Differential algebraic equations and consistent initialization — Examples

In the circuit, $f(t,v,i)=i\mathrm{/}C$ and $g(t,v,i)=Ri+v-u(t)$, so $g_i=R\mathrm{\ne }0$. Thus the selected algebraic branch is $i=(u(t)-v)\mathrm{/}R$, and the reduced ODE follows from [F3].

For $C=R=1$, $u(0)=1$, and $v_0=0$, the algebraic residual at $i_0=1$ is $1\cdot 1+0-1=0$, whereas it is $1\cdot 0+0-1=-1$ at $i_0=0$. By [F1], the first pair is consistent and the second cannot be the initial value of a solution.

Apply [F2] to the reduced ODE $v^{\mathrm{\prime }}=1-v$ from step 1.1. With $v_0=0$ and $h>0$, the endpoint equation $v_1=h(1-v_1)$ gives $v_1=h\mathrm{/}(1+h)$. Recovering the current from the algebraic branch gives $i_1=1-v_1=1\mathrm{/}(1+h)$, so $v_1-v_0=h{i}_1$ and $i_1+v_1=1$. For $h=1$, both values equal $1\mathrm{/}2$.

The exact reduced solution $v(t)=1-e^{-t}$, $i(t)=e^{-t}$ satisfies $v^{\mathrm{\prime }}=i$ and $i+v=1$, and it starts at the consistent pair $(0,1)$. Thus the example's initialization and one-step calculations satisfy both circuit equations.

Here $f(t,x,z)=z$ and $g(t,x,z)=x$, so $g_z=0$ at every point. The regular index-one hypothesis $g_z$ invertible therefore fails.

Along any solution, differentiating $g(x(t),z(t))=x(t)=0$ and substituting $x^{\mathrm{\prime }}=z$ gives $z=0$. Differentiating this relation gives $z^{\mathrm{\prime }}=0$. The original constraint does not involve $z$; the first differentiated relation determines $z$, and the second determines its first derivative. By [F2] the differentiation index is two.

For every real $M$, the candidate $(x,z)=(0,M)$ has algebraic residual $g(0,M)=0$, while step 2.1 shows that a solution has $z=0$. Taking arbitrarily small $M\mathrm{\ne }0$ rules out any finite constant $C$ for which $\mathrm{\mid }z\mathrm{\mid }\le C\mathrm{\mid }g(x,z)\mathrm{\mid }$ throughout a neighborhood of $(0,0)$.

The pair $(0,1)$ satisfies $g(0,1)=0$, but a solution would require $z(0)=0$ by step 2.1, so it is not a consistent initial pair for a solution. The pair $(1,0)$ has $g(1,0)=1\mathrm{\ne }0$, so it fails even algebraic consistency.

The constant functions $x(t)=0,z(t)=0$ satisfy both equations, so the system has a solution with the consistent initial pair $(0,0)$. The three boundary calculations above concern only this fixed index-two system and imply no index-one numerical guarantee.

Differentiating $q^{\mathsf{T}}q={\mathrm{\ell }}^2$ once and using $q^{\mathrm{\prime }}=v$ gives $q^{\mathsf{T}}v=0$, the tangency condition. Differentiating a second time gives $0=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2+q^{\mathsf{T}}{v}^{\mathrm{\prime }}=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2-g_0{q}_2-\lambda q^{\mathsf{T}}q$. Using [A1], this equation solves uniquely for $\lambda =(\mathrm{\parallel }v{\mathrm{\parallel }}_2^2-g_0{q}_2)\mathrm{/}{\mathrm{\ell }}^2$.

Differentiate $\mathrm{\parallel }v{\mathrm{\parallel }}_2^2-g_0{q}_2-\lambda {\mathrm{\ell }}^2=0$, the second-derivative constraint from step 1.1. Since $q^{\mathrm{\prime }}=v$ and $v^{\mathrm{\prime }}=-g_0{e}_2-\lambda q$, this gives $0=2v^{\mathsf{T}}{v}^{\mathrm{\prime }}-g_0{v}_2-{\lambda }^{\mathrm{\prime }}{\mathrm{\ell }}^2=-3g_0{v}_2-2\lambda q^{\mathsf{T}}v-{\lambda }^{\mathrm{\prime }}{\mathrm{\ell }}^2=-3g_0{v}_2-{\lambda }^{\mathrm{\prime }}{\mathrm{\ell }}^2,$ where the last equality uses the tangency condition in step 1.1. Since $\mathrm{\ell }>0$, this determines ${\lambda }^{\mathrm{\prime }}=-3g_0{v}_2\mathrm{/}{\mathrm{\ell }}^2$ with nonzero coefficient $-{\mathrm{\ell }}^2$ on the highest algebraic derivative.

The original constraint does not contain $\lambda$; its first derivative yields tangency but not $\lambda$, its second derivative determines $\lambda$, and its third derivative determines ${\lambda }^{\mathrm{\prime }}$. The prolonged equations are regular because the multiplier coefficients are $-{\mathrm{\ell }}^2$ at each of these latter solves. Therefore the least differentiation count in [F2] is three.

For example, $q_0=(\mathrm{\ell },0)$, $v_0=(0,s)$, and ${\lambda }_0=s^2\mathrm{/}{\mathrm{\ell }}^2$ satisfy the position and tangency constraints, and the multiplier formula in step 1.1 gives ${\lambda }_0$. Thus this tuple satisfies the initial conditions exposed by the first two differentiations.