Determine $\alpha, \beta$ and $\gamma$ so that the linear, multistep
method$$ y_{j+4} – y_j + \alpha (y_{j+3} – y_{j+1}) = h [ \beta (f_{j+3} –
f_{j+1}) + \gamma f_{j+2} ] $$for the d.e. $y' = f(x)$ has global error of order $3$. Is the
resulting method stable?
Attempt
We assume $f_{j} = f(x_j,y_j)$. At the n+1 step, the solution (exact) is $y(t_n)$ and the approximation is $y_{n+1}$. So the way we find truncation error is by using taylor expansion on $y(t_n) – y_{n+1}$ and this works fine for single-step method. But, in this situation, how would we do it?
Best Answer
The symmetry of the question suggests centering everything at $x_{j+2}$. We can translate the question by making $x_{j+2}=0$ to ease the typing. Note that $y_1=y(-h)$ and so on. First it must be exact for $f(x)=1,y(0)=0,y=x$, so plug that in. We get $$4h+2\alpha h=h\gamma\\ \gamma=4+2\alpha$$ Then it must be exact for $f(x)=x,y(x)=\frac 12x^2$, so we have $$0=h\beta(2h)\\\beta=0$$ It must be exact for $f(x)=x^2,y(x)=\frac 13x^3$, so we have $$\frac {16}3h^3+\alpha\frac 23h^3=0\\ \alpha=-8\\ \gamma=-12$$ We are only looking at the odd part of $y$ around $x_{j+2}$ and the odd part of $x$. It probably satisfies your definition of stability as the right side is zero, but it does not solve the equation.