BV solutions of the Jin-Xin model Stefano Bianchini, IAC(CNR) Roma - PowerPoint PPT Presentation

The kernels Γ 12 , Γ 22 show that only v x influences u :    e − ( x 2 − λt ) 2 / (4(1 − λ 2 )(1+ t )) Γ 12 ( t, x ) = ∂   + exp.dec, h.o. terms  , � ∂x (1 − λ 2 )(1 + t ) 2   e − ( x 2 − λt ) 2 / (4(1 − λ 2 )(1+ t )) Γ 22 ( t, x ) = ∂  ∂   + ”,”  + exp.dec . � ∂x ∂x (1 − λ 2 )(1 + t ) 2 This result is important when studying decay to an equilibrium state (¯ u, ¯ v ) = (0 , F (0) = 0), because by Duhamel formula 8

The kernels Γ 12 , Γ 22 show that only v x influences u :    e − ( x 2 − λt ) 2 / (4(1 − λ 2 )(1+ t )) Γ 12 ( t, x ) = ∂   + exp.dec, h.o. terms  , � ∂x (1 − λ 2 )(1 + t ) 2   e − ( x 2 − λt ) 2 / (4(1 − λ 2 )(1+ t )) Γ 22 ( t, x ) = ∂  ∂   + ”,”  + exp.dec . � ∂x ∂x (1 − λ 2 )(1 + t ) 2 This result is important when studying decay to an equilibrium state (¯ u, ¯ v ) = (0 , F (0) = 0), because by Duhamel formula � t � � � � � � u ( t ) u (0) 0 = Γ( t ) ∗ + 0 Γ( t − s ) ∗ ds v ( t ) v (0) F ( u ( s )) − A (0) u ( s ) � �� � ≈ G x ( t − s ) ∗ u ( s ) 2 8

• The dependence w.r.t. u 0 + �u 0 ,t can be easily seen with the example u t = � ( u xx − u tt ) . with initial data u (0) = 0, u t (0) = � − 1 . 9

• The dependence w.r.t. u 0 + �u 0 ,t can be easily seen with the example u t = � ( u xx − u tt ) . with initial data u (0) = 0, u t (0) = � − 1 . The solution is 1 − e − t/� , which converges to u ( t ) ≡ 1, t > 0. 9

• The dependence w.r.t. u 0 + �u 0 ,t can be easily seen with the example u t = � ( u xx − u tt ) . with initial data u (0) = 0, u t (0) = � − 1 . The solution is 1 − e − t/� , which converges to u ( t ) ≡ 1, t > 0. The hyperbolic limit � → 0 has the ”initial data” t → 0+ u ( t ) = 1 = lim lim � → 0 u 0 + �u t, 0 . 9

BV estimates in the conservative case 10

BV estimates in the conservative case We assume A ( u ) = D F ( u ), � = 1 and u 0 ,t ∈ L 1 . 10

BV estimates in the conservative case We assume A ( u ) = D F ( u ), � = 1 and u 0 ,t ∈ L 1 . Differentiating w.r.t. x the BGK scheme (2) 10

BV estimates in the conservative case We assume A ( u ) = D F ( u ), � = 1 and u 0 ,t ∈ L 1 . Differentiating w.r.t. x the BGK scheme (2)  f − + I − A ( u ) − I + A ( u ) f − t − f − f + =  f ± = F ± x 2 2 (8) x . I + A ( u ) f − − I − A ( u ) f + + f + f + =  x t 2 2 10

BV estimates in the conservative case We assume A ( u ) = D F ( u ), � = 1 and u 0 ,t ∈ L 1 . Differentiating w.r.t. x the BGK scheme (2)  f − + I − A ( u ) − I + A ( u ) f − t − f − f + =  f ± = F ± x 2 2 (8) x . I + A ( u ) f − − I − A ( u ) f + + f + f + =  x t 2 2 Differentiating (2) w.r.t. t we obtain 10

BV estimates in the conservative case We assume A ( u ) = D F ( u ), � = 1 and u 0 ,t ∈ L 1 . Differentiating w.r.t. x the BGK scheme (2)  f − + I − A ( u ) − I + A ( u ) f − t − f − f + =  f ± = F ± x 2 2 (8) x . I + A ( u ) f − − I − A ( u ) f + + f + f + =  x t 2 2 Differentiating (2) w.r.t. t we obtain  − I + A ( u ) g − + I − A ( u ) g − t − g − g + =  g ± = F ± x 2 2 t . (9) I + A ( u ) g − − I − A ( u ) g + t + g + g + =  x 2 2 10

BV estimates in the conservative case We assume A ( u ) = D F ( u ), � = 1 and u 0 ,t ∈ L 1 . Differentiating w.r.t. x the BGK scheme (2)  − I + A ( u ) f − + I − A ( u ) f − t − f − f + =  f ± = F ± x 2 2 x . (8) f + + f + I + A ( u ) f − − I − A ( u ) f + =  x t 2 2 Differentiating (2) w.r.t. t we obtain  − I + A ( u ) g − + I − A ( u ) g − g + t − g − =  g ± = F ± x 2 2 t . (9) g + t + g + I + A ( u ) g − − I − A ( u ) g + =  x 2 2 Our aim: � f ± (0) � L 1 , � g ± (0) � L 1 ≤ δ 0 f ± ( t ) , g ± ( t ) ∈ L 1 ( R ) . = ⇒ 10

Center manifold of travelling profiles 11

Center manifold of travelling profiles We study the ODE − σu x + A ( u ) u x = (1 − σ 2 ) u xx , 11

Center manifold of travelling profiles We study the ODE − σu x + A ( u ) u x = (1 − σ 2 ) u xx , which can be written as the first order system  u x = p   (1 − σ 2 ) p x = ( A ( u ) − σI ) p (10)  = 0  σ x 11

Center manifold of travelling profiles We study the ODE − σu x + A ( u ) u x = (1 − σ 2 ) u xx , which can be written as the first order system  u x = p   (1 − σ 2 ) p x = ( A ( u ) − σI ) p (10)  = 0  σ x Close to any equilibrium (0 , 0 , λ i (0)), one can find a center man- ifold of travelling profiles: 11

Center manifold of travelling profiles We study the ODE − σu x + A ( u ) u x = (1 − σ 2 ) u xx , which can be written as the first order system  u x = p   (1 − σ 2 ) p x = ( A ( u ) − σI ) p (10)  = 0  σ x Close to any equilibrium (0 , 0 , λ i (0)), one can find a center man- ifold of travelling profiles: ˜ p = v i ˜ r i ( u, v i , σ ) , λ i = � ˜ r i , A ( u )˜ r i � , | ˜ r i ( u ) | = 1 . (11) 11

We can parameterize by the the kinetic component f i : 12

We can parameterize by the the kinetic component f i : u t = f − − f + = − σu x = f − + f + 12

We can parameterize by the the kinetic component f i : u t = f − − f + = − σu x = f − + f + 1 1 1 − σf − = 1 + σf + . u x = 12

We can parameterize by the the kinetic component f i : u t = f − − f + = − σu x = f − + f + 1 1 1 − σf − = 1 + σf + . u x = The center manifold for the kinetic components f − , f + is (12) (13) 12

We can parameterize by the the kinetic component f i : u t = f − − f + = − σu x = f − + f + 1 1 1 − σf − = 1 + σf + . u x = The center manifold for the kinetic components f − , f + is f − = (12) (13) 12

We can parameterize by the the kinetic component f i : u t = f − − f + = − σu x = f − + f + 1 1 1 − σf − = 1 + σf + . u x = The center manifold for the kinetic components f − , f + is f − = (1 − σ ) v i ˜ r i ( u, v i , σ ) (12) (13) 12

We can parameterize by the the kinetic component f i : u t = f − − f + = − σu x = f − + f + 1 1 1 − σf − = 1 + σf + . u x = The center manifold for the kinetic components f − , f + is � � f − f − = (1 − σ ) v i ˜ r i ( u, v i , σ )= f − i i ˜ r i u, 1 − σ, σ (12) (13) 12

We can parameterize by the the kinetic component f i : u t = f − − f + = − σu x = f − + f + 1 1 1 − σf − = 1 + σf + . u x = The center manifold for the kinetic components f − , f + is � � f − f − = (1 − σ ) v i ˜ r i ( u, v i , σ )= f − = f − r − i ( u, f − i i ˜ r i u, 1 − σ, σ i ˜ i , σ ) (12) (13) 12

We can parameterize by the the kinetic component f i : u t = f − − f + = − σu x = f − + f + 1 1 1 − σf − = 1 + σf + . u x = The center manifold for the kinetic components f − , f + is � � f − f − = (1 − σ ) v i ˜ r i ( u, v i , σ )= f − = f − r − i ( u, f − i i ˜ r i u, 1 − σ, σ i ˜ i , σ ) (12) � � u, (1 + σ ) v i f + = (1 + σ )˜ r i , σ 1 + σ (13) 12

We can parameterize by the the kinetic component f i : u t = f − − f + = − σu x = f − + f + 1 1 1 − σf − = 1 + σf + . u x = The center manifold for the kinetic components f − , f + is � � f − f − = (1 − σ ) v i ˜ r i ( u, v i , σ )= f − = f − r − i ( u, f − i i ˜ r i u, 1 − σ, σ i ˜ i , σ ) (12) � � u, (1 + σ ) v i f + = (1 + σ )˜ = f + r + i ( u, f + r i , σ i ˜ i , σ ) 1 + σ (13) 12

Identification of a travelling profile: u (¯ x ), σ and v i (¯ x ), 13

Identification of a travelling profile: u (¯ x ), σ and v i (¯ x ), u σ u(x− t) x u x x − + f ,f − x x 13

x ), σ and f − Identification of a travelling profile: u (¯ i (¯ x ), u σ u(x− t) x u x x − + f ,f σ (1− )u x − x x 14

x ), σ and f + Identification of a travelling profile: u (¯ i (¯ x ), u σ u(x− t) x u x x − + σ f ,f (1+ )u x σ (1− )u x − x x 15

Decomposition in travelling profiles 16

Decomposition in travelling profiles We decompose ( f − , g − ) and ( f + , g + ) separately: 16

Decomposition in travelling profiles We decompose ( f − , g − ) and ( f + , g + ) separately: (14) 16

Decomposition in travelling profiles We decompose ( f − , g − ) and ( f + , g + ) separately: � i f − r − i ( u, f − i , σ − f − � = i ˜ i ) (14) i g − r − i ( u, f − i , σ − g − � = i ˜ i ) 16

Decomposition in travelling profiles We decompose ( f − , g − ) and ( f + , g + ) separately: � � − g − � i f − r − i ( u, f − i , σ − f − � = i ˜ i ) σ − i i = θ i (14) , i g − r − i ( u, f − i , σ − g − � f − = i ˜ i ) i 16

Decomposition in travelling profiles We decompose ( f − , g − ) and ( f + , g + ) separately: � � − g − � i f − r − i ( u, f − i , σ − f − � = i ˜ i ) σ − i i = θ i (14) , i g − r − i ( u, f − i , σ − g − � f − = i ˜ i ) i θ i λ i where θ i is the cutoff function . − − −g /f i i 16

Decomposition in travelling profiles We decompose ( f − , g − ) and ( f + , g + ) separately: � � − g − � i f − r − i ( u, f − i , σ − f − � = i ˜ i ) σ − i i = θ i (14) , i g − r − i ( u, f − i , σ − g − � f − = i ˜ i ) i θ i λ i where θ i is the cutoff function . − − −g /f i i Similarly for ( f + , g + ): 16

Decomposition in travelling profiles We decompose ( f − , g − ) and ( f + , g + ) separately: � � − g − � i f − r − i ( u, f − i , σ − f − � = i ˜ i ) σ − i i = θ i (14) , i g − r − i ( u, f − i , σ − g − � f − = i ˜ i ) i θ i λ i where θ i is the cutoff function . − − −g /f i i Similarly for ( f + , g + ):    − g + � i f + r + i ( u, f + i , σ + f + � = i ˜ i ) σ + i  , = θ i (15) i g + r + i ( u, f + i , σ + i g + � f + = i ˜ i ) i 16

To find travelling profiles, we look separately to the t , x deriva- tives of F − , F + , and try to fit n travelling profiles into F − and n into F + . 17

To find travelling profiles, we look separately to the t , x deriva- tives of F − , F + , and try to fit n travelling profiles into F − and n into F + . F − t x 17

To find travelling profiles, we look separately to the t , x deriva- tives of F − , F + , and try to fit n travelling profiles into F − and n into F + . F − t x 18

To find travelling profiles, we look separately to the t , x deriva- tives of F − , F + , and try to fit n travelling profiles into F − and n into F + . F − t φ σ (x− t) x 19

To find travelling profiles, we look separately to the t , x deriva- tives of F − , F + , and try to fit n travelling profiles into F − and n into F + . F − t φ σ (x− t) x We obtain thus 2 n travelling waves: n for F − and n for F + . 19

Equation for the components f ± i , g ± are of the form: i 20

Equation for the components f ± i , g ± are of the form: i  λ + λ − 1+˜ 1 − ˜  f − j,t − f − f − f + + ς − j j  = j + f,j ( t, x ) −  j,x j 2 2 (16) λ + λ − 1+˜ 1 − ˜  f + j,t + f + f + + ς + f − j j  = j − f,j ( t, x )  j,x j 2 2 20

Equation for the components f ± i , g ± are of the form: i  λ + λ − 1+˜ 1 − ˜  f − j,t − f − f − f + + ς − j j  = j + f,j ( t, x ) −  j,x j 2 2 (16) λ + λ − 1+˜ 1 − ˜  f + j,t + f + f + + ς + f − j j  = j − f,j ( t, x )  j,x j 2 2  λ + λ − − 1+˜ i + 1 − ˜  g − i,t − g − g − g + + ς −  i i = g,i ( t, x )  i,x 2 2 i (17) λ + λ − 1+˜ i − 1 − ˜ g + i,t + g + g + + ς + g −  i i  = g,i ( t, x )  i,x i 2 2 20

Equation for the components f ± i , g ± are of the form: i  λ + λ − 1+˜ 1 − ˜  f − j,t − f − f − f + + ς − j j  = j + f,j ( t, x ) −  j,x j 2 2 (16) λ + λ − 1+˜ 1 − ˜  f + j,t + f + f + + ς + f − j j  = j − f,j ( t, x )  j,x j 2 2  λ + λ − − 1+˜ i + 1 − ˜  g − i,t − g − g − g + + ς −  i i = g,i ( t, x )  i,x 2 2 i (17) λ + λ − 1+˜ i − 1 − ˜ g + i,t + g + g + + ς + g −  i i  = g,i ( t, x )  i,x i 2 2 with ς ± f , ς ± g sources of total variation for F ± x , F ± and g 20

Equation for the components f ± i , g ± are of the form: i  λ + λ − 1+˜ 1 − ˜  f − j,t − f − f − f + + ς − j j  = j + f,j ( t, x ) −  j,x j 2 2 (16) λ + λ − 1+˜ 1 − ˜  f + j,t + f + f + + ς + f − j j  = j − f,j ( t, x )  j,x j 2 2  λ + λ − − 1+˜ i + 1 − ˜  g − i,t − g − g − g + + ς −  i i = g,i ( t, x )  i,x 2 2 i (17) λ + λ − 1+˜ i − 1 − ˜ g + i,t + g + g + + ς + g −  i i  = g,i ( t, x )  i,x i 2 2 with ς ± f , ς ± g sources of total variation for F ± x , F ± and g   � � f + f − λ + , σ + λ − , σ − i i  . ˜ i = ˜ ˜ = ˜ λ i u, , λ i  u, 1 − σ − i i 1 + σ − i i i 20

After some computations, one obtains the source terms of the form 21

After some computations, one obtains the source terms of the form � � j | )( | f + k | + | g + j f + j g + | ς ± f,i | , | ς ± ( | f − j | + | g − | g − − f − g,i | ≤ C k | ) + C j | j j j � = k   f + � �   j | 2 + | g − j � | f − j + f + j + g + j | 2 + C χ ≇ 1 f −   j j � ( � f − L 1 + � f + L 1 ) | f − j − f + j | χ { f − j · f + j � 2 j � 2 + C < 0 } j j � L 1 + � f + j − g + j · g + ( � f − j � 2 j � 2 L 1 ) | g − j | χ { g − + C < 0 } . j j (18) 21

After some computations, one obtains the source terms of the form � � j | )( | f + k | + | g + j f + j g + | ς ± f,i | , | ς ± ( | f − j | + | g − | g − − f − g,i | ≤ C k | ) + C j | j j j � = k   f + � �   j | 2 + | g − j � | f − j + f + j + g + j | 2 + C χ ≇ 1 f −   j j � ( � f − L 1 + � f + L 1 ) | f − j − f + j | χ { f − j · f + j � 2 j � 2 + C < 0 } j j � L 1 + � f + j − g + j · g + ( � f − j � 2 j � 2 L 1 ) | g − j | χ { g − + C < 0 } . j j (18) Prove that the source terms are quadratic w.r.t. � f ± � L 1 , � g ± � L 1 . 21

Different types of source terms: 22

Different types of source terms: • interaction of different families: � ( | f − j | + | g − j | )( | f + k | + | g + k | ); j � = k 22

Different types of source terms: • interaction of different families: � ( | f − j | + | g − j | )( | f + k | + | g + k | ); j � = k • interaction of the same family: � j f + j g + | g − − f − C j | ; j j 22

Different types of source terms: • interaction of different families: � ( | f − j | + | g − j | )( | f + k | + | g + k | ); j � = k • interaction of the same family: � j f + j g + | g − − f − C j | ; j j • energy type terms: j | 2 + | g − � ( | f − j + f + j + g + j | 2 ) χ { f + j /f − j ≇ 1 } ; j 22

Different types of source terms: • interaction of different families: � ( | f − j | + | g − j | )( | f + k | + | g + k | ); j � = k • interaction of the same family: � j f + j g + | g − − f − C j | ; j j • energy type terms: j | 2 + | g − � ( | f − j + f + j + g + j | 2 ) χ { f + j /f − j ≇ 1 } ; j • L 1 decay terms: � � | f − j − f + j | χ { f − j · f + | g − j − g + j | χ { g − j · g + < 0 } + < 0 } . j j j j 22

Interaction of the same family 23

Interaction of the same family Consider the 2 × 2 system  1 −F ( u ) F − t − F − − F − =  u = F − + F + , |F ′ ( u ) | ≤ 1 − c. x 2 1+ F ( u ) F + − F + − F + =  x t 2 (19) 23

Interaction of the same family Consider the 2 × 2 system  1 −F ( u ) F − t − F − − F − =  u = F − + F + , |F ′ ( u ) | ≤ 1 − c. x 2 1+ F ( u ) F + − F + − F + =  x t 2 (19) x = f ± , F ± Let F ± = g ± , so that (same for g ± ) t � 2 f − + 1 − λ f − − 1+ λ t − f − 2 f + = x λ ( u ) = F ′ ( u ) . (20) f + + f + 2 f − − 1 − λ 1+ λ 2 f + = x t 23

Interaction of the same family Consider the 2 × 2 system  1 −F ( u ) F − t − F − − F − =  u = F − + F + , |F ′ ( u ) | ≤ 1 − c. x 2 1+ F ( u ) F + − F + − F + =  x t 2 (19) x = f ± , F ± Let F ± = g ± , so that (same for g ± ) t � 2 f − + 1 − λ f − − 1+ λ t − f − 2 f + = x λ ( u ) = F ′ ( u ) . (20) f + + f + 2 f − − 1 − λ 1+ λ 2 f + = x t Construct a functional which bounds � + ∞ � R | f − ( t, x ) g + ( t, x ) − g − ( t, x ) f + ( t, x ) | dxdt. 0 23

We can rewrite the integrand as , 24

We can rewrite the integrand as f − g + − g − f + = , 24

We can rewrite the integrand as � � f − + g + − g − f − g + − g − f + = f − f + , f + 24

We can rewrite the integrand as   � � + F +  − F − f − + g + − g − f − g + − g − f + = f − f + = F − x F + t t  , x F − f + F + x x 24

We can rewrite the integrand as   � � + F +  − F − f − + g + − g − f − g + − g − f + = f − f + = F − x F + t t  , x F − f + F + x x = strengths of waves × difference in speed . 24

We can rewrite the integrand as   � � + F +  − F − f − + g + − g − f − g + − g − f + = f − f + = F − x F + t t  , x F − f + F + x x = strengths of waves × difference in speed . This is not a Glimm functional, it is the interaction Remark. term. 24

We can rewrite the integrand as   � � + F +  − F − f − + g + − g − f − g + − g − f + = f − f + = F − x F + t t  , x F − f + F + x x = strengths of waves × difference in speed . This is not a Glimm functional, it is the interaction Remark. term. Since it holds g − + g + = f − − f + , the condition g − /f − = g + /f + implies that the solution is a travelling profile, replacing σ x = 0. 24

We can rewrite the integrand as   � � + F +  − F − f − + g + − g − f − g + − g − f + = f − f + = F − x F + t t  , x F − f + F + x x = strengths of waves × difference in speed . This is not a Glimm functional, it is the interaction Remark. term. Since it holds g − + g + = f − − f + , the condition g − /f − = g + /f + implies that the solution is a travelling profile, replacing σ x = 0. For simplicity we assume in the following λ = 0. 24