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

More documents

Recommendations

Info

10 Baeumer, Kovács, Meerschaert (see, e.g., Theorem 3.1.11 in [36]) states that if k is infinitely divisible then ˆk(λ) = e ψ(λ) where ψ(λ) = −iλa − 1 ∫ ( 2 λ2 b 2 + e −iλy − 1 + iλy ) y≠0 1 + y 2 φ(y)dy, (24) where a ∈ R, b ≥ 0 , and the jump intensity φ satisfies ∫ min{1, y 2 } φ(y)dy < ∞. y≠0 The triple [a, b, φ] is unique, and we call this the Lévy representation of the infinitely divisible density k. It follows that we can define the convolution power k t to be the infinitely divisible density with Lévy representation [ta, tb, tφ], so that k t has Fourier transform e tψ(k) for any t ≥ 0. It is well known (e.g., see [25, Example 4.1.3]) that every infinitely divisible density is associated with a strongly continuous semigroup on C 0 (R) defined by [T (t)u](x) := ∫ ∞ −∞ k t (x − y) u(y)dy. (25) The following result allows us to compute the generator of this semigroup from the Lévy representation. Proposition 2 Every function u ∈ C 0 (R) with u ′ , u ′′ ∈ C 0 (R) belongs to the domain of the generator A of the semigroup (25), and for such functions we have Au(x) = − au ′ (x) + 1 2 b2 u ′′ (x) ∫ ( ) + u(x − y) − u(x) + u′ (x)y 1 + y 2 φ(y)dy. y≠0 (26) Proof Example 4.1.12 in [25] shows that, for any u ∈ Cc ∞ (R), the space of infinitely differentiable functions u : R → R that vanish off a compact set, u belongs to the domain of A and (26) holds. Then it follows from Corollary 2.4 of [54] that the same holds for all u ∈ C 0 (R) with u ′ , u ′′ ∈ C 0 (R). ⊓⊔ Remark 2 The integral formula in the Lévy representation (24) is a slight generalisation of the formula for the Fourier transform of a compound Poisson density, and indeed any infinitely divisible density is in some sense the limiting case of a compound Poisson [36, Remark 3.1.18], a random sum with a Poisson distributed number of summands where the relative frequency of jumps of size y is given by the jump intensity φ(y). The generator formula comes from the fact that T (t)u is a convolution k t ∗ u so that its Fourier transform is e tψ(λ) û(λ), and then the generator can be computed in Fourier space by t −1 [e tψ(λ) − 1]û(λ) → ψ(λ)û(λ) so that ψ(λ)û(λ) is the Fourier transform of Au(x). Then (26) comes from inverting this Fourier transform using (24) and the fact that multiplication by iλ, e −iλy corresponds with taking the derivative or shifting in real space, respectively.
Fractional reaction-diffusion equation 11 In what follows we discuss a general abstract reaction-diffusion equation for Fisher’s equation (11) for non-negative initial data on X := C 0 (R): ˙u(t) = Au(t) + f(u(t)), u(0) = u 0 ≥ 0, (27) where f : X → X is defined via the function ˜f : R → R as ( [f(u)](x) = ˜f(u(x)) := ru(x) 1 − u(x) ) K We introduce a cut off function via the function f ˜ N : R → R as ⎧ 0 ⎪⎨ ( if u(x) < 0 [f N (u)](x) = ˜f N (u(x)) : ru(x) 1 − u(x) ) if 0 ≤ u(x) ≤ NK K ⎪⎩ rNK(1 − N) if u(x) > NK. (28) (29) Proposition 3 Let X := C 0 (R) and let f be given by (28). Then the abstract differential equation ˙u(t) = f(u(t)), u(0) = u 0 ≥ 0 (30) has a unique strong global solution given by u(t) = S(t)u 0 for each nonnegative u 0 ∈ X, and in fact we have [S(t)u 0 ](x) = [ ˜S(t)](u 0 (x)) where ˜S(t) is defined by (14). For any positive integer N ≥ 2, the abstract differential equation ˙u = f N (u), u(0) = u 0 ≥ 0 for the Lipschitz continuous function f N defined in (29) also has a unique strong global solution given by u(t) = S N (t)u 0 for each u 0 ≥ 0 in X. Furthermore, in this case, if 0 ≤ u 0 (x) ≤ NK for all x ∈ R, then we also have that 0 ≤ [S(t)u 0 ](x) = [S N (t)u 0 ](x) ≤ NK for all x ∈ R and all t ≥ 0. Proof Consider the abstract initial value problem ˙u(t) = f N (u(t)), u(0) = u 0 ≥ 0 ∈ X which has, by the Lipschitz continuity of f N , a unique global strong solution u(t) = S N (t)u 0 (see the discussion after (19)), where S N (·) is the nonlinear semigroup generated by f N . Hence, since the operator norm in this space is the supremum norm, it follows that the function u x (t) := [S N (t)u 0 ](x) is for each fixed x ∈ R the unique solution of the ordinary differential equation d dt u x(t) = ˜f N (u x (t)), u x (0) = u 0 (x) ≥ 0 ∈ R. Then it follows easily, using the uniqueness of solutions, that S N (·) is positive, i.e., if u 0 (x) ≥ 0 for all x ∈ R then u x (t) = [u(t)](x) = [S N (t)u 0 ](x) ≥ 0 for all x ∈ R, and also if u 0 (x) ≥ v 0 (x) for all x ∈ R then u x (t) = [u(t)](x) = [S N (t)u 0 ](x) ≥ v x (t) = [v(t)](x) = [S N (t)v 0 ](x) for all x ∈ R. Since ˜f N (y) < 0 for all y > K it also follows from uniqueness of solutions that, if u 0 (x) ≤ NK for all x ∈ R, then [S N (t)u 0 ](x) ≤ NK for all x ∈ R.
Page 1 and 2: Noname manuscript No. (will be inse
Page 3 and 4: Fractional reaction-diffusion equat
Page 9: Fractional reaction-diffusion equat
Page 21: Fractional reaction-diffusion equat

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

Create successful ePaper yourself

Delete template?

Save as template?