Solving a SOI MOSFET Differential Equation for Electron Distribution

[Jeppe]

Hi all!

I am trying to solve the following differential equation for the electron distribution in a SOI MOSFET structure.

The equation is:

\frac{d^2n}{dx^2} = \frac{1}{n}\left( \frac{dn}{dx}\right)^2 + A n^{2}

A is a constant.

The boundary conditions are:

n(0) = c1 (a constant)
n'(0) = 0

I know that the solution to the equation is:

n(x) = \frac{c1}{cos^2\left( \sqrt{\frac{1}{2} A \,c1}\,\,\, x \right)}

but i can not solve it myself. I have tried with both maple, mathematica, and MATLAB but none of them seems to be able to solve it.

Could anyone give me a hint on how to do it?

Thanks!

Jeppe

 

[coomast]

Hello Jeppe,

This is not a straightforward differential equation. However I am currently reading and studying the following book:

"Ordinary Differential Equations, an elementary text-book with an introduction
to Lie's theory of the group of one parameter"
written by James Morris Page, it is available on the www for free because it is very old.
It was written in 1897 and no rights are upon it anymore. So available for free.

This book is my first one the theory of Lie for solving diferential equations, some other ones will come afterwards. Now this theory helps in solving your equation. A few months back I would not be able to do it, but now I can help you. I will not go into details on the theory, just take the first step as true, which is the following. Make the substitution:

[edit]I used y for n[/edit]

u=y \qquad v=y' \qquad w=\frac{dv}{du}=\frac{y''}{y'}
the inverse being
y=u \qquad y'=v \qquad y''=v\frac{dv}{du}
After substituting this in the DE you get:
\frac{dv}{du}-\frac{v}{u}=A\frac{u^2}{v}
which is a Bernouilli differential equation and can be solved by substituting:
z=v^2
This gives a linear differential equation:
\frac{dz}{du}-\frac{2z}{u}=2Au^2
Which has the solution:
v=u\sqrt{2Au-K_1}
with K_1 an integration constant.
From this we have:
\frac{dy}{dx}=y\sqrt{2Ay-K_1}
giving now after integrating and some algebra:
y=\frac{K_1}{2A}\frac{1}{cos^2 \left(\frac{\sqrt{K_1}}{2}x+K_2\right)}
Applying the boundary conditions gives for the second one
K_2=0
and thus
y=\frac{K_1}{2A}\frac{1}{cos^2\left(\frac{\sqrt{K_1}}{2}x\right)}
The first one gives:
K_1=2Ac_1
and using this gives your solution:
y=\frac{c_1}{cos^2\left(\sqrt{\frac{Ac_1}{2}}x\right)}

coomast

 

[Jeppe]

Thanks a lot coomast! That really helped!