singular perturbation solutions of steady-state poisson-nernst

Singular Perturbation Solutions of Steady-statePoisson-Nernst-Planck Systems

Xiang-Sheng Wang

Memorial University of NewfoundlandSt. John’s, Newfoundland, Canada

joint work withDr. Dongdong He (City University of Hong Kong)

Dr. Huaxiong Huang (York University, Toronto, Canada)Dr. Jonathan J. Wylie (City University of Hong Kong)

International Conference on Approximation Theory and ApplicationsCity University of Hong Kong, May 24, 2013.

Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 1

Outline

• The singular perturbation problem

• A special case

– The outer solution– The boundary layer solutions– Asymptotic matching– Asymptotic solution

• The general case

• Numerical evidence

• Conclusion


The singular perturbation problem

• The steady-state Poisson-Nernst-Planck (PNP) system

−ε2d2φ

dx2=

n∑i=1

zici,

−Ji =dcidx

+ zicidφ

dx.

• The boundary conditions

ci(0) = ciL and φ(0) = φL,

ci(1) = ciR and φ(1) = φR.

• zi are given integers and Ji are unknown constants.

• ε > 0 is small.


The singular perturbation problem: previous works

• n = 2 with z1 = 1 and z2 = −1 (1 : −1 case):

1. V. Barcilon, D.-P. Chen, R. S. Eisenberg, and J. W. Jerome [SIAP, 1997]obtained an explicit asymptotic formula.

2. J.-H. Park and J. W. Jerome [SIAP, 1997] proved uniqueness of solution.

• n = 2 with z1 = α and z2 = β (α : β case): W. Liu [SIAP, 2005] provedexistence and uniqueness of solution.

• The general case (n ≥ 2): W. Liu [JDE, 2009] reduced the problem into solvinga system of nonlinear algebraic equations. Multiplicity of solutions for n ≥ 3was also observed.


A special case 1 : −1


−ε2d2φ

dx2= c1 − c2,

−J1 =dc1dx

+ c1dφ

dx,

−J2 =dc2dx− c2

dφ

dx.


c1(0) = c1L, c2(0) = c2L and φ(0) = φL,

c1(1) = c1R, c2(1) = c2R and φ(1) = φR.

• J1 and J2 are unknown constants.


The outer solution

The limited system in outer region:

0 = co1 − co2,

−J1 =dco1dx

+ co1dφo

dx,

−J2 =dco2dx− co2

dφo

dx.

The outer solution:

co1(x) = co2(x) = −(J1 + J2)x

2+ co1(0),

φo(x) =J1 − J2J1 + J2

log−(J1 + J2)x+ 2co1(0)

2co1(0)+ φo(0).


The boundary layer solution near x = 0

The limited system near the left boundary layer (X = x/ε):

−d2φl

dX2= cl1 − cl2,

0 =dcl1dX

+ cl1dφl

dX,

0 =dcl2dX− cl2

dφl

dX.

The boundary layer solution near x = 0:

cl1(X) = c1Le−(φl(X)−φL),

cl2(X) = c2Leφl(X)−φL,

−d2φl

dX2= c1Le

−(φl(X)−φL) − c2Leφl(X)−φL.




cl1(X) = c1Le−(φl(X)−φL),

cl2(X) = c2Leφl(X)−φL,

−d2φl

dX2= c1Le

−(φl(X)−φL) − c2Leφl(X)−φL.

Multiplying the last equation by dφl

dX and integrating it from X to ∞ yields(dφl

dX(X)

)2

−(dφl

dX(∞)

)2

= 2c1L[e−(φl(X)−φL) − e−(φ

l(∞)−φL)]

+ 2c2L[eφl(X)−φL − eφ

l(∞)−φL].



The limited system near the right boundary layer (Y = (1− x)/ε):

−d2φr

dY 2= cr1 − cr2,

0 =dcr1dY

+ cr1dφr

dY,

0 =dcr2dY− cr2

dφr

dY.


cr1(Y ) = c1Re−(φr(Y )−φR),

cr2(Y ) = c2Reφr(Y )−φR,

−d2φr

dY 2= c1Re

−(φr(Y )−φR) − c2Reφr(Y )−φR.




cr1(Y ) = c1Re−(φr(Y )−φR),

cr2(Y ) = c2Reφr(Y )−φR,

−d2φr

dY 2= c1Re

−(φr(Y )−φR) − c2Reφr(Y )−φR.

Multiplying the last equation by dφr

dY and integrating it from Y to ∞ yields(dφr

dY(Y )

)2

−(dφr

dY(∞)

)2

= 2c1R[e−(φr(Y )−φR) − e−(φ

r(∞)−φR)]

+ 2c2R[eφr(Y )−φR − eφ

r(∞)−φR].


Asymptotic matching

The unknown constants Ji and the integration constants can be determined bythe matching conditions:

φo(0) = φl(∞) and coi (0) = cli(∞),

φo(1) = φr(∞) and coi (1) = cri (∞).

Recall that X = x/ε and Y = (1− x)/ε, so we have

dφl

dX= ε

dφo

dx= O(ε)

anddφr

dY= −εdφ

o

dx= O(ε)

as ε→ 0.


Asymptotic solution

Denote a := (c1L/c2L)1/4 and b := (c1R/c2R)

1/4.

φl(X) = φL + 2 loga[1 + a+ (1− a)e−

√2(c1Lc2L)

1/4X]

1 + a− (1− a)e−√2(c1Lc2L)

1/4X,

φr(Y ) = φR + 2 logb[1 + b+ (1− b)e−

√2(c1Rc2R)

1/4Y ]

1 + b− (1− b)e−√2(c1Rc2R)

1/4Y,

φo(x) =φR − φL + log

√(c1Rc2L)/(c2Rc1L)

log√(c1Rc2R)/(c1Lc2L)

log

[(√c1Rc2Rc1Lc2L

− 1

)x+ 1

]+ φL + log

√c1Lc2L

,

φ(x) = φo(x) + φl(xε

)+ φr

(1− xε

)− φl(∞)− φr(∞) +O(ε).


Asymptotic solution

cl1(X) = c1Le−(φl(X)−φL) and cl2(X) = c2Le

φl(X)−φL,

cr1(Y ) = c1Re−(φr(Y )−φR) and cr2(Y ) = c2Re

φr(Y )−φR,

co1(x) = co2(x) = (√c1Rc2R −

√c1Lc2L)x+

√c1Lc2L,

ci(x) = coi (x) + cli

(xε

)+ cri

(1− xε

)− cli(∞)− cri (∞) +O(ε), i = 1, 2.

The constants:

J1 = −φR − φL + log(c1R/c1L)

log√

(c1Rc2R)/(c1Lc2L)(√c1Rc2R −

√c1Lc2L),

J2 =φR − φL + log

√(c2L/c2R)

log√(c1Rc2R)/(c1Lc2L)

(√c1Rc2R −

√c1Lc2L).


The general case


−ε2d2φ

dx2=

n∑i=1

zici,

−Ji =dcidx

+ zicidφ

dx.


ci(0) = ciL and φ(0) = φL,

ci(1) = ciR and φ(1) = φR.

• zi are given integers and Ji are unknown constants.

• ε > 0 is small.


The general case: previous work


The general case: previous work

The problem is reduced to a system of nonlinear algebraic equations:

“The nonlinear system is very complicated and the analysis is not carried out inthis paper.” – W. Liu [JDE, 2009]


The general case: our contributions

• By carefully maintaining the symmetries of the PNP system, we reduce theproblem to solving a single scalar transcendental equation rather than a systemof nonlinear equations found by W. Liu [JDE, 2009].

Pn−2(λ)eλ = Qn−2(λ),

where Pn−2(λ) and Qn−2(λ) are polynomials of degree n− 2.

• We confirm the observation of multiple solutions for the cases n ≥ 3 by W. Liu[JDE, 2009]. Furthermore, we prove that for the case n = 3, the oscillationsolutions are nonphysical and the (unique) non-oscillation solution is physical.

• Reference: X.-S. Wang, D. He, H. Huang and J. J. Wylie, Singular perturbationsolutions of steady-state Poisson-Nernst-Planck systems, submitted.


Numerical evidence

0.2 0.4 0.6 0.8 1.0 x

0.2

0.4

0.6

0.8

1.0

Φ

(a) Potential

c1

c2

0.2 0.4 0.6 0.8 1.0 x

1.5

2.0

2.5

3.0

c

(b) Concentrations

Figure 1: Comparison of asymptotic solutions (solid lines) with numericalsolutions (dotted lines) of the 3 : −2 case with the boundary conditions:φL = 0, c1L = 1, c2L = 1, φR = 1, c1R = 3, and c2R = 2. Here, wechoose ε = 0.1.


Numerical evidence

0.2 0.4 0.6 0.8 1.0 x

0.2

0.4

0.6

0.8

1.0

1.2

Φ

(a) Potential

c1

c2

c3

0.2 0.4 0.6 0.8 1.0 x

1

2

3

4

5

c

(b) Concentrations

Figure 2: Comparison of asymptotic solutions (solid lines) with numericalsolutions (dotted lines) of the 2 : 1 : −1 case with the boundary conditions:φL = 0, c1L = 1, c2L = 2, c3L = 3, φR = 1, c1R = 0.1, c2R = 5, and c3R = 2.Here, we choose ε = 0.1.


Numerical evidence

0.2 0.4 0.6 0.8 1.0 x

0.2

0.4

0.6

0.8

1.0

Φ

(a) Potential

c1

c2

c3

0.2 0.4 0.6 0.8 1.0 x

0.5

1.0

1.5

2.0

2.5

3.0

c

(b) Concentrations

Figure 3: Comparison of asymptotic solutions (solid lines) with numerical solutions(dotted lines) of the 2 : 1 : −1 case with the (electroneutral) boundary conditions:φL = 0, c1L = 0.1, c2L = 2, c3L = 2.2, φR = 1, c1R = 1, c2R = 1, and c3R = 3.Here, we choose ε = 0.1.


Numerical evidence

0.2 0.4 0.6 0.8 1.0 x

0.2

0.4

0.6

0.8

1.0

Φ

(a) Potential

c1

c2

c3

c4

c5

0.2 0.4 0.6 0.8 1.0 x

0.5

1.0

1.5

2.0

c

(b) Concentrations

Figure 4: Comparison of asymptotic solutions (solid lines) with numerical solutions(dotted lines) of the 3 : 2 : 1 : −1 : −2 case with the boundary conditions:φL = 0, ciL = 1, φR = 1, c1R = 0.1, c2R = 0.1, c3R = 2, c4R = 2 and c5R = 0.1.Here, we choose ε = 0.1.


Conclusion

• Explicit asymptotic solution of general PNP system.

• Uniqueness of physical solution for the case n = 3.

• Open problem: uniqueness of physical solution for the cases n ≥ 4.


Thank you!


singular perturbation solutions of steady-state poisson-nernst

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