I have a coupled set of PDEs that need to be solved as part of a larger problem. I am currently approaching this by computing spatial derivatives with finite differences and using PETSc's nonlinear conjugate gradient solver. This does converge in terms of an absolute tolerance, however, when I started thinking about the structure of my equations, I've noticed a problem.

Consider $$\partial_x F\!\left(x\right) = G\!\left(x\right)$$ where $F$, $G$ are given by $$ F\!\left(x\right) = \frac{1}{1 + f\left(x\right)^2} $$ $$ G\!\left(x\right) = \frac{f(x)}{1 + f\left(x\right)^2}. $$ We would be solving for $f\!\left(x\right)$. (This is not one of the equations I am solving, but it's similar in structure and just an illustration anyway.)

What would stop the nonlinear solver from, say, making $f\!\left(x\right)$ larger and larger until the "signal" (residual) falls below the "noise" (tolerance) level?

I've observed this happening in my own equations, and I think that my "convergence" might not be real. Yet the solver is doing exactly what it should, and what I asked it to do. Is there a way around this problem? Or is this not a problem at all?

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    $\begingroup$ For the $f,g$ you give, it's not true that $f'=g$... $\endgroup$ – Wolfgang Bangerth Jul 3 at 17:18
  • $\begingroup$ @WolfgangBangerth, Just realized that! Fixed now. Sorry for the confusion, I probably shouldn't post late at night. $\endgroup$ – emprice Jul 3 at 17:20

We can rewrite the equation as

$ \frac{-2 f f'}{(1+f^2)^2} = \frac{f}{1+f^2} $

which reduces to

$ f' = - \frac{1+f^2}{2} $

The latter does not have $1+f^2$ in the denominator, so it should not have the aforementioned numerical problem and can be easily integrated numerically.

In fact, now the equation can be easily seen to have a trivial analytic solution, which is always useful for numerical solution verification,

$ \frac{df}{1+f^2} = - \frac{dx}{2} $


$ \arctan(f) = - \frac{x}{2} + const $

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