Runge-Kutta Method: Showing Order of Accuracy is 2

[evinda]

Hello! (Wave)

The following Runge-Kutta method is given:

$$\begin{vmatrix}
0 & 0 & | 0\\
\frac{1}{2} & \frac{1}{2} & | 1 \\ \\
\frac{1}{2} & \frac{1}{2}
\end{vmatrix} \\$$

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? (Thinking)

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?

 

[jvicens]

Hello! Great job on your work so far. You have correctly applied the Runge-Kutta method and have derived the error term to be $O(h^3)$. To show that the method has an order of accuracy of $2$, we need to show that the error term can be written as $c_1h^2+c_2h^3+O(h^4)$ where $c_1$ and $c_2$ are constants.

To do this, we can use Taylor's theorem to expand $f(t^{n+1}, \zeta^{n,2})$ and $f(t^n, y(t^n))$ around the point $(t^n, y(t^n))$. We get:

$f(t^{n+1}, \zeta^{n,2})=f(t^n, y(t^n))+hf_t(t^n, y(t^n))+\frac{h^2}{2}f_{tt}(t^n, y(t^n))+\frac{h^2}{2}f_{yy}(t^n, y(t^n))+O(h^3)$

$f(t^n, y(t^n))=f(t^n, y(t^n))$

Substituting these into our error term, we get:

$\delta^n=\frac{h^2}{4} \left( hf_t(t^n, y(t^n))+\frac{h^2}{2}f_{tt}(t^n, y(t^n))+\frac{h^2}{2}f_{yy}(t^n, y(t^n))+O(h^3) \right) f_y(t^n, y(t^n))- \frac{h^2}{4} f_y(t^n, y(t^n)) \left( f(t^n, y(t^n)) \right)+O(h^3)$

$\delta^n=\frac{h^2}{4} \left( \frac{h^2}{2}f_{tt}(t^n, y(t^n))+\frac{h^2}{2}f_{yy}(t^n, y(t^n)) \right) f_y(t^n, y(t^n))+O(h^3)$

Now, we can see that $c_1=\frac{1}{4} \left( \frac{h^2}{2}f_{tt}(t^n, y(t^n))+\frac{h^2}{