Smoothing effect and Fredholm property for first-order hyperbolic - PowerPoint PPT Presentation

Smoothing effect and Fredholm property for first-order hyperbolic PDEs Irina Kmit Humboldt University of Berlin, Germany Padova, 28 June 2012

The problem under consideration n X ∂ t u j + a j ( x, t ) ∂ x u j + b jk ( x, t ) u = f j ( x, t ) , j ≤ n, ( x, t ) ∈ (0 , 1) × R k =1 u j ( x, 0) = ϕ j ( x ) , j ≤ n, x ∈ [0 , 1] u j (0 , t ) = ( Ru ) j ( t ) , 1 ≤ j ≤ m, t ∈ R + u j (1 , t ) = ( Ru ) j ( t ) , m < j ≤ n, t ∈ R + Motivation: traveling-wave models from • laser dynamics • population dynamics • chemical kinetics 1

Smoothing, fredholmness and hyperbolic PDEs • Motivation – bifurcation analysis of traveling-wave models – local investigations of semilinear and quasilinear hyperbolic problems via ∗ Implicit Function Theorem ∗ Lyapunov-Schmidt reduction – periodic synhronizations via averaging procedure ∗ • Complications and related topics – bad regularity properties of hyperbolic operator: ∗ no smoothing effect in general ∗ lost-of-smoothness effect – small denominators – non-strict hyperbolicity and constraints on lower-order terms 2

Examples 1. Smoothing effect. Heat equation: ∂u ∂t − ∆ u = 0 u ( x, 0) = ϕ ( x ) „ −| x − y | 2 1 « Z u ( x, t ) = R n exp ϕ ( y ) dy (4 πt ) n/ 2 4 t 3

Examples 1. Smoothing effect. Heat equation: ∂u ∂t − ∆ u = 0 u ( x, 0) = ϕ ( x ) „ −| x − y | 2 1 Z « u ( x, t ) = R n exp ϕ ( y ) dy (4 πt ) n/ 2 4 t 2. No smoothing effect. Wave equation: ∂ 2 u ∂t 2 − ∂ 2 u ∂x 2 = 0 Z x + t u = ϕ ( x − t ) + ϕ ( x + t ) + 1 ψ ( s ) ds 2 2 x − t 3

Smoothing property (definition) Definition. A solution u is called smoothing if for any k there exists T > 0 such that u is C k -smooth above the line t = T . Remark. Any t -periodic and smoothing solution immediately meets the C ∞ - regularity in the entire domain. Remark. The domain of influence of the initial conditions is in general infinite and is determined by the boundary operator as well as the lower-order terms . 4

Smoothing effect I: Classical boundary conditions n X ∂ t u j + a j ( x, t ) ∂ x u j + b jk ( x, t ) u = f j ( x, t ) , j ≤ n, ( x, t ) ∈ (0 , 1) × R k =1 u j ( x, 0) = ϕ j ( x ) , j ≤ n, x ∈ [0 , 1] u j (0 , t ) = h j ( t ) , 1 ≤ j ≤ m, t ∈ R + u j (1 , t ) = h j ( t ) , m < j ≤ n, t ∈ R + Assumptions: a j > 0 for all j ≤ m and a j < 0 for all j > m, x,t | a j | > 0 for all j ≤ n, inf for all 1 ≤ j � = k ≤ n there exists p jk ∈ C 1 ([0 , 1] × R ) such that b jk = p jk ( a k − a j ) and p jk = 0 in the interior of the domain { ( x, t ) : a j ( x, t ) = a k ( x, t ) } . 5

Smoothing effect I: main result Theorem. Assume that a j , b jk , f j , h j ∈ C ∞ and ϕ j ∈ C . Then any continuous solution is smoothing. Operator representation of the problem: u = CBu + Du + Ff, where ( Cu ) j ( x, t ) = c j ( x j , x, t ) u j ( x j , ω j ( x j ; x, t )) , Z ξ „ b jj « c j ( ξ, x, t ) = exp ( η, ω j ( η ; x, t )) dη, a j x ( if t > 0 , h j ( t ) ( Bu ) j ( x, t ) = ϕ j ( x ) if t = 0 , Z x j c j ( ξ, x, t ) “ ” X ( Du ) j ( x, t ) = b jk u k ( ξ, ω j ( ξ ; x, t )) dξ a j ( ξ, ω j ( ξ ; x, t )) x k � = j 6

Proof idea • 1-st step: u = CBu + Du + Ff. Hence u = ( CBu + DCBu ) + D 2 u + ( I + D ) Ff. If T 1 > 0 is sufficiently large, then u | t>T 1 = ( C + DC ) h + D 2 u + ( I + D ) Ff. The operator D 2 maps C (Π) n continuously into C 1 t (Π ∩ t > T 1 ) n . • 2-nd step: Set v = ∂ t u . Then n b jk v k − ∂ t a j X ( ∂ t + a j ∂ x ) v j + v j = G j ( f j , ∂ t f j , u ) a j k =1 Ch ′ + ˜ and we have the representation v | Π T 2 = ˜ Dv + ˜ FG ( f, ∂ t f, u ) . • Further we use induction on the order of regularity. 7

Smoothing effect II: Integral boundary conditions in population models ( ∂ t + ∂ x + µ ) u = 0 , ( x, t ) ∈ (0 , 1) × R , u ( x, 0) = ϕ ( x ) , x ∈ [0 , 1] , „Z 1 « u (0 , t ) = h γ ( x ) u ( x, t ) dx , t ∈ R , 0 ∈ C ∞ and ϕ ∈ C . Theorem. Assume that h, γ Then any continuous solution is smoothing. 8

Proof idea 1-st step: Any continuous solution for large enough T 1 is given by the formula u | t ≥ T 1 = CBu = CBCu „Z 1 « e − µx h γ ( ξ ) e − µξ u (0 , t − x − ξ ) dξ = 0 „Z t − x « e − µx h γ ( t − x − τ ) e µ ( x − t + τ ) u (0 , τ ) dτ = , t − x − 1 2-nd step: Set v = u t . Then for large enough T 2 we have „Z 1 « Z 1 e − µx h ′ γ ( ξ ) e − µξ v (0 , t − x − ξ ) dξ v | t ≥ T 2 = γ ( ξ ) u ( ξ, t − x ) dξ 0 0 „Z 1 « Z t − x e − µx h ′ γ ( t − x − τ ) e µ ( x − t + τ ) v (0 , τ ) dτ. = γ ( ξ ) u ( ξ, t − x ) dξ 0 t − x − 1 9

Smoothing effect III: Dissipative boundary conditions and periodic problems ∂ t u + a ( x, t ) ∂ x u + b ( x, t ) u = f ( x, t ) , ( x, t ) ∈ (0 , 1) × R u ( x, t + 2 π ) = u ( x, t ) , x ∈ [0 , 1] u j (0 , t ) = h j ( z ( t )) , 1 ≤ j ≤ m, t ∈ R + u j (1 , t ) = h j ( z ( t )) , m < j ≤ n, t ∈ R + with z ( t ) = ( u 1 (1 , t ) , . . . , u m (1 , t ) , u m +1 (0 , t ) , . . . , u n (0 , t )) . Theorem. Assume that a j , b jk , f j , h j ∈ C ∞ and a j , b jk , f j are 2 π - periodic in t . If (Z x j ! ) b jj − l∂ t a j ˛ exp ˛ ∂ k h ′ ˛ ˛ j ( z ) ( η, ω j ( η ; x, t )) dη < 1 a 2 a j x j for all j, k ≤ n , x ∈ [0 , 1] , t ∈ R , z ∈ R n , and l = 0 , 1 , . . . , r , then any continuous solution belongs to C r . 10

Proof idea • u = Cu + Du + Ff, u = CBu + Du + Ff ⇒ u = CBu + ( DC + D 2 ) u + ( I + D ) Ff t ) n is bijective • The operator I − CB ∈ L ( C 1 ⇒ u = ( I − CB ) − 1 h i ( DC + D 2 ) u + ( I + D ) Ff • The operator DC + D 2 is smoothing (maps C into C 1 t ) 11

Fredholmness: a problem from laser dynamics n X ∂ t u j + a j ( x ) ∂ x u j + b jk ( x ) u k = f j ( x, t ) , j ≤ n, ( x, t ) ∈ (0 , 1) × R k =1 u j ( x, t + 2 π ) = u j ( x, t ) , j ≤ n, ( x, t ) ∈ [0 , 1] × R n X r 0 u j (0 , t ) = jk u k (0 , t ) , j ≤ m, t ∈ R k = m +1 m X r 1 u j (1 , t ) = jk u k (1 , t ) , m + 1 ≤ j ≤ n, t ∈ R . k =1 12

Choice of function spaces ( u, f ) ∈ ( V γ , W γ ) Spaces for the right-hand sides: W γ = L 2 ` ´ n 0 , 1; H γ per ( R ) Spaces for the solutions: n j =1 ∈ W γ o u ∈ W γ : ∂ x u ∈ W γ − 1 , [ ∂ t u j + a j ∂ x u j ] n U γ ( a ) = 2 ‚ ‚ ‚ [ ∂ t u j + a j ∂ x u j ] n � u � 2 U γ ( a ) = � u � 2 W γ + W γ . ‚ ‚ j =1 ‚ n o V γ ( a ) = u ∈ U γ ( a ) : u fulfills refl. bound. conditions 13

Properties of the solution spaces Lemma. (i) The space U γ ( ω, a ) is complete. (ii) If γ ≥ 1 , then for any x ∈ [0 , 1] there exists a continuous trace map u ∈ U γ ( ω, a ) �→ u ( x, · ) ∈ L 2 ((0 , 2 π ); R n ) . (iii) If γ > 3 / 2 , then U γ ( ω, a ) is continuously embedded into C ([0 , 1] × [0 , 2 π ]; R n ) . 14

Operator representation of the problem A ( a, b 0 ) u + B ( b 1 ) u = f, where A ( a, b 0 ) ∈ L ( V γ ( a ); W γ ) is defined by A ( a, b 0 ) u = [ ∂ t u + a j ∂ x u j + b jj u j ] n j =1 , and B ( b 1 ) ∈ L ( W γ ) is defined by n 2 3 4X B ( b 1 ) u = b jk u k 5 j � = k j =1 15

Fredholm property Theorem. (K., Recke, 2012) Assume that a j , b jj ∈ L ∞ (0 , 1) , 0 ≤ x ≤ 1 | a j ( x ) | > 0 for all j ≤ n, inf ff for all j � = k there is c jk ∈ BV (0 , 1) such that b jk ( x ) = c jk ( x )( a j ( x ) − a k ( x )) for a.a. x ∈ [0 , 1] , n m X X R ( a, b 0 ) | r 1 jk r 0 kl | < 1 . j,l = m +1 k =1 Then (i) The operator A + B is a Fredholm operator with index zero from V γ ( a ) into W γ for all γ ≥ 1 . (ii) (smoothing effect) The subspaces ker( A + B ) and ker( ˜ A + ˜ B ) do not depend on γ , ker( A + B ) ∗ = ker( ˜ A + ˜ B ) , and n f ∈ W γ : � f, u � L 2 = 0 for all u ∈ ker “ ”o A + ˜ ˜ im ( A + B ) = B . 16

Fredholmness and non-strict hyperbolicity Fredholmness for non-strict hyperbolicity necessarily entails constraints on zero-order terms: ∂ t u 1 + ∂ x u 1 = ∂ t u 2 + ∂ x u 2 + bu 1 = 0 , u 1 ( x, t + 2 π ) − u 1 ( x, t ) = u 2 ( x, t + 2 π ) − u 2 ( x, t ) = 0 , u 1 (0 , t ) − r 0 u 2 (0 , t ) = u 2 (1 , t ) − r 1 u 1 (1 , t ) = 0 . No small denominators: r 0 r 1 < 1 . Assume that b = r 0 r 1 − 1 , r 0 covering arbitrarily large | b | . Then „ 1 « u 1 ( x, t ) = sin l ( t − x ) , u 2 ( x, t ) = b − x sin l ( t − x ) , l ∈ N , 1 − r 0 r 1 gives infinitely many linearly independent solutions. Hence, the Fredholmness is destroyed. 17

Fredholmness: proof idea Theorem (Fredholmness criterion). Let W be a Banach space, I the identity in W , and C ∈ L ( W ) with C k being compact for some k ∈ N . Then I + C is a Fredholm operator of index zero. • A ∈ L ( V γ ; W γ ) is an isomorphism, hence A + B ∈ L ( V γ ; W γ ) is Fredholm iff I + BA − 1 ∈ L ( W γ ) is Fredholm. • ( BA − 1 ) 2 ∈ L ( W γ ) is compact. • (right) parametrix or (right) regularizer: A − 1 ( I − BA − 1 ) : BA − 1 ´ 2 . A − 1 ` I − BA − 1 ´˜ ˆ ` ( A + B ) = I − 18

Thank you! 19