Interaction of weak and strong discontinuities for two-component - PowerPoint PPT Presentation

Interaction of weak and strong discontinuities for two-component mixture separation Maria Elaeva 1 , Michael Zhukov 2 , Elena Shiryaeva 2 1 Financial University under the Government of the Russian Federation, Moscow, Russia 2 Institute of Mathematica, Mechanics and Computer Science, Southern Federal University, Rostov-on-Don, Russia 12–15 September, 2016 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 1 / 40

A schematic diagram of a capillary zone electrophoresis setup anode cathode mixture buffer capillary anode cathode one sample other sample The interior of capillary is filled with an electrolyte. The sample to be analyzed is injected in the capillary. Due to electric field it travels down the capillary and forms some zones. So we have some travelling zones and we can identify a mixture. M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 2 / 40

Mathematical model of capillary zone electrophoresis System of two hyperbolic equations � � ∂ u k µ k u k ∂ t + ∂ = 0, k = 1, 2, 1 + s ∂ x s = u 1 + u 2 > − 1 Notations u k — «effective» concentrations, µ k — «effective» component mobilities, s —conductivity of the mixture System of two hyperbolic equations, written in the Riemann invariants A change of variable t → µ 1 µ 2 t ∂ R k ∂ t + λ k ( R 1 , R 2 ) ∂ R k ∂ x = 0, k = 1, 2, λ 1 ( R 1 , R 2 ) = R 1 R 1 R 2 , λ 2 ( R 1 , R 2 ) = R 2 R 1 R 2 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 3 / 40

Riemann invariants and concentrations Concentration expressed by Riemann invariants, u = u ( R ) u 1 = µ 2 ( R 1 − µ 1 )( R 2 − µ 1 ) u 2 = µ 1 ( R 1 − µ 2 )( R 2 − µ 2 ) , R 1 R 2 ( µ 1 − µ 2 ) R 1 R 2 ( µ 2 − µ 1 ) R = R ( u ) is the roots of characteristic polynomial ( 1 + u 1 + u 2 )( R ) 2 − ( µ 1 + µ 2 + u 1 µ 2 + u 2 µ 1 ) R + µ 1 µ 2 = 0 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 4 / 40

Riemann problem Initial data R 1 � t = 0 = R 1 R 2 � t = 0 = R 2 0 ( x ) , 0 ( x ) , � �  µ 1 , x < x 1 ,  µ 2 , x < x 1 ,   x 1 < x < x 2 , x 1 < x < x 2 , R 1 q 1 , R 2 q 2 , 0 ( x ) = 0 ( x ) = x 2 < x , x 2 < x , µ 1 , µ 2 ,   0 < q 1 < µ 1 < µ 2 < q 2 Rankine-Hugoniot conditions � ( µ k − R 1 )( µ k − R 2 ) � ( µ k − R 1 )( µ k − R 2 ) � � D = , k = 1, 2 µ k R 1 R 2 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 5 / 40

Hyperbolic and elliptic domains F ( u 1 , u 2 ) ≡ ( µ 1 + µ 2 + u 1 µ 2 + u 2 µ 1 ) 2 − 4 ( 1 + s ) µ 1 µ 2 F ( u 1 , u 2 ) > 0 , 1 + s > 0 — hyperbolic type, F ( u 1 , u 2 ) < 0 — elliptic type u 2 � � 0 , µ 2 − µ 1 A = µ 1 F ( u 1 , u 2 ) = 0 � � µ 1 − µ 2 B = , 0 F ( u 1 , u 2 ) < 0 µ 2 � � − µ 2 µ 1 C = µ 2 − µ 1 , µ 2 − µ 1 A C B u 1 − 1 1 − 1 + s = 0 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 6 / 40

Evolution of initial discontinuities Parameters and initial conditions x 1 = − 1, x 2 = 1 u 1 u 2 0 = 2, 0 = − 1, q 1 = 2, µ 1 = 5, µ 2 = 8, q 2 = 10 u k 2 1 x 0 2 4 6 − 1 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 7 / 40

Evolution of initial discontinuities Zone Z k = { R 1 k ( x , t ) , R 2 k ( x , t ) ; a k ( t ) , b k ( t ) } a k ( t ) , b k ( t ) — left and right border of zone, R 1 k ( x , t ) , R 2 k ( x , t ) — Riemann invariants in zone t T 6 T 3 x 2 θ Z 5 s x 1 ϕ s Z 2 Z 7 Z 6 x 2 l Z 1 Z 3 x 1 r T int Z 8 x 2 r x 1 Z 4 x l x 1 x 2 X int X 3 X 6 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 8 / 40

Evolution of initial discontinuities Solution at the moment t = + 0 Z 1 = { µ 1 , µ 2 ; − ∞ , x 1 Z 2 = { q 1 , µ 2 ; x 1 s ( t ) , x 2 s ( t ) } , l ( t ) } , Z 3 = { q 1 , R 2 a ( z 2 ) ; x 2 l ( t ) , x 2 Z 4 = { q 1 , q 2 ; x 2 r ( t ) , x 1 r ( t ) } , l ( t ) } , Z 6 = { R 1 a ( z 1 ) , q 2 ; x 1 l ( t ) , x 1 Z 7 = { µ 1 , q 2 ; x 1 r ( t ) , x 2 r ( t ) } , s ( t ) } , Z 8 = { µ 1 , µ 2 ; x 2 s ( t ) , + ∞ } , � � z 1 ( x , t ) = x − x 2 z 2 ( x , t ) = x − x 1 z 1 z 2 R 1 a ( z 1 ) = R 2 a ( z 2 ) = q 2 , , q 1 , t t Strong waves and rarefaction waves s = x 1 + q 1 µ 1 µ 2 t , s = x 2 + µ 1 µ 2 q 2 t , D 1 = q 1 µ 1 µ 2 , D 2 = µ 1 µ 2 q 2 ; x 1 x 2 l = x 2 + q 1 q 1 q 2 t , r = x 2 + µ 1 µ 1 q 2 t , x 1 x 1 l = x 1 + q 1 µ 2 µ 2 t , r = x 1 + q 1 q 2 q 2 t x 2 x 2 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 9 / 40

Evolution of initial discontinuities Evolution of initial discontinuities at the moment t = + 0 u k u k Z 2 Z 3 Z 4 Z 3 Z 7 Z 7 Z 1 Z 6 Z 8 Z 1 Z 2 Z 6 Z 8 1 1 x x 0 2 4 6 0 2 4 6 8 10 − 1 − 1 x 1 s x 2 x 1 x 2 s x 2 x 1 x 1 x 2 l r s l r s x 2 x 1 x 2 r = x 1 r l l M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 10 / 40

Evolution of initial discontinuities Interaction between two rarefaction waves The point of interaction is the solution of equation x 2 r ( t ) = x 1 l ( t ) x 2 − x 1 X int = x 1 q 1 − x 2 q 2 T int = q 1 q 2 ( q 2 − q 1 ) , q 1 − q 2 Zones Zone Z 4 disappears, new zone Z 5 appears and border of zone Z 3 and Z 6 are changed Z 5 = { R 1 ( x , t ) , R 2 ( x , t ) ; ϕ ( t ) , θ ( t ) } Z 3 = { q 1 , R 2 a ( z 2 )) ; x 2 Z 6 = { R 1 a ( z 2 ) , q 2 , θ ( t ) , x 1 l ( t ) , ϕ ( t ) } , r ( t ) } M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 11 / 40

Evolution of initial discontinuities Functions ϕ ( t ) , θ ( t ) Line ϕ ( t ) is the weak discontinuity of Riemann invariant R 1 . For determing ϕ ( t ) one have the Cauchy problem d ϕ dt = λ 1 ( q 1 , R 2 a ) , ϕ ( T int ) = X int The solution is ϕ − x 1 � 1/2 q 1 � 3/2 � X int − x 1 � 1/2 t 1/2 − T 1/2 � � � � = + int For weak discontinuity x = θ ( t ) of Riemann invariant R 2 one can find θ − x 2 � 1/2 = q 2 � 3/2 � t 1/2 − T 1/2 � X int − x 2 � 1/2 � � + � int M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 12 / 40

About the action of ϕ ( t ) and θ ( t ) Zone Z 3 disappears at the point ( T 3 , X 3 ) , when the characteristic x = ϕ ( t ) intersects the characteristic x = x 2 l ( t ) . From equation X 3 = ϕ ( T 3 ) = x 2 l ( T 3 ) we have T 3 = ( q 2 − q 1 ) 2 X 3 = x 1 + q 1 µ 2 µ 2 T 3 ( q 1 − µ 2 ) 2 T int , Zone Z 6 disappears at the point ( T 6 , X 6 ) , when the characteristic x = θ ( t ) intersects the characteristic x = x 1 r ( t ) . From equation X 6 = θ ( T 6 ) = x 1 r ( T 6 ) we have T 6 = ( q 2 − q 1 ) 2 X 6 = x 2 + µ 1 µ 1 q 2 T 6 ( q 2 − µ 1 ) 2 T int , t T 6 T 3 x 2 θ Z 5 s x 1 ϕ s Z 2 Z 7 x 2 Z 6 l Z 1 Z 3 x 1 r T int Z 8 x 2 r x 1 Z 4 x l x 1 x 2 X int X 3 X 6 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 13 / 40

The Goursat problem Zone Z 5 For equations ∂ R k ∂ t + λ k ( R 1 , R 2 ) ∂ R k ∂ x = 0, k = 1, 2, λ 1 ( R 1 , R 2 ) = R 1 R 1 R 2 , λ 2 ( R 1 , R 2 ) = R 2 R 1 R 2 we have the Goursat problem with initial data that have weak discontinuities on the characteristics x = ϕ ( t ) , x = θ ( t ) R 1 � x = ϕ ( t ) = q 1 , R 2 � x = ϕ ( t ) = R 2 a ( z 2 ( ϕ ( t ) , t )) , � � R 1 � x = θ ( t ) = R 1 a ( z 1 ( θ ( t ) , t )) , R 2 � x = θ ( t ) = q 2 � � At the point ( T int , X int ) R 1 � t = T int = q 1 , R 2 � t = T int = q 2 � � M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 14 / 40

The hodograph method For system ∂ R k ∂ t + λ k ( R 1 , R 2 ) ∂ R k ∂ x = 0, k = 1, 2, we can apply the classical hodograph method. Using the replacement of independent and dependent variables ( R 1 , R 2 ) ⇋ ( x , t ) we get x R 2 − λ 1 t R 2 = 0, x R 1 − λ 2 t R 1 = 0. Using the compatibility conditions we get 2 t R 1 R 2 + R 2 − R 1 ( t R 1 − t R 2 ) = 0 The Riemann-Green function is defined V ( r 1 , r 2 | R 1 , R 2 ) = (( R 1 + R 2 )( r 1 + r 2 ) − 2 ( R 1 R 2 + r 1 r 2 ))( r 1 − r 2 ) ( R 1 − R 2 ) 3 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 15 / 40

The hodograph method Expression for t ( R 1 , R 2 ) t ( R 1 , R 2 ) = T int V ( q 1 , q 2 | R 1 , R 2 ) = = ( x 2 − x 1 )( 2 R 1 R 2 + 2 q 1 q 2 − ( q 1 + q 2 )( R 1 + R 2 )) q 1 q 2 ( R 1 − R 2 ) 3 The initial condition R 1 � t = T int = q 1 , R 2 � t = T int = q 2 � � is satisfied t ( q 1 , q 2 ) = T int V ( q 1 , q 2 | q 1 , q 2 ) = T int M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 16 / 40

The hodograph method The equations ∂ R k ∂ t + λ k ( R 1 , R 2 ) ∂ R k ∂ x = 0, k = 1, 2, have symmetry property. After a change of variables R 1 = 1 R 2 = 1 K 1 , K 2 we get 2 x K 1 K 2 + K 2 − K 1 ( x K 1 − x K 2 ) = 0 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 17 / 40

The hodograph method Expression for x ( R 1 , R 2 ) x ( R 1 , R 2 ) = ( x 2 − x 1 )( R 1 R 2 ) 2 R 2 + R 1 − 2 ( q 1 + q 2 ) + q 1 q 2 ( R 1 − R 2 ) 3 + ( x 1 ( R 1 ) 3 − x 2 ( R 2 ) 3 ) + 3 R 1 R 2 ( R 2 x 2 − R 1 x 1 ) ( R 1 − R 2 ) 3 Functions t ( R 1 , R 2 ) , x ( R 1 , R 2 ) determine the implicit solution. t T 6 T 3 x 2 θ Z 5 s x 1 ϕ s Z 2 Z 7 Z 6 x 2 l Z 1 Z 3 x 1 r T int Z 8 x 2 r x 1 Z 4 x l x 1 x 2 X int X 3 X 6 M. Elaeva, M. Zhukov, E. Shiryaeva () 12–15 September, 2016 18 / 40