Mathematical Methods in the Applied Sciences, Vol. 12,405-412 (1990) MOS subject classification: 35 J 60, 78 A 35

Radially Symmetric Poisson-Boltzmann Equation in a Domain Expanding to Infinity

A. Krzywicki and T. Nadzieja

Institute oJ Mathematics, Wroclaw University, PI. Grunwaldzki 214, 50-384 Wroclaw, Poland

Communicated by E. Meister

The electric potential u in a solution of an electrolyte around a linear polyelectrolyte of the form of a cylinder satisfies

(ru')' = r f ( u ) , 0 < a < r < R ,

~ ' ( 0 ) = - U / U , u(R) = 0.

We study the problem when R +co.

The mathematical model of polyelectrolytes, i.e. polymeric chains which, in solu- tion, dissociate into a polymeric core and mobile counterions, is described by the following boundary value problem 121:

(ru')' = rf(u), u = u(r), r E I = [a, R ] , (1)

u'(a) = - o / a , u(R) = 0,

when

f (4 = fJ lr reIx:;::)dr -k ::;"d;;:)cir - [, r:x:(:T:) dr

and where a, N, a, p, CT are positive constants. Equation (1) presents the two-dimensional Poisson-Boltzmann equation in a sim-

plified form due to assumed radial symmetry. The equation describes the electric potential u in a solution of an electrolyte containing some number of ions with total charge N and the same number of counterions with total charge - N of the solute as well as additional ions dissociated from polyelectrolyte with total charge o. The functionf is the charge density in the solution and its particular form results from the assumed Boltzmann law.

In [l], [4] a similar physical situation considered in an exterior region is described by the following equation:

AU = Csh(u), u = ~ ( l x l ) , 1x1 > a (2)

where a, C are positive constants.

01 704214/90/05040SO8$05.00 0 1990 by B. G. Teubner Stuttgart-John Wiley & Sons, Ltd.

Received 12 April 1989 Revised 18 July 1989

406 A. Krzywicki and T. Nadzieja

We will prove rigorously that (2) can be obtained from (1) by passing to the limit

The existence and uniqueness of the solutions of the problems (1) and (2), respect- R + 00 assuming that N depends upon R such that lim N R - ’ exists.

ively, subject to the conditions

u’(a) = - C J , u ( al) = 0

are proved in [2] and [3], respectively. We begin our considerations with a few simple remarks. First of all note that when

considering a particular solution u of ( I ) , all three denominators inf(u) are constants, so u satisfies an ordinary differential equation (1) with a function f of the form

x exp (au) + y exp (Du) - z exp ( - j u ) ;

thus all arguments used for ODES may be applied in our case too. Moreover for x, y, z positive, and this is the case here,f(u) is strictly increasing with u, a fact which will be often used in the following.

Next note that the solution u of (1) satisfies an additional condition

u’(R) = 0, (3) which one obtains by integrating (1) over I.

Lemma 1. For any solution of(1)

u’ 6 0, u” > 0, f ( u ) > 0 on I .

(4)

Pro05 Consider u as a solution of an ordinary differential equation. We may simplify our reasoning by introducing a new variable F = log (r) which transforms (1) to

U ” = exp (2f ) . f ( U ), (5 ) and boundary conditions ( l ) , (3) to

P‘@) = - C J ,

P (R) = u ’ ( R ) = 0 ,

where U (F) = u(r), U ’ = du /dF = exp (F )u’, = log a < r < log R = R. We have f(0) # 0, otherwise u = 0 as a unique solution of the initial value problem (5), (7). Assumingf(0) < 0 we would have i”(J?) < 0 what implies due tof‘ > 0 and (6) the existence of Fo < R such that U ’ > 0, U < 0, .f(u) < 0 on ( ro , R), P’(ro) = 0, which contradicts the equality

- C‘(r, ,) = 1; exp(2f)f(u(F))df 1 0

following from ( 5 ) by integration. Thereforef(0) > 0 and a similar reasoning as above gives us ii ‘ < 0 for F < l?, i.e. u’ < 0 for r < R which impliesf(u) > 0. Coming back to (1) we see that

u” = - u’ /r +f(u) > 0

and, in consequence,

Radially Symmetric Poisson-Boltzmann Equation 407

The integral form of (1)

u(r) = u log (R/r) - j,R t - ' dr j: sf(u(s))ds (9)

allows us to estimate u, using u' < 0 on I

0 < u(r) < u(a) < u log ( R l a ) (10)

Theorem 1. The solution of(1) tends to zero when N + co (we keep all other constants fixed). More precisely, u < CN-'12 .

Proof: It will be convenient to writefin the form

f ( u ) = upo exp (a4 + N P + exp(84 - N P - exp( - Bu), where

p;' = IaRrexp(uu)dr, p ; ' = rexp( 8u)dr. IaR From Lemma 1, f(u) > 0; in particular, for r = R

aCco + "P+ - P-) > 0,

which may be rewritten in the form

Owing to (10) and the definition of p o , p+ the right-hand side of (1 1) is less than C N - ' . On the other hand by (8)

laR r sh( pu) dr 2 ap laR u dr 2 ap laR t dr,

where m + o(a - r)/a, for a < r < a + a m / o ,

for r 2 a + amla, t(r) =

with m = u(a) and we obtain the inequality

a2Bm2/(2u) c C N - ' ,

from which, owing to u' < 0, the required inequality follows.

We will now pass with R to infinity, that is we will expand the region of definition of u to all the exterior of the disc r < a. To have a physically interesting situation we will assume N to depend upon R in such a way that the limit concentration of ions of the solute lim N R - ~ = p 3 0 exists.

Theorem 2. If lim NR-' = p > 0 then the solutions uR of(1) haue a limit u = lim uR,

uniform on each jinite interual, satisfying the equation

R - . 1)

R-CC

(ru')'/r = 4p sh(Pu), r > a (12)

o'(a) = - o/a , u( co) = 0.

and the boundary conditions

408 A. Krzywicki and T. Nadzieja

Moreover, u satisfies the inequality

u(r) < C / r , (13)

with a constant C depending on the parameters of the problem.

Proof: The proof will be based on the fact that the family { u R } , R 2 R,, of solutions of (1) is uniformly bounded and separated from zero (see Lemma 2 below) as well as uniformly Lipschitzian due to the estimate lukl < o/a on the interval (a, R,) for any R,. Therefore, by the Arzela-Ascoli theorem, a convergent sequence {u~,}, R, + co, may be chosen having a limit function u not identically zero. The vanishing of the function u at infinity implies that both limits Np+ and Np- have the same value 2p and p, -+ 0. Passing to the limit in (9) we see that u satisfies (12). The latter boundary value problem has exactly one solution [3], therefore all the family { u R } tends to u.

To complete the proof of Theorem 2 we have (i) to establish a uniform boundedness of uR from above and from beneath, and (ii) to show (13). Owing to the monotonicity of any uR and u i > 0, to establish (i) it suffices to show

Lemma 2. The following relations hold true:

0 < inf uR(a),

sup uR(a) c co. R

R

(14)

(15)

Proof of (14). Denote mR = uR(a) (to simplify the notations we will in the following omit the subscript R and write u instead of uR and m instead of mR). We have two obvious estimates

p,, p+ < 2(R2 - a’)-’ < 3R-2 (16)

valid for large R, which yield

f(u) < f(m) < 3oR-’exp(am) + 3NR-’exp(pm) < 4pexp(ym)

for large R and y = max {a, S}. Therefore

r - l 1: sf(u)ds < 2pexp(ym)(r2 - a 2 ) / r < a/(2r)

if only

r < a + 6, 6 = [a2 + (oexp( - ym))/(4p)]’” - a .

Integrating now the equation (1) over (a, r ) we obtain

u(r) = - a/r + r - ’ 1; sf(u)ds;

therefore on the interval [a, a + 61

lu’l 2 o/(2r).

From the last inequality and from the relation

jaR ru” dr + JaR ruf(u) dr = om,

Radially Symmetric Poisson-Boltzmann Equation 409

which results from multiplying (1) by u and integrating over (a, R) we obtain, owing to the positivity off(u),

(17) 0

m >-min (log(1 + oexp( - ym)(4aZp)-’),210g(R/a)}, 8

from which (14) follows.

Proofof(15). To show (15), a more delicate estimate than (16) will be needed. Note that

(18)

for a < r < R owing to the convexity of u. So we have

u(r) < w(r) = m(R - r)/(R - a)

and by elementary calculations we get for right-hand side integrals the formulae

where the upper signs refer to the first and the lower to the second integral, respectively.

From the first inequality (19) and from N >, R2 p/2, valid for large R, we obtain

1 p(8m)’ 2 exp(Bm) - Bm - 1 + $rn[exp(Bm) - 13 ’ N p + 2-

where E = a(R - a)- ’. We know that m < 0 log (R/a); therefore &/?m < 1 for large R, which yields

Np+exp(Brn) 2 pP2m2/4.

From (19) we have also

exp(pm) - Bm - 1 + @m[exp(Bm) - 13 exp ( - Bm) + Bm - 1 + $m[ 1 - exp ( - Bm)] ’ P - / P + <

and taking R so large that @m < 1, as before we obtain

and the right-hand side goes to zero when m + 00 uniformly with respect to R. So we have for sufficiently large m

(20) “ p + exp(Bm) - P - exp( - Bm)I 3 N P + exp(Bm)/2 2 pB2m2/8

uniformly with respect to large R. The proof of (15) will be based on the equality

- au’(a)exp(/?m/2) = (p/2) rexp(jIu/2)u”dr JaR

410 A. Krzywicki and T. Nadzieja

resulting from (3) by integrating over (a, R) the equation (1) multiplied by exp(pu/2). By applying the Cauchy inequality and the boundary condition we obtain from (21) the relation

where $(u) = exp(pu)f(u). The function $ is positive together with all its derivatives, therefore

0 < $’M < $’W,

so we have

= 2a(3$’(m))-’[($(m))”’ - (I)(O))~/~].

Now $(m) = exp (Bm) Capo exp (am) + N p + exp (Pm) - Np- exp ( - Bm)l

$ ’ ( 4 = exp(Bm1 Capo(a + B)exp(am) + 2 W p + exp(pm)I,

w w $ ’ ( m ) 2 c,

+(m), $’(m) 2 Cpmzexp(pm),

($(m))3/2/$’(m) 3 Cmp1’’exp(Bm/2),

($(0))3/2/$’(m) < Cp’/’exp( - Bm),

and

from which it is seen by the first inequality of (20) that

with some positive constant C. Owing to (20) we have for sufficiently large R

which implies

where in the last inequality the estimate

$(O) = apu, + N p + - N p - < 4p

and (14) have been used. From the last inequalities and from (22) we obtain finally

from which (15) follows, and this completes the proof of Lemma 2.

Proofof(13). Take any ro > a and R =- ro and consider the equality of the form (21) where integrals are taken over ( ro , R). On its left-hand side we have now - rou’(rO)exp(/3m/2), and - rou‘(ro) is positive and less than 0, cf. (8), m meaning

now u(ro) . The final estimate (23) remains valid with a replaced by ro, which gives

m = u(ro) < C / r o ,

with some positive constant C, for any u = u R , R > ro, and this completes the proof of Theorem 2.

Radially Symmetric Poisson-Boltmann Equation 41 1

The following theorem completes the analysis of the limit passage R + co.

Theorem 3. If N R - 2 + O for R co, then lim uR = 00 and i f N R - 2 -+ co then lim uR = 0, uniformly on each finite interual.

Proof. Consider the case p = N R - ’ -+ 0. Following the first part of proof of Theorem 2 we see that (17) holds true with p arbitrarily small, which implies m + 00 and, owing to u” > 0, u + 00.

If p = N R - 2 -+ CQ, inspection of the second part of the proof of Theorem 2 shows that (23) holds true with p -+ co; therefore sup m is finite. It follows that p o / p - is bounded; therefore the right-hand side of the inequality (1 1) may be estimated by C ( p + N ) - ’ < C p - ’ . Repeating now the argument used in the proof of Theorem 1 we obtain the estimate

m < Cp-”’,

and the desired result follows.

when f in (1) is replaced by All the results of this paper may be easily extended to the more general situation

k

~ ( u ) = poexp(au) + C ~ ~ ( p : exp(&u) - pL exp( - &u)) i = 1

and the boundary conditions (1) remain unchanged, which corresponds to the case when the electrolyte contains k dissociating solutes. The positive constants Ni, p i , p: , pL denote the corresponding values of N, /I, p + , p - for the ith solute. F(u) is strictly increasing with respect to u; therefore the same arguments may be used here and the thesis of Theorem 1 holds true if only at least one of the total charges N i goes to infinity.

Consider now the case R + co. Let N i = N , ( R ) and let p i = N i R - 2 - + p i when R -+ co. The corresponding generalization of Theorems 2 and 3 reads now:

k

I. If 0 < c p i < co, the solutions uR of the problem i = 1

(rub)' = rF(uR) , a < r < R ,

tend uniformly to the solution u of k

(ru’)’ = 4r c pi sh(Biu), i = 1

u’(a) = - a/a, u ( c o ) = 0,

when R -+ co. k k

11. If 1 p i = 0, the solutions of (25) tend to infinity and if c p i = co, the i = 1

solutions tend to zero. i = l

Remark. In the three-dimensional radially symmetric case the problem corresponding to (1) takes the form

(r2u’)’ = r 2 f ( ~ ) , ~ ’ ( a ) = - ~ / a ’ , u ( R ) = 0,

412 A. Krzywicki and T. Nadzieja

with the samefonly with the integrals appearing there replaced by 1, r 2 exp (au) dr etc. All the results concerning (1) remain true without any change; moreover this time the estimates sup m < co of Lemma 2 and u 6 C/r of Theorem 2 follow trivially from the integral form of the problem

u(r) = a[(l/r) - ( l /R) ] - J,R t -2d t J:sy(u)ds.

The only exception is the first part of Theorem 3: when R + co and N R - 3 --, 0 then the corresponding solutions tend to a/r and this makes clear why in the two-dimensional case the limit was infinite.

References

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3. Krzywicki, A. and Nadzieja, T., ‘Radially symmetric solutions of the Poisson-Boltzmann equation’,

4. Lampert, M. A., The Coulomb condensate of the nonlinear Poisson-Boltzmann equations: a unified

Erscheinungen’, Phys. Ztschr., 24, 185-217 (1923).

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