# BDF2 and TR-BDF2: what is better?

What method of numerical solving ODEs is better? BDF2 or TR-BDF2?

Namely, what advantages has TR-BDF2 over BDF2?

The BDF2 method requires the values of $y_{n-1}$ and $y_n$ for computing $y_{n+1}$ but we can use, for example, the trapezoidal method for $n = 0$ and BDF2 on next steps.

The TR-BDF2 method computes an auxiliary value $y_{n+1/2}$ with the trapezoidal method and applies the BDF2 for computing $y_{n+1}$ by using $y_n$ and $y_{n+1/2}$.

TR-BDF2 for solving $y' = f(y)$ represents the following scheme: $$y_{n+1/2} = y_n + \frac{\tau}{4}(f(y_n) + f(y_{n+1/2})),$$ $$y_{n+1} = \frac{1}{3}(4y_{n+1/2} - y_n + \tau f(y_{n+1})).$$ Here $\tau$ is a step size. The both stages are implicit. The first stage is the trapezoidal method with the step size $\tau/2$ and the second stage is the BDF2 with the step size $\tau/2$.

UPD Edwards et al. in the paper Nonlinear variants of the TR/BDF2 method for thermal radiative diffusion point out that BDF2 has undesirable conservation properties. Could you explain please how can this influence the computation accuracy?

• Can you put BDF2 explicitly in your question as well? It requires one rather than two function evaluations/solutions per step, and sometimes you specifically want a one-step method (not two-step like BDF2). As to your question title: "better" is a rather ambiguous concept; it's important to be clear about what you will use it for. As far as I can tell, it just does extra work (because the second step of TR-BDF2 is the step of BDF2), but I'm not sure if that's fair. Jan 15, 2015 at 5:00

• BDF2 has (in hopefully common notation) defining polynomials $\rho(z) = \frac12-2z+\frac32z^2$ and $\sigma(z)=z^2$. Since the roots of $\sigma(z)$ are inside the unit disk, and it is A-stable, BDF2 is L-stable (same for other BDF methods with $\sigma(z)=z^s$). Are you sure what you said about L-stability is right? Jan 15, 2015 at 8:16
• A method is L-stable if it is A-stable and its stability polynomial $R$ satisfies $\lim_{z \rightarrow \infty}|R(z)| = 0$. So BDF3 through BDF6 can't be L-stable; they're not A-stable. BDF1 is backward Euler, which is A-stable and has $R(z) = (1-z)^{-1}$, so it is L-stable. I get $R(z) = (3 + z)/(2z^{2} - 5z + 3)$ for BDF2, so yes, BDF2 is L-stable also. That said, TRBDF2 does have a larger stability region, per Strang (math.mit.edu/~gs/cse/papers/optimalstability_rev6.pdf), if you take steps of slightly unequal sizes. Jan 15, 2015 at 9:18