Gradient estimates and global existence of smooth solutions to a system of reaction-diffusion equations with cross-diffusion Tuoc V. Phan University of Tennessee - Knoxville, TN Workshop in nonlinear PDES Brussels, Sept 10, 2015 Joint work with Luan T. Hoang (Texas Tech U.) and Truyen V. Nguyen (U. of Akron)
SKT cross-diffusion system Let Ω R n be open, smooth, bounded and n 2. Consider the Shigesada-Kawasaki-Teramoto system of equations u t = [(d 1 + a 11 u+a 12 v)u]+u(a 1 b 1 u c 1 v),ω (0, ), v t = [(d 2 + a 21 u+a 22 v)v]+v(a 2 b 2 u c 2 v),ω (0, ), with homogenous Newman boundary conditions and u(, 0) = u 0 ( ) 0, v(, 0) = v 0 ( ) 0 in Ω. u and v denote the population densities of two species. d k, a k, b k, c k > 0 and a ij 0 are constants; a 11, a 22 are self-diffusion coefficients and a 12, a 21 are cross-diffusion coefficients.
Divergence form The PDE SKT system can be written as { ut = [(d 1 + 2a 11 u+a 12 v) u + a 12 u v]+u(a 1 b 1 u c 1 v), v t = [(d 2 + a 21 u+2a 22 v) v + a 21 v u]+v(a 2 b 2 u c 2 v). In abstract form: where U = U t = [J(U) U]+F(U), ( ) ( ) u d1 + 2a, J(U) = 11 u+a 12 v a 12 v, v a 21 u d 2 + a 21 u+2a 22 v and F(U) = ( u(a1 b 1 u c 1 v) v(a 2 b 2 u c 2 v) ).
Local well-posedness: H. Amann Theorem Theorem (H. Amann, 1990) Let r> n and u 0, v 0 W 1,r (Ω) and non-negative. Then, there exists a maximal existence time T max > 0 such that the SKT system has unique, local non-negative solution u, v with u, v C([0, T max ); W 1,r (Ω)) C (Ω (0, T max )). Moreover, if T max < then [ u(, t) W 1,r(Ω) + ] v(, t) W 1,r(Ω) =. lim t T max
Global or finite time blow-up solution? The solution for the full STK system u t = [(d 1 + a 11 u+a 12 v)u]+u(a 1 b 1 u c 1 v), v t = [(d 2 + a 21 u+a 22 v)v]+v(a 2 b 2 u c 2 v), exists globally in time or has finite time blow up? Very few results in very special cases. We study the system when a 21 = 0, that is { ut = [(d 1 + a 11 u+a 12 v)u] +u(a 1 b 1 u c 1 v), v t = [(d 2 + a 22 v)v] +v(a 2 b 2 u c 2 v),
Known results when a 21 = 0 For a 11 > 0, the SKT System { ut = [(d 1 + a 11 u+a 12 v)u] +u(a 1 b 1 u c 1 v), v t = [(d 2 + a 22 v)v] +v(a 2 b 2 u c 2 v), with BC and initial data has global solution when n 9. Y. Lou, W.-M. Ni and J. Wu (1998) for n = 2:. D. Le, L. Nguyen, T. Nguyen (2004); Y. Choi, R. Lui, Y. Yamada (2003-2004:) for n 5: T. P. (2008) for n 9. Many other results: Restrictive conditions on the coefficients and initial data.
Today main result Theorem (L. Hoang, T. Nguyen and T. P.) Assume that a 11 > 0 and a 21 = 0. Then for any n 2, and non-negative u 0, v 0 W 1,r (Ω) with r> n. Then, the solution (u, v) of the SKT system { ut = [(d 1 + a 11 u+a 12 v)u] + u(a 1 b 1 u c 1 v), v t = [(d 2 + a 22 v)v] + v(a 2 b 2 u c 2 v), (with B.C. and initial condition) exists uniquely, globally in time solution and [ ] [ ] u, v C([0, ); W 1,r (Ω)) C (Ω (0, )).
Ideas of the proof Amann s theorem: Let T = t max > 0 the maximal time existence of the local smooth solution. Assume T<. We need to prove (for r> n): lim t T [ u(, t) W 1,r (Ω) + v(, t) W 1,r (Ω) ] <. Need to establish appropriate a priori estimates of the local solutions.
Maximum Principle The PDE SKT system: { ut = [(d 1 + a 11 u+a 12 v)u] + u(a 1 b 1 u c 1 v), v t = [(d 2 + a 22 v)v] + v(a 2 b 2 u c 2 v). The second equation can be written as v t = [d 2 + 2a 22 v] v + 2a 22 v v + v[a 2 b 2 u c 2 v] Maximum Principle implies (Lou-Ni-Wu, 98) { 0 v max sup v 0 (x), a } 2. Ω c 2 The first equation u t = [d 1 + 2a 11 u+a 12 v] u+[2a 11 u+2a 12 v] u + u[a 1 b 1 u c 1 v + a 12 v] M.P. only gives us u 0. Not upper bound of u.
Some simple a priori estimates Lemma (Lou-Ni-Wu, 98) For fixed T> 0, let Ω T = Ω [0, T]. Then, there exists C(T)>0 such that ˆ ˆ u(x, t)dx, u 2 (x, t)dxdt C(T). Ω Ω T Proof. Note that the PDE of u is u t = [(d 1 +a 12 v+2a 11 u) u+a 12 u v]+u(a 1 b 1 u c 1 v), Ω T. Also, the boundary condition give u ν = v = 0, on ν Ω (0, T). Then, the lemma follows by integrating the equation of u.
Some simple estimates (cont.) Lemma (Lou-Ni-Wu, 98) We also have v t L2 (Ω T ) + v L2 (Ω T ) + v L2 (Ω T ) C(T). Proof. Test the equation of v v t = [(d 2 + a 22 v)v]+v(a 2 b 2 u c 2 v) with [(d 2 + a 22 v)v].
Energy estimates? The PDE of u is u t = [(d 1 + 2a 11 u+a 12 v) u+ a 12 u v]+u(a 1 b 1 u c 1 v). If we multiply the eqn with u and integrate ˆ ˆ 1 d u 2 + (d 1 + 2a 11 u+a 12 v) u 2 2 dt Ω Ω = a 12ˆ u u v + other terms. Ω Note that we know u L 2 and v = (d 2 + 2a 22 v) 1 h L 2. It is therefore challenging to control Ω u u v. We need to understand the regularity of v.
Key iteration lemma The PDE of u: u t = [(d 1 + 2a 11 u+a 12 v) u+ a 12 u v]+u(a 1 b 1 u c 1 v). Lemma Let p> 2 and assume that a 11 > 0. Then for each q [ p(n + 1) ] p, (n + 2 p) + with q, we have u L q (Ω T ) C(p, q, T)[1+ v L p (Ω T )]. Main task: If u L q (Ω T ), can we derive the estimate v L q (Ω T ) C(q, p, T)[1 + u L q (Ω T )]?
Known theory on gradient estimates Consider the linear PDE w t = [a(x, t) w]+ F in Ω T := Ω (0, T) Calderón - Zygmund theory: We have v L p (Ω T ) C[1+ F L p (Ω T )] if (besides other conditions on data and Ω) (i) a is uniformly elliptic, saying: 1 a(x, t) 2, and (ii) The oscillation of a(x, t) is small: Classical results require that a(x, t) is in C(Ω T ); Recent developments require (2000s) a BMO 1. (L. Caffarreli; N. Krylov; G. Lieberman; S. Byun - L. Wang,...)
Regularity problem The PDE of v: v t = [{ d2 + 2a 22 v(x, t) } ] v +v(a 2 c 2 v) b 2 uv, in Ω (0, T). We want to establish [ v L p (Ω (0,T)) C Difficulties: 1+ u L p (Ω (0,T)) (i) Main term a(x, t) := d 2 + 2a 22 v(x, t) depends on solution. Therefore, its oscillation is not known to be small. (ii) The equation is not invariant under either of the scalings v(x, t) v(x, t) λ ]. or v(x, t) v(θx,θ2 t), λ,θ>0. θ (iii) The equation is not invariant under the change of coordinates.
Equations with scaling parameters Denote Ω T = Ω (0, T), we study the equation: w t = [(1+λαw)A w] +θ 2 w(1 λw) λθcw in Ω T, w = 0 on Ω (0, T), ν w(, 0) = w 0 ( ) in Ω. Here,θ,λ>0andα 0are constants, c(x, t) is a nonnegative measurable function, A = (a ij ) : Ω T M n n is symmetric, measurable and Λ>0 such that Λ 1 ξ 2 ξ T A(x, t)ξ Λ ξ 2 for a.e. (x, t) Ω T and for allξ R n.
Invariant scalings Let w be a solution of the PDE w t = [(1+λαw)A w] +θ 2 w(1 λw) λθcw. Then, for M>0, we see that w 1 = w/m solves t w 1 = [(1+λ 1 αw 1 )A w 1 ]+θ 2 w 1 (1 λ 1 w 1 ) λ 1 θc 1 w 1. with λ 1 =λm, c 1 (x, t) = c(x, t)/m. Also, for r> 0, w 2 (x, t) = w(rx, r 2 t)/r solves t w 2 = [(1+λ 2 αw 2 )A 2 w 2 ]+θ 2 2 w 2 (1 λ 2 w 2 ) λ 2 θ 2 c 2 w 2. with λ 2 =λr, θ 2 =θr, c 2 (x, t) = c(rx, r 2 t), A 2 (x, t) = A(rx, r 2 t).
Calderón - Zygmund type estimates Theorem (L. Hoang, T. Nguyen and T. P.) Let p> 2. Then there exists a numberδ =δ(p,λ, n,α)>0such that if Ω is a Lipschitz domain with the Lipschitz constant δand [A] BMO(ΩT ) δ, then for any weak solution w of w t = [(1+λαw)A w] +θ 2 w(1 λw) λθcw in Ω T, w = 0 on Ω (0, T), ν w(, 0) = w 0 ( ) in Ω. satisfying 0 w λ 1 in Ω T, we have ˆ {( θ ) p } w p dxdt C λ w L 2 (Ω T ) +ˆ c p dxdt Ω T Ω [ t,t] for every t (0, T). Here C> 0 is a constant depending only on Ω, t, p, Λ,α and n and independent ofθ,λ.
Main ideas (interior estimates) Perturbation technique(caffarelli Peral): Comparing the solution of w t = [(1+λαw)A w]+θ 2 w(1 λw) λθcw in Q 6 (1) with that of the reference equation h t = [(1+λαh)Ā B4 (t) h]+θ 2 h(1 λh) in Q 4, (2) where Ā B4 (t) is the average of A(, t) over B 4, that is, Ā B4 (t) := 1 ˆ A(x, t)dx. B 4 B 4 Key idea: Use the regularity of h to obtain the regularity of w when w is sufficiently close to h. Here, Q r denotes the parabolic cylinder of radius r.
Regularity of solutions for the reference equation Lemma ( De Giorgi - Nash - Moser) There exists a constant C = C(n,Λ,α) such that everyλ>0and every weak solution 0 h λ 1 solution of h t = [(1+λαh)Ā B4 (t) h]+θ 2 h(1 λh) in Q 4, we have h 2 L (Q 3 ) C(n,Λ,α) 1 ˆ h 2 dxdt. Q 4 Q 4 Note: The constant C is independent onλ.
Approximation lemma Lemma For anyε>0, there existsδ=δ(ε, n,λ,α)>0such that for all θ,λ>0, and 0<r 1, if ˆ ] 1 [ A(x, t) Ā B4r (t) 2 + c(x, t) 2 dxdt δ, Q 4r Q 4r then for any weak solution w of (1) in Q 5r satisfying ˆ 0 w λ 1 1 in Q 4r, and w 2 dxdt 1, Q 4r Q 4r and weak solution h of (2) in Q 4r satisfying h = w on p Q 4r, we have ˆ ˆ 1 w h 2 dxdt εr 2 1 and w h 2 dxdt ε. Q 4r Q 4r Q 4r Q 2r
Decay estimate of distribution of maximal function Proposition Assume c L 2 (Q 6 ). N = N(n,Λ,α)>1such that for anyε>0, we can findδ=δ(ε, n,λ)>0such that if ˆ sup sup A(x, t) ĀB 0<ρ 4 Q ρ (y, s) ρ (y)(t) 2 dxdt δ, Q ρ (y,s) (y,s) Q 1 1 then for any weak solution w of (1) satisfying 0 w λ 1 in Q 5, and {Q1 :M Q5 ( w 2 )>N} ε Q1, we have {Q 1 :M Q5 ( w 2 )>N} { {Q1 (10) n+2 ε :M Q5 ( w 2 )>1} + {Q1 :M Q5 (c2 )>δ} }.
Finishing the proof - Step I Assume for a moment that {Q 1 :M Q5 ( w 2 )>N} ε Q1. Then, by the Proposition (forǫ 1 = (10) n+2 ε) {Q 1 :M Q5 ( w 2 )>N} ǫ 1 { {Q1 :M Q5 ( w 2 )>1} + {Q1 :M Q5 (c 2 )>δ} }. Iteration: At the step k th, we use the scaling w w k := w/ N k, k = 1, 2,, we eventually obtain {Q 1 :M Q5 ( u 2 )>N k } ǫ k 1 {Q 1 :M Q5 ( u 2 )>1} k + i=1 This gives the estimate of w L p (Q 1 ). ǫ i 1 {Q 1 :M Q5 (c 2 )>δn k i }
Finishing the proof - Step II Step 2: To remove the assumption in Step 1, we just use the weak type 1 1 estimate: {Q 1 :M Q5 ( w 2 )>NM 2 } ˆ C w 2 dxdt. NM 2 Q 5 By the appropriate choice of M, we can obtain with w = w/m {Q 1 :M Q5 ( w 2 )>N} ε Q1. Since w is a weak solution, we can apply the Step I for w, to derive w L p (Q 1 ). Then we can derive the estimate of w L p (Q 1 ).
Thank you for your attention