On the Almost Axisymmetric Flows with Forcing Terms Marc Sedjro - PowerPoint PPT Presentation

On the Almost Axisymmetric Flows with Forcing Terms Marc Sedjro joint work with Michael Cullen. RWTH Aachen University October 7, 2012

Outline ◮ Analysis of the Hamiltonian of Almost Axisymmetric Flows.

Outline ◮ Analysis of the Hamiltonian of Almost Axisymmetric Flows.

Outline ◮ Analysis of the Hamiltonian of Almost Axisymmetric Flows. ◮ A Toy Model.

Outline ◮ Analysis of the Hamiltonian of Almost Axisymmetric Flows. ◮ A Toy Model. ◮ Challenges in the study of the Almost Axisymmetric Flows with Forcing Terms.

Time varying domain. The time varying domain occupied by the fluid is given by 1 := { ( λ, r , z ) | r 0 ≤ r ≤ r t Γ r t 1 ( λ, z ) , z ∈ [0 , H ] , λ ∈ [0 , 2 π ] } , For simplicity, we set r 0 = 1 in the sequel. Figure: Time varying domain.

Hamiltonian The fluid evolves with the velocity u := u ( λ, r , z ) expressed in cylindrical coordinates ( u , v , w ) . The temperature θ of the fluid inside the vortex is assumed to be greater that the ambient temperature maintained constant at θ 0 > 0 . g is the gravitational constant. The Hamiltonian of the Almost Axisymmetric Flow is ( u 2 � 2 − g θ ) rdrdzd λ. θ 0 Γ r 1 Important: The Almost Axisymmetric Flows are derived from Boussinesq’s equations with no loss of the Hamiltonian structure (George Craig).

Hamiltonian : Stable Almost axisymmetric flows Ω : Coriolis coefficient. ru + Ω r 2 : angular momentum g θ 0 θ : potential temperature. Stability condition: On each λ − section of the domain Γ r 1 , we require that → [( ru λ + Ω r 2 ) 2 , g θ λ ] ( r , z ) − θ 0 be invertible and gradient of a convex function.

Hamiltonian: Stable Almost axisymmetric flows We made crucial observation that, for stable Almost axisymetric flows for which the total mass is finite (=1), the Hamiltonian can be expressed in terms of one single measure σ : � 2 π H [ σ ] = I 0 [ σ λ ] + inf ρ ∈S I [ σ λ ]( ρ ) d λ 0 Here, σ is a probability measure such that π 1 # σ is absolutely continuous with respect to L 1 | [0 , 2 π ] . 2 − Ω √ y 1 − | y | 2 � � y 1 � I 0 [ σ λ ] = σ λ ( dy ) 2 R 2 + 2(1 − 2 x 1 ) −| x | 2 Ω 2 I [ σ λ ]( ρ ) := 1 1 � 1 � � � � 2 W 2 σ λ , (1 − 2 x 1 ) 2 χ D ρ ( x ) + (1 − 2 x 1 ) 2 dx 2 2 D ρ Here, S is the set of functions ρ : [0 , H ] → [0 , 1 / 2), D ρ := { x = ( x 1 , x 2 ) | x 1 ∈ [0 , H ] , 0 ≤ x 2 ≤ ρ ( x 1 ) }

Analysis of the Hamiltonian Assume σ 0 is a probability measure on R 2 and write I [ σ 0 ]( ρ ) = 1 1 � � 2 W 2 σ 0 , (1 − 2 x 1 ) 2 χ D ρ ( x ) + good terms 2 Existence of a minimizer. ρ ∈S is not weakly ∗ closed in L ∞ . � � Obstacle : χ D ρ

Analysis of the Hamiltonian Assume σ 0 is a probability measure on R 2 and write I [ σ 0 ]( ρ ) = 1 1 � � 2 W 2 σ 0 , (1 − 2 x 1 ) 2 χ D ρ ( x ) + good terms 2 Existence of a minimizer. ρ ∈S is not weakly ∗ closed in L ∞ . � � Obstacle : χ D ρ However, I [ σ 0 ]( ρ # ) ≤ I [ σ 0 ]( ρ ) where ρ # is the increasingly monotone rearrangement of ρ . Classical results in the direct methods of the calculus of variations ensures the existence of a minimizer.

Analysis of the Hamiltonian Uniqueness of minimizers.

Analysis of the Hamiltonian Uniqueness of minimizers. Obstacle : No convexity property for ρ → I [ σ 0 ]( ρ ) with respect to any interpolation we can think of.

Analysis of the Hamiltonian Uniqueness of minimizers. Obstacle : No convexity property for ρ → I [ σ 0 ]( ρ ) with respect to any interpolation we can think of. We use a Dual formulation of the minimization problem that yields existence and uniqueness. � H � � y 1 2 − Ω √ y 1 − Ψ( y ) � sup σ 0 ( dy )+ inf Π P ( ρ ( x 2 ) , x 2 ) dx 2 ρ ∈S R 2 { ( P , Ψ): P =Ψ ∗ , Ψ= P ∗ } 0 (1) � s 1 1 � � Π P ( x 1 , s ) = 2(1 − 2 x 1 ) − P ( x 2 , x 1 ) (1 − 2 x 2 ) 2 dx 1 for 0 ≤ x 1 < 1 . 0 (1) has a unique solution.

Analysis of the Hamiltonian. Regularity of the boundary ∂ D ρ The dual problem reveals a regularity property of ρ stronger than monotonicity. L 0 , L 0 ) × (0 , L 0 ) L 0 > 0 and P σ 0 solve the More precisely, if spt ( σ 0 ) ⊂ ( 1 variational problem (1) then the study of Euler -Lagrange equation of � H inf Π P σ 0 ( ρ ( x 2 ) , x 2 ) dx 2 ρ ∈S 0 yields C > 0 such that the minimizer ρ σ 0 satisfies ρ σ 0 (¯ x 2 ) − ρ σ 0 ( x 2 ) ≥ C (¯ x 2 − x 2 ) for all x 2 , ¯ x 2 ∈ [0 , H ] . Consequently, we obtain that ∂ D ρ σ 0 is piecewise Lipschitz continuous.

A unusual Monge-Amp` ere equation. Moreover, assume in addition, σ 0 is absolutely continuous with respect to the Lebesgue measure. If ( P σ 0 , Ψ σ 0 , ρ σ 0 ) is the variational solution(1) then P σ 0 is convex, ∇ P σ 0 is invertible (1 − 2 x 1 ) − 2 χ D ρ ( x ) L 2 a.e and (1 − 2 ∂ y 2 Ψ) 2 det ∇ 2 Ψ 1  ( i ) = σ 0       � � Ω 2 (2) ( ii ) P ρ ( x 2 ) , x 2 = on { ρ > 0 } 2(1 − 2 ρ ( x 2 ))       ( iii ) ∇ Ψ maps spt ( σ 0 ) onto D ρ .

Change of variables Let ( P λ , Ψ λ , ρ λ ) be the solution to the variational problem (1) corresponding to σ λ . Assume σ absolutely continuous with respect to Lebesgue. Define u , θ, r through g θ λ ( u λ r + Ω r 2 ) 2 = ∂ x 1 P λ , 2 x 1 = 1 − r − 2 . = ∂ x 2 P λ , (3) θ 0 and χ Γ r 1 rdrdzd λ = (1 − 2 x 1 ) − 2 χ D ρλ ( x ) dx 1 dx 2 d λ = σ dy 1 dy 2 d λ. Then, ( u , θ, r 1 ) satisfy the stability condition and ( u 2 � 2 − g θ H [ σ ] = ) rd λ drdz . θ 0 Γ r 1

Forced Axisymmetric Flows : Toy Model 2D We remove the λ dependence on the quantities involved in the Almost axisymmetric flows with forcing terms to obtain the forced axisymmetric D flows: Dt := ∂ t + v ∂ r + w ∂ z . ( ru + Ω r 2 ) 2 = r 3 ∂ r [ ϕ + Ω 2 θ 0 θ = ∂ z [ ϕ + Ω 2 2 r 2 ] , g  2 r 2 ] in Γ r 1   1  r ∂ r ( rv ) + ∂ z w = 0 in Γ r 1  ∂ t r 1 + w ∂ z r 1 = v , on { r = r 1 }   ¯  Dt ( ru + Ω r 2 ) = F , D Dt ( g D θ 0 θ ) = g in Γ r 1  θ 0 S (4) Here, Γ r t 1 := { ( r , z ) | r 1 ( t , z ) ≥ r ≥ r 0 , z ∈ [0 , H ] } , ϕ ( t , r 1 ( t , z ) , z ) = 0 , on ∂ { r 1 > r 0 } . (5) Neumann condition has been imposed on the rigid boundary. Data : F , S are prescribed functions. Unknown : u , v , w , ϕ, θ and r 1

Toy Model in “Dual Space” 2 D In view of the change of variable discussed above, existence of a variational solution to the MA equation, formal computations yield ◮ � ∂ t σ t + div ( σ t V t [ σ t ]) = 0 Toy Model ⇐ ⇒ σ | t =0 = ¯ σ 0

Toy Model in “Dual Space” 2 D In view of the change of variable discussed above, existence of a variational solution to the MA equation, formal computations yield ◮ � ∂ t σ t + div ( σ t V t [ σ t ]) = 0 Toy Model ⇐ ⇒ σ | t =0 = ¯ σ 0 ◮ Task we completed: Identify the operator σ �− → V t [ σ ].

Forced axisymmetric flows: Velocity field Regular initial data : V t [ σ ]( y ) = L t ( ∇ Ψ σ ( y ); y ) where 2 √ y 1 F t , g � (1 − 2 x 1 ) − 1 (1 − 2 x 1 ) − 1 �� 2 , x 2 2 , x 2 � � � � � L t x ; y = S t . θ 0 and Ψ σ is a solution in the variational problem (1). General initial data : Use the Riesz representation theorem to uniquely define V t [ σ ] by � � e ( x 1 ) � L t ( x , ∇ P σ ) , G ( ∇ P σ ) � dx 1 dx 2 R 2 � V t [ σ ] , G � d σ = D σ ρσ ∀ G ∈ C c ( R 2 , R 2 ) and ( P σ , ρ σ ) solves the variational problem (1).

Existence of solutions for the Forced axisymmetric flows. ◮ Appropriate conditions of the forcing terms. ◮ Continuity property in σ − → V t [ σ ] ( and σ − → σ V t [ σ ]). = ⇒ Global solution in time.

Almost Axisymmetric Flow with Forcing Terms Back to the full physical model Dt := ∂ t + u D These equations are given by (here, r ∂ λ + v ∂ r + w ∂ z ) � Du u 2  Dt + uv r + 1 D θ � r r ∂ λ ϕ + 2Ω v = F , r + 2Ω u = ∂ r ϕ, Dt = S ,      r ∂ r ( rv ) + 1 1 ∂ z ϕ − g θ r ∂ λ u + ∂ z w = 0 θ 0 = 0     ∂ t r 1 + u  r 1 ∂ λ r 1 + w ∂ z r 1 = v on { r = r 1 } (6) in the region Γ r 1 := { ( λ, r , z ) | r 1 ( λ, z ) ≥ r ≥ r 0 , z ∈ [0 , H ] , λ ∈ [0 , 2 π ] } , subject to the boundary condition ϕ ( t , λ, r 1 ( t , λ, z ) , z ) = 0 , on ∂ { r 1 > r 0 } . (7) Neumann condition has been imposed on the rigid boundary.

Almost axisymmetric Flow with Forcing Terms : Dual space 3 D The equations above can be recast as a transport equation : σ 0 << L 3 ∂ t σ t + div ( σ t X t [ σ t ]) = 0; σ | t =0 = ¯ (8) Here X t [ σ ]( y ) = L t ( ∇ Ψ σ ( y ) , y ) Ψ σ ( λ, · ) solves the Monge Amp` ere equations (2) and L t ( x , y ) = � √ y 1 √ y 1 , 2 √ y 1 F t ( λ, e √ y 1 , g � 1 1 4 ( x 1 ) , x 2 ) + 2 x 1 4 ( x 1 ) , x 2 ) − Ω − 2 x 1 S t ( λ, e r 0 θ 0 with x = ( λ, x 1 , x 2 ) , y = ( λ, y 1 , y 2 ) and e ( x 1 ) = (1 − 2 x 1 ) − 2 .

Challenges in the continuity equation ◮ Defining well the velocity X t [ σ ]. ◮ Existence and Regularity of � ∂ Ψ ∂λ , ∂ Ψ ∂ Υ , ∂ Ψ � ∇ Ψ = ∂ Z ◮ Regularity in a Monge-Ampere equation with respect to a parameter: 1 y 1 , y 2 Ψ λ = σ λ (1 − 2 ∂ y 1 Ψ λ ) 2 det ∇ 2