Report CopyRight/DMCA Form For : Lecture 1 The Poisson Boltzmann Equation
Lecture 1 The Poisson Boltzmann Equation I Background I The PB Equation Some Examples I Existence Uniqueness and Uniform Bound I Free Energy Functional Variations I Free Energy Functional Minimizers and Bounds I PB Does Not Predict Like Charge Attraction I References
Background,Coulomb s Law,I potential U21,F21 U21 r r21. Poisson s equation 0,I electrostatic potential,I charge density. I 0 vacuum permittivity,I dielectric coefficient or relative permittivity. The Poisson Boltzmann Equation,0 qj cj e qj f,Poisson s equation x 0 x x. Charge density j 1 qj cj x,Boltzmann distributions cj x cj e qj x. Charge neutrality j 1 qj cj 0,I f R given fixed charge density. I cj R concentration of jth ionic species,I cj bulk concentration of jth ionic species. I qj zj e charge of an ion of jth species zj valence e. elementary charge,I inverse thermal energy 1 kB T,PBE 0 qj cj e qj f. I The Debye Hu ckel approximation linearized PBE,Here 0 is the ionic strength or the inverse Debye. screening length 1,D defined by,I The sinh PBE for 1 1 salt q1 q2 q c2 c1 c. 0 2qc sinh q f,Some Examples, Example 1 A negatively charged plate at z 0 with constant. surface charge density 0 and with 1 1 salt solution in z 0. 00 8 ec sinh e e for z 0,The solution is,D Debye screening length. lGC Gouy Chapman length,lB Bjerrum length, Example 2 A spherical solute with a point charge at center. immersed in a solution with multiple species of ions. Debye Hu ckel approximation,0 r R w 0 2 Q O,4 m 0 r R 4 w 0 R 1 R. 4 w 0 1 R r,The Yukawa potential,Solution of,Existence Uniqueness and Uniform Bound. Consider the boundary value problem of PBE,PBE 0 B 0 f in. B 1 M e qj 1,Define Z h,I 2 f B dV,Hg H 1 g on,Theorem Li Cheng Zhang SIAP 2011. I The functional I Hg1 R has a unique minimizer,I The minimizer is bounded in L uniformly in. I The minimizer is the unique solution to the boundary value. problem of PBE,I 2 f B dV, Proof Step 1 Existence and uniqueness of minimizer. First the lower bound by Poincare inequality,I C1 k k2H 1 C2 Hg1. Let inf Hg1 I Then is finite There exist k Hg1,k 1 2 such that I k By the lower bound k is. bounded in H 1 Hence it has a subsequence not relabeled. such that k weakly in H 1 and a e in for some, Hg1 The weak convergence and Fatou s lemma lead to. Uniqueness of minimizer follows from the strict convexity of I. I u 1 v I u 1 I v 0 1,Step 2 The L bound for uniform for min max. Let g Hg1 be such that 0 g f Then g is,bounded in L uniformly in Let 0 H01 be the unique. minimizer in H01 of,J 2 B g dV,Then 0 g Prove k 0 kL C uniform in. Since B 0 there exists 0 with B 0 0 1 and,B 0 0 1 a e in Note is uniform in Define by. x 0 x if 0 x,Then H01 We have J 0 J and 0,B g 0 dV B g dV. Consequently we have by the convexity of B R R that. 0 B g 0 B g dV,B g 0 B g dV,B 0 g 0 dV,B 0 g 0 dV,Hence 0 0 and 0 a e. Step 3 The minimizer is the unique solution to the. boundary value problem of PBE,Routine calculations. I I t 0 Cc1,Since L we have,0 f B 0 dV 0 H01, So is a weak solution to the boundary value problem of PBE. Uniqueness again follows from the convexity Q E D,Free Energy Functional Variations. Electrostatic free energy functional of ionic concentrations. 1 cj ln 3 cj 1,G c j cj dV,x f x j 1 qj cj x,0 B C e g 0 on. I the thermal de Broglie wavelength,I j chemical potential for the jth ionic species. Observations,j 1 qj cj is affine in c So is linear and. quadratic in c,I G c is strictly convex in c,G u 1 v G u 1 G v 0 1. G c cj ln cj 1 j cj dV,First variations,G c tdj ej G c j dj dV. G c j qj 1 ln 3 cj j,Equilibrium conditions,G c j 0 j cj x cj e qj x j. cj 3 e j These are the Boltzmann distributions, Minimum value of G is the electrostatic free energy the PB free. energy given by note the sign,Gmin cj e qj 1 dV,Second variations. G c u v G c tv u,qj qk uj Lvk dV,In particular if u v then. 2 G c u u qj uj L qj uj dV 0,So G is convex,Free Energy Functional Minimizers and. X c c1 cM L1 RM qj cj H 1,Theorem B L SIMA 2009,I The functional G has a unique minimizer c X. I There exist constants 1 0 and 2 0 such that,1 cj x 2 a e x j 1 M. I All cj are given by the Boltzmann distributions, I The corresponding potential is the unique solution to the PBE. Remark Bounds are not physical A drawback of the PB theory.