Nonlinear transport phenomena: models, method of solving and unusual features (1)

1. Nonlinear transport phenomena: models, method of solving and unusual features Vsevolod Vladimirov AGH University of Science and technology, Faculty of Applied Mathematics ´ Krakow, August 10, 2010 Summer School: KPI, 2010 Nonlinear transport phenomena 1 / 38

2. THE AIM OF THIS LECTURE IS TO PRESENT THE SIMPLEST PATTERNS SUPPORTED BY TRANSPORT EQUATIONS WHAT ARE PATTERNS? Summer School: KPI, 2010 Nonlinear transport phenomena 2 / 38

3. THE AIM OF THIS LECTURE IS TO PRESENT THE SIMPLEST PATTERNS SUPPORTED BY TRANSPORT EQUATIONS WHAT ARE PATTERNS? Summer School: KPI, 2010 Nonlinear transport phenomena 2 / 38

4. Examples of patterns: Solution to the system of reaction-diﬀusion equation ut = Du ∆u + a(1 − u) vt = Dv ∆v + u v 2 − k v u(0, x) = f (x), v(0, x) = g(x), 2 ∂2 ∆= i=1 ∂x2 i periodically initiated cordial impulse; formation of the train of solitary waves; tsunami; tornado etc., etc. Summer School: KPI, 2010 Nonlinear transport phenomena 3 / 38

11. Our aim to discuss the patterns supported by the various transport equations in case of one spatial variable. Advantages: ability to present and analyze the analytical solutions to non-linear models the possibility to analyze the problem by means of qualitative theory methods. By patterns we temporarily mean: non-monotonic solution to the modelling system, maintaining their shape during the evolution; or the non-monotonic solutions evolving in a self-similar mode or the non-monotonic solutions tending to inﬁnity in ﬁnite time (blowing-up patterns) Summer School: KPI, 2010 Nonlinear transport phenomena 4 / 38

17. Heat transport: the balance equation d ρ u(t, x)d x = − q dσ + f (u(t, x); t, x)d x, (1) dt Ω ∂Ω Ω where q is the density of the heat ﬂux on the boundary ∂ Ω. On virtue of the Fick law, is as follows: q = −κ u(t, x); u(t, x) is the energy per unit mass (the temperature); ρ is the heat capacity per unit volume; κ(u; t, x) is the heat transport coeﬃcient; f (u(t, x); t, x) is the voluminal heat source. Summer School: KPI, 2010 Nonlinear transport phenomena 5 / 38

21. Using the Green-Gauss-Ostrogradsky theorem, we are able to write down the following identities: − q dσ ≡ κ u(t, x) dσ = [κ u(t, x)] d x. ∂Ω ∂Ω Ω So, the balance equation can be rewritten as ∂ u(t, x) ρ − [κ u(t, x)] − f [u(t, x); t, x] d x = 0. Ω ∂t Since the volume Ω is arbitrary, fulﬁllment of the balance equation is possible providing that the integrand is equal to zero In the case of one spatial variable, we get this way the equation ∂ u(t, x) ∂ ∂ u(t, x) ˜ = κ (u(t, x), t, x) ˜ + f (u(t, x), t, x) , ∂t ∂x ∂x (2) ˜ κ = κ/ρ, f = f /ρ. ˜ Summer School: KPI, 2010 Nonlinear transport phenomena 6 / 38

25. Self-similar solutions to the heat equation. Statement 1. Any physical law can be written down in the form a = ϕ (a1 , a2 , ..., an ; an+1 , ...an+m ) , where a1 , a2 , ..., an are the physical quantities expressed in the basic units, a, an+1 , ...an+m are the physical quantities expressed in derived units. Example of basic units: the length[L], the time [T ], the mass [M ] Examples of physical quantities expressed in derived units: dx L d2 x M · L V = , F =m 2 , etc., etc. dt T dt T2 Summer School: KPI, 2010 Nonlinear transport phenomena 7 / 38

30. Statement 2. Derived physical quantities a, an+1 , ...an+m can be expressed through the basic ones in the following form: a = ar1 ar2 ...arn Π, 1 2 n rj rj j an+j = a11 a22 ...arn Πj , n j = 1, 2, ...., m, where Π, Π1 , ...., Πm are dimensionless parameters. So, the physical law a = F (a1 , ...an ; ...an+j , ....) is equivalent to 1 rj rj rj Π= F a1 , a2 , ..., an ;...a11 a22 ...ann Πj ... . (3) ar1 1 ar2 2 ...arn n Summer School: KPI, 2010 Nonlinear transport phenomena 8 / 38

31. Statement 2. Derived physical quantities a, an+1 , ...an+m can be expressed through the basic ones in the following form: a = ar1 ar2 ...arn Π, 1 2 n rj rj j an+j = a11 a22 ...arn Πj , n j = 1, 2, ...., m, where Π, Π1 , ...., Πm are dimensionless parameters. So, the physical law a = F (a1 , ...an ; ...an+j , ....) is equivalent to 1 rj rj rj Π= F a1 , a2 , ..., an ;...a11 a22 ...ann Πj ... . (3) ar1 1 ar2 2 ...arn n Summer School: KPI, 2010 Nonlinear transport phenomena 8 / 38

32. The essence of Π theorem Theorem. The RHS of the formula (3) depends, at most, on the m dimensionless parameters Π1 , ...., Πm , and does not depend on the dimension quantities a1 , a2 , ..., an . 1 So, the physical law a = F (a1 , ...an ; ...an+j , ....) in dimensionless variables takes the form to Π = Ψ (Π1 , Π2 , ...Πm ) . (4) 1 A. Samarskij, A. Mikhailov, Mathematical Modelling: Ideas, Methods, Examples, Mosclw, 1997, Ch. V. Summer School: KPI, 2010 Nonlinear transport phenomena 9 / 38

35. Applications. Point explosion (linear case) u t = κ ux x , (5) ∞ u(0, x) = 0, x ∈ R, x = 0, u(t, x) d x = Q = const. −∞ The solution to (5) can be expressed as u = f (t, κ, Q; x). Basic units: E · L2 t[T ], κ , Q[E · L]. T Derived units: Q √ u = √ Π, x = κ t Π1 . κt Thus, passing to dimensionless variables we get √ κt √ Π= f (t, κ, Q; κ t Π1 ) = ϕ (Π1 ) . Q Summer School: KPI, 2010 Nonlinear transport phenomena 10 / 38

40. Corollary.Ansatz Q u = √ ϕ(ξ), ξ = Π1 κt should strongly simplify the point explosion problem (5). In fact, inserting this ansatz into the heat transport equation, we get d 2 ϕ[ξ] + ξ ϕ[ξ] + ϕ[ξ] ≡ ¨ ˙ (ξϕ[ξ] + 2 ϕ[ξ]) = 0, ˙ (6) dξ with the additional condition (as well expressed in the dimensionless variables): ∞ ϕ[ξ] dξ = 1. (7) −∞ Summer School: KPI, 2010 Nonlinear transport phenomena 11 / 38

43. Statement 3. Solution to the problem d (ξϕ[ξ] + 2 ϕ[ξ]) = 0, ˙ dξ ∞ ϕ[ξ] dξ = 1 −∞ is given by the formula 1 ξ2 ϕ[ξ] = √ e− 4 . 4π Corollary. The only solution to the linear point explosion problem (5) is given by the formula Q x2 u= √ e− 4 κ t . (8) 4κπt Summer School: KPI, 2010 Nonlinear transport phenomena 12 / 38

44. Statement 3. Solution to the problem d (ξϕ[ξ] + 2 ϕ[ξ]) = 0, ˙ dξ ∞ ϕ[ξ] dξ = 1 −∞ is given by the formula 1 ξ2 ϕ[ξ] = √ e− 4 . 4π Corollary. The only solution to the linear point explosion problem (5) is given by the formula Q x2 u= √ e− 4 κ t . (8) 4κπt Summer School: KPI, 2010 Nonlinear transport phenomena 12 / 38

45. Figure: Summer School: KPI, 2010 Nonlinear transport phenomena 13 / 38

48. Remark 1. For Q = 1 the previous formula written as 1 x2 E=√ e− 4 κ t 4πκt deﬁnes so called fundamental solution of the operator ˆ ∂ ∂2 L= − κ 2. ∂t ∂x Knowledge of this solution enables to solve the initial value problem. Theorem Solution to the initial value problem ut − κ u x x , u(0, x) = F (x) is given by the formula +∞ u(t, x) = E ∗ F (t, x) ≡ E (t, x − y) · F (y) d y −∞ provided that the integral in the RHS does exist. Summer School: KPI, 2010 Nonlinear transport phenomena 16 / 38

49. Remark 1. For Q = 1 the previous formula written as 1 x2 E=√ e− 4 κ t 4πκt deﬁnes so called fundamental solution of the operator ˆ ∂ ∂2 L= − κ 2. ∂t ∂x Knowledge of this solution enables to solve the initial value problem. Theorem Solution to the initial value problem ut − κ u x x , u(0, x) = F (x) is given by the formula +∞ u(t, x) = E ∗ F (t, x) ≡ E (t, x − y) · F (y) d y −∞ provided that the integral in the RHS does exist. Summer School: KPI, 2010 Nonlinear transport phenomena 16 / 38

50. Peculiarity of the solution Q x2 u= √ e− 4 κ t 4κπt to the linear point explosion: At any moment of time all the space ”knows ” that the point explosion took place. This is physically incorrect!!! Summer School: KPI, 2010 Nonlinear transport phenomena 17 / 38

51. Peculiarity of the solution Q x2 u= √ e− 4 κ t 4κπt to the linear point explosion: At any moment of time all the space ”knows ” that the point explosion took place. This is physically incorrect!!! Summer School: KPI, 2010 Nonlinear transport phenomena 17 / 38

52. Point explosion (nonlinear case) ut = κ [u ux ]x , (9) ∞ u(0, x) = 0, x ∈ R, x = 0, u(t, x) d x = Q = const. −∞ The solution to (9) can be expressed as u = f (t, κ, Q; x). Basic units: L2 t[T ], κ , Q[E · L]. E·T Derived units: Q2/3 u= Π, x = [κ Q t]1/3 Π1 . [κ t]1/3 Thus, passing to dimensionless variables we get [κ t]1/3 Π= f t, κ, Q; Π1 [κ Q t]1/3 = ϕ (Π1 ) . Q2/3 Summer School: KPI, 2010 Nonlinear transport phenomena 18 / 38

57. Corollary. Ansatz Q x u= ϕ(ξ), ξ = Π1 = [κ Q t]1/3 [κ Q t]1/3 should strongly simplify the point explosion problem (9). In fact, inserting this ansatz into the heat transport equation, we get d (ξϕ[ξ] + 3 ϕ[ξ] ϕ[ξ]) = 0, ˙ (10) dξ with the additional condition (as well expressed in the dimensionless variables): ∞ ϕ[ξ] dξ = 1. (11) −∞ Summer School: KPI, 2010 Nonlinear transport phenomena 19 / 38

60. Statement 4. Solution to the problem d (ξϕ[ξ] + 3 ϕ[ξ] ϕ[ξ]) = 0, ˙ dξ ∞ ϕ[ξ] dξ = 1 −∞ is given by the formula 1 6 ξΦ − ξ 2 if |ξ| < ξΦ , 2 ϕ[ξ] = 0 otherwise, where ξΦ = (9/2)1/3 . Corollary. The only solution to the nonlinear point explosion problem (9) is given by the formula 1/6   (9/2)1/3 − x2 t 2 Q  [κ Q t]2/3 if |x| < 9 (κ 2 Q) , u= 6 [κ Q t]1/3  0 otherwise  (12) Summer School: KPI, 2010 Nonlinear transport phenomena 20 / 38

67. Conclusion. Nonlinearity of the heat transport coeﬃcient causes the localization of the heat wave within the ﬁnite domain < −L(t), L(t) >, where 1/6 9 (κ t Q)2 L(t) = = C t1/3 . 2 Summer School: KPI, 2010 Nonlinear transport phenomena 25 / 38

68. Unlimited growth of temperature and the complete localization of the heat energy ut = κ [uσ ux ]x , (13) −n u(t, 0) = A (−t) , t < 0, n > 0, x ∈ R+ , lim u(t, x) = 0(14) t→−∞ In general, the problem is self-similar and the ansatz A x u= ϕ(ξ), ξ=√ (−t)n Aσ κ t1−n σ reduces the initial problem to corresponding ODE. If we assume that n = σ = 1, than ξ will not depend on t: A x u= ϕ(ξ), ξ=√ . (−t) Aκ Summer School: KPI, 2010 Nonlinear transport phenomena 26 / 38

72. The reduced equation ϕϕ + ϕ2 − ϕ = 0, ¨ ˙ together with the initial condition ϕ(0) = 1 (arising from the A condition u(t, 0) = (−t) ), has the solution √ √ (1 − ξ/ 6)2 , if 0 < ξ < √ 6, ϕ(ξ) = 0 if ξ ≥ 6, The corresponding solution to the BV problem (13) is as follows: A 2 √ 1− √ x , if 0 < x < 6 κ A, u(t, x) = −t 6κA √ 0 if x ≥ 6 κ A. Moral of the tale: [1]. Heat energy is completely localized in √ the segment 0 < x < 6 κ A0 ; [2]. The√temperature u(t, x) tends to +∞ in every point x ∈ [0, 6 κ A0 ) as t → 0−. Summer School: KPI, 2010 Nonlinear transport phenomena 27 / 38

79. Blow-up regime caused by the source term incorporation Model equation: ut = κ [u ux ]x + q0 u2 . (15) Ansatz u(t, x) = P (t) Q(x) leads, after some algebraic manipulation, to the equations P −2 (t)P (t) = κ(Q(x) Q(x) ) + q0 Q(x)2 Q−1 (x) = C1 . (16) ˙ Theorem. The equation (15) has the following solution: q0 2 4 cos2 x , if |x| < 2 π0 κ , u(t, x) = 8κ q 3 q0 (tf − t) 0 otherwise. (17) The above solution 2 π2 κ is localized within the interval [−L, L], L = q0 ; tends to +∞ ∀ x ∈ (−L, L), as t → tf − 0. Summer School: KPI, 2010 Nonlinear transport phenomena 31 / 38

89. THE ROLE OF SELF-SIMILAR SOLUTIONS Summer School: KPI, 2010 Nonlinear transport phenomena 36 / 38

90. Maximum principle. Comparison theorems ( initial value problem). Let us consider the Cauchy problem ut = [κ(u) ux ]x , (18) u(0, x) = u0 (x) ≥ 0, x ∈ R, κ(u) > 0. Theorem 1. Maximum of the solution u(t, x) at any time t > 0 does not exceed the maximum of the initial data: maxt>0, −∞<x<∞ u(t, x) ≤ max−∞<x<∞ u0 (x). 2 2 A. Samarskij, A. Mikhailov, Mathematical Modelling: Ideas, Methods, Examples, Moscow, 1997, A. Samarskij, V. Galaktionov et al, Blow-up regimes in quasilinear parabolic equations, Berlin: Walter de Gruyter, 1995. Summer School: KPI, 2010 Nonlinear transport phenomena 37 / 38

94. Corollary (comparison theorem). Let u1 (t, x), u2 (t, x), u3 (t, x), are the solutions to the equation ut = [κ(u) ux ]x , u(0, x) = u0 (x) ≥ 0, x ∈ R, κ(u) > 0, corresponding to the initial data u1 (x), u2 (x), and u3 (x) 0 0 0 such that u1 (x) ≤ u2 (x) ≤ u3 (x) 0 0 0 for −∞ < x < ∞. Then u1 (t, x) ≤ u2 (t, x) ≤ u3 (t, x) for all −∞ < x < ∞, and t > 0. Summer School: KPI, 2010 Nonlinear transport phenomena 38 / 38

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