Barenblatt\'s solution

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  mass balance  ε∂t ρ + div(ρv) = 0 µv = −k ∇p Darcy’s law  p = p0 ργ state equation

k p0 k div(ρ∇ργ ) ε∂t ρ = − div(ρv) = div(ρ∇p) = µ µ I

ρ : density

I

ε, k , µ > 0 : material constants

I

p : pressure

I

γ ≥ 1 : polytropic exponent

I

v : velocity

I

p0 : reference pressure 2 / 11

Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by  mass balance  ε∂t ρ + div(ρv) = 0 µv = −k ∇p Darcy’s law  p = p0 ργ state equation

k p0 k div(ρ∇ργ ) ε∂t ρ = − div(ρv) = div(ρ∇p) = µ µ I

ρ : density

I

ε, k , µ > 0 : material constants

I

p : pressure

I

γ ≥ 1 : polytropic exponent

I

v : velocity

I

p0 : reference pressure 2 / 11

Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by  mass balance  ε∂t ρ + div(ρv) = 0 µv = −k ∇p Darcy’s law  p = p0 ργ state equation

k p0 k div(ρ∇ργ ) ε∂t ρ = − div(ρv) = div(ρ∇p) = µ µ I

ρ : density

I

ε, k , µ > 0 : material constants

I

p : pressure

I

γ ≥ 1 : polytropic exponent

I

v : velocity

I

p0 : reference pressure 2 / 11

Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by  mass balance  ε∂t ρ + div(ρv) = 0 µv = −k ∇p Darcy’s law  p = p0 ργ state equation

k p0 k div(ρ∇ργ ) ε∂t ρ = − div(ρv) = div(ρ∇p) = µ µ I

ρ : density

I

ε, k , µ > 0 : material constants

I

p : pressure

I

γ ≥ 1 : polytropic exponent

I

v : velocity

I

p0 : reference pressure 2 / 11

Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by  mass balance  ε∂t ρ + div(ρv) = 0 µv = −k ∇p Darcy’s law  p = p0 ργ state equation

k γp0 k div(ργ ∇ρ) ε∂t ρ = − div(ρv) = div(ρ∇p) = µ µ I

ρ : density

I

ε, k , µ > 0 : material constants

I

p : pressure

I

γ ≥ 1 : polytropic exponent

I

v : velocity

I

p0 : reference pressure 2 / 11

Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by  mass balance  ε∂t ρ + div(ρv) = 0 µv = −k ∇p Darcy’s law  p = p0 ργ state equation

k γp0 k ∆(ργ+1 ) ε∂t ρ = − div(ρv) = div(ρ∇p) = µ µ(γ + 1) I

ρ : density

I

ε, k , µ > 0 : material constants

I

p : pressure

I

γ ≥ 1 : polytropic exponent

I

v : velocity

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

k k γp0 ε∂t ρ = − div(ρv) = div(ρ∇p) = ∆(ργ+1 ) µ µ(γ + 1) I

ρ : density

I

ε, k , µ > 0 : material constants

I

p : pressure

I

γ ≥ 1 : polytropic exponent

I

v : velocity

I

p0 : reference pressure 2 / 11

Scaling properties Let I I

u a classical solution of the PME in (0, ∞) × Rn α, β > 0 constants with α(m − 1) + 2β = 1

Define for λ > 0 m =0 I uλ (t, x) := λα u(λt, λβ x) ⇒ ∂t uλ − ∆uλ

3 / 11

Scaling properties Let I I

u a classical solution of the PME in (0, ∞) × Rn α, β > 0 constants with α(m − 1) + 2β = 1

Define for λ > 0 m =0 I uλ (t, x) := λα u(λt, λβ x) ⇒ ∂t uλ − ∆uλ

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 −α u(1, t −β x) =: t −α v (t −β x),

v : Rn → R

Reduction to one space variable I

0 = −t α+1 (∂t u−∆u m ) = αv (t −β x)+βDv (t −β x)·t −β x +∆v m (t −β x) 3 / 11

Scaling properties Let I I

u a classical solution of the PME in (0, ∞) × Rn α, β > 0 constants with α(m − 1) + 2β = 1

Define for λ > 0 m =0 I uλ (t, x) := λα u(λt, λβ x) ⇒ ∂t uλ − ∆uλ

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 −α u(1, t −β x) =: t −α v (t −β x),

v : Rn → R

Reduction to one space variable I

0 = −t α+1 (∂t u−∆u m ) = αv (t −β x)+βDv (t −β x)·t −β x +∆v m (t −β x) 3 / 11

Scaling properties Let I I

u a classical solution of the PME in (0, ∞) × Rn α, β > 0 constants with α(m − 1) + 2β = 1

Define for λ > 0 m =0 I uλ (t, x) := λα u(λt, λβ x) ⇒ ∂t uλ − ∆uλ

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 −α u(1, t −β x) =: t −α v (t −β x),

v : Rn → R

Reduction to one space variable I

0 = −t α+1 (∂t u − ∆u m ) = αv (y ) + βDv (y ) · y + ∆v m (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 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = αr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m

4 / 11

Derivation of a special solution Ansatz I

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

w :R→R

0 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = αr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m

4 / 11

Derivation of a special solution Ansatz I

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

w :R→R

0 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = nβr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m

4 / 11

Derivation of a special solution Ansatz I

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

w :R→R

0 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = nβr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m | {z } | {z } =β∂r (r n w)

=∂r (r n−1 ∂r w m )

4 / 11

Derivation of a special solution Ansatz I

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

w :R→R

0 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = nβr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m | {z } | {z } =β∂r (r n w)

=∂r (r n−1 ∂r w m )

= ∂r (βr n w + r n−1 ∂r w m ) = ∂r (βr n w + r n−1

m w∂r w m−1 ) m−1

4 / 11

Derivation of a special solution Ansatz I

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

w :R→R

0 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = nβr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m | {z } | {z } =β∂r (r n w)

=∂r (r n−1 ∂r w m )

= ∂r (βr n w + r n−1 ∂r w m ) = ∂r (βr n w + r n−1 I

m w∂r w m−1 ) m−1

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

m−1

m−1 2 m−1 m−1 βr ⇒ w =C− βr , C > 0 =− m 2m

4 / 11

Derivation of a special solution Ansatz I

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

w :R→R

0 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = nβr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m | {z } | {z } =β∂r (r n w)

=∂r (r n−1 ∂r w m )

= ∂r (βr n w + r n−1 ∂r w m ) = ∂r (βr n w + r n−1 I

m w∂r w m−1 ) m−1

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

m−1

m−1 2 m−1 m−1 βr ⇒ w =C− βr , C > 0 =− m 2m

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 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = nβr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m | {z } | {z } =β∂r (r n w)

=∂r (r n−1 ∂r w m )

= ∂r (βr n w + r n−1 ∂r w m ) = ∂r (βr n w + r n−1 I

m w∂r w m−1 ) m−1

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

m−1

m−1 2 m−1 m−1 βr ⇒ w =C− βr , C > 0 =− m 2m

Solution u(t, x) = t −α v (t −β 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 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = nβr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m | {z } | {z } =β∂r (r n w)

=∂r (r n−1 ∂r w m )

= ∂r (βr n w + r n−1 ∂r w m ) = ∂r (βr n w + r n−1 I

m w∂r w m−1 ) m−1

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

m−1

m−1 2 m−1 m−1 βr ⇒ w =C− βr , C > 0 =− m 2m

Solution u(t, x) = t −α w(|t −β 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 = r n−1 (αv (y ) + βDv (y ) · y + ∆v m (y ))     = nβr n−1 w + βr n ∂r w + r n−1 ∂r2 w m + (n − 1)r n−2 ∂r w m | {z } | {z } =β∂r (r n w)

n

= ∂r (βr w + r I

n−1

=∂r (r n−1 ∂r w m )

m

n

∂r w ) = ∂r (βr w + r

n−1

m w∂r w m−1 ) m−1

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

m−1

m−1 2 m−1 m−1 βr ⇒ w =C− βr , C > 0 =− m 2m

Solution u(t, x) = t −α

! 1   + m−1 β(m − 1) |x|2 C− 2m t 2β 4 / 11

Barenblatt’s solution Definition Let α =

n n(m−1)+2 ,

β = αn , C > 0. Barenblatt’s solution to the PME is

Um (t, x; C) := t −α



! 1  + m−1 β(m − 1) |x|2 C− 2m t 2β

It is also known as ZKB solution in literature.

Remarks I

Um is a smooth solution where Um > 0

I

Finite propagation speed

I

Non-smoothness on |x| = t β

I

Scaling invariance

I

Which role does C play?



C 2m β (m−1)

1 2

=: r (t) for m ≥ 2

5 / 11

Elimination of the free parameter Lemma (Mass conservation) Fix C > 0. As a map (0, ∞) → L1 (Rn ), Um is mass preserving, i.e. M := kUm (t, · ; C)kL1 (Rn ) is independent of t and is called mass of Um .

Proof t1 t2

I

Let t1 , t2 > 0 and λ =

I

Scaling invariance: Um (t1 , x; C) = λα Um (t2 , λβ x; C) kUm (t1 , · ; C)k1 = λα λ−nβ kUm (t2 , · ; C)k1 = kUm (t2 , · ; C)k1

I

6 / 11

Elimination of the free parameter Lemma (Mass conservation) Fix C > 0. As a map (0, ∞) → L1 (Rn ), Um is mass preserving, i.e. M := kUm (t, · ; C)kL1 (Rn ) is independent of t and is called mass of Um .

Proof t1 t2

I

Let t1 , t2 > 0 and λ =

I

Scaling invariance: Um (t1 , x; C) = λα Um (t2 , λβ x; C) kUm (t1 , · ; C)k1 = λα λ−nβ kUm (t2 , · ; C)k1 = kUm (t2 , · ; C)k1

I

Definition (Mass as parameter) Let γ =

1 m−1

+ n2 . The mass M and the free parameter C are related by

M = a(m, n) · C γ Write Um (t, x; M) for Barenblatt’s solution with mass M. 6 / 11

Elimination of the free parameter Lemma (Mass conservation) Fix C > 0. As a map (0, ∞) → L1 (Rn ), Um is mass preserving, i.e. M := kUm (t, · ; C)kL1 (Rn ) is independent of t and is called mass of Um .

Proof t1 t2

I

Let t1 , t2 > 0 and λ =

I

Scaling invariance: Um (t1 , x; C) = λα Um (t2 , λβ x; C) kUm (t1 , · ; C)k1 = λα λ−nβ kUm (t2 , · ; C)k1 = kUm (t2 , · ; C)k1

I

Definition (Mass as parameter) Let γ =

1 m−1

+ n2 . The mass M and the free parameter C are related by

M = a(m, n) ·

Cγ

n 2

=π ·


n mα−m − 2 2mn

·

m Γ( m−1 ) m Γ( m−1 + n2 )

·C γ

Write Um (t, x; M) for Barenblatt’s solution with mass M. 6 / 11

Comparison to the heat equation Note I

For m = 1 the PME becomes the heat equation ∂t u − ∆u = 0

I

(HE)

Fundamental solution for the HE given by the Gaussian kernel |x|2 − n2 G(t, x) = (4πt) exp(− 4t )

x

x

7 / 11

Comparison to the heat equation Note I

For m = 1 the PME becomes the heat equation ∂t u − ∆u = 0

I

(HE)

Fundamental solution for the HE given by the Gaussian kernel |x|2 − n2 G(t, x) = (4πt) exp(− 4t )

x

7 / 11

Comparison to the heat equation Note I

For m = 1 the PME becomes the heat equation ∂t u − ∆u = 0

I

(HE)

Fundamental solution for the HE given by the Gaussian kernel |x|2 − n2 G(t, x) = (4πt) exp(− 4t )

x

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

Asymptotics of Barenblatt’s solution Theorem Let Um (t, x; M) Barenblatt’s solution with mass M. We have the limits lim Um (t, · M) = Mδ0

in the sense of distributions

t→0

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

m→1

Proof 

C 2m β (m−1)

1
2

I

supp Um (t, · ; M) ⊆ B 0, t β

I

By mass preservation: lim Um (t, · M) = Mδ0 t→0

8 / 11

Asymptotics of Barenblatt’s solution Theorem Let Um (t, x; M) Barenblatt’s solution with mass M. We have the limits lim Um (t, · M) = Mδ0

in the sense of distributions

t→0

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

m→1

Proof 

C 2m β (m−1)

1
2

I

supp Um (t, · ; M) ⊆ B 0, t β

I

By mass preservation: lim Um (t, · M) = Mδ0

I

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

t→0

8 / 11

The Cauchy Dirichlet problem (CDP) Let I

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   ∂t u − ∆(Φ(u)) = f u(0, x) = u0 (x) (CDP)  u(t, x) = 0

in QT in Ω on [0, T ) × ∂Ω

9 / 11

The Cauchy Dirichlet problem (CDP) Let I

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   ∂t u − ∆(Φ(u)) = f u(0, x) = u0 (x) (CDP)  u(t, x) = 0 I

in QT in Ω on [0, T ) × ∂Ω

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

9 / 11

Weak solutions for the CDP Definition A weak solution of CDP in QT is a function u ∈ L1 (QT ) s.t. 1

2

w := Φ(u) ∈ L1 (0, T ; W01,1 (Ω)) RR RR R (∇w · ∇η − u∂t η) dx dt = u0 (x)η(0, x) dx + f η dx dt QT

Ω

QT

holds for any η ∈ C 1 (QT ) which vanishes on [0, T ) × ∂Ω and for t = T

Remarks I I

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

10 / 11

Weak solutions for the CDP Definition A weak solution of CDP in QT is a function u ∈ L1 (QT ) s.t. 1

2

w := Φ(u) ∈ L1 (0, T ; W01,1 (Ω)) RR RR R (∇w · ∇η − u∂t η) dx dt = u0 (x)η(0, x) dx + f η dx dt QT

Ω

QT

holds for any η ∈ C 1 (QT ) which vanishes on [0, T ) × ∂Ω and for t = T

Remarks I I

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

10 / 11

Weak solutions for the CDP Definition A weak solution of CDP in QT is a function u ∈ L1 (QT ) s.t. 1

2

w := Φ(u) ∈ L1 (0, T ; W01,1 (Ω)) RR RR R (∇w · ∇η − u∂t η) dx dt = u0 (x)η(0, x) dx + f η dx dt QT

Ω

QT

holds for any η ∈ C 1 (QT ) which vanishes on [0, T ) × ∂Ω and for t = T

Remarks I I

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

10 / 11

Weak solutions for the CDP Definition A weak solution of CDP in QT is a function u ∈ L1 (QT ) s.t. 1

2

w := Φ(u) ∈ L1 (0, T ; W01,1 (Ω)) RR RR R (∇w · ∇η − u∂t η) dx dt = u0 (x)η(0, x) dx + f η dx dt Ω

QT

QT

holds for any η ∈ C 1 (QT ) which vanishes on [0, T ) × ∂Ω and for t = T

Remarks I I I

Integration by parts shows: smooth solutions are weak solutions What about initial data...? satisfied in the sense that for any ϕ ∈ C 1 (Ω) with ϕ = 0 on ∂Ω Z Z lim u(t)ϕ dx = u0 ϕ dx t→0

Ω

Ω 10 / 11

A well-known weak solution Modify Barenblatt’s solution I

Take x0 ∈ Ω, τ > 0

I

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

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

Proof I

v has the stated regularity

I

Let P := {(t, x) ∈ QT | v (t, x) > 0}

I

v is smooth solution within P and v m is C 1 up to |x| = r (t)

I

Integration by parts yields the integral equality (2) 11 / 11