I am trying to solve a numerical analysis dealing with Runge Kutta methods. The problem is in solving the differential equation:
$$\frac{d \vec{y}(x)}{dx} = \vec{F}(x,\vec{y}).$$
Defining the error of the discrete gradient:
$$\vec{e}(x) = \vec{y}(x) - \vec{y}(x+h) + \alpha \vec{K}_0 + \beta \vec{K}_1,$$
with $\vec{K}_0 = h \vec{F}(x,\vec{y})$ and $\vec{K}_1= h\vec{F}(x+\frac{h}{2},\vec{y} + \frac{1}{2} \vec{K}_0)$; how can we pick $\alpha$ and $\beta$ to get $\vec{e}(x) \sim O(h^3)$.
My solution is:
Using $$\vec{y}(x+h) = \vec{y}(x) + h \vec{y}'(x) + \frac{h^2}{2} \vec{y}''(x) + O(h^3) = \vec{y}(x) + h \vec{F} + \frac{h^2}{2} \frac{d\vec{F}}{dx} + O(h^3)$$
and $$\vec{K}_1=h\bigg(\vec{F}(x,\vec{y}) + \frac{h}{2}\frac{d\vec{F}}{dx}(x,\vec{y})+O(h^2)\bigg)$$
I find:
$$-1+\alpha+\beta =0 \qquad {\rm and}\qquad \beta-1 =0$$
I am bothered to find $\alpha=0$, is there something wrong with my solution.
No, there is nothing wrong, you have just computed the modified/improved Euler or explicit midpoint method.