Discussion
For the initial value problem
\(y' = f(x, y),\ y(x_{0}) = y_{0}\),
generate approximations \(y_{n}\) to \(y(x_{n})\), \(x_{n} = x_{0} + nh\), for a fixed \(h \gt 0\) and \(n = 1, 2, 3, \ldots\), using the recursion formula
\(y_{n} = y_{n-1} + ak_{1} + bk_{2}\), where
\(k_{1} = hf(x_{n-1}, y_{n-1})\) and \(k_{2} = hf(x_{n-1} + \alpha h, y_{n-1} + \beta k_{1})\).
Which one of the following choices of \(a, b, \alpha, \beta\) for the above recursion formula gives the Runge-Kutta method of order 2 ?
Answer: C
The second-order Runge-Kutta choice is \(a = 0.25, b = 0.75, \alpha = 2/3, \beta = 2/3\).
Matching the Taylor expansion of \(y(x_{n})\) to second order requires
\[ \begin{aligned} a + b &= 1 \\ b\alpha &= \tfrac{1}{2},\quad b\beta = \tfrac{1}{2} \end{aligned} \]
For \(a = 0.25, b = 0.75, \alpha = \beta = 2/3\): \(a + b = 1\) and \(b\alpha = b\beta = 0.5\). This is Ralston's method.
\(a = b = 1\) gives \(a + b = 2\). \(a = b = 0.5, \alpha = \beta = 2\) gives \(b\alpha = 1\). \(a = b = 0.5, \alpha = 1, \beta = 2\) gives \(b\beta = 1\). Each fails a condition.
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