The Porous Medium Equation egert/Barenblatt.pdf · PDF file...

Click here to load reader

  • date post

    30-Sep-2020
  • Category

    Documents

  • view

    0
  • download

    0

Embed Size (px)

Transcript of The Porous Medium Equation egert/Barenblatt.pdf · PDF file...

  • The Porous Medium Equation

    Moritz Egert, Samuel Littig, Matthijs Pronk, Linwen Tan

    Coordinator: Jürgen Voigt

    18 June 2010

    1 / 11

  • Physical motivation

    Flow of an ideal gas through a homogeneous porous medium can be described by

     ε∂tρ+ div(ρv) = 0 mass balance

    µv = −k ∇p Darcy’s law p = p0 ργ state equation

    ε∂tρ = − div(ρv) = k µ

    div(ρ∇p) = kp0 µ

    div(ρ∇ργ)

    I ρ : density I p : pressure I v : velocity

    I ε, k , µ > 0 : material constants I γ ≥ 1 : polytropic exponent I p0 : reference pressure

    2 / 11

  • Physical motivation

    Flow of an ideal gas through a homogeneous porous medium can be described by

     ε∂tρ+ div(ρv) = 0 mass balance

    µv = −k ∇p Darcy’s law p = p0 ργ state equation

    ε∂tρ = −div(ρv) = k µ

    div(ρ∇p) = kp0 µ

    div(ρ∇ργ)

    I ρ : density I p : pressure I v : velocity

    I ε, k , µ > 0 : material constants I γ ≥ 1 : polytropic exponent I p0 : reference pressure

    2 / 11

  • Physical motivation

    Flow of an ideal gas through a homogeneous porous medium can be described by

     ε∂tρ+ div(ρv) = 0 mass balance

    µv = −k ∇p Darcy’s law p = p0 ργ state equation

    ε∂tρ = − div(ρv) = k µ

    div(ρ∇p) = kp0 µ

    div(ρ∇ργ)

    I ρ : density I p : pressure I v : velocity

    I ε, k , µ > 0 : material constants I γ ≥ 1 : polytropic exponent I p0 : reference pressure

    2 / 11

  • Physical motivation

    Flow of an ideal gas through a homogeneous porous medium can be described by

     ε∂tρ+ div(ρv) = 0 mass balance

    µv = −k ∇p Darcy’s law p = p0 ργ state equation

    ε∂tρ = − div(ρv) = k µ

    div(ρ∇p) = kp0 µ

    div(ρ∇ργ)

    I ρ : density I p : pressure I v : velocity

    I ε, k , µ > 0 : material constants I γ ≥ 1 : polytropic exponent I p0 : reference pressure

    2 / 11

  • Physical motivation

    Flow of an ideal gas through a homogeneous porous medium can be described by

     ε∂tρ+ div(ρv) = 0 mass balance

    µv = −k ∇p Darcy’s law p = p0 ργ state equation

    ε∂tρ = − div(ρv) = k µ

    div(ρ∇p) = kγp0 µ

    div(ργ∇ρ)

    I ρ : density I p : pressure I v : velocity

    I ε, k , µ > 0 : material constants I γ ≥ 1 : polytropic exponent I p0 : reference pressure

    2 / 11

  • Physical motivation

    Flow of an ideal gas through a homogeneous porous medium can be described by

     ε∂tρ+ div(ρv) = 0 mass balance

    µv = −k ∇p Darcy’s law p = p0 ργ state equation

    ε∂tρ = − div(ρv) = k µ

    div(ρ∇p) = kγp0 µ(γ + 1)

    ∆(ργ+1)

    I ρ : density I p : pressure I v : velocity

    I ε, k , µ > 0 : material constants I γ ≥ 1 : polytropic exponent I p0 : reference pressure

    2 / 11

  • Physical motivation

    Flow of an ideal gas through a homogeneous porous medium can be described by

    Definition The porous medium equation (PME) is

    ∂tu(t , x) = ∆xum(t , x), u ≥ 0, m > 1 (t , x) ∈ (0,∞)× Rn

    ε∂tρ = − div(ρv) = k µ

    div(ρ∇p) = kγp0 µ(γ + 1)

    ∆(ργ+1)

    I ρ : density I p : pressure I v : velocity

    I ε, k , µ > 0 : material constants I γ ≥ 1 : polytropic exponent I p0 : reference pressure

    2 / 11

  • Scaling properties Let

    I u a classical solution of the PME in (0,∞)× Rn

    I α, β > 0 constants with α(m − 1) + 2β = 1

    Define for λ > 0 I uλ(t , x) := λαu(λt , λβx) ⇒ ∂tuλ −∆umλ = 0

    Idea Scaling u → uλ maps solutions of the PME to other solutions. Find a scaling invariant solution, that is uλ = u for all λ > 0.

    Ansatz I u(t , x) = t−αu(1, t−βx) =: t−αv(t−βx), v : Rn → R

    Reduction to one space variable I 0 = −tα+1(∂tu −∆um) = αv() + βDv() · + ∆vm()

    3 / 11

  • Scaling properties Let

    I u a classical solution of the PME in (0,∞)× Rn

    I α, β > 0 constants with α(m − 1) + 2β = 1

    Define for λ > 0 I uλ(t , x) := λαu(λt , λβx) ⇒ ∂tuλ −∆umλ = 0

    Idea Scaling u → uλ maps solutions of the PME to other solutions. Find a scaling invariant solution, that is uλ = u for all λ > 0.

    Ansatz I u(t , x) = t−αu(1, t−βx) =: t−αv(t−βx), v : Rn → R

    Reduction to one space variable I 0 = −tα+1(∂tu−∆um) = αv(t−βx)+βDv(t−βx)·t−βx +∆vm(t−βx)

    3 / 11

  • Scaling properties Let

    I u a classical solution of the PME in (0,∞)× Rn

    I α, β > 0 constants with α(m − 1) + 2β = 1

    Define for λ > 0 I uλ(t , x) := λαu(λt , λβx) ⇒ ∂tuλ −∆umλ = 0

    Idea Scaling u → uλ maps solutions of the PME to other solutions. Find a scaling invariant solution, that is uλ = u for all λ > 0.

    Ansatz I u(t , x) = t−αu(1, t−βx) =: t−αv(t−βx), v : Rn → R

    Reduction to one space variable I 0 = −tα+1(∂tu−∆um) = αv(t−βx)+βDv(t−βx)·t−βx +∆vm(t−βx)

    3 / 11

  • Scaling properties Let

    I u a classical solution of the PME in (0,∞)× Rn

    I α, β > 0 constants with α(m − 1) + 2β = 1

    Define for λ > 0 I uλ(t , x) := λαu(λt , λβx) ⇒ ∂tuλ −∆umλ = 0

    Idea Scaling u → uλ maps solutions of the PME to other solutions. Find a scaling invariant solution, that is uλ = u for all λ > 0.

    Ansatz I u(t , x) = t−αu(1, t−βx) =: t−αv(t−βx), v : Rn → R

    Reduction to one space variable I 0 = −tα+1(∂tu −∆um) = αv(y) + βDv(y) · y + ∆vm(y)

    3 / 11

  • Derivation of a special solution Ansatz

    I v is radial, i.e. v(y) = w(|y |) := w(r), w : R→ R

    0 = rn−1 (αv(y) + βDv(y) · y + ∆vm(y))

    = ( αrn−1w + βrn∂r w

    ) + (

    rn−1∂2r w m + (n − 1)rn−2∂r wm

    )

    I w , ∂r w → 0 as r →∞ and w ≥ 0 seems appropriate

    ∂r wm−1 = − m − 1

    m βr ⇒ wm−1 = C − m − 1

    2m βr2, C > 0

    Solution

    u(t , x)

    4 / 11

  • Derivation of a special solution Ansatz

    I v is radial, i.e. v(y) = w(|y |) := w(r), w : R→ R

    0 = rn−1 (αv(y) + βDv(y) · y + ∆vm(y))

    = ( αrn−1w + βrn∂r w

    ) + (

    rn−1∂2r w m + (n − 1)rn−2∂r wm

    )

    I w , ∂r w → 0 as r →∞ and w ≥ 0 seems appropriate

    ∂r wm−1 = − m − 1

    m βr ⇒ wm−1 = C − m − 1

    2m βr2, C > 0

    Solution

    u(t , x)

    4 / 11

  • Derivation of a special solution Ansatz

    I v is radial, i.e. v(y) = w(|y |) := w(r), w : R→ R

    0 = rn−1 (αv(y) + βDv(y) · y + ∆vm(y))

    = (

    nβrn−1w + βrn∂r w )

    + (

    rn−1∂2r w m + (n − 1)rn−2∂r wm

    )

    I w , ∂r w → 0 as r →∞ and w ≥ 0 seems appropriate

    ∂r wm−1 = − m − 1

    m βr ⇒ wm−1 = C − m − 1

    2m βr2, C > 0

    Solution

    u(t , x)

    4 / 11

  • Derivation of a special solution Ansatz

    I v is radial, i.e. v(y) = w(|y |) := w(r), w : R→ R

    0 = rn−1 (αv(y) + βDv(y) · y + ∆vm(y))

    = (

    nβrn−1w + βrn∂r w )

    ︸ ︷︷ ︸ =β∂r (rnw)

    + (

    rn−1∂2r w m + (n − 1)rn−2∂r wm

    ) ︸ ︷︷ ︸

    =∂r (rn−1∂r wm)

    I w , ∂r w → 0 as r →∞ and w ≥ 0 seems appropriate

    ∂r wm−1 = − m − 1

    m βr ⇒ wm−1 = C − m − 1

    2m βr2, C > 0

    Solution

    u(t , x)

    4 / 11

  • Derivation of a special solution Ansatz

    I v is radial, i.e. v(y) = w(|y |) := w(r), w : R→ R

    0 = rn−1 (αv(y) + βDv(y) · y + ∆vm(y))

    = (

    nβrn−1w + βrn∂r w )

    ︸ ︷︷ ︸ =β∂r (rnw)

    + (

    rn−1∂2r w m + (n − 1)rn−2∂r wm

    ) ︸ ︷︷ ︸

    =∂r (rn−1∂r wm)

    = ∂r (βrnw + rn−1∂r wm) = ∂r (βrnw + rn−1 m

    m − 1 w∂r wm−1)

    I w , ∂r w → 0 as r →∞ and w ≥ 0 seems appropriate

    ∂r wm−1 = − m − 1

    m βr ⇒ wm−1 = C − m − 1

    2m βr2, C > 0

    Solution

    u(t , x)

    4 / 11

  • Derivation of a special solution Ansatz

    I v is radial, i.e. v(y) = w(|y |) := w(r), w : R→ R

    0 = rn−1 (αv(y) + βDv(y) · y + ∆vm(y))

    = (

    nβrn−1w + βrn∂r w )

    ︸ ︷︷ ︸ =β∂r (rnw)

    + (

    rn−1∂2r w m + (n − 1)rn−2∂r wm

    ) ︸ ︷︷ ︸

    =∂r (rn−1∂r wm)

    = ∂r (βrnw + rn−1∂r wm) = ∂r (βrnw + rn−1 m

    m − 1 w∂r wm−1)

    I w , ∂r w → 0 as r →∞ and w ≥ 0 seems appropriate

    ∂r wm−1 = − m − 1

    m βr ⇒ wm−1 = C − m − 1

    2m βr2, C > 0

    Solution

    u(t , x)

    4 / 11

  • Derivation of a special solution Ansatz

    I v is radial, i.e. v(y) = w(|y |) := w(r), w : R→ R

    0 = rn−1 (αv(y) + βDv(y) · y