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What kind of approaches can be used to study the following quasilinear parabolic pde for a scalar function $u=u(x,t)$ ? $$ u_t= u \ u_{x x} $$

The physical problem where this pde comes from dictates that the Cauchy problem of interest corresponds to an initial condition which is a third-order polynomial with no-constant term $$ u(x,0) = x \ (x^2-s x + p), \ \ \ x \in \mathbb{R}^+, \ \ \ s p \neq 0 $$ Note that $u(x,0)$ is initially $\propto x$, and that $u(x,t)$ will remain so during the evolution as long as the pde makes sense. Indeed, and this is the crux of the matter, numerical experiments and heuristic arguments strongly indicate that there exists a blow-up time $0< T=T(s,p) < \infty$, where the solution explodes to infinity, with $$ u(x,t) \sim \frac{f(x)}{T-t}, \ \ \ t \to T^- $$

  • Can one prove this, and relate $T$ and the amplitude $f(x)$ to the parameters $s,p$ entering the initial condition ?
  • It is possible (and how !) to continue the solution after the blow-up time ?
  • A colleague of mine suggested that I could exchange the role of the dependent and independent variables, and/or to look at a paper by Clarkson, Fokas, \& Ablowitz, "Hodograph transformations of linearizable partial differential equations", where they study similar equations and relate them to the Harry-Dym equation and "solvable" (by inverse scattering transform or Painlevé reductions) variants. Needless to say, the functional form of the terms in my pde do not fall under the conditions of applicability of Clarkson-fokas-Ablowitz' main theorem, while the hodograph transformation: $$ (t,x,u(x,t)) \longrightarrow (t,y=u,x=v(y,t)) $$ so that (using $u_x = 1/v_y$, $u_{x x}=-v_y^{-3}v_{y y}$, $u_t= -v_t/v_y$) $$ v_t + y (v_y^{-1})_y =0 $$ does not seem to give a pde much easier to solve (or to recognize, such as the porous medium equation)

Edit: an alternative perhaps preferable form (closer to the porous medium equation!) is for the function $w=\frac{1}{2}{\rm Log}(u^2)$, which obeys $$ w_t = \left(w_x e^w\right)_x $$ The Lie-point symmetries of the latter are recalled in this review (section 5), but none would match the initial condition I have to deal with.

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    $\begingroup$ I always get nervous when my solutions grow at spatial infinity. One possibility would be to make the transformation v = (x+1)^{-4}u so that v is, at least initially, well behaved at spatial infinity. The resulting equation has a v^2 term on the right hand side. This gives bounds for finite-time blow up at least for initial data with negative minima. For positive initial data, this form of the equation at least highlights the competition between the curvature and amplitude of a positive maximum. $\endgroup$ Commented Feb 6, 2012 at 13:57
  • $\begingroup$ Thanks Aaron for the comment. Do you think that such a method would also entail a better-behaved numerical scheme ? $\endgroup$ Commented Feb 6, 2012 at 14:57
  • $\begingroup$ The solution (x^2/2+ax+b)/(T-t) blows up at T, but has the wrong initial value. $\endgroup$ Commented Feb 7, 2012 at 21:49
  • $\begingroup$ @Bob. Yes, I had seen this, and it also turns out that the separable solution you mention comes out through an ODE reduction after studying the Lie-point symmetries for the related function $w=\frac{1}{2}{\rm Log}(u^2)$ obeying the perhaps somewhat better-looking $w_t=(w_x e^w)_x$. I have edited my question accordingly, and I have given a ref. concerning the Lie symmetry-analysis. $\endgroup$ Commented Feb 9, 2012 at 11:27
  • $\begingroup$ There is a sort of Backlund transformation. If v_x = log(uv) and v_t = U_x+v_x, then u_t = u u_{xx} iff v_t = v v_{xx}. For example if u=1, then v is a traveling wave with profile satisfying v_x=log v. But I don't see any better examples so far. $\endgroup$ Commented Feb 9, 2012 at 20:55

2 Answers 2

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By dividing with $u$, you may formally write the equation as $$\tag 1 \partial_t \log{u} = \Delta u. $$ For the log-diffusion equation $$ \tag 2 \partial_t u = \Delta \log u $$ the trick is to write $$ \log u = \lim_{m \to 0} \frac{u^m-1}{m} $$ and then study the corresponding porous medium equation. The issue is of course the convergence of the equation, but here is a recent preprint concerning such issues for equation (2):

http://www.vanderbilt.edu/math/people/liao/sm_files/m_limit_112311.pdf

Now, a short while ago I actually discussed with some of the authors of the above paper whether similar methods could be used for equations of type (1). I gathered that there is more or less no literature for this equation (although you might still want to search a bit). However, something similar might still work and it is an reasonable research question, I think. However, this probably won't help you too much.

In any case, substituting $v=u^m/m$ into the corresponding approximative equation $$ \tag 3 \partial_t \frac{u^m-1}{m}= \Delta u $$ yields $$ \partial_t v = m^{1/m} \Delta v^{1/m}, \quad m \in (0,1) $$ which is a porous medium equation for $v$. Here the problem is that you would like to take the limit $m \to 0$ and of course the convergence issues need to be studied. But at least a lot is known for the porous medium equation so maybe there is some hope to get something out of it.

Just out of curiosity, what might be the applications for such equations?

edit: Let me just add, that there is also some literature for equation (3) directly. In particular, see this paper and the references therein:

http://arxiv.org/pdf/1208.0621v2.pdf

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I don't know if this is useful, but anyway: Let's compute $$ \partial_t \partial_x u=\partial_x u\partial_x^2 u+u\partial_x^3u, $$ and $$ \partial_t \partial_x^2 u=\partial_x^2 u\partial_x^2 u+\partial_x u\partial_x^3u+\partial_x u\partial_x^3 u+u\partial_x^4u. $$ Now notice that if $u_0(x_0)=0$ then $u(t,x_0)=0$ as long as the solution remains $C^2$. Assume that we have a non-negative initial data such that there exists $x_0$ such that $u_0(x_0)=0$. Then $x_0$ is a point of minimum ($\partial_x u(x_0,t)=0$) and $u(x_0,t)=0$. The equation evaluated at this point is $m(t)=\partial_x^2 u(x_0,t)$ with ODE $$ m'(t)=m(t)^2, $$ and this ODE has finite time lifespan. Thus, the classical solution is not global.

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