Solving Nernst-Planck Equations-Project Report-

Leo Kim23163058

Physics 210

05 December 2007

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1 Introduction

The first and ultimate goal of this project is solving Nernst-Planck equation numerically. I de-veloped and expanded this topic from my old project topic, the simple diffusion equation. To behonest with you, as you already mentioned, it is very interesting but very ambitious. However, Idid learn a lot from this project although I didn’t accomplish what I intended to do in the firstplace. I used the various types of programs to do this project, including Maple, Mathematica,Excel, and many different kind of image tools. It is really hard to edit and produce the postscriptfiles on the Window without any pro-programs such as Photoshop.

Not only I learned and used various type of computer programs to fulfill the computationalpart of the course, but I also read different books to understand the actual physics behind it.This is the computational PHYSICS course, so I believe I actually need to understand the physicsof my project. Moreover, I also believe that it is really important to do the physics topic whichyou feel interest about. For such a reason, this project gives me that bittersweet taste of thechallenging but fun task.

Basically, I did start from the scratch. The heart of my project is finite difference methodwhich will solve the PDE numerically. I did sample examples for the certain wave and heatproblems with finite difference method in Excel. Then I used Maple and Mathematica to writea code to solve PDE in such a way and plot them. For the last part, I actually tried to solveNernst-Planck equation with Excel which turns out to be interesting.

2 Theory

This section explains briefly about the Nernst-Planck equation, and finite difference method forPDE. Also,

2.1 Nernst-Planck equation

Nernst-Planck equation can be thought as the advanced version of the simple diffusion equationbecause it describes the concentration change due to the properties of diffusion and effect of theelectrostatics. And you can say it is more realistic as well because pretty much everything inour body has a net charge as an ion. Although, I did find the article which explains how tocompute exact solution but even with that article[1] I did have some difficult time to understandthe solution fully. As I already mention, PDE needs certain boundary conditions to be solved. Iwill choose closed system where no extra ions can be diffused in and out. Then I will use equationto find flux of anions and cations over long time and eventually they will be in equilibrium state.For this particular case, the Nernst-Planck equations[1] will be

Φa(x, t) = −Da(∂na/∂x)− (qDana/kT )(∂φ/∂x), (1)Φb(x, t) = −Db(∂nb/∂x) + (qDbnb/kT )(∂φ/∂x), (2)

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∂na/∂t = −∂Φa/∂x, (3)∂nb/∂t = −∂Φb/∂x, (4)

where na is the number of cations per unit volume, Φa is the flux of cations, nb is the numberof anions per unit volume, Φb is the flux of anions, φ is the electric potential, k is Boltzmann’sconstant, T is the absolute temperature, and Da and Db are the diffusion coefficients of thecations and the anions. q and -q are charge of cation and anion. However, this PDE is notsimple to solve even though you use the computer program because it is non-linear PDE. But,you can apply the physical conditions, which is just described previously, you can get the linearPDE.

∂2Φ/∂x2 = −(4/ε)(na − nb), (5)

Equation (5) is the Poisson’s equation which will take care of the potential term of the originalNernst-Planck equation. Then you need the boundary conditions.

∂na(0, t)/∂x = 0, ∂nb(0, t)/∂x = 0, (6)∂na(L, t)/∂x = 0, ∂nb(L, t)/∂x = 0, (7)

After sub-in the equation (5) to the original Nernst-Planck equation and applied equation (6)and (7) produce,

∂na/∂t = Da(2na/∂x2) + (qDan

∗/kT )(∂2Φ/∂x2), (8)∂nb/∂t = Db(2nb/∂x

2) + (qDbn∗/kT )(∂2Φ/∂x2), (9)

where the n∗ is the number of the ions in the equilibrium. However, you can neglect the effectof the potential terms due to the effect of the ionic screening[2]. Effect of the ionic screeningintroduces the co-ions around the charge which will make the electronic interaction be valid forwithin nano length so the potential terms have neglecting value. And finally, you get the equa-tions in form,

Equations to be solved by numerical method

∂na/∂t = Da(∂2na/∂x2)− αDa(na − nb), (11)

∂nb/∂t = Db(∂2nb/∂x2) + αDb(na − nb),(12)

where na and nb are function of x and t. Da, Da, and α are constants.

2.2 The Finite Difference Method

[3] Finite Difference Method is the main tool of my project. Its concept is quite simple butactual computation might take long and boring. There are two different ways to solve the heator wave equation with this method. But, I will show only the heat equation since the simplifiedNernst-planck equation indeed looks like the heat equation but Nernst-Planck equation has theextra ”coupled term”.

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First you have to choose the time and distance steps for the grid in which the the numericalsolution of the certain equation will be found. Of course, you can decide the how big is the gridwill be. And, yes if you can produce more grid, that will be better. It is same as you take moreN for your normal numerical solution. Your discretization expression will find the value of thenext layer of the grid then you can use that value to find the next upper layer value of the gridand so on.

Now, this is time to do the discretization of the heat equation. The basic and general heatequations have first partial derivative respect to time and second partial derivative respect to thedistance. Then discretization of each of term will be,

ut(ih, jk) ≈ 1/k(ui,j+1 − uij), (13)uxx(ih, jk) ≈ 1/h2(ui+1,j − 2uij + ui−1.j), (14)

where k is the time step and h is the distance step.

And finally, you can produce the discretization of the heat equation,

ui,j+1 = (1− 2s)uij + s(ui+1,j + ui−1,j), (15)s = k/h2, (16)

Similarly, you can find the discretization of the wave equation,

ui,j+1 = 2(1− s)uij + s(ui+1,j + ui−1,j)− ui,j−1, (17)s = k2/h2, (18)

I work out sample wave equation for practice in Excel and you can find out them in the Resultsection.

3 Results and Discussion

In this section, there will be results of work and discussion.

3.1 Practice: Numerical solution for wave function on Excel

For this particular example, I used the Excel to find the numerical values. Excel provides theexcellent grid for me so all I need to do is that determine and set the constants for the time anddistance steps. Then I need to carefully type in the expression of the discretization of the waveequation.

Wave function is given,

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utt = 4uxx, 0x30, 0t30, (19)u(0, t) = 0, (20)u(L, t) = 0, (21)

u(x, 0) = p(x), (22)ut(x, 0) = 0, (23)

where p(x) is

Although my Nernst-Planck equation is similar to heat equation, I did the wave equationbecause if I plot the numerical solution then I know if that numerical solution is wrong or rightby just looking at the graph. If my numerical solution for the wave is correct, it should be peri-odic function. But, I can’t be sure that by just looking at the 1-D graph of the heat equation,the numerical solution of the heat equation is right or wrong. And, indeed, the graph of thenumerical solution of wave looks promising.The size and resolution of the image are poor but you can see the full size version in my Excel

Figure 1: Graph of the numerical solution of the wave function

file(wave function.xls). You just need to change the worksheets by clicking which are located atthe left corner of the excel documents. However, the shape of graph is drawn as expected. Asyou can see, every time the steps are added, the shape of the graph becomes more clear. Thus, Iuse the smaller time step so I can improve the numerical solution which is also done in the sameexcel file.

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3.2 Practice:Wave equation solving code in Mathematica

Originally, I was trying this job with the Maple but something terrible things happened. TheMaple was too heavy for my computer so long time usage of the Maple crushed my system,exactly its own works. So I can’t use any Maple data or code for this project even if there weresome. I restart everything with the Mathematica and I think I did make working code.

Actual code:[4],[5],[6].FDgrid[n ,m ]:=Module

[{i}, u = Table[0, {n}, {m}]; For

[i = 1, i ≤ n, i++, u[[i,1]] = f [i];

];FDgrid[n ,m ]:=Module

[{i}, u = Table[0, {n}, {m}]; For

[i = 1, i ≤ n, i++, u[[i,1]] = f [i];

];FDgrid[n ,m ]:=Module

[{i}, u = Table[0, {n}, {m}]; For

[i = 1, i ≤ n, i++, u[[i,1]] = f [i];

];

For[i = 2, i ≤ n− 1, i++, u[[i,2]] =

(1− r2

)f [i] + k g[i] + r2

2 (f [i+ 1] + f [i− 1]);]

;]

;For[i = 2, i ≤ n− 1, i++, u[[i,2]] =

(1− r2

)f [i] + k g[i] + r2

2 (f [i+ 1] + f [i− 1]);]

;]

;For[i = 2, i ≤ n− 1, i++, u[[i,2]] =

(1− r2

)f [i] + k g[i] + r2

2 (f [i+ 1] + f [i− 1]);]

;]

;

This code will generate the grid.

FDsolve[n ,m ]:=FDsolve[n ,m ]:=FDsolve[n ,m ]:=Module[{i, j},For[j = 3, j ≤ m, j++,Module[{i, j},For[j = 3, j ≤ m, j++,Module[{i, j},For[j = 3, j ≤ m, j++,For[i = 2, i ≤ n− 1, i++,For[i = 2, i ≤ n− 1, i++,For[i = 2, i ≤ n− 1, i++,u[[i,j]] =

(2− 2r2

)u[[i,j−1]] + r2

(u[[i+1,j−1]] + u[[i−1,j−1]]

)−u[[i,j]] =

(2− 2r2

)u[[i,j−1]] + r2

(u[[i+1,j−1]] + u[[i−1,j−1]]

)−u[[i,j]] =

(2− 2r2

)u[[i,j−1]] + r2

(u[[i+1,j−1]] + u[[i−1,j−1]]

)−

u[[i,j−2]];]

;]

;]

;u[[i,j−2]];]

;]

;]

;u[[i,j−2]];]

;]

;]

;

This code will actually solve the wave equation way I did in Excel.

Now I will set up the wave equation for this particular example.a = 1.0;a = 1.0;a = 1.0;b = 1.0;b = 1.0;b = 1.0;c = 2.0;c = 2.0;c = 2.0;n = 11;n = 11;n = 11;m = 21;m = 21;m = 21;F [x ] = Sin[πx] + Sin[2πx];F [x ] = Sin[πx] + Sin[2πx];F [x ] = Sin[πx] + Sin[2πx];G[x ] = 0.0;G[x ] = 0.0;G[x ] = 0.0;h = a

n−1 ;h = an−1 ;h = an−1 ;

k = bm−1 ;k = bm−1 ;k = bm−1 ;

f [i ] = F [h(i− 1)];f [i ] = F [h(i− 1)];f [i ] = F [h(i− 1)];g[j ] = G[k(k − 1)];g[j ] = G[k(k − 1)];g[j ] = G[k(k − 1)];

F[] and G[] are initial conditions and boundary conditions are zero.

Now, let’s solve the equation and plot it.

r = ckh ;r = ckh ;r = ckh ;

FDgrid[n,m];FDgrid[n,m];FDgrid[n,m];FDsolve[n,m];FDsolve[n,m];FDsolve[n,m];ListPlot3D[u,AxesLabel→ {"t(j)", "x(i)", "u"}];ListPlot3D[u,AxesLabel→ {"t(j)", "x(i)", "u"}];ListPlot3D[u,AxesLabel→ {"t(j)", "x(i)", "u"}];Print["The numerical solution to the wave P.D.E."];Print["The numerical solution to the wave P.D.E."];Print["The numerical solution to the wave P.D.E."];Print ["utt(x,t)=4 uxx(x,t)"] ;Print ["utt(x,t)=4 uxx(x,t)"] ;Print ["utt(x,t)=4 uxx(x,t)"] ;

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Print[" u(x,0) = f(x) = ", F [x]];Print[" u(x,0) = f(x) = ", F [x]];Print[" u(x,0) = f(x) = ", F [x]];Print ["ut(0,x) = g(x) = ", G[x]] ;Print ["ut(0,x) = g(x) = ", G[x]] ;Print ["ut(0,x) = g(x) = ", G[x]] ;Print["c = ", c];Print["c = ", c];Print["c = ", c];Print["h = ", h];Print["h = ", h];Print["h = ", h];Print["k = ", k];Print["k = ", k];Print["k = ", k];Print

[" ck

h = ", ckh

];Print

[" ck

h = ", ckh

];Print

[" ck

h = ", ckh

];

Figure 2: The numerical solution to the wave P.D.E.

The numerical solution to the wave P.D.E.utt(x,t)=4 uxx(x,t)u(x,0) = f(x) = Sin[πx] + Sin[2πx]ut(0,x) = g(x) = 0.c = 2.h = 0.1k = 0.05ckh = 1.

Let’s compare with the exact solution. The code below will plot the real wave function. Inorder to compare both graphs easily, I set up the grid of this graph as m and n.U [x , t ] = Sin[πx]Cos[2πt] + Sin[2πx]Cos[4πt];U [x , t ] = Sin[πx]Cos[2πt] + Sin[2πx]Cos[4πt];U [x , t ] = Sin[πx]Cos[2πt] + Sin[2πx]Cos[4πt];SetOptions[Plot3D,PlotPoints→ {m,n}];SetOptions[Plot3D,PlotPoints→ {m,n}];SetOptions[Plot3D,PlotPoints→ {m,n}];Plot3D[U [x, t], {t, 0, 1}, {x, 0, 1},AxesLabel→ {"t(j)", "x(i)", "u"}];Plot3D[U [x, t], {t, 0, 1}, {x, 0, 1},AxesLabel→ {"t(j)", "x(i)", "u"}];Plot3D[U [x, t], {t, 0, 1}, {x, 0, 1},AxesLabel→ {"t(j)", "x(i)", "u"}];Print["The analytic solution to the wave P.D.E."];Print["The analytic solution to the wave P.D.E."];Print["The analytic solution to the wave P.D.E."];Print ["utt(x,t)=4 uxx(x,t)"] ; Print[" u(x,0) = f(x) = ", F [x]]; Print ["ut(0,x) = g(x) = ", G[x]] ;Print ["utt(x,t)=4 uxx(x,t)"] ; Print[" u(x,0) = f(x) = ", F [x]]; Print ["ut(0,x) = g(x) = ", G[x]] ;Print ["utt(x,t)=4 uxx(x,t)"] ; Print[" u(x,0) = f(x) = ", F [x]]; Print ["ut(0,x) = g(x) = ", G[x]] ;Print[" u[x,t] = ", U [x, t]];Print[" u[x,t] = ", U [x, t]];Print[" u[x,t] = ", U [x, t]];

The analytic solution to the wave P.D.E.utt(x,t)=4 uxx(x,t)u(x,0) = f(x) = Sin[πx] + Sin[2πx]

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Figure 3: The analytic solution to the wave P.D.E.

ut(0,x) = g(x) = 0.u[x,t] = Cos[2πt]Sin[πx] + Cos[4πt]Sin[2πx]u[x,t] = Cos[2πt]Sin[πx] + Cos[4πt]Sin[2πx]

As you can see, both graphs are exactly same. Code works great. Now, let’s make the tableof numbers which is similar to the one in the Excel.

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This code will generate the table.Print["The numerical solution to the wave P.D.E."];Print["The numerical solution to the wave P.D.E."];Print["The numerical solution to the wave P.D.E."];Print ["utt(x,t)=4 uxx(x,t)", "\n"] ;Print ["utt(x,t)=4 uxx(x,t)", "\n"] ;Print ["utt(x,t)=4 uxx(x,t)", "\n"] ;Print[NumberForm[TableForm[Transpose[Chop[u]]], 3]];Print[NumberForm[TableForm[Transpose[Chop[u]]], 3]];Print[NumberForm[TableForm[Transpose[Chop[u]]], 3]];The numerical solution to the wave P.D.E.utt(x,t)=4 uxx(x,t)

0 0.897 1.54 1.76 1.54 1. 0.363 −0.142 −0.363 −0.279 00 0.769 1.33 1.54 1.38 0.951 0.429 0 −0.21 −0.182 00 0.432 0.769 0.948 0.951 0.809 0.588 0.361 0.182 0.0684 00 0 0.0516 0.182 0.377 0.588 0.741 0.769 0.639 0.363 00 −0.38 −0.588 −0.519 −0.182 0.309 0.769 1.02 0.951 0.571 00 −0.588 −0.951 −0.951 −0.588 0 0.588 0.951 0.951 0.588 00 −0.571 −0.951 −1.02 −0.769 −0.309 0.182 0.519 0.588 0.38 00 −0.363 −0.639 −0.769 −0.741 −0.588 −0.377 −0.182 −0.0516 0 00 −0.0684 −0.182 −0.361 −0.588 −0.809 −0.951 −0.948 −0.769 −0.432 00 0.182 0.21 0 −0.429 −0.951 −1.38 −1.54 −1.33 −0.769 00 0.279 0.363 0.142 −0.363 −1. −1.54 −1.76 −1.54 −0.897 00 0.182 0.21 0 −0.429 −0.951 −1.38 −1.54 −1.33 −0.769 00 −0.0684 −0.182 −0.361 −0.588 −0.809 −0.951 −0.948 −0.769 −0.432 00 −0.363 −0.639 −0.769 −0.741 −0.588 −0.377 −0.182 −0.0516 0 00 −0.571 −0.951 −1.02 −0.769 −0.309 0.182 0.519 0.588 0.38 00 −0.588 −0.951 −0.951 −0.588 0 0.588 0.951 0.951 0.588 00 −0.38 −0.588 −0.519 −0.182 0.309 0.769 1.02 0.951 0.571 00 0 0.0516 0.182 0.377 0.588 0.741 0.769 0.639 0.363 00 0.432 0.769 0.948 0.951 0.809 0.588 0.361 0.182 0.0684 00 0.769 1.33 1.54 1.38 0.951 0.429 0 −0.21 −0.182 00 0.897 1.54 1.76 1.54 1. 0.363 −0.142 −0.363 −0.279 0

3.3 Real1: Solving Nernst-Planck equation in the Excel1

Finally, we come to this far, my original goal to solve the Nernst-Planck equation. As I did forthe wave-function, I hope that Nernst-Planck equation can be done in Excel so I can carry ontothe Mathematica.

For the simplicity sake, I will introduce easier notations and constant values: [1].

a = uij , (24)b = ui,j+1, (25)c = ui+1,j , (26)d = ui−1,j), (27)s = Dak/h

2, (28)w = αDak, (29)

α = 4πq2n∗/εKbT , (30)

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Da = 1.0× 10−5cm2sec−1 , (31)Db = 0.5× 10−5cm2sec−1, (32)

L = 1.0cm, (33)T = 298.18◦K, (34)ε = 78.53, (35)

q = 4.803× 10−10, (36)n∗ = 3.0× 108cm−3, (37)

Kb=Boltzmann constant, (38)and of course, s and w will have Db when they are dealing with nb.

Everything is calculated in the Excel worksheet(Nernst-Planck equation.xls). Since equa-tion(11) and (12) are coupled, their discretization is somewhat different with the heat or waveequation.

For na(in Excel, this is expressed as u family), discretization isb = (1− 2s− w)a+ s(c+ d) + wvi,j (vi,j is the ui,j of nb), andfor nb(in Excel, this is expressed as v family), discretization is

b = (1− 2s− w)a+ s(c+ d) + wui,j .

As you can see both have extra ending term which is not their own function. Thus, I decide touse the cross reference in the Excel. Basically, I will make two work sheets which are one forna or u and one for nb or v. Then I will construct the layers one by one but with each other’sone-lower layer value. That is what the discretization equation says! In such a mechanism, Ionly build the one layer by time since both layers are building at same time. If you don’t havethe layers of another, you can’t find the next value of your own.

Figure 4: Graph of u

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Figure 5: Graph of v

However, look at these graphs. They are graph of worksheet ”u” and ”v” in the Excel file.They don’t tell much about the concentration behavior. They just describe the initial distribu-tion of the concentration which I create in the first place. Originally, I plan to give the Gaussiandistribution for the initial state of concentration in the box(I said box, because it is the isolatedsystem). Furthermore, the behavior of both u and v should be almost exactly proportional to thecos(πx/L), but they don’t seem to follow the rule of the equation. However, they are inversedeach other, that can be explained by the charge of the ion. u is supposed to describe the motionof the cations and v is supposed to destine the motion of the anions so it can be reasonable theymay have inverse relationship. And due the difference of the diffusion constant,D their max valueof u and min value of v don’t perfectly cancel each others. That is a good feature of these graphssince their behaviors are expected as difference of D.

This time, I use smaller distance step so I can get more accurate graphs. Now, the distancesteps are about twice as much as compare to the first one. And all of the work is in the Uu andVv worksheets in the Excel file.

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Figure 6: Graph of u for uu worksheet

Figure 7: Graph of v for vv worksheet

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They are not also telling much about the motion of the concentration. However, at least theyhave better shape of the graph which will support the as the steps get smaller, you will get moreaccurate answers. That is the main concepts of all of the numerical methods so I believe, myapproach and method to this certain problem is not that bad so far. Or my first assumption andset up and property of the equation can be very wrong because as you can see, the shape of thegraph doesn’t change over time, static state. So problem can be arise from that equation tellsconcentration to move but my given conditions and set up of the discretization tell concentrationnot to move.

3.4 Real2: Solving Nernst-Planck equation in the Excel2

I can’t give up just like that, so I tried little bit different approach this time. There must beNtotal in the system and I set that Ntotal = na + nb = constant = 2n∗. And I am still using themore smaller space steps. These works are doen in the worksheet ”uuu” and ”vvvv” in the Excelfile.

Figure 8: Graph of u for uuu worksheet

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Figure 9: Graph of v for vvv worksheet

The results look very interesting. The graph of the v is totally in the negative side and itsmax value is 0 which suggests all the anions are attracted to the cation side. That can be possiblesince electrons are pretty much only movable and want to move somewhere if there is attraction.Moreover, the shapes of graphs are more like cos which is the expected behavior for the t=longtime. However, this is not answer for sure and this can’t satisfy me due to the face that thisgraph is also having no effect over time. Their shapes are almost same and don’t change overthe time. May be the point about the wrong relationship between the property of the equationand my condition are right or there are just simple problems which I can’t find right now. But,one thing is for sure, this is wrong about the logic and theory which I build for this method andproject.

4 Conclusion and Future plan

FINALLY, time for the conclusion. This project was such a bittersweet chocolate for me. Theequations are evil and no one really can help me. But, this is MY project and MY topic so I didwhat can I do for the best. Definitely this is not easy project for general. I didn’t realize thattype in the LaTex form is such a hard job. It feels really great if I did something which I thinkcan’t do that especially for writing the code for the Mathematica shows the lessons of the P210.I can do the programming without any previous experience because I have the documentationsand GOOGLE! And it really feels sad and bad for the last part which I really want to solve.However, I think it is just a small step which I will take closer to the my goal like space andtime steps are building each layers of the numerical solution. I will probably research about thisrelated topic for my honors thesis for 4th year if I am still doing Biophysics in next year whichis the my future plan. I will master the Nernst-Planck equation and let all the world know howEASY that equation should be. For that purpose, I won’t make this project public. This isbetween you and me, Jess! Thank you to teach me such a great strength and knowledge!

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References

[1] S.B. Malvadkar, M.D. Kostin, “Solutions of the Nernst-Planck Equations forIonic Diffusion for Conditions near Equilibrium”, The Journal of ChemicalPhysics, Oct. 1972, vol 57, is. 8, pp. 3263-3265.

[2] Philip Nelson,Biological Physics, pp.265-272.

[3] One of the my Math 316 notes on the finite difference method.

[4] http://www.math.buffalo.edu/ pitman/courses/mth438/na/node15.htmlSECTION000132000000000000000.

[5] http://www.me.rochester.edu/courses/ME223/webexamp/findiffexamp1nb.pdf.

[6] http://www.math.su.se/ lambe/nsc/nsc.html.

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