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I'm not able to give a general solution of this pde:

$$ {\partial \over \partial t} {\Phi(x,t)}={k^2\Phi(x,t)^2}{\partial^2 \over \partial x^2}{\Phi(x,t)}$$ Can someone help me?

Riccardo.Alestra
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    This looks rather difficult, as it is a nonlinear PDE. – Fabian Feb 09 '12 at 16:08
  • I found one! how about $\Phi=ax+b$? – yohBS Feb 09 '12 at 16:23
  • Thanks yohBS, but I don't see any dependence from the time t – Riccardo.Alestra Feb 09 '12 at 16:30
  • @yohBS: and it is far from a general solution ;-) – Fabian Feb 09 '12 at 18:28
  • @Riccardo.Alestra: nonlinear PDEs are pretty hard (apart from some well-known integrable examples). Thus, I would recommend to give some additional motivation: where does this equation arise? why are you interested in it? – Fabian Feb 09 '12 at 18:31
  • @Fabian: it's a Fick diffusion equation with a diffusion coefficient proportional to the square of the concentration itself. – Riccardo.Alestra Feb 09 '12 at 20:20
  • @Riccardo.Alestra: rather unfortunately your description ("Fick diffusion equation with a diffusion coefficient proportional to the square of the concentration") does not give any more information than the equation itself. – Willie Wong Feb 09 '12 at 22:05
  • Have you tried checking one of the references like Handbook of Differential Equations? – Willie Wong Feb 09 '12 at 22:19
  • A side comment: I don't like describing that as a diffusion equation. Diffusion should conserve total particle number, which is $N(t) = \int \Phi(x,t) dx$. In your model $\partial_t N(t) < 0$ strictly assuming the particle density $\Phi \geq 0$. Somehow you are losing particles. – Willie Wong Feb 09 '12 at 22:26
  • @Willie Wong: You are right. In fact this is a diffusion equation of atoms doping a particular kind of material in which part of the atoms form a bound with the material itself. – Riccardo.Alestra Feb 10 '12 at 20:47

1 Answers1

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Here's my try. It's a standard trick that is used in physics, but I can't guarantee that the solution is the most general. Actually, I'm certain that it is not.

First of all, define $\tilde t = t k^2$ so that the equation does not have $k$ anymore. Now guess an ansatz of the form $\Phi=X(x)T(t)$, where $X$ is a function of $x$ only and $T$ is a function of $T$ only. Plugging that into the equation yields

$$X(x) T'(t)-T(t)^3 X(x)^2 X''(x)=0$$

which can be transformed, assuming all $T,X$ do not vanish to

$$\frac{T'(t)}{T(t)^3}=X(x)''X(x)$$

Since the left hand side depends on $t$ only, and the right one on $x$, they both must be a constant, which we'll denote by $g$. So you need to solve the two ODE's:

$$\begin{align} T'&=g T^3\\ X''&=\frac{1}{X} \end{align} $$

The solution of the first equation is $$T=\pm\frac{1}{\sqrt{-2(g t+C)}}$$ where $C$ is an integration constant. If you choose it negative enough then $T$ will be real. The solution of the $X$ equation is a bit more difficult, but Mathematica finds one: $$X=A^{-2} e^{ \operatorname{InverseErf}\left[\pm i\sqrt{A \frac{2g}{\pi } (x+B)^2}\right]^2}$$ where $A,B$ are integration constants. Is this any good?

yohBS
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    The problem with separation of variables is that, while for linear equations one can often end up writing the general solution as a linear superposition of special separation of variables solutions, for nonlinear equations this class of solutions is only representative of some special cases. – Willie Wong Feb 09 '12 at 22:03
  • I am well aware of that, and I wrote explicitly that this is not the most general solution. – yohBS Feb 10 '12 at 08:11
  • @yohBs: Ok. I accept your answer, but as you said this solution is not the general solution. Thanks – Riccardo.Alestra Feb 10 '12 at 20:51
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    @Riccardo.Alestra Thx. By the way, if your equation really describes "diffusion with a diffusion coefficient proportional to the square of the concentration itself", then it should read $$\partial_t \Phi=\nabla\cdot\left(\Phi^2\vec\nabla\Phi\right)$$ Written like that it will conserve the particle number. – yohBS Feb 12 '12 at 06:58