Finite-time Blowup of Semilinear PDEs via the Feynman-Kac - PowerPoint PPT Presentation

Finite-time Blowup of Semilinear PDEs via the Feynman-Kac Representation E A LFREDO L ´ J OS ´ OPEZ -M IMBELA C ENTRO DE I NVESTIGACI ´ ON EN M ATEM ´ ATICAS G UANAJUATO , M EXICO jalfredo@cimat.mx

Introduction and backgrownd Consider a reaction-diffusion equation of the form ∂w ∂t = Lw + w 1+ β , w (0 , x ) = ϕ ( x ) , x ∈ E, (1) where L is the generator of a strong Markov process in a locally compact space E , β > 0 is constant, ϕ ≥ 0 is bounded and measurable. Well-known facts: • ∀ ϕ ≥ 0 ∃ T ϕ ∈ (0 , ∞ ] such that (1) has a unique solution w on R d × [0 , T ϕ ) . • w is bounded on R d × [0 , T ] for any 0 < T < T ϕ . • If T ϕ < ∞ , then � w ( · , t ) � L ∞ ( R d ) → ∞ as t ↑ T ϕ . When T ϕ = ∞ we say that w is a global solution When T ϕ < ∞ we say that w blows up in finite time or that w is non-global. 1

The study of blow up properties of (1) goes back to the fundamental work of Fujita (1966): For E = R d and L = ∆ • If d < 2 /β , then any non-trivial positive solution blows up in finite time. • If d > 2 /β , then Equation (1) admits a global solution for all sufficiently small initial values ϕ . Later on, Hayakawa (1973) and Aronson and Weinberger (1978) proved that the critical dimension d = 2 β also pertains to the finite-time blowup regime. The case of L = ∆ α := − ( − ∆) α/ 2 in E = R d was settled by Sugitani (1975), who showed that if d ≤ α/β, then for any non-vanishing initial condition the solution blows up in finite time. 2

An example with an integral power non-linearity ∂w ∂t = Lw + w 2 , w (0) = ϕ Blowup or stability of solutions? Recall ∂f ∂t = f 2 , f = K 1 has solution f ( t ) = 1 K − t Thus: Blowup in the absence of motion For a (small) initial ϕ with bounded support, the motion tends to smear out u , hence counteract the blowup. Where is the border? 3

Probabilistic representation In case of integer exponents β ≥ 1 it was proved by McKean (1975) that    �  , w ( t, x ) = E ϕ ( y ) x ∈ E, t ≥ 0 . y ∈ B x ( t ) solves the equation ∂w ∂t = Lw − w + w 1+ β , w (0) = ϕ Here ( B x ( t )) t ≥ 0 is a branching particle system in E • starting from an ancestor at x ∈ E , • with exponential (mean–one) individual lifetimes, • branching numbers 1 + β • particle motions with generator L . How to remove the “Feynman-Kac” term − w ? 4

FACTS: • w is nonglobal provided that � 1 � 1 /β T ( t ) ϕ ( x ) ≥ for some x ∈ E and t ≥ 0 , tβ where { T ( t ) } t ≥ 0 is the semigroup with generator L . • The condition � ∞ y ∈ E [ T ( s ) ϕ ( y )] β ds < 1 sup β 0 ensures sup w ( t, x ) < ∞ for all t ≥ 0 , x ∈ E i.e. w is a global solution. (e.g. Nagasawa and Sirao (1969) and Weissler (1981)) 5

Constructing Subsolutions by the Feynman-Kac Formula We now consider the equation ∂w ( t ) = ∆ α w ( t ) + w ( t ) 1+ β , w ( x, 0) = ϕ ( x ) , x ∈ R d . (2) ∂t Recall that the solution u of the IVP ∂u = ∆ α u ( t ) + u ( t ) v ( t ) , (3) ∂t ( t, x ) ∈ [0 , T ) × R d u (0 , x ) = ϕ ( x ) , with v : [0 , T ) × R d → R + locally bounded, has by the Feynman-Kac formula a proba- bilistic representation as the density of the measure � � � � t E x 1 ( W t ∈ dy ) exp v ( s )( W s ) ϕ ( x ) dx = u ( t, y ) dy (4) 0 where E x denotes expectation with respect to the symmetric α -stable process ( W t ) started at W 0 = x . The representation (4) shows in particular that any solution � u of (3) with v replaced by � v ≤ v and � u 0 ≤ u 0 fulfills � u ≤ u . 6

Consider the initial value problems ∂f t = ∆ α f t , f 0 = ϕ ( v t ≡ 0) ∂t ∂g t = ∆ α g t + f t g t , g 0 = ϕ ( v t ≡ f t ) ∂t ∂h t = ∆ α h t + g t h t , h 0 = ϕ ( v t ≡ g t ) . ∂t Then, by the previous remark f t ≤ g t ≤ h t ≤ w t , i.e., f t , g t and h t are subsolutions of ∂w t ∂t = ∆ α w t + w 1+ β , w 0 = ϕ. 7

Basic Estimates We denote by B r the ball in R d with radius r centered at the origin, and write p t ( x ) for the transition density of ( W t ) . Let � f t ( y ) := p t ( y − x ) ϕ ( x ) dx = E y [ ϕ ( W t )] . Lemma 1. For all t ≥ 1 we have the inequality � f t ( y ) ≥ c 0 t − d/α 1 B 1 ( t − 1 /α y ) ϕ ( x ) dx B 1 for some c 0 > 0 . Indeed, let y ∈ B t 1 /α . Then, by the scaling property of W t f t ( y ) = E y [ ϕ ( W t )] = E 0 [ ϕ ( W t + y )] � � �� t 1 /α ( W 1 + t − 1 /α y ) = ϕ E 0 � p 1 ( x − t − 1 /α y ) ϕ ( t 1 /α x ) dx ≥ B 1 � ϕ ( t 1 /α x ) dx ≥ c 0 B 1 � c 0 t − d/α = ϕ ( x ) dx B t 1 /α � c 0 t − d/α 1 B 1 ( t − 1 /α y ) ≥ ϕ ( x ) dx B 1 8

This argument also shows that, for sufficiently large t , f t ( y ) ≥ c ′ 0 t − d/α 1 B 1 ( t − 1 /α y ) . for some c ′ 0 > 0 . In the same way one can prove the following Lemma 2. There exists a c > 0 such that for all t ≥ 2 , y ∈ B t 1 /α , x ∈ B 1 and s ∈ [1 , t/ 2] , P x { W s ∈ B s 1 /α | W t = y } ≥ c. Proof. Using self-similarity, continuity and strict positivity of stable densities, we have that for all s ∈ [1 , t/ 2] , � p s ( z − x ) p t − s ( y − z ) dz p t ( y − x ) B s 1 /α � s − d/α p 1 ( s − 1 /α ( z − x ))( t − s ) − d/α p 1 (( t − s ) − 1 /α ( y − z )) = dz t − d/α p 1 ( t − 1 /α ( y − x )) B s 1 /α � · (inf w ∈ B 2 p 1 ( w )) 2 s − d/α ( t − s ) − d/α ≥ dz t − d/α p 1 (0) B s 1 /α s − d/α Vol( B s 1 /α )(inf w ∈ B 2 p 1 ( w )) 2 ≥ . 2 p 1 (0) 9

Let g t solve ∂g t ∂t = ∆ α g t + g t f β t , g 0 = ϕ, (5) where f t is defined in Lemma 1. Proposition 1 Let d < α/β . Then g t grows to ∞ uniformly on the unit ball as t → ∞ , i.e., t →∞ inf lim x ∈ B 1 g t ( x ) = ∞ . Proof. From the Feynman-Kac representation we know that g t is given by � t � � � � f s ( W s ) β ds � g t ( y ) = ϕ ( x ) p t ( y − x ) E x exp � W t = y dx. � 0 Using Lemma 1 and Jensen’s inequality, it follows that for y ∈ B t 1 /α , g t ( y ) � t/ 2 � � � � � c 1 s − βd/α 1 B s 1 /α ( W s ) ds ≥ ϕ ( x ) p t ( y − x ) E x exp � W t = y dx � 1 � t/ 2 � � � s − βd/α P x { W s ∈ B s 1 /α | W t = y } ds ≥ ϕ ( x ) p t ( y − x ) exp c 2 dx 1 � t/ 2 � � c 3 t − d/α exp s − βd/α ds ≥ c 4 , (6) 1 where we have used Lemma 2 to obtain the last inequality, and where c i , i = 1 , 2 , 3 , 4 , are positive constants. The result follows from the condition d < α/β . 10

Completion of the proof of blowup Since w t ≥ g t , from Proposition 1 it follows that K ( t ) := inf w t ( x ) → ∞ as t → ∞ . (7) x ∈ B 1 We re-start (2) with the initial condition w t 0 , with a suitable choice of t 0 given below. Writing u t = w t 0 + t , the equation becomes ∂u t ∂t = ∆ α u t + u 1+ β x ∈ R d . , u 0 ( x ) = w t 0 ( x ) , t Its integral form is � t � � p t − s ( y − x ) u s ( y ) 1+ β dy. u t ( x ) = p t ( y − x ) u 0 ( y ) dy + ds 0 Noting that ζ := min 0 ≤ s ≤ 1 P x { W s ∈ B 1 } > 0 , min we deduce that, for every t ∈ [0 , 1] , x ∈ B 1 � t � 1+ β � min u t ( x ) ≥ ζK ( t 0 ) + ζ min u s ( y ) ds. x ∈ B 1 y ∈ B 1 0 Moreover, min u t ( x ) ≥ v ( t ) , where x ∈ B 1 � t v ( s ) 1+ β ds, v ( t ) = ζK ( t 0 ) + ζ 0 and v ( t ) blows up at time 1 τ = βζ 1+ β K ( t 0 ) β . Due to (7), we can choose t 0 so big that τ < 1 . This yields min u 1 ( x ) = ∞ , which shows blowup of w t 0 +1 . x ∈ B 1 11

Remark By a second application of the Feynman-Kac formula, one can easily prove that the subsolution h t ∂h t ∂t = ∆ α h t + g β t h t , h 0 = ϕ, (where g t is the subsolution obtained above), is such that x ∈ B 1 h t ( x ) → ∞ inf t → ∞ as even if d = α/β . Hence, our equation has no positive global solutions if d ≤ α/β . 12

Other Generators Consider the one-dimensional semilinear equation ∂w t ∂t = Γ w t + νt σ w 1+ β , w 0 ( x ) = ϕ ( x ) , x ∈ R + , (8) t where ν, σ, β ≥ 0 , ϕ ≥ 0 and ∞ [ f ( x + y ) − f ( x )] e − y � Γ f ( x ) = y dy, 0 i.e. Γ is the generator of the standard Gamma process . • The Γ -process enjoys no self-similarity or symmetry, nor dimensional-dependent behavior. • The transition densities of the Γ -motion process are explicitly given, • The bridges of the Γ -process are beta distributed. 13

In this case, we need to impose conditions on the decay of the initial value ϕ ( x ) as x → ∞ . By adapting the Feynman-Kac approach one can prove [LM, N. Privault]: • Every bounded, measurable initial condition ϕ ≥ 0 satisfying c 1 x − a 1 ≤ ϕ ( x ) , x > x 0 for some positive constants x 0 , c 1 , a 1 , where a 1 β < 1 + σ , produces a non-global solution. • On the other side, if ϕ fulfils ϕ ( x ) ≤ c 2 x − a 2 , x > x 0 , where x 0 , c 2 , a 2 are positive numbers and a 2 β > 1 + σ , then the solution w t to (8) is global and, moreover, 0 ≤ w t ( x ) ≤ Ct − a 2 , x ≥ 0 , for some constant C > 0 . 14