From Relaxation to Slow Erosion Wen Shen Department of Mathematics, Penn State University SISSA, Italy, June 16, 2016 Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 1 / 28
List of some joint works: Uniqueness for discontinuous ODEs and conservation laws; BV estimate for a relaxation model for multicomponent chromatography; Differential games: non-cooperative and semi-cooperative differential games; Optimality conditions for solutions to hyperbolic balance laws; Differential games related to fish harvesting; Slow erosion of granular flow: a semigroup approach; Growth model using PDEs – ongoing work. Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 2 / 28
Chromatography: a relaxation model u t + u x = − 1 δ ( F ( u ) − v ) v t = 1 δ ( F ( u ) − v ) A fluid flow with unit speed travels over a solid bed. u , v ∈ R n : concentration of n components of chemicals in the fluid and the solid bed. Equilibrium state: If v = F ( u ), then no exchange of chemicals will happen. δ : relaxation time, how quickly the equilibrium configuration is reached. Zero relaxation limit: as δ → 0, we get v → F ( u ), and ( u + F ( u )) t + u x = 0 . ⇒ An n × n system of conservation laws for u . Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 3 / 28
Chromatography: a relaxation model u t + u x = − 1 δ ( F ( u ) − v ) v t = 1 δ ( F ( u ) − v ) A fluid flow with unit speed travels over a solid bed. u , v ∈ R n : concentration of n components of chemicals in the fluid and the solid bed. Equilibrium state: If v = F ( u ), then no exchange of chemicals will happen. δ : relaxation time, how quickly the equilibrium configuration is reached. Zero relaxation limit: as δ → 0, we get v → F ( u ), and ( u + F ( u )) t + u x = 0 . ⇒ An n × n system of conservation laws for u . Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 3 / 28
Chromatography: a relaxation model u t + u x = − 1 δ ( F ( u ) − v ) v t = 1 δ ( F ( u ) − v ) A fluid flow with unit speed travels over a solid bed. u , v ∈ R n : concentration of n components of chemicals in the fluid and the solid bed. Equilibrium state: If v = F ( u ), then no exchange of chemicals will happen. δ : relaxation time, how quickly the equilibrium configuration is reached. Zero relaxation limit: as δ → 0, we get v → F ( u ), and ( u + F ( u )) t + u x = 0 . ⇒ An n × n system of conservation laws for u . Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 3 / 28
Chromatography: a relaxation model u t + u x = − 1 δ ( F ( u ) − v ) v t = 1 δ ( F ( u ) − v ) A fluid flow with unit speed travels over a solid bed. u , v ∈ R n : concentration of n components of chemicals in the fluid and the solid bed. Equilibrium state: If v = F ( u ), then no exchange of chemicals will happen. δ : relaxation time, how quickly the equilibrium configuration is reached. Zero relaxation limit: as δ → 0, we get v → F ( u ), and ( u + F ( u )) t + u x = 0 . ⇒ An n × n system of conservation laws for u . Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 3 / 28
Chromatography: a relaxation model u t + u x = − 1 δ ( F ( u ) − v ) v t = 1 δ ( F ( u ) − v ) A fluid flow with unit speed travels over a solid bed. u , v ∈ R n : concentration of n components of chemicals in the fluid and the solid bed. Equilibrium state: If v = F ( u ), then no exchange of chemicals will happen. δ : relaxation time, how quickly the equilibrium configuration is reached. Zero relaxation limit: as δ → 0, we get v → F ( u ), and ( u + F ( u )) t + u x = 0 . ⇒ An n × n system of conservation laws for u . Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 3 / 28
Goal of the study: • Given δ > 0, establish existence and uniqueness of the the solution for the relaxation system. • Zero relaxation limit as δ → 0. The key estimate: A compactness estimate. A bound on the total variations of the solutions for the relaxation system, uniform w.r.t. the relaxation parameter δ . Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 4 / 28
Goal of the study: • Given δ > 0, establish existence and uniqueness of the the solution for the relaxation system. • Zero relaxation limit as δ → 0. The key estimate: A compactness estimate. A bound on the total variations of the solutions for the relaxation system, uniform w.r.t. the relaxation parameter δ . Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 4 / 28
Goal of the study: • Given δ > 0, establish existence and uniqueness of the the solution for the relaxation system. • Zero relaxation limit as δ → 0. The key estimate: A compactness estimate. A bound on the total variations of the solutions for the relaxation system, uniform w.r.t. the relaxation parameter δ . Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 4 / 28
Key feature: Langmuir isotherm, F = ( F 1 , · · · , F n ) k i u i F i ( u ) = 1 + k 1 u 1 + · · · + k n u n The Jacobian matrix A ( u ) = DF ( u ) has n distinct real eigen-values, each family is genuinely nonlinear. Furthermore, the integral curves of each family are straight lines and coincide with the shock curves. ⇒ a type of Temple class Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 5 / 28
Key feature: Langmuir isotherm, F = ( F 1 , · · · , F n ) k i u i F i ( u ) = 1 + k 1 u 1 + · · · + k n u n The Jacobian matrix A ( u ) = DF ( u ) has n distinct real eigen-values, each family is genuinely nonlinear. Furthermore, the integral curves of each family are straight lines and coincide with the shock curves. ⇒ a type of Temple class Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 5 / 28
Key feature: Langmuir isotherm, F = ( F 1 , · · · , F n ) k i u i F i ( u ) = 1 + k 1 u 1 + · · · + k n u n The Jacobian matrix A ( u ) = DF ( u ) has n distinct real eigen-values, each family is genuinely nonlinear. Furthermore, the integral curves of each family are straight lines and coincide with the shock curves. ⇒ a type of Temple class Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 5 / 28
l i , r i : the left and right normalized eigenvectors of A ( u ) = DF ( u ). � i u i u i u x = � x r i ( u ) , x = l i ( u ) · u x i v i v i v x = � x r i ( u ) , x = l i ( u ) · v x Define the directional derivative: φ ( u + h � v ) − φ ( u ) φ ( u ) • � v = lim h h → 0 Key property: r i ( u , v ) • r i ( u , v ) ≡ 0 , for all i , for all ( u , v ) Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 6 / 28
l i , r i : the left and right normalized eigenvectors of A ( u ) = DF ( u ). � i u i u i u x = � x r i ( u ) , x = l i ( u ) · u x i v i v i v x = � x r i ( u ) , x = l i ( u ) · v x Define the directional derivative: φ ( u + h � v ) − φ ( u ) φ ( u ) • � v = lim h h → 0 Key property: r i ( u , v ) • r i ( u , v ) ≡ 0 , for all i , for all ( u , v ) Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 6 / 28
l i , r i : the left and right normalized eigenvectors of A ( u ) = DF ( u ). � i u i u i u x = � x r i ( u ) , x = l i ( u ) · u x i v i v i v x = � x r i ( u ) , x = l i ( u ) · v x Define the directional derivative: φ ( u + h � v ) − φ ( u ) φ ( u ) • � v = lim h h → 0 Key property: r i ( u , v ) • r i ( u , v ) ≡ 0 , for all i , for all ( u , v ) Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 6 / 28
� u xt + u xx = − A ( u ) u x + v x v xt = A ( u ) u x − v x In components: ( u i x ) t + ( u i x ) x = − λ i ( A ) u i x + v i j G ij u j x − � x ( v i x ) t = λ i ( A ) u i x − v i j G ij u j j , k H ijk v j x u k x + � x + � x where G ij = l i · (( F ( u ) − v ) • r j ) , H ijk ( u , v ) = l i · ( r j • r k ) . Need to show that these terms are integrable over ( t , x ) ∈ [0 , ∞ ) × ( −∞ , ∞ ). Thanks to the key feature, one can show – H ijk = 0 for j = k – G ij includes only terms u i x with i � = j . Need to show terms v j x u k x , u j x u k x with j � = k are integrable. Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 7 / 28
� u xt + u xx = − A ( u ) u x + v x v xt = A ( u ) u x − v x In components: ( u i x ) t + ( u i x ) x = − λ i ( A ) u i x + v i j G ij u j x − � x ( v i x ) t = λ i ( A ) u i x − v i j G ij u j j , k H ijk v j x u k x + � x + � x where G ij = l i · (( F ( u ) − v ) • r j ) , H ijk ( u , v ) = l i · ( r j • r k ) . Need to show that these terms are integrable over ( t , x ) ∈ [0 , ∞ ) × ( −∞ , ∞ ). Thanks to the key feature, one can show – H ijk = 0 for j = k – G ij includes only terms u i x with i � = j . Need to show terms v j x u k x , u j x u k x with j � = k are integrable. Wen Shen (Penn State) From Relaxation to Slow Erosion SISSA, Italy, June 16, 2016 7 / 28
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