Are the sequences $\frac{ih}{2}((1-ih)^{k}-(1+ih)^{k})$ and $\frac{1}{(1+h^{2})^{k}}\frac{ih}{2}((1-ih)^{k}-(1+ih)^{k})$ bounded?

Question

I want to understand the explicit and the implicit Euler method better. Assume I have the initial value problem $y''+y=0$, $y(0)=0$, $y'(0)=1$, which is of course solved by $y=\sin(x)$, and I convert this into a system of first-degree linear equations \begin{equation*} \left(\begin{matrix}y'\\z'\end{matrix}\right)=\left(\begin{matrix}0 & 1\\-1 & 0\end{matrix}\right)\left(\begin{matrix}y\\z\end{matrix}\right)\text{.} \end{equation*} When I apply the explicit and the implicit Euler methods to the differential equation $y''+y=0$, with stepwidth $h>0$, I get approximation points that satisfy: \begin{eqnarray*} y_{k} & = & \frac{ih}{2}((1-ih)^{k}-(1+ih)^{k}) & \text{for the explicit Euler method}\\ \tilde{y}_{k} & = & \frac{ih}{2(1+h^{2})^{k}}((1-ih)^{k}-(1+ih)^{k}) & \text{for the implicit Euler method} \end{eqnarray*} I am interested in seeing how fast these sequences grow with $k$. My guess is that the first sequence is unbounded, whereas the second converges to zero. At least that is what I get for , but I have no idea how to prove this. Any suggestions?

Answer 1

Let $\theta=\arctan h, r=\sqrt{1+h^2}$ and then $$ 1+ih=r(\cos\theta+i\sin\theta), 1-ih=\sqrt{1+h^2}(\cos\theta-i\sin\theta). $$ So \begin{eqnarray*} y_{k} & = & \frac{ih}{2}((1-ih)^{k}-(1+ih)^{k}) \\ &=& hr^k\sin(k\theta). \end{eqnarray*} Since $h>0$, you have $r>1$ and $\theta\in(0,\frac\pi2)$ and hence $\{y_k\}$ diverges. Similarly \begin{eqnarray*} \tilde{y}_{k} & = & \frac{ih}{2(1+h^{2})^{k}}((1-ih)^{k}-(1+ih)^{k}) \\ &=&hr^{-k}\sin(k\theta). \end{eqnarray*} So $\{\tilde{y}_k\}$ converges to $0$ since $r>1$.