Order of accuracy of Runge-Kutta method

Question

The following Runge-Kutta method is given by the Butcher tableau:

$$ \begin{array}{c|ccccc} \tau_1 =0 & a_{11}=0 & a_{12} = 0\\ \tau_2 =1 & a_{21} = \frac{1}{2} & a_{22} = \frac{1}{2}\\ \hline & b_1 = \frac{1}{2} & b_2 = \frac{1}{2} & \ \end{array} $$

and I want to show that its order of accuracy is $2$.

I have tried the following:

$$t^{n,1}=t^n \\ t^{n,2}=t^n+ \frac{h}{2} \\ \zeta^{n,1}=y(t^n) \\ \zeta^{n,2}=y(t^n)+ \frac{h}{2} f(t^n, y(t^n))$$

$$\delta^n=y(t^n+h f \left( t^n+ \frac{h}{2}, y(t^n)+ \frac{h}{2} f(t^n, y(t^n))\right))-y(t^n+h) \\ = -\frac{h}{2} f(t^n,y(t^n))+ \frac{h}{2} f(t^{n+1}, \zeta^{n,2})- \frac{h^2}{2} f_t(t^n,y(t^n))-\frac{h^2}{2} f_y(t^n, y(t^n)) f(t^n,y(t^n))+O(h^3)$$

$$f(t^{n+1}, \zeta^{n,2})=f \left( t^n+h, y(t^n)+ \frac{h}{2} (f(t^n,y(t^n))+f(t^{n+1}, \zeta^{n,2})) \right)=f(t^n,y(t^n))+hf_t(t^n,y(t^n))+ \frac{h}{2} (f(t^n,y(t^n))+ f(t^{n+1}, \zeta^{n,2})) f_y(t^n, y(t^n))+O(h^2)$$

So we have:

$\delta^n=\frac{h^2}{4} f(t^{n+1}, \zeta^{n,2}) f_y(t^n,y(t^n))- \frac{h^2}{4} f_y(t^n,y(t^n)) f(t^n, y(t^n))+O(h^3)$

Is it right so far or have I done something wrong?

If it is right then it should hold $f(t^{n+1}, \zeta^{n,2}) f_y(t^n,y(t^n))= f_y(t^n,y(t^n)) f(t^n, y(t^n))$ so that we get $\delta^n=O(h^3)$. But does this hold? If so, how could we show this?

Answer 1

This is the trapezoidal method, one scheme among the Lobatto family. The scheme could be written as follows \begin{align*} k_1&=f(t_{n-1},y_{n-1})\\ k_2&=f(t_n,y_{n-1}+\frac{h}{2}k_{1}+\frac{h}{2}k_{2})\\ y_{n}&=y_{n-1}+\frac{h}{2}k_{1}+\frac{h}{2}k_{2} \end{align*} Now to show order of consistency, we insert our exact solution in the numerical scheme, as shown below. \begin{align*} \tilde{k_1}&=f(t_{n-1},y(t_{n-1}))\\ \tilde{k_2}&=f(t_n,y(t_{n-1})+\frac{h}{2}\tilde{k_{1}}+\frac{h}{2}\tilde{k_{2}})\\ \end{align*} Note $\tilde{k_1}$ and $\tilde{k_2}$ have the exact solution $y(t_{n-1})$ as argument. Now we do Taylor expansion around $(t_{n-1},y(t_{n-1})).$ Note that $\tilde{k_1}=f$. Expanding $\tilde{k_2}$ we obtain $$ \tilde{k_2}=f+hf_t+f_y\frac{h}{2}\left(\tilde{k_1}+\tilde{k_2}\right)+O(h^2) $$ From this we can see that $\tilde{k_2}=f+O(h)$. Hence we can rewrite the above expression as $$ \tilde{k_2}=f+hf_t+hff_y $$ Plug this into $y(t_n)$ to obtain $$ y(t_n)=y(t_{n-1})+\frac{h^2}{2}f_t+hf+\frac{h^2}{2}ff_y+O(h^3) $$ This exactly matches the expansion of $f(t,y)$ upto the second order. Hence the method is consistent of order $2$.