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(First time asking question on this forum so please kindly let me know if this is out of scope/inappropriate etc.)

I have a problem that leads me to the following quadratic system of PDEs:-

$ c_1 w_q (\partial_tw_q) + c_2 w_q (\partial_xw_q) + c_3 (\partial_{x}w_q)^2 +c_4 w_q (\partial_{xx}w_q) + c_5 (w_q)^2 + c_6 w_q w_{q-1} = 0\; \forall q \in \mathbb{N} $

with $w_q(T,x)=1$ and $w_0(t,x) = 1$, $t \in [0,T] \subset \mathbb{R}$ and $c_1 = 1;\;\; c_2,c_5 \in \mathbb{R};\;\; c_3<0;\;\;c_4,c_6>0$

I would like to find the series of solutions $w_q(t,x) \;\forall q \in \mathbb{N}$.

I would have really liked this to work out into a linear system of PDEs but i'm having no luck at all. I've read something about Quadratic Differential Systems but all the sources seem to be dealing with quadratic systems of ODEs.

Has anyone encountered anything similar or is my system too exotic to have been considered in the literature? (or just my lack of erudition!) In addition to solutions, would appreciate pointers to relevant materials, keywords or even wild ideas that I should look at.

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    $\begingroup$ >>> I would like to find the series of solutions $w(t,x)_q$ <<< Under what additional constraints? I doubt you are after $0,1,0,2,0,239,0,c_2t-c_1x,...$ etc. $\endgroup$
    – fedja
    Commented Jul 6, 2014 at 9:01
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    $\begingroup$ There is not even an equation! $\endgroup$
    – Matthias Ludewig
    Commented Jul 6, 2014 at 11:17
  • $\begingroup$ Thanks Kofi and @fedja for pointing out the mistakes. I'm sorry for not mentioning the $c_n$ even though that might help in making the problem more specific. However, my point was just to ask for general comments and direction (e.g. book references, have you seen this before somewhere etc), which I should perhaps have made more clear in the question. My problem is that I've never seen such types of equations before. $\endgroup$
    – rrz
    Commented Jul 8, 2014 at 8:48
  • $\begingroup$ OK, but it is still unclear if you want to find just one solution (which is trivial: look for functions that depend on $t$ only) or you want to show that nothing else is there, or something else. $\endgroup$
    – fedja
    Commented Jul 8, 2014 at 9:54
  • $\begingroup$ Dear @fedja, in fact the above problem was motivated by exactly such a "trivial" $w_q(t)$ series of ODE which I wished to extend to $w_q(t,x)$. I am looking for non-trivial solutions (non-constant [not possible anyway I think], at least first order function of x and t). Showing non-existence is fine too although I suspect it exists as I have an approximation for the underlying motivating problem. $\endgroup$
    – rrz
    Commented Jul 8, 2014 at 10:34

1 Answer 1

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Anyway, let's comb it a bit. Since we start at $1$ at $T$, I'll assume that $w_q>0$ at least for $t$ close to $T$. Once $w_0$ is fixed everywhere, we have no choice but to solve left to right.

Put $W=w_q^p$. Then $(W)_{xx}=pw^{p-2}(ww_{xx}+(p-1)w_x^2)$, so we can choose the power to combine two second order terms into one ($p=0$ should be understood as $W=\log w_q$). Then we get $$ c_1W_t+c_2W_x+c_3W_{xx}+c_4w_{q-1}W^{(p-1)/p}=0. $$ This is almost standard heat except it is written in a skewed coordinate system instead of the rectangular one (which can be trivially adjusted) plus you have an extra term. Since you start at $1$, in any decent function space you can hum the usual Duhamel-Gronwall tune to prove local uniqueness. Of course, your setup may be so exotic that even the unperturbed heat equation has many solutions, so if you want me or anyone else to go beyond this point, pose your problem with full and clear details.

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  • $\begingroup$ Thanks so much, I was suspecting diffusion - everything in the motivating field ends up being diffusion so this makes a lot of intuitive sense. I'm going to add some extra clarification to $c_i$s above and also add an extra term $w^2$ that I missed (yes bad, very sorry). I'm going to study this a bit before accepting as answer (is this correct protocol?). $\endgroup$
    – rrz
    Commented Jul 8, 2014 at 15:41
  • $\begingroup$ $w^2$ changes nothing. Of course, there is no reason to accept until you get sure the argument works. $\endgroup$
    – fedja
    Commented Jul 8, 2014 at 16:37
  • $\begingroup$ I can confirm your working, but turns out that I had another ansatz that took me to a similar solution but I abandoned it precisely because the extra term $W^{(p-1)/p}$ was not linear. While this is interesting in and of itself, I think this extra term makes it even more hairy than the formulation I show above at least for numerical work. I'm going to accept the answer now, but I may come back later on to revisit and offer more clarifications (or provide another solution!) $\endgroup$
    – rrz
    Commented Jul 10, 2014 at 16:58
  • $\begingroup$ Well, you certainly cannot make it perfectly linear. Also, from the viewpoint of numerics a fractional power is just as bad as the function in front of derivatives: once $w$ hits $0$, you are screwed either way. :-) $\endgroup$
    – fedja
    Commented Jul 11, 2014 at 4:29

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