 0 and α 0 1/(δ 1). Then the boundary value problem has a unique solution u with u C 1 [0, 1] C q+1 (0, 1] (q depends on b, c, f ), and for all x (0, 1] there exists a constant C, which is independent of x but depends on the problem data including δ, such that { C u (i) if i = 0, 1, (x) Cx δ i if i = 2, 3,..., q + 1.

7 Existence, and bounds on derivatives of the solution [M. Stynes & JLG, IMA J. Numer. Anal. 2015] Let b, c, f be sufficiently smooth. Assume that c 0, α 1 0 and α 0 1/(δ 1). Then the boundary value problem has a unique solution u with u C 1 [0, 1] C q+1 (0, 1] (q depends on b, c, f ), and for all x (0, 1] there exists a constant C, which is independent of x but depends on the problem data including δ, such that { C u (i) if i = 0, 1, (x) Cx δ i if i = 2, 3,..., q + 1. but, how do the constants C i depend on δ? From now on, the constants are independent of x and δ.

8 A sample problem Recall that 1 < δ < 2. Consider the particular problem D δ u(x) + 2u (x) = 1 for x (0, 1), u(0) 1 δ 1 u (0) = 0.4, u(1) = 1.7 Exact solution of this problem can be computed using the Laplace transform

9 Exact solutions for δ = 1.6, Exact solution x Exact solution

10 Exact solutions for δ = 1.3, Exact solution x Exact solution

11 Exact solution for δ = Exact solution u x

12 A heuristic argument explaining the occurrence of a layer? (imitating an argument from singularly perturbed ODEs) Because lim δ 1 + Dδ u(x) = u (x) u (0), the ODE approaches a 1st-order ODE as δ 1 +, and then there are too many boundary conditions, so a boundary layer develops...

13 1D Convection-diffusion problems Example: εu + u = x, x (0, 1), u(0) = 0, u(1) = 1. u R (x) = x, x (0, 1), u R(0) = 0, u R (x) = x 2 / Computed solution x Solution for ε = 1, 10 1, 10 2 and 10 6

14 An example with no boundary layer This argument is incorrect as we now show by another constant-coefficient example whose exact solution can be computed.

15 An example with no boundary layer This argument is incorrect as we now show by another constant-coefficient example whose exact solution can be computed. Consider the particular problem D δ u(x) u (x) = 1 for x (0, 1), u(0) 1 δ 1 u (0) = 0.4, u(1) = 1.7

16 Solutions for δ = 1.6, Exact solution u x Exact solution u x

17 Solutions for δ = 1.2, Exact solution u x 4 Exact solution u x

18 Constant coefficient problem Consider the particular BVP D δ u(x) + bu (x) = f for x (0, 1), u(0) α 0 u (0) = γ 0, u(1) + α 1 u (1) = γ 1, where 1 < δ < 2 and the constants α 0, α 1, γ 0, γ 1, b and f are given, with α 1 0 and α 0 1/(δ 1). We solve the BVP exactly by applying a Laplace transform. The argument can be extended to the case of a variable rhs f (x). Inversion of Laplace transform solution leads naturally to Mittag-Leffler functions: E α,β (z) = k=0 z k Γ(αk + β) which is well defined for α, β, z C with Re(α) > 0.

19 Constant coefficient problem Applying L to the differential equation and observing that L{D δ u} = s δ 2 [ s 2 L{u} su(0) u (0) ] we obtain f L{u} = s 2 (s δ 1 b) + u(0) s + sδ 3 u (0) s δ 1 b f = s 2 (s δ 1 b) + γ 0 + α 0 u (0) s where we used the boundary condition at x = 0. To invert the Laplace transform, use that L { x β 1 E α,β (±ax α ) } = k!sα β s α a. + sδ 3 u (0) s δ 1 b,

20 Constant coefficient problem Thus, u(x) = γ 0 + α 0 u (0) + fx [ b + u (0) f ] b xe δ 1,2 (bx δ 1 ). Differentiating this expression, we obtain that u (x) = f [ b + u (0) f ] E δ 1,1 (bx δ 1 ). b Imposing the boundary condition u(1) + α 1 u (1) = γ 1, we get u (0) = f b + γ 1 γ 0 (1 + α 0 + α 1 )f /b α 0 + E δ 1,2 (b) + α 1 E δ 1,1 (b).

21 Exact solution of particular BVP The solution of the BVP with c 0 and b and f constant is u(x) =γ 0 + (α 0 + x)f b ( ) α 0 + xe δ 1,2 (bx δ 1 ) + γ 1 γ 0 (1 + α 0 + α 1 ) f ; α 0 + E δ 1,2 (b) + α 1 E δ 1,1 (b) b and u (x) = f b + E δ 1,1 (bx δ 1 ) α 0 + E δ 1,2 (b) + α 1 E δ 1,1 (b) ( γ 1 γ 0 (1 + α 0 + α 1 ) f b ). Recall that the layer appeared at x = 1 when δ 1 +. The only possible causes of this layer are E δ 1,n (bx δ 1 ).

22 Some estimates for E δ 1,n (t), t R Recall that 1 < δ < 2 and n is a positive integer. Elementary manipulations of the infinite series show that 1 if t 0, (n 1)! 2 E δ 1,n (t) if 0 t < 1, 1 t δe if t = 1. [Factor 1/(δ 1) here is sharp] δ 1 Conclusion: as δ 1 +, the value of E δ 1,n (t) does not blow up for t < 1 but does blow up for t = 1.

23 Some estimates for E δ 1,n (t), t R Recall that 1 < δ < 2 and n is a positive integer. Elementary manipulations of the infinite series show that 1 if t 0, (n 1)! 2 E δ 1,n (t) if 0 t < 1, 1 t δe if t = 1. [Factor 1/(δ 1) here is sharp] δ 1 Conclusion: as δ 1 +, the value of E δ 1,n (t) does not blow up for t < 1 but does blow up for t = 1. In addition, E δ 1,n ( ) is an increasing function so it blows up as δ 1 + for t > 1 but much more spectacularly... E δ 1,n (t) 1 δ 1 t(1 n)/(δ 1) exp for each fixed t > 1 as δ 1 +. (t 1/(δ 1))

24 Contour for integrating M-L function [I. Podlubny, 1999] For z G + (1, ϕ), 0 < α < 2, and πα/2 < ϕ min{π, πα}, E α,β (z) = 1 α z(1 β)/α exp(z 1/α )+ 1 ζ (1 β)/α exp(ζ 1/α ) 2απi ζ z γ(1,ϕ) where γ(1, ϕ) is the following contour in the complex plane: dζ, y (1, ) 1 (1, ) O - x (1, )

25 Unlike the case b > 1 where u (x) blows up only at x = 1 as δ 1 +, when b = 1 the derivative u (x) blows up at every point x in (0, 1] as δ approaches its limiting value! Case b > 1 [M. Stynes & JLG, Lecture Notes in Comput. Sci. Engineering, 2015] We have that lim δ 1 u (x) = + { ±, if x = 1, f b for each x (0, 1). Case b = 1 [M. Stynes & JLG, Lecture Notes in Comput. Sci. Engineering, 2015]. We have that lim δ 1 u (x) = ± for each x (0, 1]. +

26 A sample problem with b > 1 Recall the particular problem D δ u(x) + 2u (x) = 1 for x (0, 1), u(0) 1 δ 1 u (0) = 0.4, u(1) = Exact solution u x Figure: Solution for δ = 1.11

27 A sample problem with b = 1 Consider now the particular problem D δ u(x) + 1u (x) = 1 for x (0, 1), u(0) 1 δ 1 u (0) = 0.4, u(1) = Exact solution u Exact solution u x x Figure: Solution for δ = 1.1 and δ = 1.05

28 A sample problem with b = 1 Consider now the particular problem D δ u(x) + 1u (x) = 1 for x (0, 1), u(0) 1 δ 1 u (0) = 0.4, u(1) = Exact solution u Exact solution u x x Figure: Solution for δ = 1.01 and δ = 1.001

29 Variable-coefficient problem We now consider the following two-point boundary value problem D δ u(x) + b(x)u (x) + c(x)u(x) = f (x) for x (0, 1), u(0) α 0 u (0) = γ 0, u(1) + α 1 u (1) = γ 1, where 1 < δ < 2 and D δ is the Caputo fractional derivative. The constants α 0, α 1, γ 0, γ 1 and functions b, c and f are given. Assume that α 1 0 and α 0 1 δ 1. We will explain when boundary layers at x = 1 occur in the general case where the differential operator has variable coefficients and f is any smooth function.

30 Comparison principle and Barrier functions [M. Stynes & JLG, IMA J. Numer. Anal. 2015] Assume that g C 2 (0, 1], g (x) Cx θ, 0 < x 1 and b, c C[0, 1] with c(x) 0 for all x (0, 1). If D δ g + bg + cg 0 on (0, 1), g(0) α 0 g (0) 0 and g(1) + α 1 g (1) 0, where α 0 1 δ 1 and α 1 0, then g 0 on [0, 1].

31 Comparison principle and Barrier functions [M. Stynes & JLG, IMA J. Numer. Anal. 2015] Assume that g C 2 (0, 1], g (x) Cx θ, 0 < x 1 and b, c C[0, 1] with c(x) 0 for all x (0, 1). If D δ g + bg + cg 0 on (0, 1), g(0) α 0 g (0) 0 and g(1) + α 1 g (1) 0, where α 0 1 δ 1 and α 1 0, then g 0 on [0, 1]. Counterexample Set g(x) = x x x. Then D 1.2 g(x) 0 for 0 < x < 1, g(0) = g(1) = 0, but g(0.8) < 0, so the Dirichlet boundary conditions do not justify a comparison principle. Note that g(0) α 0 g (0) < 0 if α 0 > 0.

32 Comparison principle and Barrier functions [M. Stynes & JLG, IMA J. Numer. Anal. 2015] Assume that g C 2 (0, 1], g (x) Cx θ, 0 < x 1 and b, c C[0, 1] with c(x) 0 for all x (0, 1). If D δ g + bg + cg 0 on (0, 1), g(0) α 0 g (0) 0 and g(1) + α 1 g (1) 0, where α 0 1 δ 1 and α 1 0, then g 0 on [0, 1]. Comparison principle and barrier functions Thus, if ( D δ + b(x)d + c(x)i )(B ± g) 0, x (0, 1) (I α 0 D)(B ± g)(0) 0 (I + α 1 D)(B ± g)(1) 0, = g B.

33 Summary of the variable-coefficient problem In our class of problems define D δ u(x) + b(x)u (x) + c(x)u(x) = f (x) for x (0, 1), u(0) α 0 u (0) = γ 0, u(1) + α 1 u (1) = γ 1, M = max x [0,1] b(x). u u (0) u (1) c > 0 Bounded No Layer No layer (Robin) Layer (Dirichlet) c 0 M < 0 Bounded No Layer No Layer 0 M < 1 Bounded No Layer No Layer M = 1 Unbounded No Layer Layer M > 1 & b > 0 Unbounded No Layer Layer Table: [M. Stynes & JLG, Comput. Methods Appl. Math. 2015]

34 An example: Consider the problem with Then, D δ u(x) + b(x)u (x) + c(x)u(x) = f (x) for x (0, 1), u(0) 1 δ 1 u (0) = γ 0, u(1) + α 1 u (1) = γ 1, M = max b(x) > 1, b(x) > 0, c(x) 0, x [0, 1]. x [0,1] (i) u C δ 1, (ii) u (0) C, (iii) u (1) C δ 1, if α 1 > 0, (iv) (Robin) u (1) CM1/(δ 1), if α 1 = 0 (Dirichlet) δ 1

35 Example: M > 1, b > 0 and α 1 > 0 Example D δ u(x) + (x + 0.2)u (x) = 2(3 + x) for x (0, 1), u(0) 1 δ 1 u (0) = 0.4, u(1) + u (1) = 1.7. The exact solution of this problem is unknown. We compute a numerical solution with a finite difference scheme on a uniform mesh with N = This scheme uses L2 approximation for D δ [K.B Oldham, J. Spanier, 1974] upwinding for u Thus, this scheme is monotone [JLG & M. Stynes, J. Comput. Math. Appl. 2015].

36 Example: M > 1, b > 0 and α 1 > 0 Example Observe that D δ u(x) + (x + 0.2)u (x) = 2(3 + x) for x (0, 1), u(0) 1 δ 1 u (0) = 0.4, u(1) + u (1) = 1.7. M = = 1.2 > 1 and b 0.2 > 0. Table: Verifying sharpness of estimates (i) u u (1) C δ 1 C δ 1 and (iii) δ = 1.01 δ = δ = δ = (δ 1) max 0 j N u j (δ 1) u N u N 1 x N x N

37 Figure: Plot of b(x) = 8x 2 + 6x + 1: b(x) > 1 on part of the interval but b < 0 near x = 1. Example: An interior layer Example D δ u(x) + ( 8x 2 + 6x + 1)u (x) = 1 for x (0, 1), u(0) 1 δ 1 u (0) = 0.4, u(1) = b(x)= 8x 2 +6x x

38 Example: An interior layer 8 90 Computed solution x Computed solution x Figure: Test problem with δ = 1.1 (left) and δ = 1.01 (right)

39 Example: An interior layer Computed solution Computed solution x x Figure: Test problem with δ = (left) and δ = (right)

40 Riemann-Liouville fractional derivative problem Finally, we consider the following BVP D δ RL u(x) + bu (x) = f for x (0, 1), u(0) = 0, u(1) + α 1 u (1) = γ 1, with b and f constants and DRL δ is the Riemann-Liouville fractional derivative of order δ and it is defined by DRL δ g(x) = d 2 [ 1 x ] dx 2 (x t) 1 δ g(t) dt for 0 < x 1. Γ(2 δ) t=0 Existence and uniqueness of solution is discussed in [N. Kopteva, M. Stynes, submitted].

41 Riemann-Liouville fractional derivative problem The Laplace transform L of the Riemann-Liouville fractional derivative is L{D δ RL u} = sδ L{u} C 1 sc 2, with C 1 = [ ] [ ] D δ 1 RL u (0) and C 2 = D δ 2 RL u (0). taking the Laplace transform, one obtains L{u} = Using again that C 1 s(s δ 1 b) + C 2 s δ 1 b f s 2 (s δ 1 b). L { x β 1 E α,β (±ax α ) } = k!sα β s α a,

42 Riemann-Liouville fractional derivative problem we deduce that u(x) = C 1 x δ 1 E δ 1,δ (bx δ 1 ) + C 2 x δ 2 E δ 1,δ 1 (bx δ 1 ) fx δ E δ 1,δ+1 (bx δ 1 ). In this formula the constants C 1 and C 2 must be chosen to satisfy the boundary conditions. The boundary condition u(0) = 0 forces C 2 = 0. Differentiating this expression, we get u (x) = C 1 x δ 2 E δ 1,δ 1 (bx δ 1 ) fx δ 1 E δ 1,δ (bx δ 1 ), and hence, imposing that u(1) + α 1 u (1) = γ 1, C 1 = γ 1 + f [E δ 1,δ+1 (b) + α 1 E δ 1,δ (b)] E δ 1,δ (b) + α 1 E δ 1,δ 1 (b).

43 Exact solution constant-coefficient problem Consequently, E δ 1,1 (bx δ 1 ) 1 u(x) = γ 1 E δ 1,1 (b) 1 + α 1 E δ 1,0 (b) + x f ( ) 1 E δ 1,2 (bx δ 1 ) b + f ( ) Eδ 1,2 (b) 1 + α 1 [E δ 1,1 (b) 1] ( E δ 1,1(bx δ 1 ) 1) b E δ 1,1 (b) 1 + α 1 E δ 1,0 (b) and u (x) = γ 1 x 1 E δ 1,0 (bx δ 1 ) E δ 1,1 (b) 1 + α 1 E δ 1,0 (b) + f ( ) 1 E δ 1,1 (bx δ 1 ) b + f ( ) b x 1 Eδ 1,2 (b) 1 + α 1 [E δ 1,1 (b) 1] E δ 1,0 (bx δ 1 ). E δ 1,1 (b) 1 + α 1 E δ 1,0 (b) Observe that for each δ (1, 2) u (x) as x 0 +

44 Summary of the RL constant-coefficient problem Recall that D δ RL u(x) + bu (x) = f for x (0, 1), u(0) = 0, u(1) + α 1 u (1) = γ 1. u u (1) b < 1 Bounded No layer b = 1 Unbounded Layer: u (1) C δ 1 b > 1 Unbounded Layer: u (1) C (Robin BC) (δ 1) 2 Layer: u (1) Cb1/(δ 1) (δ 1) 2 (Dirichlet BC) Table: [JLG & M. Stynes, proceedings BAIL 2015]

45 An example Consider the test problem: DRL δ u(x) + u (x) = 1 for x (0, 1), u(0) = 0, u(1) = Exact solution u Exact solution u x x Figure: Exact solution for δ = 1.01 (left fig.) and δ = (right fig.).

46 Figure: Exact solution for δ = Another example Consider the test problem: DRL δ u(x) + 1.1u (x) = 1 for x (0, 1), u(0) = 0, u(1) = Exact solution u x

47 Conclusions: We considered a 2pt BVP whose leading term is a Caputo fractional derivative of order δ with 1 < δ < 2. We first considered the special case of a constant-coefficient operator and we showed that when δ is near 1, the solution of the BVP may exhibit a boundary layer at x = 1. Moving on to the general case of a variable-coefficient differential operator, we then determined conditions on the data of the problem under which boundary layers at each endpoint can occur. This analysis showed that a crucial parameter in the presence or absence of a boundary layer at x = 1 is the quantity M := max x [0,1] b(x). We gave an example, where the sign of b is variable, and we observed that the solution can also exhibit an interior layer. We finally considered a Riemann-Liouville fractional derivative 2pt BVP and we determined when a layer at x = 1 can occur.

48 References: [GS15a] [GS15b] [KS] [OS74] J.L. Gracia, M. Stynes, Central difference approximation of convection in Caputo fractional derivative two-point boundary value problems, J. Comput. Math. Appl. 273 (2015) J.L. Gracia, M. Stynes, Boundary layers in a Riemann-Liouville fractional derivative two-point boundary value problem. Accepted in Lecture Notes in Comput. Sci. Engineering (proceedings BAIL 2014). N. Kopteva, M. Stynes, Analysis and numerical solution of a Riemann-Liouville fractional derivative two-point boundary value problem. (Submitted for publication). K.B. Oldham, J. Spanier, The fractional calculus, Academic Press, 1974.

49 References: [P99] I. Podlubny, Fractional Differential Equations. An Introduction to Fractional Derivatives, Fractional Di.erential Equations, to Methods of Their Solution and some of Their Applications, Math. Sci. Eng. 198, Academic Press, San Diego, [SG15a] M. Stynes, J.L. Gracia, Blow-up of solutions and interior layers in a Caputo two-point boundary value problem. Accepted in Lecture Notes in Comput. Sci. Engineering (proceedings BAIL 2014). [SG15b] M. Stynes, J.L. Gracia, A finite difference method for a two-point boundary value problem with a Caputo fractional derivative, IMA J. Numer. Anal. 35 (2015) [SG15c] M. Stynes, J.L. Gracia, Boundary layers in a two-point boundary value problem with a Caputo fractional derivative, Comput. Methods Appl. Math. 15 (2015)

50 Thank you very much for your attention

51 Thank you very much for your attention Boundary layers in a two-point boundary value problem with a Caputo fractional derivative J.L. Gracia and M. Stynes Institute of Mathematics and Applications (IUMA) and Department of Applied Mathematics, University of Zaragoza (Spain) National University of Ireland (Cork) and Beijing Computational Science Research Center (China) Third Fractional Calculus, Probability and Non-Local Operators: Applications and Recent Developments. Bilbao Nov Research partially supported by IUMA, project MTM P and Diputación General de Aragón