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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?

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  • $\begingroup$ Inserting Taylor expansions centered at $x_{j+2}$ should give simple enough equations to determine the parameters. Did you do that? $\endgroup$ – LutzL May 13 at 12:19
  • $\begingroup$ I was doing centered at $x_j$ probaby that why I got mistake $\endgroup$ – ILoveMath May 13 at 16:18
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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.

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  • $\begingroup$ @LutzL: you are right. Fixed. I confused myself with the centering on $0$ $\endgroup$ – Ross Millikan May 23 at 14:35
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Another test problem to get more coefficient identities at once it the classical $f(t,y)=y$ with solution $y(t)=Ce^x$. Select the constant so that $y_{j+2}=1$, then the equation reads $$ 2\sinh(2h)+2α\sinh(h)=h(2β\sinh(h)+γ)+O(h^{p+1}) $$ Now the left side is odd in $h$, so the right side can not have even terms, requiring $β=0$. Inserting the power series of the exponential and comparing coefficients of equal degree gives next \begin{align} \left[2h+\frac{(2h)^3}{6}+\frac{(2h)^5}{5!}+...\right]+α\left[h+\frac{h^3}{6}+\frac{h^5}{5!}+...\right]&=\frac{hγ}2+O(h^{p+1}) \\[1em]\hline 2+α&=\frac{γ}2\\ \frac{8+α}6&=0 \end{align} so that $α=-8$, $γ=-12$ and $p=4$.


As for stability, look at the characteristic polynomial for the left side, $$ z^4-8z^3+8z-1=(z^2-1)(z^2-8z+1)=(z-1)(z+1)(z-4-\sqrt{15})(z-4+\sqrt{15}) $$ The root $z=-1$ gives a weakly unstable component, the root $z=4+\sqrt{15}$ makes the method strongly unstable, errors due to the initialization of the multi-step method and floating point truncations will be rapidly magnified.


possible typo in given equation?

Assuming the first component of the right side should lead to a non-trivial contribution, assume that there is a typo and it is a sum, not a difference. Then the test equation leads to $$ 2\sinh(2h)+2α\sinh(h)=h(2β\cosh(h)+γ)+O(h^{p+1}) $$ with the power series expansion \begin{align} \sum_k(2^{2k+1}+α)\frac{h^{2k+1}}{(2k+1)!} &=hβ\sum_k\frac{h^{2k}}{(2k)!}+\frac{hγ}2+O(h^{p+1}) \\[1em]\hline 2+α&=β+\frac{γ}2\\ \frac{8+α}6&=\frac{β}2\\ \frac{32+α}{120}&=\frac{β}{24} \end{align} giving $β=12$, $α=28$, $γ=36$ with $p=6$ but even more instability.

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