Analysis of PDE models of molecular motors. Benoît Perthame based on works with P. E. Souganidis

Transcription

1 Analysis of PDE models of molecular motors Benoît Perthame based on works with P. E. Souganidis

2 Asymmetry and motion Except human constructions movement always uses asymmetry

3 Asymmetry and motion See M. Golubitsky ; F. Alouges, A. DeSimone, A. Lefebvre,

4 Asymmetry and motion At small scales, movement uses more elementary devices K[f] = t f(t, x, ξ) + ξ xf }{{} run K(c; ξ, ξ )f(ξ )dξ = K[f] }{{} tumble, K(c; ξ, ξ)dξ f, It is enough that K(ξ, ξ ) IS NOT symmetric to create a drift

5 Molecular motors : motivation Eukariotic cells are structured by various filaments, microtubules... the cytoskeleton

6 Molecular motors : motivation One can observ trafficking of molecules organized along these filaments (more than by diffusion). This is because some proteins can link to these filaments and move along. A very general effect. There should be a simple universal explanation for this motor effect. A. Ajdari, J. Prost, J.-F. Johanny Julicher, Ermentrout, Oster, Peskine, Majda, Howard, Astumina, Hänggi...

7 Molecular motors : motivation One can observ trafficking of molecules organized along these filaments (more than by diffusion). This is because some proteins can link to these filaments and move along. A very general effect. There should be a simple universal explanation for this motor effect. A. Ajdari, J. Prost, J.-F. Joanny Julicher, Ermentrout, Oster, Peskine, Majda, Howard, Astumina, Hänggi...

8 Molecular motors : motivation The conclusion : Molecules diffusing in an asymmetric potential move in a certain direction What is an asymmetric potential? Is just non-symmetric enough like for E Coli?

9 Molecular motors : motivation Molecules diffusing in an asymmetric potential move The framework proposed by Hastings, Kinderlehrer, Kowalczyk... is the Fokker-Planck-Kolmogorov eigenfunction equation ε 2 u x 2 ( ψ (x)u ) x = 0, 0 < x < 1, ε u x + ψ (x)u(x) = 0 at x = 0, 1 (zero flux). Does the density concentrate on one hand : 0 or 1.

10 Molecular motors : motivation ε 2 u x 2 ( ψ (x)u ) x = 0, 0 < x < 1, ε u x + ψ (x)u(x) = 0 at x = 0, 1 (zero flux). Take the potential ψ( ) periodic (with a small period) 1 u(x) = e ψ(x) ε / e ψ(x) ε 0 The mass is exponentially located at minimum of ψ and thus uniformly distributed! A non-symmetric potential is clearly not enough! Physicists knew it well! dx.

11 Molecular motors : motivation ε 2 u x 2 ( ψ (x)u ) x = 0, 0 < x < 1, ε u x + ψ (x)u(x) = 0 at x = 0, 1 (zero flux). Take the potential ψ( ) periodic (with a small period) 1 u(x) = e ψ(x) ε / e ψ(x) ε 0 The mass is exponentially located at minimum of ψ and thus uniformly distributed! A non-symmetric potential is clearly not enough! Physicists knew it well! dx.

12 Molecular motors : motivation ε 2 u x 2 ( ψ (x)u ) x = 0, 0 < x < 1, ε u x + ψ (x)u(x) = 0 at x = 0, 1 (zero flux). Take the potential ψ( ) periodic (with a small period) 1 u(x) = e ψ(x) ε / e ψ(x) ε 0 The mass is exponentially located at minimum of ψ and thus uniformly distributed! A non-symmetric potential is clearly not enough! Physicists knew it well! dx.

13 OUTLINE OF THE LECTURE I. A pure transport case : conformation changes II. A motor using diffusion III. Flashing rachets

14 I. Conventional kinesis : pure transport motors Molecules can reach two conformations u 1, u 2 ε 2 u 1 x 2 ( ψ 1 (x)u 1 ε 2 u 2 x 2 ( ψ 2 (x)u 2 ) ) x + ν 1u 1 = ν 2 u 2, 0 < x < 1, x + ν 2u 2 = ν 1 u 1, ε u i x + ψ i (x)u i(x) = 0 at x = 0, 1 (zero flux). Still an eigenvalue problem The adjoint has 0 a dominant eigenvalue Still mass conservative : 1 0 [u 1 + u 2 ] = 1

15 I. Conventional kinesis : pure transport motors Molecules can reach two conformations u 1, u 2 ε 2 u 1 x 2 ( ψ 1 (x)u 1 ε 2 u 2 x 2 ( ψ 2 (x)u 2 ) ) x + ν 1u 1 = ν 2 u 2, 0 < x < 1, x + ν 2u 2 = ν 1 u 1, ε u i x + ψ i (x)u i(x) = 0 at x = 0, 1 (zero flux). A family of particular solutions is ν i = α(x)e ψ i (x) ε, u i = e ψ i (x) ε. Again there is no motion possible whatever the asymmetry of the potential.

16 I. Conventional kinesis : pure transport motors Definition We say that (ψ 1, ψ 2, ν 1, ν 2 ) is asymmetric if lim i,ε = δ(x) ε 0 or δ(x 1). and more acurately u i,ε = e R i,ε/ε, R i,ε (x) ε 0 R(x), max R(x) = R(0) or R(1) x (0,1) The intuition is δ(x) 1 2πε e x 2 2ε, R ε = x ε ln(2πε) R is an effective potential. For ν i constants, this may occur

17 I. Conventional kinesis : pure transport motors ! !0.02 0!0.03!0.2!0.4!0.04!0.6!0.05!0.8! Left : potentials ψ i,! Right : Effective potential R i,ε

18 I. Conventional kinesis : pure transport motors Theorem (CHK, PS) With ν i constant, ψ are asymmetric if max(ψ 1 (x), ψ 2 (x)) > 0 x (0, 1), min(ψ 1 (x), ψ 2 (x)) > 0 on I 1 and concentration occurs at x = !0.2!0.4!0.6!0.8!

19 II. Conventional kinesis : homogenization case One can use more general asymmetric potentials by juxtaposing them many times : ε 2 u 1 x 2 ( ψ 1 (x ε ) u ) 1 x + ν 1( x ε ) u 1 = ν 2 ( x ε ) u 2, 0 < x < 1, ε 2 u 2 x 2 + ν 2 ( x ε ) u 2 = ν 1 ( x ε ) u 1, u 2 x = ε u 1 x + ψ 1 (x ε )u 1(x) = 0 at x = 0, 1 (zero flux). Set like this it is an homogenization problem The mathematical theory is mainly born with the works of F. Murat and L. Tartar Developed further in Lab. J.-L. Lions (Cioranescu, Allaire)

20 II. Conventional kinesis : homogenization case One can use more general asymmetric potentials by juxtaposing them many times : ε 2 u 1 x 2 ( ψ 1 (x ε ) u ) 1 x + ν 1( x ε ) u 1 = ν 2 ( x ε ) u 2, 0 < x < 1, ε 2 u 2 x 2 + ν 2 ( x ε ) u 2 = ν 1 ( x ε ) u 1, u 2 x = ε u 1 x + ψ 1 (x ε )u 1(x) = 0 at x = 0, 1 (zero flux). New feature : diffusion length interact with the period ε of the potential and coefficients for conformation changes! We still have counter-exemples : u 2 = 1, u 1 = e ψ(x ε )/ε, ν 1 = ν 2 e ψ(x ε ) ψ(y), ν i (y) even

21 II. Conventional kinesis : homogenization case

22 II. Conventional kinesis : homogenization case ε 2 u 1 x 2 ( ψ 1 (x ε ) u ) 1 x + ν 1( x ε ) u 1 = ν 2 ( x ε ) u 2, 0 < x < 1, ε 2 u 2 x 2 + ν 2 ( x ε ) u 2 = ν 1 ( x ε ) u 1, u 2 x = ε u 1 x + ψ 1 (x ε )u 1(x) = 0 at x = 0, 1 (zero flux). We keep the same notion of asymmetric potential Definition We say that (ψ 1, ν 1, ν 2 ) is asymmetric if concentration occurs lim u i,ε = δ(x) or δ(x 1). ε 0 u i,ε = e R i,ε/ε, R i,ε (x) ε 0 R(x), max R(x) = R(0) or R(1) x (0,1)

23 II. Conventional kinesis : homogenization case The effective hamiltonian H(p) is defined by the cell problem 2 ϕ 1 (y) y 2 ( ψ 1 (y) ϕ 1 2 ϕ 2 (y) y 2 ϕ 2 (y)e py, ϕ 1 (y)e py are 1 periodic. Properties H(0) = 0, H(p) is convex, ) y + ν 1(y) ϕ 1 ν 2 (y) ϕ 2 = H(p)ϕ 1 (y), 0 < y < 1, + ν 2 (y) ϕ 2 ν 1 (y) ϕ 1 = H(p)ϕ 2 (y) H(p) = F (p)(e p 1) with F (p) the total flux (independent of y) F (p) = ( ) ϕ 1 y ψ 1 (y) ϕ 1 ( ) y ϕ 2 y + H(p) [ϕ 1 + ϕ 2 ] 0

24 II. Conventional kinesis : homogenization case The effective hamiltonian H(p) defines the motor effect : Theorem The potential is asymmetric if one of the equivalent properties are true p 0, such that H( p) = 0, F (0) 0, and then F ( p) = 0. Conclusion1 A single number measures the efficiency of the motor Conclusion 2 Potentials are generically asymmetric. Or conversely, it is exceptional that F (0) = 0 Conclusion 3 By perturbations of a symmetric potential one can create a asymmetric potential

25 II. Conventional kinesis : homogenization case The effective hamiltonian H(p) defines the motor effect : Theorem The potential is asymmetric if one of the equivalent properties are true p 0, such that H( p) = 0, F (0) 0, and then F ( p) = 0. Conclusion 1 A single number measures the efficiency of the motor Conclusion 2 Potentials are generically asymmetric. Or conversely, it is exceptional that F (0) = 0 Conclusion 3 By perturbations of a symmetric potential one can create a asymmetric potential

26 II. Conventional kinesis : homogenization case Remark For neutron transport, similar effects are known (asymmetry of micro-structures change growth rates) (Allaire, Capdebosq). The model is set with Dirichlet and the quantity of importance is min H(p) = H( p)

27 II. Conventional kinesis : homogenization case R i,ε Theorem For asymmetric potentials u i,ε = e ε and R i,ε ε 0 R, x (0, 1), R(x) < R(0) or R(1) and the sign of p = R decides between 0 and 1. Proof ε 2 R 1,ε x 2 ε 2 R 2,ε x 2 ( ) R1,ε x ( ) R2,ε 2 x 2 ( ψ 1 (y) R 1,ε ) x R 2,ε R 1,ε + ν 1(y) = ν 2 (y) e ε, + ν 2 (y) = ν 1 (y) ϕ 1 e Use specific a priori bounds, the method of viscosiy solutions (Crandall, Lions), the modulated test function method (L. C. Evans), some duality arguments. R 1,ε R 2,ε ε

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30 III. Flashing rachets u t ε 2 u x 2 ( ψ ( ε t, x ε ) u) = 0, x 0 < x < 1, ε u x + ψ ( ε t, x ε )u(t, x) = 0 at x = 0, 1 (zero flux), u(t, x) ε periodic in time By Floquet theory, this time-periodic solution (up to multiplication by a constant) attracts all the solutions. The same theory applies and asymmetry is characterized by an effective hamiltonian H(p). The limiting effective potential R is independent of time.

31 III. Flashing rachets The effective hamiltonian H(p) is defined by the cell problem Properties ϕ(τ,y) t 2 ϕ(y) y 2 ( ψ (τ, y) ϕ ) = H(p)ϕ(τ, y), 0 < y < 1, y ϕ(τ, y)e py are 1 periodic in y and τ. H(0) = 0, H(p) is convex, H(p) = 1 0 F (τ, p)dτ (e p 1) with F (p) the total flux F (τ, p) = ϕ y (τ, 0) ψ (τ, 0) ϕ(τ, 0) Asymmetric potentials means F (0) 0, p s.t. H( p) = 0

32 Conclusion Conclusion A single number F (0) characterizes if asymmetry and motor effect hold Effciency is given by the number p = R Open questions Conventional kinesis : study the limits t, ε 0 Flashing rachets : other scaling in time Pulsating waves for x R (see Blanchet, Dolbeault, Kowalczyk) Prove rigorously that the set of asymmetric potential is open and dense (see P. Collet & S. Martinez)

33 Conclusion Laboratoire J.-L. Lions UPMC