Problem in a Taylor expansion of Runge-Kutta method

Question

I'm stuck with this problem:

Determine all three-stage, thre-order RK methods

I'm reading some papers about this and they do:

$$ y_{i} = y_{i-1} + \sum_{i=0}^m b_{j}k_{j} $$

$$ \begin{cases} k_{1} = hf(t_{i-1},y_{i-1}) \\ k_{2} = hf(t_{i-1}+c_{2}h,y_{i-1}+a_{21}k_{1}) \\ k_{3} = hf(t_{i-1}+c_{3}h,y_{i-1}+a_{31}k_{1}++a_{32}k_{2}) \\ \end{cases} $$

Now, they do a third order Taylor expansion of $y(t;t_{i-1},y_{i-1})$. The point is here:

$$ y_{i} = y_{i-1} + f(t_{i-1},y_{i-1})h + \cfrac{1}{2}\left[ \frac{\partial }{\partial t} f(t_{i-1},y_{i-1}) + f(t_{i-1},y_{i-1}) \frac{\partial }{\partial y} f(t_{i-1},y_{i-1}) \right]h^2 + ... $$

I'm trying to get this expression but I don't know to do it:

$$ \frac {d}{dh} f(t_{i-1},y_{i-1}) = \frac {\partial }{\partial t} f(t_{i-1},y_{i-1})\frac {dt}{dh} + \frac {\partial }{\partial y} f(t_{i-1},y_{i-1})\frac {dy}{dh} $$

So that, it means that:

$$ \begin{cases} \frac {dt}{dh} = 1 \\ \frac {dy}{dh} = f(t_{i-1},y_{i-1}) \\ \end{cases} $$

But I don't understand it (in case this is right).

Please, anyone could help me to understand it? Thanks in advance.

Answer 1

In the derivative $\frac{dy}{dh}$, you are considering the function $h\mapsto y(t_{i-1}+h)$ for an exact solution of the ODE with initial condition $y(t_{i-1})=y_{i-1}$. In consequence you get indeed $$ \frac{dy}{dh}(t_{i-1})=y'(t_{i-1})=f(t_{i-1},y_{i-1}). $$

Any ODE problem can be transformed into an autonomous first order system $u'(t)=F(u(t))$. In application this can then be reduced back to the non-autonomous case. In the autonomous case the propagation of the exact solution has the Taylor expansion $$\begin{multline} u(t+h)=u(t)+h·F(u(t))+\frac{h^2}2F'[F]+\frac{h^3}6\Bigl(F''[F,F]+F'[F'[F]]\Bigr) \\+\frac{h^4}{24}\Bigl(F'''[F,F,F]+3F''[F'[F],F]+F'[F''[F,F]]+F'[F'[F'[F]]]\Bigr)+O(h^5) \end{multline}$$ For the stages of the method one gets the expansions $$\begin{align} k_1&=hF\\ k_2&=hF+a_{21}h^2F'[F]+\tfrac12 a_{21}^2h^3 F''[F,F]+O(h^4)\\ k_3&=hF+hF'[a_{31}hF+a_{32}h(F+a_{21}hF'[F]]+\tfrac12hF''[a_{31}hF+a_{32}hF]^2+O(h^4)\\ &=hF + (a_{31}+a_{32})h^2F'[F] + a_{32}a_{21}h^3F'[F'[F]] + \tfrac12(a_{31}+a_{32})^2h^3F''[F,F]+O(h^4) \end{align}$$ and assembled $$\begin{multline} u_{+1} = u+(b_1+b_2+b_3)hF+(b_2a_{21}+b_3(a_{31}+a_{32}))h^2F'[F] \\+ \tfrac12(b_2a_{21}^2+b_3(a_{31}+a_{32})^2)h^3F''[F,F] + b_3a_{32}a_{21}h^3F'[F'[F]] + O(h^4) \end{multline}$$ Now using $a_{21}=c_2$ and $a_{31}+a_{32}=c_3$ and comparing both expansions gives the conditions \begin{align} \text{order $1$}&:& b_1+b_2+b_3&=1\\ \text{order $2$}&:& b_2c_2+b_3c_3&=\tfrac12\\ \text{order $3$}&:& b_2c_2^2+b_3c_3^2&=\tfrac13\\ & & b_3a_{32}c_2&=\tfrac16\\ \end{align} See also Butcher 2008: RK introduction and low order methods for a more systematic approach and examples of methods.