Consider the first order intial value problem $y'(x)=-y(x),x>0,y(0)=1$ and the corresponding numerical scheme $$4 \bigg( \frac{y_{n+1}-y_{n-1}}{2h}\bigg)-3\bigg( \frac{y_{n+1}-y_{n}}{h}\bigg)=-y_n,$$ with $y_0=1$ and $y_1=e^{-h},$ where $h$ is the step size. Then which of the following statements are true ?
- The order of the scheme is $1$
- The order of the scheme is $2$
- $|y_n-y(nh)|\to \infty$ as $n \to \infty$
- $|y_n-y(nh)|\to 0$ as $n \to \infty$
My attempt:
The exact solution for this problem is $y(x)=e^{-x}.$
After solving for $y_{n+1}$ in the given iterative scheme, I get $$y_{n+1}=(3+h)y_n-2y_{n-1}.$$
So, we have that $y_2=(3+h)y_1-2y_0=(3+h)e^{-h}-2.$
I am not able to proceed any further.
In general, I want to know how to calculate the order given any numerical scheme for a first order IVP.
Please help. Thanks in advance.
To determine the order of a linear multi-step method (with equidistant steps) you insert the exact solution, for example set $y_n=e^{-hn}$. After dividing out the common factor, the equation reads $$ 4\frac{e^{-h}-e^h}{2h}-3\frac{e^{-h}-1}h=-1+O(h^p). $$ Inserting the exponential Taylor series, this gives $$ -4\left(1+\frac{h^2}{3!}+\frac{h^4}{5!}+...\right)+3\left(1-\frac{h}{2!}+\frac{h^2}{3!}-\frac{h^3}{4!}+\frac{h^4}{5!}+...\right)=-1+O(h^p) $$ Cancel terms and factors $$ -\frac32h-\frac16h^2+...=O(h^p) $$ so that the order is $p=1$.