Fractional reaction-diffusion equation for species ... - ResearchGate

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4 Baeumer, Kovács, Meerschaert 2 Problem outline The classical reaction-diffusion equation (1) and its fractional analogue (2) are both special cases of a general form ∂u ∂t (x, t) = (Au)(x, t) + f(u(x, t)), u(x, 0) = u 0(x) (5) where A is a pseudo-differential operator [25] that appears as the generator of some continuous convolution semigroup [1,44]. Our goal in this section is to explore the connection between this continuous time evolution equation and its discrete time analogue, the integro-difference equation u n+1 (x) = ∫ ∞ −∞ k τ (x, y)g τ (u n (y)) dy (6) where u n (x) = u(x, nτ) with some τ > 0 fixed. Our general approach is based operator theory for abstract differential equations [1,20,44] and infinitely divisible probability distributions [21,36]. Example 1 To demonstrate the basic idea let us consider the classical reactiondiffusion equation (1) with a growth term f(u) = ru for some real r ≥ 0. It is well know that if r = 0 then the solution is given by u(x, t) = ∫ ∞ −∞ 1 √ e − (x−y)2 4Dt u 0 (y) dy := [T (t)u 0 ](x) 4πDt where the dispersal operator T (t) maps the initial condition at time t = 0, the function u 0 (x), to the solution of the diffusion equation at time t > 0. Likewise, if D = 0 then the solution is given by u(x, t) = e rt u 0 (x) := [S(t)u 0 ](x). where the growth operator S(t) maps the initial condition to the solution of the growth equation at time t > 0. Clearly, [S(t+s)u 0 ](x) = [S(t)S(s)u 0 ](x) = [S(s)S(t)u 0 ](x) and [T (t + s)u 0 ](x) = [T (t)T (s)u 0 ](x) = [T (s)T (t)u 0 ](x) for all t, s ≥ 0 (the semigroup property). Using the product rule its is easy to see that the solution to (1) is of the form u(x, t) = ∫ ∞ −∞ √ 1 4πDt e − (x−y)2 4Dt e rt u 0 (y) dy = [T (t)S(t)u 0 ](x) (7) = [S(t)T (t)u 0 ](x) = e rt ∫ ∞ −∞ √ 1 4πDt e − (x−y)2 4Dt u 0 (y) dy, (8) that is; the solution of the complex problem (1) can be written using the solution of the simpler sub-problems. Now if we fix τ > 0, then by (7) and (8) we have u n+1 (x) = u(x, (n + 1)τ) = [T ((n + 1)τ)S((n + 1)τ)u 0 ](x) = [(T (τ)) n+1 (S(τ)) n+1 u 0 ](x) = [T (τ)S(τ)(T (τ)) n (S(τ)) n u 0 ](x) = [T (τ)S(τ)T (nτ)S(nτ)u 0 ](x) = [T (τ)S(τ)u n ](x) = ∫ ∞ −∞ √ 1 4πDτ e − (x−y)2 4Dτ e rτ u n (y) dy := ∫ ∞ −∞ k τ (x, y)g τ (u n (y)) dy.
Fractional reaction-diffusion equation 5 This means that the exact solution at time t = nτ, for any positive integer n = 1, 2, . . . satisfies an iteration of the form (6) with k τ given by (4) and g τ (u n (y)) = e rτ u n (y). Iterations of the form u n+1 = S(τ)T (τ)(u n ) are called sequential splitting. We now try to follow the same basic idea in the case when both the differential operator and the growth term are more complicated. Assume that the continuous model is well posed in the sense that given u 0 we obtain a unique solution on R × [0, T ]. In this case, at least formally, the solution can be written as u(x, t) = [W (t)u 0 ](x) and we call W (t) the solution operator of (5). In Example 1 the solution operator is given by (7) and since T (t) and S(t) commute W (t) = T (t)S(t). If we would be able to construct W (t) as in Example 1, then the solution of (5) would satisfy the natural iteration u n+1 (x) = [W (τ)u n ](x). Unfortunately if the function f is not just a multiplication by a constant, then it is usually impossible to obtain a closed form of W (t). Therefore, we might want to follow the idea of sequential splitting here, too, that is; instead of looking at (5) we look at two separate differential equations which are easier to solve and construct an iteration based on their solutions. Of course we then have to investigate what this has to do with W (t), the solution operator of (5). Let us denote by T (t) the solution operator to the pseudo-differential equation and by S(t) the solution operator to ∂u (x, t) = (Au)(x, t) (9) ∂t ∂u (x, t) = f(u(x, t)). (10) ∂t The idea behind sequential splitting is that for τ small, we solve first (10) on [0, τ] and then using the obtained solution at time t = τ as initial data for (9) we solve that on [0, τ]. Then hope that this will be close to the result if we would do both at the same time. Mathematically, instead of computing [W (τ)u 0 ](x) we compute [T (τ)S(τ)u 0 ](x). Now if S(t) and T (t) commute, then this is the same thing. If not, then it is usually different. Assume that we are interested in the solution of (5) at time t = nτ. Then, by the above described idea, we set up an iteration u n+1 (x) = [T (τ)S(τ)u n ](x) with starting value u 0 , or equivalently, u n+1 (x) = [(T (τ)S(τ)) n u 0 ] (x). Usually, u n (x) ≠ u(x, nτ) (except for the commuting case), but is it true that u n (x) is close to u(x, nτ) in some sense For example, if we fix t and choose finer and finer time steps τ = t n , then does u n(x) converge to u(x, t) in some sense as n → ∞, in other words, does [ T ( t n )S( t n )] n u0 converge to W (t)u 0 as n → ∞ Example 2 Let us consider the fractional reaction-diffusion equation with Fisher’s growth term ( ) ∂u ∂t (x, t) = ru(x, t) u(x, t) 1 − + D ∂α u K ∂x α (x, t); u(x, 0) = u 0(x) (11)
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