singular perturbation solutions of steady-state poisson-nernst
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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
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 2
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.
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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.
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 4
A special case 1 : −1
• The steady-state Poisson-Nernst-Planck (PNP) system
−ε2d2φ
dx2= c1 − c2,
−J1 =dc1dx
+ c1dφ
dx,
−J2 =dc2dx− c2
dφ
dx.
• The boundary conditions
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.
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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).
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 6
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.
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The boundary layer solution near x = 0
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.
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].
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The boundary layer solution near x = 1
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.
The boundary layer solution near x = 1:
cr1(Y ) = c1Re−(φr(Y )−φR),
cr2(Y ) = c2Reφr(Y )−φR,
−d2φr
dY 2= c1Re
−(φr(Y )−φR) − c2Reφr(Y )−φR.
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 9
The boundary layer solution near x = 1
The boundary layer solution near x = 1:
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].
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 10
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.
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 11
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).
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 12
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).
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The general case
• 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.
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 14
The general case: previous work
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 15
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]
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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.
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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.
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 18
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.
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 19
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.
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 20
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.
Singular Perturbation Solutions of Steady-state Poisson-Nernst-Planck Systems First Previous Next Last 21
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.
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Thank you!
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