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

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)