# Determine the variables so as to make differential equation has global error of order 3

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?

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

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.

• @LutzL: you are right. Fixed. I confused myself with the centering on $0$ – Ross Millikan May 23 at 14:35

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.