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In the frame of semiconductor physics, I find myself in front of a smart but difficult second-order nonlinear ODE : \begin{equation} \tag{$E$-ODE} \label{eq:E-ODE} \boxed{ \phi_T \frac{\mathrm{d}^2 E}{\mathrm{d}{x}^2} + E \frac{\mathrm{d}{E}}{\mathrm{d}{x}} - \frac{e N_D}{\varepsilon} E = \kappa \text{.} } \end{equation} I prefer to consider general boundary conditons, as I am looking for an general analytical solution.

Equation can be rewritten as

\begin{equation*} \frac{\mathrm{d}^2 E}{\mathrm{d}{x}^2} = -\frac{1}{\phi_T} E \frac{\mathrm{d}{E}}{\mathrm{d}{x}} + \frac{e N_D}{\varepsilon \phi_T} E + \frac{\kappa}{\phi_T} \text{.} \end{equation*} Setting \begin{equation*} \begin{aligned} & y \left( x \right) \equiv E \left( x \right) \\ & a \equiv -\frac{1}{\phi_T} \\ & b \equiv \frac{e N_D}{\varepsilon \phi_T} \\ & c \equiv \frac{\kappa}{\phi_T} \end{aligned} \end{equation*} \begin{equation*} \frac{\mathrm{d}^2 y}{\mathrm{d}{x}^2} = a y \frac{\mathrm{d}{y}}{\mathrm{d}{x}} + b y + c \end{equation*} Using shortened notations for derivatives: \begin{equation} \tag{$y$-ODE} \label{eq:y-ODE} \boxed{ y'' = a y y' + b y + c \text{.} } \end{equation}

Here are my attempts :

  • it is a second-order nonlinear differential equation;
  • it is an autonomous equation : $y'' = F \left( y, y' \right)$;
  • it is a Liénard equation : \begin{equation} \tag{Liénard} \label{eq:Liénard} y'' + f \left( y \right) y' + g \left( y \right) = 0 \text{,} \end{equation} with $f \left( y \right) = - a y$ and $g \left( y \right) = - b y - c$.
  • with the substitution $w = y'$, it is an Abel equation of the second kind : \begin{equation} \tag{Abel} \label{eq:Abel} w w'_{y} + f \left( y \right) w + g \left( y \right) = 0 \end{equation}

I have tried searching in dedicated textbooks, for instance in Polyanin Handbook of Exact Solutions for Ordinary Differential Equations, but I have got the impression that my ODE has no analytical solution...

What do you think? Does one among you see or know one way to solve this equation?

Under request, I might provide boundary conditions for some specific problem.

Remark: I am familiar with numerical solving of ODE's, but this is not what I am looking for here.

Many thanks forward,

Léopold

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  • $\begingroup$ i think your equation is an object for studies, i haven't found an analytical solution $\endgroup$ – Dr. Sonnhard Graubner Dec 20 '17 at 17:38
  • $\begingroup$ Thanks a lot for your quick answer, I will keep that in mind! Which way(s) have you (quickly) tried? You searched in your favorite ODE textbook or explored some usual method to deal with such kind of ODE? $\endgroup$ – Léopold Van Brandt Dec 20 '17 at 19:46
  • $\begingroup$ One thing to wonder about is the possibility of periodic solutions, indeed, the possibility of just one periodic solution. Along with the fourth edition of Nonlinear Ordinary Differential Equations, by Jordan and Smith, there is a problem book with solutions, Nonlinear Ordinary Differential Equations: Problems and Solutions: A Sourcebook for Scientists and Engineers (Oxford Texts in Applied and Engineering Mathematics) D. W. Jordan; Peter Smith, seems published 2007. $\endgroup$ – Will Jagy Dec 20 '17 at 22:40
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    $\begingroup$ Thank you Will for your answer. Well, in this case, I know from physics that the solution of this ODE for electric field is hardly ever periodic. $\endgroup$ – Léopold Van Brandt Dec 21 '17 at 8:44
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$$\frac{d^2y}{dx^2}=ay\frac{dy}{dx}+by+c$$ This is autonomous ODE. One can reduce it to the first order, thanks to the usual change :

$\frac{dy}{dx}=u(y) \quad\to\quad \frac{d^2y}{dx^2}=u\frac{du}{dy}\quad\to\quad u\frac{du}{dy}=a\,y\,u(y)+b\,y+c$

This is an Abel's ODE of the second kind.

The change $\quad u(y)=\frac{1}{v(y)}\quad$ leads to an Abel's ODE of the first kind : $$\frac{dv}{dy}=-(b\,y+c)v^3-a\,y\,v^2$$ For further progress, see :

https://www.hindawi.com/journals/ijmms/2011/387429/#sec2

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  • $\begingroup$ Many thanks, your link seems very serious and this is already quite advanced calculus for me. I take a look at it tomorrow with a clear head, and I will get back to you. $\endgroup$ – Léopold Van Brandt Dec 20 '17 at 20:00
  • $\begingroup$ Hello. I quickly get stuck with the calculus. Especially, do you know how I can extract, obtain $x(\xi)$ and $\xi(x)$ from $$x = \int f_3 E^2 \mathrm{d}\xi \quad E = \exp\left( \int \left( f_1 - \frac{f_2^2}{f_3} \right) \mathrm{d}x \right)$$? Note that $\xi \equiv y$, and that their $x$ is different from our $x$. $\endgroup$ – Léopold Van Brandt Dec 21 '17 at 9:36
  • $\begingroup$ and that $f_3 (\xi) = - b \xi - c$, $f_2 (\xi) = - a \xi$ and $f_1 = f_0 = 0$. $\endgroup$ – Léopold Van Brandt Dec 21 '17 at 9:44
  • $\begingroup$ $$\int \left(f_1-\frac{f_2^2}{f_3} \right)d\xi = \int \frac{a^2\xi^2}{b\xi+c}d\xi =\frac{a^2}{2b^3}\left(2c^2\ln(b\xi+c)+b\xi(b\xi-2c) \right) $$ $$E=(b\xi+c)^{a^2c^2/b^3}\exp\left(\frac{a^2}{2b}(b\xi+c)\xi \right)$$ $$x=-\int (b\xi+c)^{1+2a^2c^2/b^3}\exp\left(\frac{a^2}{b}(b\xi+c)\xi \right)d\xi$$ As in the general case, we are stuck by an integral without closed form. All depends on the kind of functions $f_1(\xi),f_2(\xi),f_3(\xi)$. The method is of interest on theoretical view point. But in the majority of practical cases, the next steps are implicit. $\endgroup$ – JJacquelin Dec 21 '17 at 11:55
  • $\begingroup$ In fact, the key question is : what is the aim of the calculus? Suppose that, in a favourable case, an analytical solution be obtained (on the form of implicit or parametric equations). These non-explicit equations require numerical calculus to be used. Then, why not directly solve the ODE thanks to numerical methods? The interest to an analytical solving of the ODE(when it is possible) is purely theoretical. It is a common fact that many ODEs are not solvable on closed form. I am afraid that it is the case here. As you know, other empirical approches are usual especially in physics. $\endgroup$ – JJacquelin Dec 21 '17 at 11:56

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