method of frozen coefficients and its relation to von Neumann stability analysis

I am considering two equations $$u_t=a(x)u_{xx}$$ and $$v_t=b(x)v_x$$ as classical representatives of the parabolic and hyperbolic family of equations. If $a(x)=a$ and $b(x)=b$ were constants, to show stability of any finite difference scheme I use for those equations, I could use the Fourier or so called Von Neumann stability analysis, i.e. would have to calculate the magnitude of the amplification factor.

In the case of variable coefficients though, where $a=a(x)$ and $b=b(x)$, I am somewhat confused. From the book by Strikwerda (Finite Difference Schemes and PDEs) on page 59 it says that one can use method of frozen coefficients, i.e. freeze the values at every grid point, use the Fourier analysis and deduce stability for variable coefficients. On the other hand, in the book of Ascher (Numerical methods for Evolutionary Differential Equations), on page 153 it says that even ff the method is stable with constant coefficients then it doesn't necessary imply stability for variable coefficients.

Thus I see a contradiction between two books. My problem is that I see how the von Neumann analysis works and I wonder if I can use that with variable-coefficient PDEs as well. If I can, is that the same for both of the equations I am looking at or there are some differences between the two in terms of the method of frozen coefficients? Basically, where is the limitation of the Von Neumann stability analysis for parabolic and hyperbolic equations?

• Are the variable coefficients bounded and positive? If so, you may be able to use these bounds in your von neumann stability analysis. See Sewell's book, chapters 2 & 3 for examples of analysis of parabolic and hyperbolic equations with variable coefficients.
– Paul
Apr 15, 2013 at 13:22
• Well, they are bounded on any finite interval $I\in \mathbb{R}$, but I am ready to make such an assumption and then see if I can extend it. Fourier is considered without any boundary conditions, and is supposed to be on the whole real line, so let's assume it is bounded everywhere. Apr 15, 2013 at 22:52
• Without any loss of generality, you can restrict the problem onto your specific domain so that you can bound the variable coefficients by its maximum and minimum value (even if you don't know what it is, you can assume they exist). Then, these bounds form part of your stability criteria in the end. You can generalize this argument onto any arbitrary domain by assuming that a maximum and minimum value of $a(x)$ or $b(x)$ exists and repeating the same analysis. Think of it as if you're analyzing the worst case scenario and analyzing its stability.
– Paul
Apr 16, 2013 at 14:40
• The coefficients are like polynomials, say $x^2$, so they are bounded on any bounded set. However, to apply the Fourier analysis, I see that one has either consider the problem on the whole real line, or consider a periodic problem, so it can be extended there. In my case, I am solving that numerically, so I would have to put boundary condition there and they don't have to be periodic. One can transform the equation to the one with zero boundary condition, but the equation would get an additional term. Any other approaches to handle boundary conditions that are not periodic? Apr 17, 2013 at 0:12

1 Answer

One author is being practical, while the other is being rigorous. The short answer is that stability of the frozen-coefficient problems is proven to guarantee stability of the variable-coefficient problem if the coefficients and solution are smooth enough.

If you want to know how much smoothness is needed, prepare to get your hands dirty. The fundamental reference on the parabolic side is

Fritz, John. On integration of parabolic equations by difference methods: I. Linear and quasi-linear equations for the infinite interval. Communications on Pure and Applied Mathematics, 5(2):155--211 (1952).

For hyperbolic problems, see

Strang, Gilbert. Accurate partial difference methods II. Numerische Mathematik 6, 37-46 (1964).

One additional note: Ascher claims that his examples 5.1 and 5.6 demonstrate how schemes that are von Neumann stable in the smooth case can be unstable in the non-smooth case. But both examples are misleading. Example 5.1, involving the KdV equation, actually has a smooth solution for all time. Meanwhile, Example 5.6 is still a linear constant-coefficient problem; only the solution is discontinuous. The oscillations that appear there are due to numerical dispersion and not instability. So these two examples do not demonstrate the pathologies that can arise. Unfortunately, I don't know of one that does.