Global ODE and Log-density Integration Error — Examples

Extend the given field to the open domain ${\mathbb{R}}^2$ by the same formula $f(t,x)=x$; its restriction to $[0,1]\times \mathbb{R}$ is unchanged. Define ${\mathrm{\Phi }}_t(x_0):=\exp (t)x_0$ for $t\in [0,1]$. By [F3] and [F9], ${\mathrm{\partial }}_t{\mathrm{\Phi }}_t(x_0)=\exp (t)x_0=f(t,{\mathrm{\Phi }}_t(x_0))$ and ${\mathrm{\Phi }}_0(x_0)=x_0$; [F4] makes every slice a C¹ diffeomorphism $\mathbb{R}\to \mathbb{R}$ with inverse $y\mapsto y\mathrm{/}\exp (t)$. Thus $x(t)={\mathrm{\Phi }}_t(1)=\exp (t)$ is the exact path. Euler satisfies $x_{n+1}=(1+h)x_n$, so induction gives $x_n=(1+h{)}^n$. Since $h>0$, [F5] gives $\exp (h)-1=\exp (c)h$ for some $c\in (0,h)$, and [F6] with $\exp (0)=1$ gives $\exp (c)>1$, hence $1+h<\exp (h)$. By [F8], $(1+h{)}^N<\exp (h{)}^N$; [F7] gives $\exp (h{)}^N=\exp (Nh)=\exp (1)$, so $\mathrm{\mid }{x}_N-x(1)\mathrm{\mid }=\exp (1)-(1+h{)}^N>0$. For $1\le n\le N$, [F8] and [F7] give $1\le x_n\le \exp (nh)\le \exp (1)$, while $x_0=1$; exact states lie in $[1,\exp (1)]$ by [F6]. Therefore $K=[0,\exp (1)]$ contains the exact path and Euler nodes. The bounded convex open set $U=(-1,2)\times (-1,\exp (1)+1)$ contains $[0,1]\times K$, and $f(t,x)=x$ is C² with bounded derivatives there, so [F2] applies.

Here $D{\mathrm{\Phi }}_1(1)=\exp (1)$, so [F1] gives $p_1(\exp (1))\mathrm{/}{p}_0(1)=1\mathrm{/}\exp (1)=\exp (-1)$ by [F4]. Since $p_0(1)>0$, [F10] and [F11] show that the exact log-density change is $\log (p_1(\exp (1)))-\log p_0(1)=-1$. Euler subtracts $h{\mathrm{div}}_{x}f=h$ on each of the $N$ steps, so its computed change is $-Nh=-1$; with the given exact initialization, the log-density error is zero despite the positive state error from step 1.1.

Define ${\mathrm{\Phi }}_t(x_0):=\exp (t^2\mathrm{/}2)x_0$ for $t\in [0,1]$. By [F5], ${\mathrm{\partial }}_t{\mathrm{\Phi }}_t(x_0)=t\exp (t^2\mathrm{/}2)x_0=t{\mathrm{\Phi }}_t(x_0)=f(t,{\mathrm{\Phi }}_t(x_0))$, and ${\mathrm{\Phi }}_0(x_0)=x_0$; [F6] makes each slice a C¹ diffeomorphism $\mathbb{R}\to \mathbb{R}$ with inverse $y\mapsto y\mathrm{/}\exp (t^2\mathrm{/}2)$. Thus $x(t)={\mathrm{\Phi }}_t(0)=0$ is the exact path. Euler has $x_{n+1}=x_n+h{t}_n{x}_n$, hence $x_n=0$ for every $n$ by induction. The compact convex state set $K=[-1,1]$ contains the exact and Euler states, and the bounded convex open set $U=(-1,2)\times (-2,2)$ contains $[0,1]\times K$; $f(t,x)=tx$ is C² with bounded derivatives there, so [F3] applies to its left-Euler density update.

Since $D{\mathrm{\Phi }}_1(0)=\exp (1\mathrm{/}2)$, [F1] gives $p_1(0)\mathrm{/}{p}_0(0)=1\mathrm{/}\exp (1\mathrm{/}2)=\exp (-1\mathrm{/}2)$ by [F6]. Since $p_0(0)>0$, [F7] and [F8] give the exact log-density change $\log p_1(0)-\log p_0(0)=-1\mathrm{/}2$. The left-Euler update gives $-h{\sum }_{n=0}^{N-1}{t}_n=-h^2{\sum }_{n=0}^{N-1}n=-h^{2}N(N-1)\mathrm{/}2=-1\mathrm{/}2+h\mathrm{/}2$: the sum formula holds at $N=1$, and if it holds at $N$ then adding the next term $N$ gives $(N+1)N\mathrm{/}2$. Thus its absolute log-density error is $h\mathrm{/}2>0$. For midpoint quadrature, $-h{\sum }_{n=0}^{N-1}(t_n+h\mathrm{/}2)=-h^2(N(N-1)\mathrm{/}2+N\mathrm{/}2)=-h^2{N}^2\mathrm{/}2=-1\mathrm{/}2$, so its error is zero with the state solver unchanged. In [F2], state sensitivity vanishes because the integrand $t$ is independent of $x$, and midpoint quadrature has zero local defect because $t$ is affine.