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 mediumcan be described by

ε∂tρ+ div(ρv) = 0 mass balance

µv = −k ∇p Darcy’s lawp = p0 ρ

γ state equation

ε∂tρ = − div(ρv) =kµ

div(ρ∇p) =kp0

µdiv(ρ∇ργ)

I ρ : densityI p : pressureI v : velocity

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

2 / 11

Physical motivation

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

ε∂tρ+ div(ρv) = 0 mass balance

µv = −k ∇p Darcy’s lawp = p0 ρ

γ state equation

ε∂tρ = −div(ρv) =kµ

div(ρ∇p) =kp0

µdiv(ρ∇ργ)

I ρ : densityI p : pressureI v : velocity

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

2 / 11

Physical motivation

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

ε∂tρ+ div(ρv) = 0 mass balance

µv = −k ∇p Darcy’s lawp = p0 ρ

γ state equation

ε∂tρ = − div(ρv) =kµ

div(ρ∇p) =kp0

µdiv(ρ∇ργ)

I ρ : densityI p : pressureI v : velocity

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

2 / 11

Physical motivation

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

ε∂tρ+ div(ρv) = 0 mass balance

µv = −k ∇p Darcy’s lawp = p0 ρ

γ state equation

ε∂tρ = − div(ρv) =kµ

div(ρ∇p) =kp0

µdiv(ρ∇ργ)

I ρ : densityI p : pressureI v : velocity

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

2 / 11

Physical motivation

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

ε∂tρ+ div(ρv) = 0 mass balance

µv = −k ∇p Darcy’s lawp = p0 ρ

γ state equation

ε∂tρ = − div(ρv) =kµ

div(ρ∇p) =kγp0

µdiv(ργ∇ρ)

I ρ : densityI p : pressureI v : velocity

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

2 / 11

Physical motivation

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

ε∂tρ+ div(ρv) = 0 mass balance

µv = −k ∇p Darcy’s lawp = p0 ρ

γ state equation

ε∂tρ = − div(ρv) =kµ

div(ρ∇p) =kγp0

µ(γ + 1)∆(ργ+1)

I ρ : densityI p : pressureI v : velocity

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

2 / 11

Physical motivation

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

DefinitionThe porous medium equation (PME) is

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

ε∂tρ = − div(ρv) =kµ

div(ρ∇p) =kγp0

µ(γ + 1)∆(ργ+1)

I ρ : densityI p : pressureI v : velocity

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

2 / 11

Scaling propertiesLet

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 > 0I uλ(t , x) := λαu(λt , λβx) ⇒ ∂tuλ −∆um

λ = 0

IdeaScaling u → uλ maps solutions of the PME to other solutions. Find ascaling invariant solution, that is uλ = u for all λ > 0.

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

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

3 / 11

Scaling propertiesLet

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 > 0I uλ(t , x) := λαu(λt , λβx) ⇒ ∂tuλ −∆um

λ = 0

IdeaScaling u → uλ maps solutions of the PME to other solutions. Find ascaling invariant solution, that is uλ = u for all λ > 0.

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

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

3 / 11

Scaling propertiesLet

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 > 0I uλ(t , x) := λαu(λt , λβx) ⇒ ∂tuλ −∆um

λ = 0

IdeaScaling u → uλ maps solutions of the PME to other solutions. Find ascaling invariant solution, that is uλ = u for all λ > 0.

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

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

3 / 11

Scaling propertiesLet

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 > 0I uλ(t , x) := λαu(λt , λβx) ⇒ ∂tuλ −∆um

λ = 0

IdeaScaling u → uλ maps solutions of the PME to other solutions. Find ascaling invariant solution, that is uλ = u for all λ > 0.

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

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

3 / 11

Derivation of a special solutionAnsatz

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 wm + (n − 1)rn−2∂r wm

)

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

∂r wm−1 = −m − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x)

4 / 11

Derivation of a special solutionAnsatz

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 wm + (n − 1)rn−2∂r wm

)

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

∂r wm−1 = −m − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x)

4 / 11

Derivation of a special solutionAnsatz

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 wm + (n − 1)rn−2∂r wm

)

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

∂r wm−1 = −m − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x)

4 / 11

Derivation of a special solutionAnsatz

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 wm + (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 − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x)

4 / 11

Derivation of a special solutionAnsatz

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 wm + (n − 1)rn−2∂r wm

)︸ ︷︷ ︸

=∂r (rn−1∂r wm)

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

w∂r wm−1)

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

∂r wm−1 = −m − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x)

4 / 11

Derivation of a special solutionAnsatz

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 wm + (n − 1)rn−2∂r wm

)︸ ︷︷ ︸

=∂r (rn−1∂r wm)

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

w∂r wm−1)

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

∂r wm−1 = −m − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x)

4 / 11

Derivation of a special solutionAnsatz

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 wm + (n − 1)rn−2∂r wm

)︸ ︷︷ ︸

=∂r (rn−1∂r wm)

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

w∂r wm−1)

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

∂r wm−1 = −m − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x)

4 / 11

Derivation of a special solutionAnsatz

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 wm + (n − 1)rn−2∂r wm

)︸ ︷︷ ︸

=∂r (rn−1∂r wm)

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

w∂r wm−1)

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

∂r wm−1 = −m − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x) = t−αv(t−βx)

4 / 11

Derivation of a special solutionAnsatz

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 wm + (n − 1)rn−2∂r wm

)︸ ︷︷ ︸

=∂r (rn−1∂r wm)

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

w∂r wm−1)

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

∂r wm−1 = −m − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x) = t−αw(|t−βx |)

4 / 11

Derivation of a special solutionAnsatz

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 wm + (n − 1)rn−2∂r wm

)︸ ︷︷ ︸

=∂r (rn−1∂r wm)

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

w∂r wm−1)

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

∂r wm−1 = −m − 1m

βr ⇒ wm−1 = C − m − 12m

βr2, C > 0

Solution

u(t , x) = t−α((

C − β(m − 1)

2m|x |2

t2β

)+) 1

m−1

4 / 11

Barenblatt’s solution

DefinitionLet α = n

n(m−1)+2 , β = αn , C > 0. Barenblatt’s solution to the PME is

Um(t , x ; C) := t−α((

C − β(m − 1)

2m|x |2

t2β

)+) 1

m−1

It is also known as ZKB solution in literature.

RemarksI Um is a smooth solution where Um > 0I Finite propagation speed

I Non-smoothness on |x | = tβ(

Cβ

2m(m−1)

) 12

=: r(t) for m ≥ 2

I Scaling invarianceI Which role does C play?

5 / 11

Elimination of the free parameterLemma (Mass conservation)

Fix C > 0. As a map (0,∞)→ L1(Rn), Um is mass preserving, i.e.M := ‖Um(t , · ; C)‖L1(Rn) is independent of t and is called mass of Um.

ProofI Let t1, t2 > 0 and λ = t1

t2

I Scaling invariance: Um(t1, x ; C) = λαUm(t2, λβx ; C)

I ‖Um(t1, · ; C)‖1 = λαλ−nβ ‖Um(t2, · ; C)‖1 = ‖Um(t2, · ; C)‖1

Definition (Mass as parameter)

Let γ = 1m−1 + n

2 . The mass M and the free parameter C are related by

M = a(m,n) · Cγ

= πn2 ·(mα−m

2mn

)− n2 · Γ( m

m−1 )

Γ( mm−1 + n

2 ) ·Cγ

Write Um(t , x ; M) for Barenblatt’s solution with mass M.

6 / 11

Elimination of the free parameterLemma (Mass conservation)

Fix C > 0. As a map (0,∞)→ L1(Rn), Um is mass preserving, i.e.M := ‖Um(t , · ; C)‖L1(Rn) is independent of t and is called mass of Um.

ProofI Let t1, t2 > 0 and λ = t1

t2

I Scaling invariance: Um(t1, x ; C) = λαUm(t2, λβx ; C)

I ‖Um(t1, · ; C)‖1 = λαλ−nβ ‖Um(t2, · ; C)‖1 = ‖Um(t2, · ; C)‖1

Definition (Mass as parameter)

Let γ = 1m−1 + n

2 . The mass M and the free parameter C are related by

M = a(m,n) · Cγ

= πn2 ·(mα−m

2mn

)− n2 · Γ( m

m−1 )

Γ( mm−1 + n

2 ) ·Cγ

Write Um(t , x ; M) for Barenblatt’s solution with mass M.

6 / 11

Elimination of the free parameterLemma (Mass conservation)

Fix C > 0. As a map (0,∞)→ L1(Rn), Um is mass preserving, i.e.M := ‖Um(t , · ; C)‖L1(Rn) is independent of t and is called mass of Um.

ProofI Let t1, t2 > 0 and λ = t1

t2

I Scaling invariance: Um(t1, x ; C) = λαUm(t2, λβx ; C)

I ‖Um(t1, · ; C)‖1 = λαλ−nβ ‖Um(t2, · ; C)‖1 = ‖Um(t2, · ; C)‖1

Definition (Mass as parameter)

Let γ = 1m−1 + n

2 . The mass M and the free parameter C are related by

M = a(m,n) · Cγ = πn2 ·(mα−m

2mn

)− n2 · Γ( m

m−1 )

Γ( mm−1 + n

2 ) ·Cγ

Write Um(t , x ; M) for Barenblatt’s solution with mass M.

6 / 11

Comparison to the heat equationNote

I For m = 1 the PME becomes the heat equation

∂tu −∆u = 0 (HE)

I Fundamental solution for the HE given by the Gaussian kernelG(t , x) = (4πt)−

n2 exp(− |x |

2

4t )

x x

What happens to Barenblatt’s solution in the limit m→ 1?

7 / 11

Comparison to the heat equationNote

I For m = 1 the PME becomes the heat equation

∂tu −∆u = 0 (HE)

I Fundamental solution for the HE given by the Gaussian kernelG(t , x) = (4πt)−

n2 exp(− |x |

2

4t )

x

What happens to Barenblatt’s solution in the limit m→ 1?

7 / 11

Comparison to the heat equationNote

I For m = 1 the PME becomes the heat equation

∂tu −∆u = 0 (HE)

I Fundamental solution for the HE given by the Gaussian kernelG(t , x) = (4πt)−

n2 exp(− |x |

2

4t )

x

What happens to Barenblatt’s solution in the limit m→ 1?7 / 11

Asymptotics of Barenblatt’s solution

TheoremLet Um(t , x ; M) Barenblatt’s solution with mass M. We have the limits

limt→0

Um(t , · M) = Mδ0 in the sense of distributions

limm→1

Um(t , x ; M) = MG(t , x) pointwise on (0,∞)× Rn

Proof

I supp Um(t , · ; M) ⊆ B(0, tβ

(Cβ

2m(m−1)

) 12 )

I By mass preservation: limt→0

Um(t , · M) = Mδ0

I Limit for m→ 1 is a truly marvelous calculation but this margin isto narrow to contain it

8 / 11

Asymptotics of Barenblatt’s solution

TheoremLet Um(t , x ; M) Barenblatt’s solution with mass M. We have the limits

limt→0

Um(t , · M) = Mδ0 in the sense of distributions

limm→1

Um(t , x ; M) = MG(t , x) pointwise on (0,∞)× Rn

Proof

I supp Um(t , · ; M) ⊆ B(0, tβ

(Cβ

2m(m−1)

) 12 )

I By mass preservation: limt→0

Um(t , · M) = Mδ0

I Limit for m→ 1 is a truly marvelous calculation but this margin isto narrow to contain it

8 / 11

The Cauchy Dirichlet problem (CDP)

LetI Ω ⊆ Rn bounded with ∂Ω smooth, T ∈ (0,∞]

I Q := R+ × Ω, QT := (0,T )× Ω

I u0 ∈ L1(Ω), f ∈ L1(Q)

I Φ ∈ C(R) strictly increasing with Φ(±∞) = ±∞, Φ(0) = 0

Consider

(CDP)

∂tu −∆(Φ(u)) = f in QT

u(0, x) = u0(x) in Ωu(t , x) = 0 on [0,T )× ∂Ω

I Choose Φ(u) = |u|m−1u and f = 0 for the PME

9 / 11

The Cauchy Dirichlet problem (CDP)

LetI Ω ⊆ Rn bounded with ∂Ω smooth, T ∈ (0,∞]

I Q := R+ × Ω, QT := (0,T )× Ω

I u0 ∈ L1(Ω), f ∈ L1(Q)

I Φ ∈ C(R) strictly increasing with Φ(±∞) = ±∞, Φ(0) = 0

Consider

(CDP)

∂tu −∆(Φ(u)) = f in QT

u(0, x) = u0(x) in Ωu(t , x) = 0 on [0,T )× ∂Ω

I Choose Φ(u) = |u|m−1u and f = 0 for the PME

9 / 11

Weak solutions for the CDPDefinitionA weak solution of CDP in QT is a function u ∈ L1(QT ) s.t.

1 w := Φ(u) ∈ L1(0,T ; W 1,10 (Ω))

2∫∫QT

(∇w · ∇η − u∂tη) dx dt =∫Ω

u0(x)η(0, x) dx +∫∫QT

fη dx dt

holds for any η ∈ C1(QT ) which vanishes on [0,T )× ∂Ω and for t = T

RemarksI Integration by parts shows: smooth solutions are weak solutionsI What about initial data...?

I satisfied in the sense that for any ϕ ∈ C1(Ω) with ϕ = 0 on ∂Ω

limt→0

∫Ω

u(t)ϕdx =

∫Ω

u0ϕdx

10 / 11

Weak solutions for the CDPDefinitionA weak solution of CDP in QT is a function u ∈ L1(QT ) s.t.

1 w := Φ(u) ∈ L1(0,T ; W 1,10 (Ω))

2∫∫QT

(∇w · ∇η − u∂tη) dx dt =∫Ω

u0(x)η(0, x) dx +∫∫QT

fη dx dt

holds for any η ∈ C1(QT ) which vanishes on [0,T )× ∂Ω and for t = T

RemarksI Integration by parts shows: smooth solutions are weak solutionsI What about initial data...?

I satisfied in the sense that for any ϕ ∈ C1(Ω) with ϕ = 0 on ∂Ω

limt→0

∫Ω

u(t)ϕdx =

∫Ω

u0ϕdx

10 / 11

Weak solutions for the CDPDefinitionA weak solution of CDP in QT is a function u ∈ L1(QT ) s.t.

1 w := Φ(u) ∈ L1(0,T ; W 1,10 (Ω))

2∫∫QT

(∇w · ∇η − u∂tη) dx dt =∫Ω

u0(x)η(0, x) dx +∫∫QT

fη dx dt

holds for any η ∈ C1(QT ) which vanishes on [0,T )× ∂Ω and for t = T

RemarksI Integration by parts shows: smooth solutions are weak solutionsI What about initial data...?

I satisfied in the sense that for any ϕ ∈ C1(Ω) with ϕ = 0 on ∂Ω

limt→0

∫Ω

u(t)ϕdx =

∫Ω

u0ϕdx

10 / 11

Weak solutions for the CDPDefinitionA weak solution of CDP in QT is a function u ∈ L1(QT ) s.t.

1 w := Φ(u) ∈ L1(0,T ; W 1,10 (Ω))

2∫∫QT

(∇w · ∇η − u∂tη) dx dt =∫Ω

u0(x)η(0, x) dx +∫∫QT

fη dx dt

holds for any η ∈ C1(QT ) which vanishes on [0,T )× ∂Ω and for t = T

RemarksI Integration by parts shows: smooth solutions are weak solutionsI What about initial data...?I satisfied in the sense that for any ϕ ∈ C1(Ω) with ϕ = 0 on ∂Ω

limt→0

∫Ω

u(t)ϕdx =

∫Ω

u0ϕdx

10 / 11

A well-known weak solution

Modify Barenblatt’s solutionI Take x0 ∈ Ω, τ > 0I Set v(t , x) := Um(t + τ, x − x0; M)

I Let T > 0 be small enough so that v = 0 on [0,T )× ∂Ω

TheoremDefine v(t , x) as above. Then v is a weak solution of the CDP for thePME in QT . If m ≥ 2, then v is not a classical solution of that problem.

ProofI v has the stated regularityI Let P := (t , x) ∈ QT | v(t , x) > 0I v is smooth solution within P and vm is C1 up to |x | = r(t)I Integration by parts yields the integral equality (2)

11 / 11