1. Nov 4, 2016

### Langam

As a part of my research work, I need to find the number of charged particles at a given time 't', at a distance 'x' from anode. I derived a set of PDEs as per my requirement and assumptions which needs to be solved analytically.

\frac{\partial{N_e}}{\partial{t}} = \alpha N_e |W_e| - \beta N_eN_p -\eta N_e |W_e|-W_e\frac{\partial{N_e}}{\partial{x}} + D_e\frac{\partial^2{N_e}}{\partial{x}^2}

\frac{\partial{N_p}}{\partial{t}} = \alpha N_e |W_e| - \beta N_eN_p -W_p\frac{\partial{N_p}}{\partial{x}}

\frac{\partial{N_n}}{\partial{t}} = \eta N_e |W_e| - W_n\frac{\partial{N_n}}{\partial{x}}

where,
N – no of particles, W- drift velocity, \alpha - ionization coefficient, \beta - recombination coefficient, \eta - attachment coefficient, e - electrons, p - positive ions, n - negative ions.
The initial electron and positive ion distributions are equal and given by:

\begin{aligned}
N_e(t=0,x) & = N_p(t=0,x)\\
& = 10^8 \exp(-[\frac{(x-d/2)^2}{(4.883*10^-3)^2}])
\end{aligned}

Assuming, the anode is perfect absorber of electrons, i.e. derivative is zero, while cathode is a non-emitter of negative ions, as boundary conditions:

\begin{aligned}
\frac{\partial{N_e}}{\partial{t}}\vert_{x=0,t} & = 0\\
N_n(x=d,t)&= 0
\end{aligned}

Here, W- drift velocity, $\alpha$ - ionization coefficient, $\eta$ - attachment coefficient depend on field(E) given by:

-\epsilon _r* \frac{\partial{E}}{\partial{x}} + \frac{e(N_p-N_e-N_n)}{\epsilon _0} =0

Boundary conditions for field:

E(x=0) = E_{app}

Since, the time step I have to consider is very small (less than nanoseconds), I assume that the \alpha, W, \eta and field (E) is constant for each time step. Similarly, N_p is taken to be constant while computing $N_e$ and likewise.

I looked up some textbooks and materials on net and have tried solving these equations. I am not confident about the solution and approach though.

My solution:
For N_e

While solving for $N_e$, since I expect an exponential increase, I assume the solution of the form:

N_e(x,t)= e^{(Xx+Yt)}u(x,t)

Then I try to simplify it to get it into a heat equation of the form,

\frac{\partial{u}}{\partial{t}}= D_e\frac{\partial^2{u}}{\partial{x^2}}

I used separable variables form on this equation and applied initial and boundary conditions to arrive at the solution:

\begin{aligned}
N_e(x,t)&= e^{[\frac{W_e}{2D_e}]x+[A - \frac{W_e^2}{4D_e}]t}e^{\lambda t} [10^8 e^{-d^2(10.48*10^3)}] \sinh{\sqrt\frac{\lambda}{D_e}x}\\
\text{where,}\ \lambda &= \frac{D_e}{x^2}[ln(e^{-ax^2+bx} + \sqrt(e^{-2ax^2+2bx}+1))]^2\\
a & = \frac{1}{M},\ b = \frac{d}{M}-\alpha, \ M =(4.883*10^{-3})^2
\end{aligned}

For N_p
I define new variables

\begin{aligned}
\zeta & = Ax+Bt\\
\eta & = Cx+Dt\\
\text{such that}, N_p & = f(\zeta , \eta)
\end{aligned}

I try to get it into the linear homogeneous equation of form,

\frac{\partial{N_p}}{\partial{\zeta}}+N_p N_e \beta = \alpha N_e |W_e|

I solve this using integrating factors and arrive at solution :

N_p(x,t)=\frac{\alpha W_p |W_e|}{\beta x}+e^{N_e\beta t}[10^8 e^{-\frac{(x-W_pt-d/2)^2}{(4.883*10^{-3})^2}}-\frac{\alpha W_p |W_e|}{\beta (x-W_pt)}]\\

For N_n:
I define new variables such that,

(x,t)=(\zeta=x-W_nt, \eta= x+W_nt)

Then on simplification and applying boundary conditions, I arrive at

N_n = \frac{N_e\eta |W_e| (x-d)}{2W_n}

For E:

E(x) = E_{app} + \frac{e(N_p-N_e-N_n)}{\epsilon _0 * \epsilon _r}

I tried in Mathematica. It could not list the answer to these equations. Please let me know whether the approach and the solution is correct for my boundary and initial conditions.

Langam

2. Nov 9, 2016