Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
This looks rather difficult, as it is a nonlinear PDE. –  Fabian Feb 9 '12 at 16:08
    
I found one! how about $\Phi=ax+b$? –  yohBS Feb 9 '12 at 16:23
    
Thanks yohBS, but I don't see any dependence from the time t –  Riccardo.Alestra Feb 9 '12 at 16:30
    
@yohBS: and it is far from a general solution ;-) –  Fabian Feb 9 '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 9 '12 at 18:31
show 5 more comments

1 Answer 1

up vote 2 down vote accepted

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?

share|improve this answer
    
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 9 '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 8:11
    
@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
2  
@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 6:58
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.