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• 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