On a low Mach number limit for Supernovae Donatella Donatelli joint work with E. Feireisl ( arXiv:1604.05860, [math.AP] ) Dipartimento di Ingegneria e Scienze dell’Informazione e Matematica Universit` a degli Studi dell’Aquila 67100 L’Aquila, Italy donatella.donatelli@univaq.it
Motivation A broad range of interesting phenomena in science and engineering occur in a low Mach number regime, in which the fluid velocity is much less than the speed of sound Ma = fluid velocity sound speed 2
What are supernovae? A supernova is the explosion of a star. It is the largest explosion that takes place in space. A supernova happens where there is a change in the core, or center, of a star. For example if we have two stars that orbit the same point (binary star systems) and one of the stars, a carbon-oxygen white dwarf, steals matter from its companion star. Eventually, the white dwarf accumulates too much matter. Having too much matter causes the star to explode, resulting in a supernova. 3
What are supernovae? But supernovas are difficult to see in our own Milky Way galaxy because dust blocks our view. In 1604, Johannes Kepler discovered the last observed supernova in the Milky Way. NASA’s Chandra telescope discovered the remains of a more recent supernova. A supernova burns for only a short period of time, but it can tell scientists a lot about the universe. When the star explodes, it shoots elements and debris into space. Many of the elements we find here on Earth are made in the core of stars. 4
Cassiopeia A Cassiopeia A is a supernova remnant in the constellation Cassiopeia. The supernova occurred approximately 11,000 light-years away within the Milky Way. Discovery date: 1947. 5
Supernova 1987A Credits: Nasa - Hubble Space Telescope 6
Why construct a mathematical model? to track the final steps that leads from a white dwarf to a supernova ⇓ to know more about how the supernova ignite 7
Range of timescale 100 years convection that precedes ignition last few hours convection becomes more vigorous as the heat release intensifies and convection can no longer carry away the heat. 1 second duration of the explosion � � � In the last minutes of the convective phase, velocity reach ∼ 1% of the sound speed � � � low Mach number regime 8
Our Goal Understand the separation of scales that occurs in a low Mach number regime Physically, one can think of the solution to a low Mach number model as supporting infinitely fast acoustic equilibration rather than finite-velocity acoustic wave propagation Mathematically , this is manifest in the addition of a constraint on the velocity field to the system of evolution equations. 9
The mathematical model Compressible viscous fluid ̺ = ̺ ( x, t ) density u = u ( x, t ) velocity vector Θ = Θ( x, t ) temperature X k = X k ( x, t ) mass fraction of the species ω k = ˙ ˙ ω k ( x, t ) production rates Viscous stress tensor � � x u − 2 ∇ x u + ∇ t S ( ∇ x u ) = 3div x u I + λ div x u I , λ ≥ 0 , 10
The mathematical model Mass conservation ∂ t ̺ + div( ̺ u ) = 0 Momentum balance ∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + ∇ x p = µ div S ( ∇ x u ) + ̺f Energy balance � ∂ t ( ̺E ) + div( ̺ u E + p u ) = − ̺q k ˙ ω k k Combustion equation ∂ t ( ̺X k ) + div( ̺ u X k ) = ̺ ˙ ω k E = e + | u | 2 / 2 , total energy p = p ( ̺, Θ) pressure 11
Nondimensional equations T ref = reference time L ref = reference lenght U ref = reference velocity ρ ref = reference density p ref = reference pressure 12
Nondimensional equations T ref = reference time L ref = reference lenght U ref = reference velocity ρ ref = reference density p ref = reference pressure X Nondimensional variables = X ′ = X ref L ref U ref Re = ρ ref U ref L ref √ Sr = Ma = T ref U ref T ref µ ref p ref /ρ ref U ref √ Fr = L ref f ref 12
Nondimensional equations T ref = reference time L ref = reference lenght U ref = reference velocity ρ ref = reference density p ref = reference pressure X Nondimensional variables = X ′ = X ref L ref U ref Re = ρ ref U ref L ref √ Sr = Ma = T ref U ref T ref µ ref p ref /ρ ref U ref √ Fr = L ref f ref ∂ t ̺ + div( ̺ u ) = 0 ∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + ∇ x p = div S ( ∇ x u ) + ̺f 12
Nondimensional equations T ref = reference time L ref = reference lenght U ref = reference velocity ρ ref = reference density p ref = reference pressure X Nondimensional variables = X ′ = X ref L ref U ref Re = ρ ref U ref L ref √ Sr = Ma = T ref U ref T ref µ ref p ref /ρ ref U ref √ Fr = L ref f ref Sr∂ t ̺ + div( ̺ u ) = 0 Ma 2 ∇ p = 1 1 1 Sr∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + Re div S ( ∇ x u ) + Fr 2 ̺f 12
Nondimensional equations Sr∂ t ̺ + div( ̺ u ) = 0 Ma 2 ∇ p = 1 1 1 Sr∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + Re div S ( ∇ x u ) + Fr 2 ̺f 13
Nondimensional equations Sr∂ t ̺ + div( ̺ u ) = 0 Ma 2 ∇ p = 1 1 1 Sr∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + Re div S ( ∇ x u ) + Fr 2 ̺f L ref Sr = = 1 T ref U ref ( the characteristic time scale T ref equals the convection time scale L ref /U ref ) 13
Nondimensional equations Sr∂ t ̺ + div( ̺ u ) = 0 Ma 2 ∇ p = 1 1 1 Sr∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + Re div S ( ∇ x u ) + Fr 2 ̺f L ref Sr = = 1 T ref U ref ( the characteristic time scale T ref equals the convection time scale L ref /U ref ) Ma = ε 13
Nondimensional equations Sr∂ t ̺ + div( ̺ u ) = 0 Ma 2 ∇ p = 1 1 1 Sr∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + Re div S ( ∇ x u ) + Fr 2 ̺f L ref Sr = = 1 T ref U ref ( the characteristic time scale T ref equals the convection time scale L ref /U ref ) Ma = ε strong stratification Fr ∼ Ma ∼ ε 13
Nondimensional equations Sr∂ t ̺ + div( ̺ u ) = 0 Ma 2 ∇ p = 1 1 1 Sr∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + Re div S ( ∇ x u ) + Fr 2 ̺f L ref Sr = = 1 T ref U ref ( the characteristic time scale T ref equals the convection time scale L ref /U ref ) Ma = ε high Reynolds number Re ∼ ε − α strong stratification Fr ∼ Ma ∼ ε 13
Nondimensional equations Sr∂ t ̺ + div( ̺ u ) = 0 Ma 2 ∇ p = 1 1 1 Sr∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + Re div S ( ∇ x u ) + Fr 2 ̺f L ref Sr = = 1 T ref U ref ( the characteristic time scale T ref equals the convection time scale L ref /U ref ) Ma = ε high Reynolds number Re ∼ ε − α strong stratification Fr ∼ Ma ∼ ε ∂ t ̺ + div( ̺ u ) = 0 ∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + 1 ε 2 ∇ p = ε α div S ( ∇ x u ) + 1 ε 2 ̺f 13
Low Mach number limit ∂ t ̺ + div( ̺ u ) = 0 ∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + 1 ε 2 ∇ p = ε α div S ( ∇ x u ) + 1 ε 2 ̺f 14
Low Mach number limit ∂ t ̺ + div( ̺ u ) = 0 ∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + 1 ε 2 ∇ p = ε α div S ( ∇ x u ) + 1 ε 2 ̺f � � � ∂ t V + V · ∇ x V + ∇ x Π = − ̺ 0 f where ̺ 0 is the unique solution of the static equation ∇ x p ( ̺ 0 ) = ̺ 0 f and.... div( ̺ 0 V ) = 0 14
Low Mach number limit ∂ t ̺ + div( ̺ u ) = 0 ∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + 1 ε 2 ∇ p = ε α div S ( ∇ x u ) + 1 ε 2 ̺f � � � ∂ t V + V · ∇ x V + ∇ x Π = − ̺ 0 f where ̺ 0 is the unique solution of the static equation ∇ x p ( ̺ 0 ) = ̺ 0 f and.... div V = −∇ x ̺ 0 V ̺ 0 constrain on the velocity field 14
Main problem: fast oscillating acoustic waves Linearized system Fourier Transform � 0 � � ˆ ρ t + u x = 0 � ˆ � iξ � ρ t ρ u t + 1 + iξ = 0 ε 2 ρ x = 0 u t ˆ ˆ u 0 ε 2 15
Main problem: fast oscillating acoustic waves Linearized system Fourier Transform � 0 � � ˆ ρ t + u x = 0 � ˆ � iξ � ρ t ρ u t + 1 + iξ = 0 ε 2 ρ x = 0 u t ˆ ˆ u 0 ε 2 Solutions � ˆ � � ˆ � � ˆ � ρ ( ξ, t ) ρ + ( ξ ) ρ − ( ξ ) θ ( ξ ) = ξ e iθ ( ξ ) t + e − iθ ( ξ ) t = u ( ξ, t ) ˆ u + ( ξ ) ˆ u − ( ξ ) ˆ ε 15
Main problem: fast oscillating acoustic waves Where do we see these waves? H ⊥ [ u ] u = H [ u ] + ���� � �� � soleinodal part gradient part H ⊥ [ u ] = ∇ Ψ div H [ u ] = 0 16
Main problem: fast oscillating acoustic waves Where do we see these waves? H ⊥ [ u ] u = H [ u ] + ���� � �� � soleinodal part gradient part H ⊥ [ u ] = ∇ Ψ div H [ u ] = 0 ρ t + div u = 0 ∂ t ρ + ∆Ψ = 0 u t + 1 ∂ t ∇ Ψ + 1 ε 2 ∇ r = 0 ε 2 ∇ ρ = 0 16
The model (a simplified) ∂ t ̺ + div( ̺ u ) = 0 ∂ t ( ̺ u ) + div( ̺ u ⊗ u ) + ∇ x p = µ div S ( ∇ x u ) + ̺f � ∂ t ( ̺E ) + div( ̺ u E + p u ) = − ̺q k ˙ ω k k ∂ t ( ̺X k ) + div( ̺ u X k ) = ̺ ˙ ω k E = e + | u | 2 / 2 , total energy p = p ( ̺, Θ) pressure e = e (Θ , ρ ) internal energy p = p ( ̺, Θ) = p ion + p rad , 17
The model (a simplified) ∂ t ̺ + div( ̺ u ) = 0 ∂ t ( ̺ u ) + div( ̺ u ⊗ u )+ ∇ x p = µ div S ( ∇ x u ) + ̺f � ❍❍❍❍ ✟ ∂ t ( ̺E ) + div( ̺ u E + p u ) = − ̺q k ˙ ω k ✟✟✟✟ k ❍ ❤❤❤❤❤❤❤❤❤❤❤❤❤ ✭ ✭✭✭✭✭✭✭✭✭✭✭✭✭ ∂ t ( ̺X k ) + div( ̺ u X k ) = ̺ ˙ ω k ❤ E = e + | u | 2 / 2 , total energy p = p ( ̺, Θ) pressure e = e (Θ , ρ ) internal energy p = p ( ̺, Θ) = p ion + p rad , 17
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