The Porous Medium Equation

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1 The Porous Medium Equation Moritz Egert, Samuel Littig, Matthijs Pronk, Linwen Tan Coordinator: Jürgen Voigt 18 June / 11

2 Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by ε t ρ + div(ρv) = 0 mass balance µv = k p Darcy s law p = p 0 ρ γ state equation ε t ρ = div(ρv) = k µ div(ρ p) = kp 0 µ div(ρ ργ ) ρ : density p : pressure v : velocity ε, k, µ > 0 : material constants γ 1 : polytropic exponent p 0 : reference pressure 2 / 11

3 Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by ε t ρ + div(ρv) = 0 mass balance µv = k p Darcy s law p = p 0 ρ γ state equation ε t ρ = div(ρv) = k µ div(ρ p) = kp 0 µ div(ρ ργ ) ρ : density p : pressure v : velocity ε, k, µ > 0 : material constants γ 1 : polytropic exponent p 0 : reference pressure 2 / 11

4 Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by ε t ρ + div(ρv) = 0 mass balance µv = k p Darcy s law p = p 0 ρ γ state equation ε t ρ = div(ρv) = k µ div(ρ p) = kp 0 µ div(ρ ργ ) ρ : density p : pressure v : velocity ε, k, µ > 0 : material constants γ 1 : polytropic exponent p 0 : reference pressure 2 / 11

5 Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by ε t ρ + div(ρv) = 0 mass balance µv = k p Darcy s law p = p 0 ρ γ state equation ε t ρ = div(ρv) = k µ div(ρ p) = kp 0 µ div(ρ ργ ) ρ : density p : pressure v : velocity ε, k, µ > 0 : material constants γ 1 : polytropic exponent p 0 : reference pressure 2 / 11

6 Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by ε t ρ + div(ρv) = 0 mass balance µv = k p Darcy s law p = p 0 ρ γ state equation ε t ρ = div(ρv) = k µ div(ρ p) = kγp 0 µ div(ργ ρ) ρ : density p : pressure v : velocity ε, k, µ > 0 : material constants γ 1 : polytropic exponent p 0 : reference pressure 2 / 11

7 Physical motivation Flow of an ideal gas through a homogeneous porous medium can be described by ε t ρ + div(ρv) = 0 mass balance µv = k p Darcy s law p = p 0 ρ γ state equation ε t ρ = div(ρv) = k µ div(ρ p) = kγp 0 µ(γ + 1) (ργ+1 ) ρ : density p : pressure v : velocity ε, k, µ > 0 : material constants γ 1 : polytropic exponent p 0 : reference pressure 2 / 11

8 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, ) R n ε t ρ = div(ρv) = k µ div(ρ p) = kγp 0 µ(γ + 1) (ργ+1 ) ρ : density p : pressure v : velocity ε, k, µ > 0 : material constants γ 1 : polytropic exponent p 0 : reference pressure 2 / 11

9 Scaling properties Let u a classical solution of the PME in (0, ) R n α, β > 0 constants with α(m 1) + 2β = 1 Define for λ > 0 u λ (t, x) := λ α u(λt, λ β x) t u λ uλ m = 0 3 / 11

10 Scaling properties Let u a classical solution of the PME in (0, ) R n α, β > 0 constants with α(m 1) + 2β = 1 Define for λ > 0 u λ (t, x) := λ α u(λt, λ β x) t u λ uλ m = 0 Idea Scaling u u λ maps solutions of the PME to other solutions. Find a scaling invariant solution, that is u λ = u for all λ > 0. Ansatz u(t, x) = t α u(1, t β x) =: t α v(t β x), v : R n R Reduction to one space variable 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

11 Scaling properties Let u a classical solution of the PME in (0, ) R n α, β > 0 constants with α(m 1) + 2β = 1 Define for λ > 0 u λ (t, x) := λ α u(λt, λ β x) t u λ uλ m = 0 Idea Scaling u u λ maps solutions of the PME to other solutions. Find a scaling invariant solution, that is u λ = u for all λ > 0. Ansatz u(t, x) = t α u(1, t β x) =: t α v(t β x), v : R n R Reduction to one space variable 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

12 Scaling properties Let u a classical solution of the PME in (0, ) R n α, β > 0 constants with α(m 1) + 2β = 1 Define for λ > 0 u λ (t, x) := λ α u(λt, λ β x) t u λ uλ m = 0 Idea Scaling u u λ maps solutions of the PME to other solutions. Find a scaling invariant solution, that is u λ = u for all λ > 0. Ansatz u(t, x) = t α u(1, t β x) =: t α v(t β x), v : R n R Reduction to one space variable 0 = t α+1 ( t u u m ) = αv(y) + βdv(y) y + v m (y) 3 / 11

13 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) 4 / 11

14 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) 4 / 11

15 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) 4 / 11

16 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) }{{}}{{} =β r (r n w) = r (r n 1 r w m ) 4 / 11

17 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) }{{}}{{} =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 m 1 w r w m 1 ) 4 / 11

18 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) }{{}}{{} =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 m 1 w r w m 1 ) w, r w 0 as r and w 0 seems appropriate r w m 1 = m 1 m βr w m 1 = C m 1 2m βr 2, C > 0 4 / 11

19 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) }{{}}{{} =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 m 1 w r w m 1 ) w, r w 0 as r and w 0 seems appropriate r w m 1 = m 1 m βr w m 1 = C m 1 2m βr 2, C > 0 Solution u(t, x) 4 / 11

20 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) }{{}}{{} =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 m 1 w r w m 1 ) w, r w 0 as r and w 0 seems appropriate r w m 1 = m 1 m βr w m 1 = C m 1 2m βr 2, C > 0 Solution u(t, x) = t α v(t β x) 4 / 11

21 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) }{{}}{{} =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 m 1 w r w m 1 ) w, r w 0 as r and w 0 seems appropriate r w m 1 = m 1 m βr w m 1 = C m 1 2m βr 2, C > 0 Solution u(t, x) = t α w( t β x ) 4 / 11

22 Derivation of a special solution Ansatz 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 r 2 w m + (n 1)r n 2 r w m) }{{}}{{} =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 m 1 w r w m 1 ) w, r w 0 as r and w 0 seems appropriate r w m 1 = m 1 m βr w m 1 = C m 1 2m βr 2, C > 0 Solution u(t, x) = t α ( ( C β(m 1) 2m x 2 t 2β ) + ) 1 m 1 4 / 11

23 Barenblatt s solution Definition Let α = n n(m 1)+2, β = α n, C > 0. Barenblatt s solution to the PME is U m (t, x; C) := t α ( ( C β(m 1) 2m x 2 t 2β ) + ) 1 m 1 It is also known as ZKB solution in literature. Remarks U m is a smooth solution where U m > 0 Finite propagation speed Non-smoothness on x = t β ( C β Scaling invariance Which role does C play? ) 1 2m 2 (m 1) =: r(t) for m 2 5 / 11

24 Elimination of the free parameter Lemma (Mass conservation) Fix C > 0. As a map (0, ) L 1 (R n ), U m is mass preserving, i.e. M := U m (t, ; C) L 1 (R n ) is independent of t and is called mass of U m. Proof Let t 1, t 2 > 0 and λ = t 1 t2 Scaling invariance: U m (t 1, x; C) = λ α U m (t 2, λ β x; C) U m (t 1, ; C) 1 = λ α λ nβ U m (t 2, ; C) 1 = U m (t 2, ; C) 1 6 / 11

25 Elimination of the free parameter Lemma (Mass conservation) Fix C > 0. As a map (0, ) L 1 (R n ), U m is mass preserving, i.e. M := U m (t, ; C) L 1 (R n ) is independent of t and is called mass of U m. Proof Let t 1, t 2 > 0 and λ = t 1 t2 Scaling invariance: U m (t 1, x; C) = λ α U m (t 2, λ β x; C) U m (t 1, ; C) 1 = λ α λ nβ U m (t 2, ; C) 1 = U m (t 2, ; C) 1 Definition (Mass as parameter) Let γ = 1 m 1 + n 2. The mass M and the free parameter C are related by M = a(m, n) C γ Write U m (t, x; M) for Barenblatt s solution with mass M. 6 / 11

26 Elimination of the free parameter Lemma (Mass conservation) Fix C > 0. As a map (0, ) L 1 (R n ), U m is mass preserving, i.e. M := U m (t, ; C) L 1 (R n ) is independent of t and is called mass of U m. Proof Let t 1, t 2 > 0 and λ = t 1 t2 Scaling invariance: U m (t 1, x; C) = λ α U m (t 2, λ β x; C) U m (t 1, ; C) 1 = λ α λ nβ U m (t 2, ; C) 1 = U m (t 2, ; C) 1 Definition (Mass as parameter) Let γ = 1 m 1 + n 2. The mass M and the free parameter C are related by M = a(m, n) C γ = π n 2 ( mα m 2mn ) n 2 Γ( m m 1 ) Γ( m m 1 + n 2 ) Cγ Write U m (t, x; M) for Barenblatt s solution with mass M. 6 / 11

27 Comparison to the heat equation Note For m = 1 the PME becomes the heat equation t u u = 0 (HE) Fundamental solution for the HE given by the Gaussian kernel G(t, x) = (4πt) n 2 exp( x 2 4t ) x x 7 / 11

28 Comparison to the heat equation Note For m = 1 the PME becomes the heat equation t u u = 0 (HE) Fundamental solution for the HE given by the Gaussian kernel G(t, x) = (4πt) n 2 exp( x 2 4t ) x 7 / 11

HE given by the Gaussian kernel G(t, x) = (4πt) n 2 exp( x 2 4t

29 Comparison to the heat equation Note For m = 1 the PME becomes the heat equation t u u = 0 (HE) Fundamental solution for the HE given by the Gaussian kernel G(t, x) = (4πt) n 2 exp( x 2 4t ) x What happens to Barenblatt s solution in the limit m 1? 7 / 11

30 Asymptotics of Barenblatt s solution Theorem Let U m (t, x; M) Barenblatt s solution with mass M. We have the limits Proof lim U m(t, M) = Mδ 0 in the sense of distributions t 0 lim U m(t, x; M) = MG(t, x) pointwise on (0, ) R n m 1 supp U m (t, ; M) B ( 0, t β ( C β ) 1 2m 2 ) (m 1) By mass preservation: lim t 0 U m (t, M) = Mδ 0 8 / 11

31 Asymptotics of Barenblatt s solution Theorem Let U m (t, x; M) Barenblatt s solution with mass M. We have the limits Proof lim U m(t, M) = Mδ 0 in the sense of distributions t 0 lim U m(t, x; M) = MG(t, x) pointwise on (0, ) R n m 1 supp U m (t, ; M) B ( 0, t β ( C β ) 1 2m 2 ) (m 1) By mass preservation: lim t 0 U m (t, M) = Mδ 0 Limit for m 1 is a truly marvelous calculation but this margin is to narrow to contain it 8 / 11

32 The Cauchy Dirichlet problem (CDP) Let Ω R n bounded with Ω smooth, T (0, ] Q := R + Ω, Q T := (0, T ) Ω u 0 L 1 (Ω), f L 1 (Q) Φ C(R) strictly increasing with Φ(± ) = ±, Φ(0) = 0 Consider (CDP) t u (Φ(u)) = f in Q T u(0, x) = u 0 (x) in Ω u(t, x) = 0 on [0, T ) Ω 9 / 11

33 The Cauchy Dirichlet problem (CDP) Let Ω R n bounded with Ω smooth, T (0, ] Q := R + Ω, Q T := (0, T ) Ω u 0 L 1 (Ω), f L 1 (Q) Φ C(R) strictly increasing with Φ(± ) = ±, Φ(0) = 0 Consider (CDP) t u (Φ(u)) = f in Q T u(0, x) = u 0 (x) in Ω u(t, x) = 0 on [0, T ) Ω Choose Φ(u) = u m 1 u and f = 0 for the PME 9 / 11

34 Weak solutions for the CDP Definition A weak solution of CDP in Q T is a function u L 1 (Q T ) s.t. 1 w := Φ(u) L 1 (0, T ; W 1,1 0 (Ω)) 2 ( w η u t η) dx dt = Ω Q T u 0 (x)η(0, x) dx + Q T f η dx dt holds for any η C 1 (Q T ) which vanishes on [0, T ) Ω and for t = T Remarks Integration by parts shows: smooth solutions are weak solutions What about initial data...? 10 / 11

35 Weak solutions for the CDP Definition A weak solution of CDP in Q T is a function u L 1 (Q T ) s.t. 1 w := Φ(u) L 1 (0, T ; W 1,1 0 (Ω)) 2 ( w η u t η) dx dt = Ω Q T u 0 (x)η(0, x) dx + Q T f η dx dt holds for any η C 1 (Q T ) which vanishes on [0, T ) Ω and for t = T Remarks Integration by parts shows: smooth solutions are weak solutions What about initial data...? 10 / 11

36 Weak solutions for the CDP Definition A weak solution of CDP in Q T is a function u L 1 (Q T ) s.t. 1 w := Φ(u) L 1 (0, T ; W 1,1 0 (Ω)) 2 ( w η u t η) dx dt = Ω Q T u 0 (x)η(0, x) dx + Q T f η dx dt holds for any η C 1 (Q T ) which vanishes on [0, T ) Ω and for t = T Remarks Integration by parts shows: smooth solutions are weak solutions What about initial data...? 10 / 11

37 Weak solutions for the CDP Definition A weak solution of CDP in Q T is a function u L 1 (Q T ) s.t. 1 w := Φ(u) L 1 (0, T ; W 1,1 0 (Ω)) 2 ( w η u t η) dx dt = Ω Q T u 0 (x)η(0, x) dx + Q T f η dx dt holds for any η C 1 (Q T ) which vanishes on [0, T ) Ω and for t = T Remarks 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 u(t)ϕ dx = u 0 ϕ dx lim t 0 Ω Ω 10 / 11

38 A well-known weak solution Modify Barenblatt s solution Take x 0 Ω, τ > 0 Set v(t, x) := U m (t + τ, x x 0 ; M) 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 Q T. If m 2, then v is not a classical solution of that problem. Proof v has the stated regularity Let P := {(t, x) Q T v(t, x) > 0} v is smooth solution within P and v m is C 1 up to x = r(t) Integration by parts yields the integral equality (2) 11 / 11

Nonlinear Control. Nonlinear Control Lecture # 8 Time Varying and Perturbed Systems

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