Revisions to How to make a function depend on the solution of an integration?

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edited Apr 2, 2019 at 0:09

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I have two functions which are the linear and non-linear solution of an equation. I'm trying to calculate where does the linear solution is a good solution of my equation, when compared with the non-linear. To do so I am using

$$\int_{\tfrac{a}{2}}^\infty\left(\sigma^{NL}(x)-\sigma^{L}(x)\right)^2\mathrm dx$$

The variables I'm using are

T = 298.15(*K*);
k1 = 1.38064852*10^-23 (*J/K*);
ϵ0 = 8.85418781761*10^-12 (*C^2/N m^2||F/m||C/V m*);
ϵ = 78.5;
ϵR = ϵ ϵ0 (*C^2/N m^2||F/m||C/Vm*);
c = 2.99792458*10^8(*m/s*);
z = 1;
e1 = 1.60217733*10^-19 (*C*);
NA = 6.022140857*10^23 (*mol^-1*);
a = 4.25(*Å*);
a1 = a*10^-10 (*m*);
x1 = x*10^-10 (*m*);

And the main code is

nMKS[ρ0_] = ρ0 *10^3 NA (*m^-3*);
κMKS[ρ0_] = Sqrt[(2 nMKS[ρ0] (e1^2) (z^2) )/(ϵR k1 T)](*m^-1*);
LσMMKS[ρ0_, ψH_] = ϵR κMKS[ρ0] ψH/1000  (*C/m^2*);
LψPBMKS[ρ0_, ψH_, x_] = ψH Exp[κMKS[ρ0] (a1/2 - x1)] (*mV*);
LσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {ϵR κMKS[ρ0] LψPBMKS[ρ0, ψH, x]/1000, x1 >= a1/2}
  }](*C/m^2*);
NLσMMKS[ρ0_, ψH_] = (2 ϵR κMKS[ρ0] k1 T)/( z e1) Sinh[(z e1 )/(2 k1 T) ψH/1000] (*C/m^2*);
ZMKS[ψH_] = (z e1 ψH/1000)/(2 k1 T);
NLσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {(4 ϵR κMKS[ρ0] k1 T)/(z e1) (((Exp[2 ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])/(
    (Exp[ZMKS[ψH]] + 1)^2 - ((Exp[ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])^2)), x1 >= a1/2}
  }](*C/m^2*);

Hence, when I try to integrate my equation with

Dif2σ[ρ0_, ψH_, x_] = FullSimplify[(NLσMKS[ρ0, ψH, x] - LσMKS[ρ0, ψH, x])^2]
Vσ[ρ0_, ψH_] = FullSimplify[Sqrt[Integrate[Dif2σ[ρ0, ψH, x], {x, a/2, Infinity}]]]

Mathematica is not able to solve it. How can I force mathematicaMathematica to actually solve this and thereafter define a function from its solution?

I have two functions which are the linear and non-linear solution of an equation. I'm trying to calculate where does the linear solution is a good solution of my equation, when compared with the non-linear. To do so I am using

The variables I'm using are

T = 298.15(*K*);
k1 = 1.38064852*10^-23 (*J/K*);
ϵ0 = 8.85418781761*10^-12 (*C^2/N m^2||F/m||C/V m*);
ϵ = 78.5;
ϵR = ϵ ϵ0 (*C^2/N m^2||F/m||C/Vm*);
c = 2.99792458*10^8(*m/s*);
z = 1;
e1 = 1.60217733*10^-19 (*C*);
NA = 6.022140857*10^23 (*mol^-1*);
a = 4.25(*Å*);
a1 = a*10^-10 (*m*);
x1 = x*10^-10 (*m*);

And the main code is

nMKS[ρ0_] = ρ0 *10^3 NA (*m^-3*);
κMKS[ρ0_] = Sqrt[(2 nMKS[ρ0] (e1^2) (z^2) )/(ϵR k1 T)](*m^-1*);
LσMMKS[ρ0_, ψH_] = ϵR κMKS[ρ0] ψH/1000  (*C/m^2*);
LψPBMKS[ρ0_, ψH_, x_] = ψH Exp[κMKS[ρ0] (a1/2 - x1)] (*mV*);
LσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {ϵR κMKS[ρ0] LψPBMKS[ρ0, ψH, x]/1000, x1 >= a1/2}
  }](*C/m^2*);
NLσMMKS[ρ0_, ψH_] = (2 ϵR κMKS[ρ0] k1 T)/( z e1) Sinh[(z e1 )/(2 k1 T) ψH/1000] (*C/m^2*);
ZMKS[ψH_] = (z e1 ψH/1000)/(2 k1 T);
NLσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {(4 ϵR κMKS[ρ0] k1 T)/(z e1) (((Exp[2 ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])/(
    (Exp[ZMKS[ψH]] + 1)^2 - ((Exp[ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])^2)), x1 >= a1/2}
  }](*C/m^2*);

Hence, when I try to integrate my equation with

Dif2σ[ρ0_, ψH_, x_] = FullSimplify[(NLσMKS[ρ0, ψH, x] - LσMKS[ρ0, ψH, x])^2]
Vσ[ρ0_, ψH_] = FullSimplify[Sqrt[Integrate[Dif2σ[ρ0, ψH, x], {x, a/2, Infinity}]]]

Mathematica is not able to solve it. How can I force mathematica to actually solve this and thereafter define a function from its solution?

I have two functions which are the linear and non-linear solution of an equation. I'm trying to calculate where does the linear solution is a good solution of my equation, when compared with the non-linear. To do so I am using

$$\int_{\tfrac{a}{2}}^\infty\left(\sigma^{NL}(x)-\sigma^{L}(x)\right)^2\mathrm dx$$

The variables I'm using are

T = 298.15(*K*);
k1 = 1.38064852*10^-23 (*J/K*);
ϵ0 = 8.85418781761*10^-12 (*C^2/N m^2||F/m||C/V m*);
ϵ = 78.5;
ϵR = ϵ ϵ0 (*C^2/N m^2||F/m||C/Vm*);
c = 2.99792458*10^8(*m/s*);
z = 1;
e1 = 1.60217733*10^-19 (*C*);
NA = 6.022140857*10^23 (*mol^-1*);
a = 4.25(*Å*);
a1 = a*10^-10 (*m*);
x1 = x*10^-10 (*m*);

And the main code is

nMKS[ρ0_] = ρ0 *10^3 NA (*m^-3*);
κMKS[ρ0_] = Sqrt[(2 nMKS[ρ0] (e1^2) (z^2) )/(ϵR k1 T)](*m^-1*);
LσMMKS[ρ0_, ψH_] = ϵR κMKS[ρ0] ψH/1000  (*C/m^2*);
LψPBMKS[ρ0_, ψH_, x_] = ψH Exp[κMKS[ρ0] (a1/2 - x1)] (*mV*);
LσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {ϵR κMKS[ρ0] LψPBMKS[ρ0, ψH, x]/1000, x1 >= a1/2}
  }](*C/m^2*);
NLσMMKS[ρ0_, ψH_] = (2 ϵR κMKS[ρ0] k1 T)/( z e1) Sinh[(z e1 )/(2 k1 T) ψH/1000] (*C/m^2*);
ZMKS[ψH_] = (z e1 ψH/1000)/(2 k1 T);
NLσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {(4 ϵR κMKS[ρ0] k1 T)/(z e1) (((Exp[2 ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])/(
    (Exp[ZMKS[ψH]] + 1)^2 - ((Exp[ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])^2)), x1 >= a1/2}
  }](*C/m^2*);

Hence, when I try to integrate my equation with

Dif2σ[ρ0_, ψH_, x_] = FullSimplify[(NLσMKS[ρ0, ψH, x] - LσMKS[ρ0, ψH, x])^2]
Vσ[ρ0_, ψH_] = FullSimplify[Sqrt[Integrate[Dif2σ[ρ0, ψH, x], {x, a/2, Infinity}]]]

Mathematica is not able to solve it. How can I force Mathematica to actually solve this and thereafter define a function from its solution?

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edit approved Apr 1, 2019 at 19:41

mjw

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I have two functions which are the linear and non-linear solution of an equation. I'm trying to calculate where does the linear solution is a good solution of my equation, when compared with the non-linear. To do so I am using [![Formula used][1]][1] [1]: https://i.sstatic.net/KCLED.png

The variables I'm using are

T = 298.15(*K*);
k1 = 1.38064852*10^-23 (*J/K*);
ϵ0 = 8.85418781761*10^-12 (*C^2/N m^2||F/m||C/V m*);
ϵ = 78.5;
ϵR = ϵ ϵ0 (*C^2/N m^2||F/m||C/Vm*);
c = 2.99792458*10^8(*m/s*);
z = 1;
e1 = 1.60217733*10^-19 (*C*);
NA = 6.022140857*10^23 (*mol^-1*);
a = 4.25(*Å*);
a1 = a*10^-10 (*m*);
x1 = x*10^-10 (*m*);

And the main code is

nMKS[ρ0_] = ρ0 *10^3 NA (*m^-3*);
κMKS[ρ0_] = Sqrt[(2 nMKS[ρ0] (e1^2) (z^2) )/(ϵR k1 T)](*m^-1*);
LσMMKS[ρ0_, ψH_] = ϵR κMKS[ρ0] ψH/1000  (*C/m^2*);
LψPBMKS[ρ0_, ψH_, x_] = ψH Exp[κMKS[ρ0] (a1/2 - x1)] (*mV*);
LσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {ϵR κMKS[ρ0] LψPBMKS[ρ0, ψH, x]/1000, x1 >= a1/2}
  }](*C/m^2*);
NLσMMKS[ρ0_, ψH_] = (2 ϵR κMKS[ρ0] k1 T)/( z e1) Sinh[(z e1 )/(2 k1 T) ψH/1000] (*C/m^2*);
ZMKS[ψH_] = (z e1 ψH/1000)/(2 k1 T);
NLσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {(4 ϵR κMKS[ρ0] k1 T)/(z e1) (((Exp[2 ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])/(
    (Exp[ZMKS[ψH]] + 1)^2 - ((Exp[ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])^2)), x1 >= a1/2}
  }](*C/m^2*);

Hence, when I try to integrate my equation with

Dif2σ[ρ0_, ψH_, x_] = FullSimplify[(NLσMKS[ρ0, ψH, x] - LσMKS[ρ0, ψH, x])^2]
Vσ[ρ0_, ψH_] = FullSimplify[Sqrt[Integrate[Dif2σ[ρ0, ψH, x], {x, a/2, Infinity}]]]

Mathematica is not able to solve it. How can I force mathematica to actually solve this and thereafter define a function from its solution? Thanks in advance.

I have two functions which are the linear and non-linear solution of an equation. I'm trying to calculate where does the linear solution is a good solution of my equation, when compared with the non-linear. To do so I am using [![Formula used][1]][1] [1]: https://i.sstatic.net/KCLED.png

The variables I'm using are

T = 298.15(*K*);
k1 = 1.38064852*10^-23 (*J/K*);
ϵ0 = 8.85418781761*10^-12 (*C^2/N m^2||F/m||C/V m*);
ϵ = 78.5;
ϵR = ϵ ϵ0 (*C^2/N m^2||F/m||C/Vm*);
c = 2.99792458*10^8(*m/s*);
z = 1;
e1 = 1.60217733*10^-19 (*C*);
NA = 6.022140857*10^23 (*mol^-1*);
a = 4.25(*Å*);
a1 = a*10^-10 (*m*);
x1 = x*10^-10 (*m*);

And the main code is

nMKS[ρ0_] = ρ0 *10^3 NA (*m^-3*);
κMKS[ρ0_] = Sqrt[(2 nMKS[ρ0] (e1^2) (z^2) )/(ϵR k1 T)](*m^-1*);
LσMMKS[ρ0_, ψH_] = ϵR κMKS[ρ0] ψH/1000  (*C/m^2*);
LψPBMKS[ρ0_, ψH_, x_] = ψH Exp[κMKS[ρ0] (a1/2 - x1)] (*mV*);
LσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {ϵR κMKS[ρ0] LψPBMKS[ρ0, ψH, x]/1000, x1 >= a1/2}
  }](*C/m^2*);
NLσMMKS[ρ0_, ψH_] = (2 ϵR κMKS[ρ0] k1 T)/( z e1) Sinh[(z e1 )/(2 k1 T) ψH/1000] (*C/m^2*);
ZMKS[ψH_] = (z e1 ψH/1000)/(2 k1 T);
NLσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {(4 ϵR κMKS[ρ0] k1 T)/(z e1) (((Exp[2 ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])/(
    (Exp[ZMKS[ψH]] + 1)^2 - ((Exp[ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])^2)), x1 >= a1/2}
  }](*C/m^2*);

Hence, when I try to integrate my equation with

Dif2σ[ρ0_, ψH_, x_] = FullSimplify[(NLσMKS[ρ0, ψH, x] - LσMKS[ρ0, ψH, x])^2]
Vσ[ρ0_, ψH_] = FullSimplify[Sqrt[Integrate[Dif2σ[ρ0, ψH, x], {x, a/2, Infinity}]]]

Mathematica is not able to solve it. How can I force mathematica to actually solve this and thereafter define a function from its solution? Thanks in advance.

I have two functions which are the linear and non-linear solution of an equation. I'm trying to calculate where does the linear solution is a good solution of my equation, when compared with the non-linear. To do so I am using

The variables I'm using are

T = 298.15(*K*);
k1 = 1.38064852*10^-23 (*J/K*);
ϵ0 = 8.85418781761*10^-12 (*C^2/N m^2||F/m||C/V m*);
ϵ = 78.5;
ϵR = ϵ ϵ0 (*C^2/N m^2||F/m||C/Vm*);
c = 2.99792458*10^8(*m/s*);
z = 1;
e1 = 1.60217733*10^-19 (*C*);
NA = 6.022140857*10^23 (*mol^-1*);
a = 4.25(*Å*);
a1 = a*10^-10 (*m*);
x1 = x*10^-10 (*m*);

And the main code is

nMKS[ρ0_] = ρ0 *10^3 NA (*m^-3*);
κMKS[ρ0_] = Sqrt[(2 nMKS[ρ0] (e1^2) (z^2) )/(ϵR k1 T)](*m^-1*);
LσMMKS[ρ0_, ψH_] = ϵR κMKS[ρ0] ψH/1000  (*C/m^2*);
LψPBMKS[ρ0_, ψH_, x_] = ψH Exp[κMKS[ρ0] (a1/2 - x1)] (*mV*);
LσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {ϵR κMKS[ρ0] LψPBMKS[ρ0, ψH, x]/1000, x1 >= a1/2}
  }](*C/m^2*);
NLσMMKS[ρ0_, ψH_] = (2 ϵR κMKS[ρ0] k1 T)/( z e1) Sinh[(z e1 )/(2 k1 T) ψH/1000] (*C/m^2*);
ZMKS[ψH_] = (z e1 ψH/1000)/(2 k1 T);
NLσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {(4 ϵR κMKS[ρ0] k1 T)/(z e1) (((Exp[2 ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])/(
    (Exp[ZMKS[ψH]] + 1)^2 - ((Exp[ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])^2)), x1 >= a1/2}
  }](*C/m^2*);

Hence, when I try to integrate my equation with

Dif2σ[ρ0_, ψH_, x_] = FullSimplify[(NLσMKS[ρ0, ψH, x] - LσMKS[ρ0, ψH, x])^2]
Vσ[ρ0_, ψH_] = FullSimplify[Sqrt[Integrate[Dif2σ[ρ0, ψH, x], {x, a/2, Infinity}]]]

Mathematica is not able to solve it. How can I force mathematica to actually solve this and thereafter define a function from its solution?

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edited Apr 1, 2019 at 17:52

MarcoB

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T = 298.15(*K*);
k1 = 1.38064852*10^-23 (*J/K*);
\[Epsilon]0ϵ0 = 8.85418781761*10^-12 (*C^2/N m^2||F/m||C/V m*);
\[Epsilon]ϵ = 78.5;
\[Epsilon]RϵR = \[Epsilon]ϵ \[Epsilon]0ϵ0 (*C^2/N m^2||F/m||C/Vm*);
c = 2.99792458*10^8(*m/s*);
z = 1;
e1 = 1.60217733*10^-19 (*C*);
NA = 6.022140857*10^23 (*mol^-1*);
a = 4.25(*\[Angstrom]**Å*);
a1 = a*10^-10 (*m*);
x1 = x*10^-10 (*m*);

nMKS[\[Rho]0_]nMKS[ρ0_] = \[Rho]0ρ0 *10^3 NA (*m^-3*);
\[Kappa]MKS[\[Rho]0_]κMKS[ρ0_] = Sqrt[(
  2 nMKS[\[Rho]0]nMKS[ρ0] (e1^2) (z^2) )/(\[Epsilon]RϵR k1 T)](*m^-1*);
L\[Sigma]MMKS[\[Rho]0_LσMMKS[ρ0_, \[Psi]H_]ψH_] = \[Epsilon]RϵR \[Kappa]MKS[\[Rho]0]κMKS[ρ0] \
\[Psi]HψH/1000  (*C/m^2*);
L\[Psi]PBMKS[\[Rho]0_LψPBMKS[ρ0_, \[Psi]H_ψH_, 
   x_] = \[Psi]HψH Exp[\[Kappa]MKS[\[Rho]0]Exp[κMKS[ρ0] (a1/2 - x1)] (*mV*);
L\[Sigma]MKS[\[Rho]0_LσMKS[ρ0_, \[Psi]H_ψH_, x_] = 
  Piecewise[{{0, 
    {0, 0 <= x1 < a1/
      2}, 
   {\[Epsilon]RϵR \[Kappa]MKS[\[Rho]0]κMKS[ρ0] L\[Psi]PBMKS[\[Rho]0LψPBMKS[ρ0, \
\[Psi]HψH, x]/1000, x1 >= a1/2}
  }](*C/m^2*);
NL\[Sigma]MMKS[\[Rho]0_NLσMMKS[ρ0_, \[Psi]H_]ψH_] = (
   2 \[Epsilon]RϵR \[Kappa]MKS[\[Rho]0]κMKS[ρ0] k1 T)/( z e1)
    Sinh[(z e1 )/(2 k1 T) \[Psi]HψH/1000] (*C/m^2*);
ZMKS[\[Psi]H_]ZMKS[ψH_] = (z e1 \[Psi]HψH/1000)/(2 k1 T);
NL\[Sigma]MKS[\[Rho]0_NLσMKS[ρ0_, \[Psi]H_ψH_, x_] = 
  Piecewise[{{0, 
    {0, 0 <= x1 < a1/2}, 
   {(4 \[Epsilon]RϵR \[Kappa]MKS[\[Rho]0]κMKS[ρ0] k1 T)/( 
      z e1) (((Exp[2 ZMKS[\[Psi]H]]ZMKS[ψH]] - 
            1) Exp[\[Kappa]MKS[\[Rho]0]Exp[κMKS[ρ0] (a1/2 - x1)])/((Exp[
             ZMKS[\[Psi]H]](Exp[ZMKS[ψH]] + 
            1)^2 - ((Exp[ZMKS[\[Psi]H]]Exp[ZMKS[ψH]] - 
              1) Exp[\[Kappa]MKS[\[Rho]0]Exp[κMKS[ρ0] (a1/2 - x1)])^2)), 
     x1 >= a1/2}
  }](*C/m^2*);

Dif2\[Sigma][\[Rho]0_Dif2σ[ρ0_, \[Psi]H_ψH_, x_] = 
 FullSimplify[(NL\[Sigma]MKS[\[Rho]0NLσMKS[ρ0, \[Psi]HψH, x] - 
    L\[Sigma]MKS[\[Rho]0LσMKS[ρ0, \[Psi]HψH, x])^2]
V\[Sigma][\[Rho]0_Vσ[ρ0_, \[Psi]H_]ψH_] = 
 FullSimplify[Sqrt[
  Integrate[Dif2\[Sigma][\[Rho]0FullSimplify[Sqrt[Integrate[Dif2σ[ρ0, \[Psi]HψH, x], {x, a/2, Infinity}]]]

T = 298.15(*K*);
k1 = 1.38064852*10^-23 (*J/K*);
\[Epsilon]0 = 8.85418781761*10^-12 (*C^2/N m^2||F/m||C/V m*);
\[Epsilon] = 78.5;
\[Epsilon]R = \[Epsilon] \[Epsilon]0 (*C^2/N m^2||F/m||C/Vm*);
c = 2.99792458*10^8(*m/s*);
z = 1;
e1 = 1.60217733*10^-19 (*C*);
NA = 6.022140857*10^23 (*mol^-1*);
a = 4.25(*\[Angstrom]*);
a1 = a*10^-10 (*m*);
x1 = x*10^-10 (*m*);

nMKS[\[Rho]0_] = \[Rho]0 *10^3 NA (*m^-3*);
\[Kappa]MKS[\[Rho]0_] = Sqrt[(
  2 nMKS[\[Rho]0] (e1^2) (z^2) )/(\[Epsilon]R k1 T)](*m^-1*);
L\[Sigma]MMKS[\[Rho]0_, \[Psi]H_] = \[Epsilon]R \[Kappa]MKS[\[Rho]0] \
\[Psi]H/1000  (*C/m^2*);
L\[Psi]PBMKS[\[Rho]0_, \[Psi]H_, 
   x_] = \[Psi]H Exp[\[Kappa]MKS[\[Rho]0] (a1/2 - x1)] (*mV*);
L\[Sigma]MKS[\[Rho]0_, \[Psi]H_, x_] = 
  Piecewise[{{0, 
     0 <= x1 < a1/
      2}, {\[Epsilon]R \[Kappa]MKS[\[Rho]0] L\[Psi]PBMKS[\[Rho]0, \
\[Psi]H, x]/1000, x1 >= a1/2}}](*C/m^2*);
NL\[Sigma]MMKS[\[Rho]0_, \[Psi]H_] = (
   2 \[Epsilon]R \[Kappa]MKS[\[Rho]0] k1 T)/( z e1)
    Sinh[(z e1 )/(2 k1 T) \[Psi]H/1000] (*C/m^2*);
ZMKS[\[Psi]H_] = (z e1 \[Psi]H/1000)/(2 k1 T);
NL\[Sigma]MKS[\[Rho]0_, \[Psi]H_, x_] = 
  Piecewise[{{0, 
     0 <= x1 < a1/2}, {(4 \[Epsilon]R \[Kappa]MKS[\[Rho]0] k1 T)/( 
      z e1) (((Exp[2 ZMKS[\[Psi]H]] - 
            1) Exp[\[Kappa]MKS[\[Rho]0] (a1/2 - x1)])/((Exp[
             ZMKS[\[Psi]H]] + 
            1)^2 - ((Exp[ZMKS[\[Psi]H]] - 
              1) Exp[\[Kappa]MKS[\[Rho]0] (a1/2 - x1)])^2)), 
     x1 >= a1/2}}](*C/m^2*);

Dif2\[Sigma][\[Rho]0_, \[Psi]H_, x_] = 
 FullSimplify[(NL\[Sigma]MKS[\[Rho]0, \[Psi]H, x] - 
    L\[Sigma]MKS[\[Rho]0, \[Psi]H, x])^2]
V\[Sigma][\[Rho]0_, \[Psi]H_] = 
 FullSimplify[Sqrt[
  Integrate[Dif2\[Sigma][\[Rho]0, \[Psi]H, x], {x, a/2, Infinity}]]]

T = 298.15(*K*);
k1 = 1.38064852*10^-23 (*J/K*);
ϵ0 = 8.85418781761*10^-12 (*C^2/N m^2||F/m||C/V m*);
ϵ = 78.5;
ϵR = ϵ ϵ0 (*C^2/N m^2||F/m||C/Vm*);
c = 2.99792458*10^8(*m/s*);
z = 1;
e1 = 1.60217733*10^-19 (*C*);
NA = 6.022140857*10^23 (*mol^-1*);
a = 4.25(*Å*);
a1 = a*10^-10 (*m*);
x1 = x*10^-10 (*m*);

nMKS[ρ0_] = ρ0 *10^3 NA (*m^-3*);
κMKS[ρ0_] = Sqrt[(2 nMKS[ρ0] (e1^2) (z^2) )/(ϵR k1 T)](*m^-1*);
LσMMKS[ρ0_, ψH_] = ϵR κMKS[ρ0] ψH/1000  (*C/m^2*);
LψPBMKS[ρ0_, ψH_, x_] = ψH Exp[κMKS[ρ0] (a1/2 - x1)] (*mV*);
LσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {ϵR κMKS[ρ0] LψPBMKS[ρ0, ψH, x]/1000, x1 >= a1/2}
  }](*C/m^2*);
NLσMMKS[ρ0_, ψH_] = (2 ϵR κMKS[ρ0] k1 T)/( z e1) Sinh[(z e1 )/(2 k1 T) ψH/1000] (*C/m^2*);
ZMKS[ψH_] = (z e1 ψH/1000)/(2 k1 T);
NLσMKS[ρ0_, ψH_, x_] = 
  Piecewise[{
   {0, 0 <= x1 < a1/2}, 
   {(4 ϵR κMKS[ρ0] k1 T)/(z e1) (((Exp[2 ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])/(
    (Exp[ZMKS[ψH]] + 1)^2 - ((Exp[ZMKS[ψH]] - 1) Exp[κMKS[ρ0] (a1/2 - x1)])^2)), x1 >= a1/2}
  }](*C/m^2*);

Dif2σ[ρ0_, ψH_, x_] = FullSimplify[(NLσMKS[ρ0, ψH, x] - LσMKS[ρ0, ψH, x])^2]
Vσ[ρ0_, ψH_] = FullSimplify[Sqrt[Integrate[Dif2σ[ρ0, ψH, x], {x, a/2, Infinity}]]]

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asked Apr 1, 2019 at 17:21

Adrián Silva Caballero

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