Forchheimer Equations in Porous Media - Part III

Forchheimer Equations in Porous Media - Part III Luan Hoang, Akif Ibragimov Department of Mathematics and Statistics, Texas Tech niversity http://www.math.umn.edu/ lhoang/ luan.hoang@ttu.edu Applied Mathematics Seminar Texas Tech niversity September 15&22, 21

Outline 1 Introduction 2 Bounds for the solutions 3 Dependence on the boundary data 4 Dependence on the Forchheimer polynomials

Introduction Darcy s Law: the two term law the power law the three term law αu = p, αu + β u u = p, c n u n 1 u + au = p, Au + B u u + C u 2 u = p. Here α, β, c, A, B, and C are empirical positive constants.

General Forchheimer equations Generalizing the above equations as follows g( u )u = p. Let G(s) = sg(s). Then G( u ) = p u = G 1 ( p ). Hence where p u = g(g 1 ( p )) = K( p ) p, K(ξ) = K g (ξ) = 1 g(s) = 1 g(g 1, sg(s) = ξ. (ξ)) We derived non-linear Darcy equations from Forchheimer equations.

Equations of Fluids Let ρ be the density. Continuity equation dρ = (ρu), dt For slightly compressible fluid it takes dρ dp = 1 κ ρ, where κ 1. Substituting this into the continuity equation yields dρ dp dp dt = ρ u dρ dp u p, dp = κ u u p. dt Since κ 1, we neglect the second term in continuity equation dp dt = κ u.

Combining the equation of pressure and the Forchheimer equation, one gets after scaling: dp dt = (K( p ) p ). Consider the equation on a bounded domain in R 3. The boundary of consists of two connected components: exterior boundary Γ 2 and interior (accessible) boundary Γ 1. Dirichlet condition on Γ 1 : p(x, t) = ψ(x, t) which is known for x Γ i. On Γ 2 : u N = p N =.

Class FP(N, α) We introduce a class of Forchheimer polynomials Definition A function g(s) is said to be of class FP(N, α) if g(s) = a s α + a 1 s α 1 + a 2 s α 2 +... + a N s α N = N a j s α j, j= where N >, = α < α 1 < α 2 <... < α N, and a, a N >, a 1,..., a N 1. Notation α N = deg(g).

Let N 1 and α be fixed. Let R(N) = { a = (a, a 1,..., a N ) R N+1 : a, a N >, a 1,..., a N 1 }. Let g = g(s, a) be in FP(N, α) with a R(N). We denote a = α N a (, 1), b = α N + 1 2 a = α N (, 1), α N + 2 { χ( a) = max a, a 1,..., a N, 1 1 }, 1. a a N Many estimates below will have constants depending on χ( a).

Lemma Let g(s, a) be in class FP(N, α). One has for any ξ that and for any m 1, δ > that C 1 χ( a) 1 a (1 + ξ) a K(ξ, a) C χ( a) 1+a (1 + ξ) a, C 1 χ( a) 1 a δ a (1 + δ) a (ξm a δ m a ) K(ξ, a)ξ m C χ( a) 1+a ξ m a, where C = C (N, α N ) depends on N, α N only. In particular, when m = 2, δ = 1, one has 2 a C 1 χ( a) 1 a (ξ 2 a 1) K(ξ, a)ξ 2 C χ( a) 1+a ξ 2 a.

The Monotonicity Proposition (i) For any y, y R n, one has ( K( y, a)y K( y, a)y ) (y y ) ak( y y, a) y y 2. (ii) For any functions p 1 and p 2 one has ( ) ( ) K( p1, a) p 1 K( p 2, a) p 2 p1 p 2 dx ( a C 5 ( ) K( p 1 p 2, a) p 1 p 2 2 dx ) 2 p 1 p 2 2 a 2 a ( ) a, dx 1 + p1 L 2 a () p 2 L 2 a () where C 5 = C 5 (N, deg(g), χ( a)).

Poincaré Sobolev inequality Let f (x) and ξ(x) be two functions on with f (x) vanishing on Γ 1 and ξ(x). Then ( ) 2 f (x) (2 a) (2 a) dx C 6 ( ) K(ξ(x), a) f (x) 2 dx ( 1 + ) a 2 a H(ξ(x), a)dx, where (2 a) = n(2 a)/(n (2 a)) and C 6 = C 6 (N, deg(g), χ( a), ). Subsequently, when deg(g) 4/(n 2) one has ( )( ) a f (x) 2 dx C 7 K(ξ(x), a) f (x) 2 2 a dx 1 + H(ξ(x), a)dx, where C 7 = C 7 (N, deg(g), χ( a), ).

Degree Condition (DC) deg(g) 4 n 2 2 (2 a) = n(2 a) n (2 a).

Extension of the boundary data Let Ψ(x, t), x, t [, ) be an extension of ψ(x, t) from Γ 1 to. For instance, one can use the following harmonic extension Ψ of ψ: Ψ = on, Ψ = ψ and Ψ =. Γ1 Γ2 We denote such Ψ by H(ψ). Then the Sobolev norms of H(ψ) on can be bounded by the Sobolev or Besov norms of ψ on Γ 1. In particular, we have k t k H(ψ) W C(k, r) k r,2 () t k ψ W, r,2 (Γ 1 ) for k =, 1, 2, r =, 1.

Shift of solutions Shifted solution: Let p = p Ψ, then p satisfies Define: p t = (K( p ) p) Ψ t on (, ), p = on Γ 1 (, ). H(ξ, a) = ξ 2 K( s, a)ds. We denote H(x, t) = H[p](x, t) = H( p(x, t), a). Constants in estimates below depend on χ( a).

A priori Estimates - I(a) Lemma One has where G 1 (t) = 1 d 2 dt p 2 (x, t)dx C H(x, t)dx + CG 1 (t), ( ) 2 a ( ) 1 Ψ(x, t) 2 dx+ Ψ t (x, t) r r dx (1 a) + Ψ t (x, t) r r dx with r denoting the conjugate exponent of (2 a), thus explicitly having the value n(2 a) r = (2 a)(n + 1) n = n(2 + α N). n + 2 + α N

A priori Estimates - I(b) Corollary One has for t that p 2 (x, t)dx where Λ 1 (t) = t p 2 (x, )dx + CΛ 1 (t), G 1 (τ)dτ. In the case deg(g) 4/(n 2) one has p 2 (x, t)dx p 2 (x, )dx + CΛ 2 (t), where Λ 2 (t) = (1 + Env(G 1 )(t)) 2 2 a, with Env(G1 )(t) being a continuous, increasing envelop of the function G 1 (t) on [, ).

A priori Estimates - I: Proofs No Degree Condition: minimal information With Degree Condition: Sobolev-Poincare inequality and non-linear differential inequality (weaker than the usual Gronwall s)

Non-linear differential inequality Lemma Suppose y Ay α + f (t), for all t >, with A, α > and y(t), f (t). Let F (t) be a continuous, increasing envelop of f (t) on [, ). Then one has y(t) y() + A 1/α F (t) 1/α, t.

A priori Estimates - II(a) Lemma For any ε >, one has d H(x, t)dx + dt where C ε is positive and G 2 (t) = p 2 t (x, t)dx ε H(x, t)dx + C ε G 2 (t), Ψ t (x, t) 2 dx + Ψ t (x, t) 2 dx. Consequently, one has [ ] d H(x, t) + p 2 (x, t)dx + p 2 t (x, t)dx C H(x, t)dx +CG 3 (t), dt where G 3 (t) = G 1 (t) + G 2 (t).

A priori Estimates - II(b) Corollary (i) Given δ >, there is C δ > such that for all t one has t H(x, t)dx e δt H(x, )dx + C δ e δ(t τ) G 2 (τ)dτ. (ii) For t, one has H(x, t) + p 2 (x, t)dx + t p 2 t (x, τ)dxdτ H(x, ) + p 2 (x, )dx + C t G 3 (τ)dτ.

A priori Estimates - II(c) Corollary For t, one has H(x, t)dx e C 1t + C t H(x, )dx + C p 2 (x, )dx ( ) e C 1(t τ) Λ 1 (τ) + G 3 (τ) dτ. In case deg(g) 4/(n 2), one has for t that H(x, t)dx e C 1t H(x, )dx + C + C t p 2 (x, )dx ( ) e C 1(t τ) Λ 2 (τ) + G 3 (τ) dτ.

A priori Estimates - II(d) Lemma Suppose deg(g) 4 n 2. Then ( ) p 2 (x, t)dx Ch(t) H(x, t)dx + Ψ(x, t) 2 dx, ( ) 2 a ( ) p 2 2 (x, t)dx C 1 + H(x, t)dx + C Ψ(x, t) 2 dx, where h(t) = ( ) a 2 a 1 + H(x, t)dx.

A priori Estimates - II(e) Proposition Suppose deg(g) 4 n 2. One has the following two estimates H(x, t) + p 2 (x, t)dx e C ( t ) 1 h 1 (τ)dτ H(x, ) + p 2 (x, )dx and H(x, t)+p 2 (x, t)dx + C t e C 1 t τ h 1 (θ)dθ G 3 (τ)dτ, H(x, )+p 2 (x, )dx +C ( 1+Env(G 3 )(t) ) 2 2 a.

A priori Estimates - III(a) Let q = p t, q = p t. Lemma One has for t > that d q 2 dx C K( p ) q 2 dx + C dt Ψ t 2 dx + qψ tt dx. Note: May happen lim sup p 2 t (x, t)dx =. t

A priori Estimates - III(b) Proposition One has for t t > that [ p 2 t (x, t)dx t 1 H(x, ) + p 2 (x, )dx + C t { ( + C Ψ t (x, τ) 2 dx + h(τ) If deg(g) 4 n 2, then for t t > one has + C t ] G 3 (τ)dτ Ψ tt (x, τ) r dx ) 2 r } dτ. p 2 t (x, t)dx t 1 [ t 1 e C 1 t h(τ) dτ t H(x, )+p 2 (x, )dx+c t e C t 1 τ 1 dθ( h(θ) Ψ t (x, τ) 2 dx + h(τ) Ψ tt (x, τ) 2 dx G 3 (τ)dτ ) dτ.

A priori Estimates - IV(a) Proposition Suppose deg(g) 4/(n 2). (i) For any t t > one has H(x, t) + p 2 t (x, t) + p 2 (x, t)dx (1 + t 1 [ t )e C 1 t h 1 (τ)dτ H(x, ) + p 2 (x, )dx + C t where G 4 (t) = G 3 (t) + Ψ2 tt(x, t)dx. ] t G 3 (τ)dτ + C e C t 1 τ h 1 (θ)dθ G 4 (τ)dτ,

A priori Estimates - IV(b) Proposition (continued) (ii) Assume, in addition, that M def == ( ) e C 1(t τ) Λ 2 (t) + G 3 (t) dτ <. Then there is d > depending on the initial data of the solution and the value M above so that H(x, t) + p 2 t (x, t) + p 2 (x, t)dx (1 + t 1 + p 2 (x, )dx + C for all t t >. t )e d (t t ) [ H(x, ) ] t G 3 (τ)dτ + C e d(t τ) G 4 (τ)dτ,

A priori Estimates - IV(c): No condition on h(t) Proposition Suppose deg(g) 4 n 2. One has for t t > that H(x, t) + p 2 t (x, t) + p 2 (x, t)dx (1 + t 1 ) H(x, ) + p 2 (x, )dx + C ( 1 + Env(G 4 )(t) ) 2 2 a.

A priori Estimates - V(a): niform bounds Theorem Suppose deg(g) 4/(n 2). Assume that Then sup [, ) sup ψ(, t) W 1,2 (Γ 1 ), sup ψ t (, t) W 1,2 (Γ 1 ) <. [, ) [, ) ( H(x, t) + p 2 (x, t)dx 4 H(x, ) + p 2 (x, )dx + C 1+ sup ψ(, t) 2 W 1,2 (Γ 1 ) + sup [, ) [, ) 2 a 1 a ψ t (, t) L 2 (Γ 1 ) + sup ψ t (, t) 2 W 1,2 (Γ 1 ) [, ) ) 2 2 a.

A priori Estimates - V(b): niform bounds Theorem (continued) If, in addition, then for any t > one has sup ψ tt (, t) L 2 (Γ 1 ) <, [, ) sup H(x, t) + pt 2 (x, t) + p 2 (x, t)dx C(1 + t 1 [t, ) + p 2 (x, )dx + Ct 1 + sup [, ) Γ 1 ψ(x, ) 2 dσ + C ) H(x, ) ( 1 + sup ψ(, t) 2 W 1,2 (Γ 1 ) [, ) 2 a 1 a ψ t (, t) L 2 (Γ 1 ) + sup ψ t (, t) 2 W 1,2 (Γ 1 ) + sup ψ tt (, t) 2 L 2 (Γ 1 ) [, ) [, ) ) 2 2 a.

Dependence on the boundary data Let p 1 (x, t) and p 2 (x, t) be two solutions of the IBVP with the boundary profiles ψ 1 (x, t) and ψ 2 (x, t), respectively. Let Ψ k (x, t) be an extension of ψ k (x, t), for k = 1, 2. We denote z(x, t) = p 1 (x, t) p 2 (x, t), Ψ(x, t) = Ψ 1 (x, t) Ψ 2 (x, t), p k = p k Ψ k, k = 1, 2, z = p 1 p 2 = z Ψ. Let H k (x, t) = H[p k ](x, t) = H( p k (x, t), a) for k = 1, 2. Recall that a = deg(g) deg(g)+1 and b = a 2 a = deg(g) deg(g)+2. We will establish various estimates for Z(t) def == z2 (x, t)dx, for t.

First, we derive a general differential inequality for Z(t). Lemma One has for all t >, 1 d ( ) 2 z 2 dx C z 2 a 2 a dx (1 + H 1 2 dt L 1 + H 2 L 1) b + C(1 + H 1 L 1 + H 2 L 1) b Ψ 2 L 2 a + C( H 1 L 1 + H 2 L 1) 1/2 Ψ L 2 where r is the conjugate of (2 a). + C (1 + H 1 L 1 + H 2 L 1) b Ψ t 2 L r,

Let G j [Ψ k ], k = 1, 2, j = 1, 2, 3, 4, denote the quantity G j for corresponding solution p k with boundary data extension Ψ k. Similarly, let Λ j [Ψ k ], k = 1, 2, j = 1, 2, denote the corresponding quantity Λ j defined for Ψ k. Let m(t) = m 1 (t) + m 2 (t), where for k = 1, 2, m k (t) = e C 1t H k (x, )dx + p 2 k (x, )dx t ( ) + e C 1(t τ) Λ 1 [Ψ k ](τ) + G 3 [Ψ k ](τ) dτ.

Estimate for finite time interval Theorem One has for all t that z 2 (x, t)dx z 2 (x, )dx + C t ( Ψ(, τ) 2 L 2 a + m(τ) 1/2 Ψ(, τ) L 2 + (1 + m(τ)) b Ψ t (, τ) 2 L r Consequently, for any given T >, ) dτ. sup z 2 (x, t)dx 4 z 2 (x, )dx+6 sup Ψ(, t) 2 L +CT sup Ψ(, t) 2 [,T ] [,T ] [,T ] + CT (1 + A + D (T )) δ( sup Ψ(, t) L 2 + sup Ψ t (, t) 2 L r [,T ] [,T ] ),

Above, δ = max{1/2, b}, A = H 1 (x, )dx + D (T ) = 2 sup k=1 [,T ] t p 2 1(x, )dx + H 2 (x, )dx + p 2 2(x, )dx, ( ) e C 1(t τ) Λ 1 [Ψ k ](τ) + G 3 [Ψ k ](τ) dτ.

Corollary Suppose for both k = 1, 2 and t one has Ψ k (, t) 2 2 a L + (Ψ 2 1 a k ) t (, t) L r C(1 + t) r 1 and (Ψ k ) t (, t) 2 L 2 + (Ψ k ) t (, t) 2 L 2 C(1 + t) r 2, where r 1, r 2 >. Let r 3 = 1 + max{r 1 + 1, r 2 }. Then z 2 (x, t)dx + C(1 + A ) t z 2 (x, )dx where A is defined previously. (1 + τ) r 3/2 Ψ(, τ) L 2 + (1 + τ) r 3b Ψ t (, τ) 2 L r dτ,

Estimate for all time under the Degree Condition sing appropriate estimates of H k(x, t)dx, we denote Also, let m k (t) = e C 1t H k (x, )dx + p 2 k (x, )dx t ( + e C 1(t τ) Λ 2 [Ψ k ](τ) + G 3 [Ψ k ](τ) m(t) = m 1 (t) + m 2 (t) and S(t, t) = t ) dτ, k = 1, 2. t (1 + m(τ)) b dτ.

Theorem Suppose deg(g) 4/(n 2). Then for all t one has z 2 (x, t)dx e C 1S(,t) z 2 (x, )dx + C t e C 1S(τ,t) ( Ψ(, τ) 2 L 2 a + m(τ) 1/2 Ψ(, τ) L 2 + (1 + m(τ)) b Ψ t (, τ) 2 L r ) dτ.

Corollary Suppose deg(g) 4/(n 2). Assume that 2 ( k=1 sup [, ) Then one has z 2 (x, t)dx 4 sup [, ) where + C sup [, ) A = 1+ ( 2 ( k=1 ) Λ 2 [Ψ k ](t) + sup G 3 [Ψ k ](t) <. [, ) A b Ψ(, t) 2 L 2 a + A b+ 1 2 z 2 (x, )dx + C sup Ψ(, t) 2 L 2 [, ) Ψ(, t) L 2 + A 2b Ψ t (, t) 2 L r p k (x, ) 2 a +p 2 k (x, )dx+ sup Λ 2 [Ψ k ](t)+ sup G 3 [Ψ k ](t) [, ) [, ) ),

Corollary Suppose deg(g) 4/(n 2). Assume that lim S(, t) =, t lim (1 + t m(t))b+ 1 2 Ψ(, t) L 2 =, Then lim t z2 (x, t)dx =. lim (1 + m(t)) b Ψ t (, t) L r =. t

Dependence on the Forchheimer polynomials Let the boundary data ψ(x, t) on Γ 1 be fixed. For each coefficient vector a, we denote by p(x, t; a) the solution of p p t = (K( p ) p), p Γ1 = ψ, =, N Γ2 with K = K(ξ, a). Let g 1 = g(s, a (1) ) and g 2 = g(s, a (2) ) be two functions of class FP(N, α). Let p k = p k (x, t; a (k) )) for k = 1, 2. Let p = p 1 p 2, then p t = (K( p 1, a (1) ) p 1 ) (K( p 2, a (2) ) p 2 ). Multiplying this equation by p and integrating by parts over the domain yield 1 d 2 dt p 2 dx = (K( p 1, a (1) ) p 1 K( p 2, a (2) ) p 1 ) ( p 1 p 2 )dx

Perturbed Monotonicity Let a and a be two arbitrary vectors. We denote by a a and a a the maximum and minimum vectors of the two, respectively, with components ( a a ) j = max{a j, a j} and ( a a ) j = min{a j, a j}. Then component-wise one has a a a, a a a. Define χ( a, a ) = max{χ( a), χ( a )}. Note that χ( a a ), χ( a a ) χ( a, a ), χ(t a + (1 t) a ) χ( a, a ) t [, 1].

Perturbed Monotonicity Lemma Let g(s, a) and g(s, a ) belong to class FP(N, α). Then for any y, y in R n, one has (K( y, a)y K( y, a )y ) (y y ) (1 a)k( y y, a a ) y y 2 where a (, 1). Nχ( a, a ) a a K( y y, a a )( y y ) y y,

Let M k (t) = 1 + e C 1t p k (x, ) 2 a dx + p 2 k (x, )dx t + e C 1(t τ) ( Λ 1 (τ) + G 3 (τ) ) dτ, k = 1, 2, and M = M 1 + M 2. One has K( p k, a (1) a (2) ) p k 2 dx C + C where C depends on χ( a (1) a (2) ) and χ( a (k) ). Denote H k (x, t) = H( p k (x, t), a (k) ) for k = 1, 2. p k 2 a dx CM k,

Finite time estimate First, we obtain the continuous dependence for finite time. Proposition For t one has p 1 (x, t) p 2 (x, t) 2 dx p 1 (x, ) p 2 (x, ) 2 dx + C a (1) a (2) t M(τ)dτ, where C > depends on N, α N, and χ( a (1), a (2) ). Consequently, the solution p(x, t; a) depends continuously (in finite time intervals) on the initial data and the coefficient vector a R(N).

Large time estimate under the Degree Condition nder the Degree Condition, p k(x, t) 2 a dx is bounded by CM k (t) where M k (t) = 1 + e C 1t p k (x, ) 2 a dx + p 2 k (x, )dx t + e C 1(t τ) ( Λ 2 (τ) + G 3 (τ) ) dτ, k = 1, 2, and M(t) = M 1 (t) + M 2 (t).

Proposition Suppose α N 4/(n 2). Then for t, p 1 (x, t) p 2 (x, t) 2 dx e C t 1 M(τ) b (τ)dτ + C 2 a (1) a (2) t e C 1 t τ M(θ) b (θ)dθ M(τ)dτ. Assume, in addition, that M def == sup [, ) M(t) < then p 1 (x, t) p 2 (x, t) 2 dx e C 1t/M b p 1 (x, ) p 2 (x, ) 2 dx p 1 (x, ) p 2 (x, ) 2 dx + C 2 C 1 M 1+b a (1) a (2) for all t, and consequently lim sup p 1 (x, t) p 2 (x, t) 2 dx C 2 C 1 M 1+b a (1) a (2). t

niform estimate in the coefficients Let D be a compact set in R(N). Define ˆχ(D) = max{χ( a), a D}. Note that ˆχ(D) (, ) and for any a D, one has ˆχ(D) 1 χ( a) 1 χ( a) ˆχ(D). Let a (k) belong to D for k = 1, 2. Set 2 ( A = 2 + p k (x, ) 2 a dx + k=1 λ(t) = 1 + λ(t) = 1 + t t where C 1 depends on N, α N and ˆχ(D). Then M(t) A λ(t), M(t) A λ(t). p 2 k (x, )dx ) e C 1(t τ) ( Λ 1 (τ) + G 3 (τ) ) dτ, e C 1(t τ) ( Λ 2 (τ) + G 3 (τ) ) dτ,

Theorem Assume, additionally, that α N 4/(n 2). (i) One has for t that p 1 (x, t) p 2 (x, t) 2 dx e C 1A b t λ(τ) b dτ + C 2 A a (1) a (2) t e C 1A b t τ λ(θ) bdθ λ(τ)dτ. (ii) Assume, in addition, that M def == sup [, ) λ(t) <. Then p 1 (x, t) p 2 (x, t) 2 dx e C 1t/(A M ) b Consequently lim sup p 1 (x, ) p 2 (x, ) 2 dx p 1 (x, ) p 2 (x, ) 2 dx + C 2 C 1 1 (A M ) 1+b a (1) a (2). p 1 (x, t) p 2 (x, t) 2 dx C 2 C 1 1 (A M ) 1+b a (1) a (2). t

Next Semester - Spring 211 Forchheimer Equations in Porous Media - Part IV Boundary Flux condition Refined asymptotic estimates: limsup estimates for nonlinear differential inequalities, uniform Gronwall inequality,... Continuous dependence for pressure gradients and velocity

Next Semester - Spring 211 Forchheimer Equations in Porous Media - Part IV Boundary Flux condition Refined asymptotic estimates: limsup estimates for nonlinear differential inequalities, uniform Gronwall inequality,... Continuous dependence for pressure gradients and velocity THANK YO!