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Does anyone know what it means for a one-step method to be 'consistent'?

I've seen it written that if $a + b = 1$, then the RK2 method is consistent. How can I show that if $a + b = 1$ then RK2 is consistent? RK2 being

$$\begin{aligned} y_{i+1} &= y_i + h(ak_1 + bk_2)\\ k_1 &= f(x_i, y_i)\\ k_2 &= f(x_i + \alpha h, y_i + \beta hk_1) \end{aligned} $$

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I think there was a typing mistake in your post. I changed a capital $H$ in the formula for $k_2$ to $h$. I also fixed up the formatting a little bit, since I was editing the question anyhow. – Harald Hanche-Olsen Nov 10 '12 at 9:10
up vote 2 down vote accepted

If $y$ is an exact solution to the ODE $$y'=f(x,y)$$ and $y_i=y(x_i)$ for each $i$, where $x_{i+1}=x_i+h$, then Taylor's formula implies $$ y_{i+1}=y_i+hf(x_i,y_i)+o(h)\qquad\text{as $h\to0$}. $$ (If you're not familiar with little-oh and its big brother big-oh notation, see the wikipedia article on it.) For reasonably smooth functions $f$, the $o(h)$ term can be replaced by $O(h^2)$.

If now $(y_i)$ comes from a numerical method, you say the method is consistent if the same formula holds. Think of it as a minimal requirement for the method to give approximations to the exact solution: Without consistency, the method is no good at all.

Along with a consistency proof you often get better estimates on the $o(h)$ term that can yield a convergence proof.

In your case, just knowing that $k_2=f(x_i,y_i)+O(h)$ (where I assume that $f$ is smooth) gives you $$ y_{i+1}=y_i+h\cdot(a+b)f(x_i,y_y)+O(h^2), $$ and so the method is consistent if and only if $a+b=1$.

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I can't see how you are deriving $$ y_{i+1}=y_i+h\cdot(a+b)f(x_i,y_y)+O(h^2), $$. And how you are then concluding that it's only consistent if $a + b = 1$. Say for sake of argument $a + b = 0.5$...then the method wouldn't be consistent..why? I have no idea what $a + b$ is supposed to signfify. – sonicboom Nov 10 '12 at 11:33
For the first bit, just insert the definition of $k_1$ and the stated fact about $k_2$ in the definition of $y_{i+1}$ and get $y_{i+1}=y_i+h(a+b)f(x_i,y_i)+hbO(h)$, after collecting terms. The magic of big-O notation allows you to absorb the constant $b$ into the big O, and also $hO(h)$ really is the same as $O(h^2)$. It does take a bit of getting used to, admittedly. I don't have time for more right now, will come back with more later if nobody beats me to it. – Harald Hanche-Olsen Nov 10 '12 at 12:32
Cheers, right after I posted I copped what you were doing and I think I get it now. Is this correct - you are saying $k_2$ 'is' $k_1$ with an error term? So then when you use that version of $k_2$ in the equation it leads to the equation you gave in your answer and $a + b$ must equal $1$ for the equation to match the true value of the function $y$ at position $i+1$, ie the first two terms of the taylor series for $y$? – sonicboom Nov 10 '12 at 16:05
Yep, that's pretty much it. It can be dressed up in more mathematical language, the main point being that if $a+b\ne1$ then the mismatch is proportional to $h$, and that is not an $o(h)$-term. – Harald Hanche-Olsen Nov 10 '12 at 16:19

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