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How (or if) a comparison principle works in the case of equations singular at some point? For example, I am analyzing a partial differential equation

$$ \partial_{t}u=\partial_{rr}u+\frac{2}{r}\partial_{r}u-\frac{2}{r^{2}}u\quad u(0,t)=0,\quad \partial_{t}u(1,t)=0 $$

There is a family of stationary solutions $u_{a}(r)=a\cdot r$ that could possibly function as sub or super solutions with suitably chosen $a$. But, to my understanding, comparison principle only works when coefficients in the equation are well behaved, which is not the case here: at $r=0$ the coefficients are diverging. How can I rigorously apply the comparison principle in this case?

A somewhat related question is, what do we know about $\partial_{r}u$? In particular, can we show that $\partial_{r}u$ is bounded when $a_1r\le u(r)\le a_2r$?

Ultimately, I would like to use comparison principle to give a bound on $u$ and, in consequence, a bound on $\partial_{r}u$ with $u$ solving a quasilinear equation

$$ \partial_{t}u=\partial_{rr}u+\frac{2}{r}\partial_{r}u-\frac{2}{r^{2}}f(u)\quad u(0,t)=0. $$

Here $f$ is smooth and $f(u)=u+\mathcal O(u^{2})$.

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  • $\begingroup$ I would try applying the comparison principle on a compact domain in the interior, and see if one can approach the boundary. $\endgroup$
    – timur
    Commented Dec 20, 2013 at 21:12
  • $\begingroup$ Could you please specify the domain on which you want to solve the final quasilinear pde, the regularity required on the solution and if the domain is unbounded, the growth conditions on the solution. Cheers $\endgroup$
    – JCM
    Commented Jan 2, 2014 at 21:42
  • $\begingroup$ Here, the singular coef. in your diff. eq. doesn't concern whether or not a comparison principle can be obtained (these depend on existence of parabolic weak maximum principles). This is due to the sign of the zeroth order term in the diff. eq. (the coef. of the first order term in the diff. eq. does not effect a max. principle argument). Provided that you specify appropriate boundary conditions on the solutions, notably for $(x,t)\in ([0,1]\times\{ 0\})\cup (\{ 0\}\times (0,\infty ))\cup (\{ 1\}\times (0,\infty ))$, as well as regularity on the solutions, then a comparison principle exists. $\endgroup$
    – JCM
    Commented Aug 7, 2015 at 19:22

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