MHB Finding an Expression for Y in a Linear Multistep Method

[shen07]

consider the following LMM

$$y_n-\frac{3}{2}y_{n-1}+\frac{1}{2}y_{n-2}=h(\frac{1}{2}f_n+\frac{1}{4}f_{n-1}-\frac{1}{4}f_{n-2})$$

which is applied to the initial value problem $$y'(t)=y(t),0\leq{t}\leq{1}\\ and\\ y(0)=1$$

How do i find an expression for $$y_n$$, if the starting values are $$y_0=1\\ and\\ y_1=1$$

 

[zzephod]

consider the following LMM

$$y_n-\frac{3}{2}y_{n-1}+\frac{1}{2}y_{n-2}=h(\frac{1}{2}f_n+\frac{1}{4}f_{n-1}-\frac{1}{4}f_{n-2})$$

which is applied to the initial value problem $$y'(t)=y(t),0\leq{t}\leq{1}\\ and\\ y(0)=1$$

How do i find an expression for $$y_n$$, if the starting values are $$y_0=1\\ and\\ y_1=1$$

Usually you have a problem in the form:

$$\frac{dy}{dt}=f(t,y)$$

with an initial condition $y(0)=y_0$. So here $f_n=f(t_n,y_n)=y_n$ and your method reduces to:

$$y_n-\frac{3}{2}y_{n-1}+\frac{1}{2}y_{n-2}=h\left(\frac{1}{2}y_n+\frac{1}{4}y_{n-1}-\frac{1}{4}y_{n-2}\right)$$

which can be rearranged to give $y_n$ in terms of $y_{n-1}$ and $y_{n-2}$

Reassuringly this integrates up nicely to give a reasonable approximation to the analytic solution after allowance is made for the poor choice of $y_1$. A much better choice for $y_1$ would be $y_0+h\times f(t_0,y_0)=y_0\times (1+h)$, effectively taking an Euler step to get your multi-step method started.

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