Euler-scheme discretization for 2 dimensional vector field ODE

Question

Let $V: \mathbb{R}^m \rightarrow \mathbb{R}^m$ be a vector field on $\mathbb{R}^m$. An ODE is the equation: $$X_t = X_0 + \int_0^t V(X_s)ds$$ where $X: \mathbb{R}_{+} \rightarrow \mathbb{R}^m, X_0\in \mathbb{R}^m$. Euler scheme to solve this ODE is as follows:

1. Choose time points $0=t_0<t_1<\dots<t_n$
2. $\tilde{X}_{t_0} :=X_0$
3. $\tilde{X}_{t_{k+1}} := \tilde{X}_{t_k} + V(\tilde{X}_{t_k})(t_{k+1} - t_{k})$

Now to the exercise:

Consider a 2-dimensional vector field $V(x) = (-x_2,x_1)$ and implement Euler scheme with $X_0 = (1,0)$ on interval $[0,7]$. Use different $n$s.

The solution to this should be a circle for some reason, however my solution does definitely not remind a circle:

n <- 1000
tt <- seq(0,7,length.out = n)
xx <- matrix(0,n,2)
xx[1,] <- c(1,0)

vv <- function(x){
  return(c(-x[1],x[2]))
}

for(i in 1:(n-1)){
  xx[i+1,] <- xx[i,]+ vv(xx[i,])*(tt[i+1]-tt[i])
} 

What did I do wrong? This is quite important question, as there are more of such in the assignment.

Answer 1

For the specific given task, you can easily simulate the solution manually using complex numbers. If $z=x_1+ix_2$, then $v(z)=iz$. This gives the exact solution as $z(t)=e^{it}z(0)=\cos t+i\sin t$ which indeed traces a circle. The Euler method computes $$ z_{n}=(1+ih)z_{n-1}=(1+ih)^nz_0=(1+h^2)^{n/2}e^{i\phi n} $$ where $\phi=\arg(1+ih)=\arctan(h)$. Or in another way $$ z_n=e^{n(ih+\frac12h^2-\frac13ih^3-\frac14h^4+...)}=e^{\frac12ht_n(1-\frac12h^2+...)}e^{it_n(1-\frac13h^2+..)} $$ This gives an outwards spiral that gets tighter the smaller $h$ is. As a secondary effect one gets that the angular speed is slower than the exact solution, this error also reducing for falling $h$.