What is the order of linear multi step method $u_{j+1}=(1-a)u_j+au_{j-1}+h/4\{(a+3)u'_{j+1}+(3a+1)u'_{j-1}\}$ for solving $u'=f(x,u)~?$

Question

Problem: The order of linear multi step method $~u_{j+1}=(1-a)u_j+au_{j-1}+h/4\{(a+3)u'_{j+1}+(3a+1)u'_{j-1}\}~$ for solving $~u'=f(x,u)~$ is
$(1)~~~2~~~~$ if $~~a=-1$
$(2)~~~2~~~~$ if $~~a=-2$
$(3)~~~3~~~~$ if $~~a=-1$
$(4)~~~3~~~~$ if $~~a=-2$

My work: The general multi-step (k-step) method for the solution of $u'=f(x,u)~,~~u(x_0)=u_0~$ may be written as $$u_{j+1}=a_1u_j+a_2u_{j-1}+\cdots+a_ku_{j-k+1}+h\left(b_0u'_{j+1}+b_1u'_j+\cdots+b_ku'_{j-k+1}\right)$$ and $~T_{j+1}~$ which is the local truncation error is written as $$T_{j+1}=c_0u(x_j)+c_1hu'(x_j)+c_2h^2u''(x_j)+\cdots+c_ph^pu^{(p)}(x_j)+T_{p+1}$$where $~c_0=1-\sum_{i=1}^ka_i~,$
$c_q=\frac{1}{q!}\left[1-\sum_{i=1}^ka_i(1-i)^q\right]-\frac{1}{(q-1)!}\sum_{i=0}^kb_i(1-i)^{q-1}~,~~~q=1,2,\cdots,p$

The linear multi-step method is said to be of order $~p~$ if $~c_0=c_1=c_2=\cdots=c_p=0~$ and $~c_{p+1}\ne0~.$

Given that $~u_{j+1}=(1-a)u_j+au_{j-1}+h/4\{(a+3)u'_{j+1}+(3a+1)u'_{j-1}\}~$

Comparing $~a_1=1-a~,~~a_2=a~,~~b_0=(a+3)/4~,~~b_1=0~,~~b_2=(3a+1)/2~$ and $~k=2~.$

So $~c_0=1-\sum_{i=1}^2a_i=1-a_1-a_2=1-1+a-a=0~$
$c_1=\frac{1}{1!}\left[1-\sum_{i=1}^2a_i(1-i)^1\right]-\frac{1}{(0)!}\sum_{i=0}^2b_i(1-i)^{0}=1+a_2-b_0-b_2=0$
Similarly, $~c_2=0~,~~c_3=-1/3(1+a)$
if $~a=-1~,~$ then $~c_3=0~$
if $~a=-2~,~$ then $~c_3=1/3\ne0~$
Hence if $~a=-2~,~$ then $~c_0=c_1=c_2=0~$ but $~c_3\ne0~.$
Therefore the order is $~2~$ for $~a=-2~$ and then option $\bf (2)$ is correct.

Also by the similar process, $~a_4=-1/24(1-a)~$
if $~a=-1~,~c_4=-1/12\ne0~.$
Therefore the order is $~3~$ for $~a=-1~$ and then option $\bf (3)$ is correct.

My Query: Please anyone help me to check if there be any wrong. Or let me inform, if there be any alternative process to solve it. Thank you.

Answer 1

For linear multistep methods to detect the order it is sufficient to insert $$u(x)=e^x,~~~ u_j=e^{jh},$$ at $j=0$ or $j=1$ into the step equation and to compute the defect between both sides. After cancelling the common power it remains $$ E(h)=e^h-(1-a)e^0-ae^{-h} - \frac{h}{4}((a+3)e^h+(3a+1)e^{-h})~ $$ Now insert the power series for the exponential and collect the coefficients of equal degree. If the defect has order $E(h)=O(h^{p+1})$, then $p$ is the order of the method.

Using a computer algebra system (like the CAS Magma this can be done automatically and gives

> A<a>:=FunctionField(Integers());                          
> P<h>:=PowerSeriesRing(A);
> Exp(h)-(1-a)-a*Exp(-h)-h/4*((a+3)*Exp(h)+(3*a+1)*Exp(-h));

(-a - 1)/3*h^3 + (a - 1)/24*h^4 + (-a - 1)/30*h^5 
    + (a - 1)/360*h^6 + (-a - 1)/840*h^7 + ...

So order $p=2$ for $a\ne -1$, order $p=3$ for $a=-1$.