I'd like to find analytical solutions of that kind of differential equations :

$$\partial_t c = \partial_x (D(c)\partial_x c) $$

with $D(c)$ a polynomial. The trivial cas $D(c)=a$ with $a$ a constant is the classical diffusion equation in 1d.

For polynomial such as : $D(c)= (a+bc)^n$ with $n$ an integer, the solution can be infered by a change of variable from the simple case $D(c)= C^n$.

In this article the author finds analytical solutions for $D(c)=D_0 (c/C_0)^n$ with $D_0 \; ,\; C_0$ some constants for a diffusion from a point. The method to find his result is to look for scaling functions like what I'm writing there.

I'd like to know if there are analytical solution for the case $D(c)=(c-a)(c-b)$ with $a\neq b$ some constants. I'm interested even in particular cases.

So I'd be happy if you have some references or some technics to tackle this problem, and it's even better if you have the result for a particular case :)

I'm not even sure the solution exists all the time so if you know articles or results about that thank you in advance.


1 Answer 1


Using Maple, I find implicit solutions for the case $D(c)= (c-a)(c-b)$:

$$\eqalign{&{k_{{1}}}^{2}{k_{{2}}}^{2}{c}^{2} + \left( 2\,{k_{{1}}}^{4}k_{{2}}k_{{3 }}-2\,{k_{{1}}}^{2}{k_{{2}}}^{2}a-2\,{k_{{1}}}^{2}{k_{{2}}}^{2}b \right) c\cr &+ \left( 2\,{k_{{1}}}^{6}{k_{{3}}}^{2}-2\,a{k_{{1}}}^{4}k_{{ 2}}k_{{3}}-2\,b{k_{{1}}}^{4}k_{{2}}k_{{3}}+2\,ab{k_{{1}}}^{2}{k_{{2}}} ^{2} \right) \ln \left( -{k_{{1}}}^{2}k_{{3}}+ck_{{2}} \right)\cr & -2\,{k _{{2}}}^{4}t-2\,k_{{1}}{k_{{2}}}^{3}x-2\,{k_{{2}}}^{3}k_{{3}}-2\,k_{{4 }}{k_{{2}}}^{3} =0} $$

  • $\begingroup$ thx but could you be a bit more explicit (lol) plz ? what are the $k_n$'s ? Constants to determine ? $\endgroup$
    – J.A
    Commented Aug 6, 2019 at 15:23
  • $\begingroup$ It's interesting that Mathematica doesn't manage to do anything with the equations $\endgroup$
    – J.A
    Commented Aug 6, 2019 at 15:59
  • $\begingroup$ And would you have an idea how to deal with that kind of formula ? I don't manage to find back some simple results... And do you know what kind of method maple uses to find this result ? $\endgroup$
    – J.A
    Commented Aug 6, 2019 at 16:09
  • 1
    $\begingroup$ The $k_i$ are arbitrary constants. Unfortunately it's not going to be possible to solve the equation for $c$ in closed form. On the other hand, notice that the only dependence on $t$ and $x$ is through the terms $-2 k_2^4 t -2 k_1 k_2^3 x$, so you can solve for that combination of $t$ and $x$ in terms of $c$. $\endgroup$ Commented Aug 6, 2019 at 18:21
  • 1
    $\begingroup$ Yes: take partial derivatives of the implicit equation with respect to $t$, $x$ and $(x,x)$, and you can solve for $\partial c/\partial t$, $\partial c/\partial x$ and $\partial^2 c/\partial x^2$ in terms of $c$, $x$ and $t$. Substitute those expressions into the difference of right and left sides of the PDE and simplify, and you get $0$. $\endgroup$ Commented Aug 7, 2019 at 21:06

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