Suppose I had a boundary value problem:

$$\frac{d^2u}{dx^2} + \frac{dv}{dx}=f \text{ in } \Omega$$ $$\frac{du}{dx} +\frac{d^2v}{dx^2} =g \text{ in } \Omega$$ $$u=h \text{ in } \partial\Omega$$

My goal is to decompose the solution of this coupled problem into a sequence of uncoupled PDE's. To decouple the system, I'm applying a fixed point iteration over a sequence of approximations $(u^k,v^k)$ such that

$$\frac{d^2u^k}{dx^2} + \frac{dv^{k-1}}{dx}=f $$ $$\frac{du^{k-1}}{dx} +\frac{d^2v^k}{dx^2} =g $$

Theoretically, this would allow me to solve both equations as a purely elliptic PDE's. However, I have never seen fixed point iterations applied to PDE's in this way. I've seen fixed point iterations applied to the numerically discretized equations (finite difference method, finite element method, etc.), but never to the continuous equations directly.

Am I violating any blatant mathematical principle by doing this? Is this mathematically valid? Could I solve the coupled PDE as a sequence of uncoupled PDE's by using fixed point iteration applied to the CONTINUOUS variable problem, rather than the DISCRETE variable problem?

At this point, I'm not really concerned with whether it is practical to use this method, but rather whether it is theoretically plausible. Any feedback would be greatly appreciated!

  • 6
    $\begingroup$ In the hyperbolic PDE literature, fractional step and operator splitting methods sort of do what you describe above. $\endgroup$ Commented May 16, 2013 at 19:59
  • $\begingroup$ Did you mean $(u^k, v^k)$ instead of $(u^k, p^k)$? $\endgroup$
    – Bill Barth
    Commented May 16, 2013 at 20:06
  • $\begingroup$ @BillBarth: Yes! I just corrected it. $\endgroup$
    – Paul
    Commented May 16, 2013 at 20:40
  • $\begingroup$ @GeoffOxberry: I find operator splitting to be very different in character. $\endgroup$
    – anonymous
    Commented Jan 14, 2014 at 23:49
  • $\begingroup$ @Paul: I can think of at least one other problem where "coupled PDEs" are solved through a fixed point iteration (and not just formulated as fixed point problems): domain decomposition, see e.g. the Neumann–Dirichlet method. (the difference here being that you have two PDEs but they live on different domains and the coupling is only through an interface). $\endgroup$
    – anonymous
    Commented Jan 14, 2014 at 23:53

1 Answer 1


You define a sequence in, say, $C^{\infty}(\Omega)\times C^{\infty}(\Omega)$ by $$ \frac{\text{d}^2u^k}{\text{d}x^2} + \frac{\text{d} v^{k-1}}{\text{d} x} =f\\ \frac{\text{d}^2v^k}{\text{d}x^2} + \frac{\text{d} u^{k-1}}{\text{d} x} =g\\ $$ (plus boundary conditions).

It is clear that if this sequence converges, it will be a solution of your original set of PDEs.

As for proving whether the sequence converges in the Banach space, Banach's fixed point theorem may come in handy. It says that if your mapping $x^k\to x^{k+1}$ is a contraction, you can be sure to approach the solution given any initial guess $u^0$, $v^0$. Hence, what you'd have to check is if $$ \left\| \begin{pmatrix} u^k\\ v^k \end{pmatrix} - \begin{pmatrix} \hat{u}^k\\ \hat{v}^k \end{pmatrix} \right\| \le q \left\| \begin{pmatrix} u^{k-1}\\ v^{k-1} \end{pmatrix} - \begin{pmatrix} \hat{u}^{k-1}\\ \hat{v}^{k-1} \end{pmatrix} \right\| $$ with a fixed $|q|<1$ for any given $(u^{k-1}, v^{k-1})$, $(\hat{u}^{k-1}, \hat{v}^{k-1})$.

This logic works both in the continuous and the discrete space.

  • 3
    $\begingroup$ Shouldn't $|q|<1$? $\endgroup$
    – Paul
    Commented May 20, 2013 at 13:25

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