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I am trying to proof directly that the Radau IA method \begin{align} k_1 &= f\left(t_0, x_o+h\left(\frac14 k_1 - \frac14 k_2\right)\right) \\ k_2 &= f\left(t_0+\frac23 h, x_o+h\left(\frac14 k_1 + \frac5{12} k_2\right)\right) \\ x_1 &= x_0 + h\left(\frac14 k_1 + \frac34 k_2\right) \end{align} is of order 3 (so the local order of convergence is of order 4).

I think the idea is to compute the derivatives (this is where I am struggeling) of $$\varphi(h,t,x)=x_0 + h\left(\frac14 k_1 + \frac34 k_2\right)$$ w.r.t $h$ up to order 4 and then using Taylor expansion at $h=0$ and compare it with the expansion of the flow. Is that more or less correct? How do I actually do this? Or is there a better way for a direct proof?

leiferz
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1 Answers1

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The order conditions for $$ \begin{array}{c|cc} c_1 & a_{11}& a_{12}\\ c_2 & a_{21}& a_{22}\\ \hline & b_1& b_2 \end{array} ~~=~~ \begin{array}{c|cc} 0 & \frac14& -\frac14\\ \frac23 & \frac14& \frac5{12}\\ \hline & \frac14& \frac34 \end{array} $$ are \begin{align} \text{order 1}\\ 1&=b_1+b_2&&=\frac14+ \frac34\\ \hline \text{order 2}\\ \frac12&=b_1c_1+b_2c_2&&=\frac14·0+ \frac34·\frac23\\ \hline \text{order 3}\\ \frac13&=b_1c_1^2+b_2c_2^2&&=\frac34·\frac49\\ \frac16&=\sum b_ia_{ij}c_j&&=(b_1a_{12}+b_2a_{22})c_2&&=\left(\frac14·-\frac14+ \frac34·\frac5{12}\right)·\frac23\\ \hline \text{order 4?}\\ \frac14&\overset{?}{=}\sum b_ic_i^3=\frac34·\frac{8}{27}&&\text{No}\\ \frac18&=\sum b_ic_ia_{ij}c_j\\ \frac1{12}&=\sum b_ia_{ij}c_j^2\\ \frac1{24}&=\sum b_ia_{ij}a_{jk}c_k \end{align} So indeed, order 3 holds and in order 4 the interpolation condition can not be satisfied.

So the goal is to get a match for as many terms as possible in $y'=F(y)$, \begin{multline} y(x+h)=y(x)+Fh+\tfrac12F'Fh^2+\tfrac16(F''[F,F]+F'F'F)h^3 \\+\tfrac1{24}(F'''[F,F,F]+3F''[F'F,F]+F'F''[F,F]+F'F'F'F)h^4+... \end{multline} It seems obvious that some kind of shorthand for the expressions in $F$ is needed. The order of the derivative corresponds to the number of arguments and there is only one function in play, so $F'[F''[F,F]]$ can be conveniently replaced by $[[*,*]]$ or $[[,]]$ or more graphical representations of rooted trees. Order 3 is still low enough that this is not needed, if one were to compute the order 4 error coefficients, this shorthand would substantially reduce space.

The order 1 consistency conditions should be trivial to motivate, so $c_i=\sum a_{ij}$ and $1=\sum b_i$. For the stages one gets implicit equations \begin{align} k_i&=F+F'[a_{ij}k_j]h+\tfrac12F''[a_{ij}k_j,a_{il}k_l]h^2+...\\ &=F+a_{ij}F'[k_j]h+\tfrac12a_{ij}a_{il}F''[k_j,k_l]h^2+... \end{align} This means that $k_i=F+O(h)$, inserting this gives immediately $k_i=F+c_iF'[F]h+O(h^2)$, and then inserting this $$ k_i=F+c_iF'[F]h+a_{ij}c_jF'[F'[F]]h^2+\tfrac12c_i^2F''[F,F]h^2+O(h^3) $$ Now put this into the assembled formula $$ y_{+1}=y+b_ik_ih=y+Fh+b_ic_iF'Fh^2+b_ia_{ij}c_jF'[F'[F]]h^3+\tfrac12b_ic_i^2F''[F,F]h^3+O(h^4) $$ Comparing both step-$h$ expressions results in the cited order conditions.

Lutz Lehmann
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  • Clearly, this is correct, but one has to know the order condition beforehand. I am struggling to proof it directly. – leiferz Mar 18 '21 at 17:23
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    It is probably easiest to derive the order condition for the generic 2-stage method first. Butcher trees where invented for that, to shorten the derivative constructs. From that you get an implicit system for the series expression of the $k_i$ that you have then to compare to the expansion of the exact step. – Lutz Lehmann Mar 18 '21 at 18:04
  • Very nice, could you clarify the notation with the brackets? – leiferz Mar 19 '21 at 12:16
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    This is an ASCII version of B-trees, see https://tex.stackexchange.com/questions/283343/butcher-trees-in-tikz for how that would look graphically. Essentially, if $\tau=[\tau_1,...,\tau_k]$, then $F{\tau}=F^{(k)}[F{\tau_1},...,F{\tau_k}]$ and $F{*}$ is simply $F$. – Lutz Lehmann Mar 19 '21 at 12:22
  • See also https://math.stackexchange.com/questions/4002723/problem-in-a-taylor-expansion-of-runge-kutta-method for the same dance for explicit order-3 methods. – Lutz Lehmann Mar 22 '21 at 22:25