Interpretation of Differential Algebraic Equation

Question

Considering the Differential Algebraic Equation (DAE) of the form $$\dot{x}=f(x,y)$$ $$0=g(x,y)$$ We can use the Implicit function theorem to conclude that as long as the Jacobian $ \frac{\partial g(x,y)}{\partial y}$ is non-singular, $\dot{y}$ can be written as a function of $x$ and $y$, and thus, we can use the local equivalent ODE version of this DAE.

(i)- Under what condition we can interpret the above DAE as an ODE on the manifold $S:=\{ g(x,y)=0 \}$?

(ii)- What if the DAE is of the form $\dot{x}=f(x)$, $g(x)=0$ (i.e., when we have a complete ODE plus a set of algebraic equations)?

Answer 1

So, we are given that

$\dot x = f(x, y), \tag 1$

and

$g(x, y) = 0; \tag 2$

I assume that $f(x, y)$ and $g(x, y)$ are (at least) twice continuously differentiable functions in both $x$ and $y$; that is, of class $\mathcal C^2$. We may then differentiate (2) with respect to $t$ and obtain

$g_x(x, y) \dot x + g_y(x, y) \dot y = 0, \tag 3$

where the subscripts denote derivatives:

$g_x(x, y) = \dfrac{\partial g(x, y)}{\partial x}, \tag 4$

and so forth; we can substitute (1) for $\dot x$ in (3):

$g_x(x, y) f(x, y) + g_y(x, y) \dot y = 0, \tag 5$

and since we assume

$g_y(x, y) = \dfrac{\partial g(x, y)}{\partial y} \tag 6$

is non-singular, we may invert it in (5) and write

$g_y^{-1}(x, y)g_x(x, y)f(x, y) + \dot y = 0, \tag 7$

or

$\dot y = -g_y^{-1}(x, y)g_x(x, y)f(x, y); \tag 8$

(1) and (8) together form an ordinary differential equation for the pair $(x, y)$; presumably, if we set

$x(t_0) = x_0, \tag 9$

and

$y(t_0) = y_0, \tag{10}$

then there will be an integral curve $\gamma(t) = (x(t), y(t))$ of the vector field $X(x, y)$,

$X(x, y) = \begin{pmatrix} f(x, y) \\ -g_y^{-1}(x, y)g_x(x, y)f(x, y) \end{pmatrix}, \tag{11}$

such that

$\gamma(t_0) = (x(t_0), y(t_0)) = (x_0, y_0); \tag{12}$

the existence of such a solution curve $\gamma(t)$ through any point $(x_0, y_0)$ follows from the differentiability of the vector field $X(x, y)$, and is the reason we hypothesized $f(x, y), g(x, y) \in \mathcal C^2$; for then, $\nabla g(x, y) \in \mathcal C^1$, as is

$\dot y = -g_y^{-1}(x, y)g_x(x, y) f(x,, y); \tag{13}$

it is well-known that continuous differentiability implies Lipschitz continuity, at least locally; see the answer to this question; furthermore, local Lipschitz continuity yields existence and uniqueness of a local solution through any point, by the Picard-Lindeloef theorem.

The above remarks show how the vector field $X(x, y)$ such that $\dot \gamma(t) = X(\gamma(t))$ may be constructed in the event that $g_y^{-1}(x, y)$ exists, and that integral curves of $X(x, y)$ exist and are unique in the sense that at most one satisfies a given set of initial conditions (9)-(10).

Now, as for point

(i), we note that the assumption that $g_y(x, y)$ is invertible allows invocation of the implicit function theorem to conclude that, locally, $y$ may be expressed as a differentiable function $y(x)$ of $x$ such that $g(x, y(x)) = 0$; from this we see that the graph of $y(x)$ is indeed a manifold with local coordinates given by $x$; the fact that $S = \{(x, y) \mid g(x, y) = 0 \}$ is a manifold is essential because it makes the tangent bundlle $TS$ a meaningful concept, so that sections of it can be defined; it also ensures that $\nabla g(x, y)$ may legitimately regarded as a vector normal to $S$, and this promotes our next observation that (3) may be written

$\nabla g(x, y) \cdot \dot \gamma(t) = g_x(x, y)\dot x + g_y(x, y) \dot y = 0; \tag{14}$

since $\nabla g(x, y)$ is normal to the manifold $g(x, y) = 0$, this equation tells us that $\dot \gamma(t)$ is tangent to this manifold, i.e., is locally given by a section of the tangent bundle of the manifold given by $g(x, y) = 0$; therefore we see that the vector field $X(x, y) = \dot \gamma$ may in fact be regarded as giving a differential equation on $S = \{ (x, y) \mid g(x, y) = 0\}$. We may also see directly that $X(x, y) = \dot \gamma(t)$ is locally a section of the tangent bundle to $S$ by observing that (11) implies

$\nabla g(x, y) \cdot X(x, y) = g_x(x, y) f(x, y) + g_y(x, y)(-g_y^{-1}(x, y) g_x(x, y) f(x, y)$ $= g_x(x, y) f(x, y) - g_x(x, y) f(x, y) = 0, \tag{15}$

which directly shows the vector field $X(x, y)$ is tangent to $S$, being normal to $\nabla g(x, y)$.

So the condition $\exists g_y^{-1}(x, y)$ is sufficient both to grant a manifold structure to $S$ and to define $X(x, y)$ as a section of $TS$. We can thus interpret $X$ as a differential equation on $S$ in this case.

As for

(ii), we've practically answered it already at this point; indeed, if

$\dot x = f(x), \tag{16}$

and

$g(x(t)) = c, \; \text{a constant}, \tag{17}$

then

$\dfrac{dg(x(t))}{dt} = 0; \tag{18}$

a result similar to that in (i) may be obtained if we hypothesize that $\nabla g(x) \ne 0$ in a region of interest; then the sets $S_c = \{x \mid g(x) = c \}$ will be well-defined submanifolds, and

$\nabla g(x(t)) \cdot \dot x(t) = \dfrac{dg(x(t))}{dt} = 0, \tag{19}$

which again shows that $f(x) = \dot x(t)$ is tangent to the set $S_c$ for constant $c$; thus in particular, $\dot x(t)$ is a vector field tangent to $\{x \mid g(x) = 0 \}$.

Answer 2

I think there is an important question regarding this kind of systems. At least when one wants to solve the resulting ODE numerically.

As the solution should stay in the manifold $g(x,y)=0$, then $[\dot{x},\dot{y}]'$ must be orthogonal to $\nabla_{[x,y]} g(x,y)$, and as a consequence $\nabla_{[x,y]} [\dot{x},\dot{y}]'$, that is the ODE Jacobian, is singular, meaning that it has a 0 eigenvalue.

So the resulting ODE is marginally stable, meaning that the use of a standard ODE numerical integrator will drift away from the manifold $g(x,y)=0$

Proof:

Taylor expansion of the ODE around $\begin{bmatrix}x_0\\ y_0\end{bmatrix}$ so that $g(x_0,y_0)=0$ $$\begin{bmatrix}\dot{x}\\\dot{y}\end{bmatrix}=\begin{bmatrix}\dot{x}_0\\\dot{y}_0\end{bmatrix}+\nabla_{[x,y]} \begin{bmatrix}\dot{x}\\ \dot{y}\end{bmatrix} (\begin{bmatrix}x\\y\end{bmatrix}-\begin{bmatrix}x_0\\y_0\end{bmatrix})+\cal{O}(\begin{bmatrix}{\Delta x}^2\\ {\Delta y}^2\end{bmatrix})$$

Left multiply by $\nabla_{[x,y]} g(x=x_0,y=y_0) \cdot$ (scalar product)

$$\nabla_{[x,y]} g(x_0,y_0) \cdot \begin{bmatrix}\dot{x}\\\dot{y}\end{bmatrix}=\nabla_{[x,y]} g(x_0,y_0) \cdot \begin{bmatrix}\dot{x}_0\\\dot{y}_0\end{bmatrix}+\nabla_{[x,y]} g(x_0,y_0) \cdot\nabla_{[x,y]} \begin{bmatrix}\dot{x}\\ \dot{y}\end{bmatrix} \begin{bmatrix}{\Delta x}\\ {\Delta y}\end{bmatrix}+\nabla_{[x,y]} g(x_0,y_0) \cdot\cal{O}(\begin{bmatrix}{\Delta x}^2\\ {\Delta y}^2\end{bmatrix})$$

So as $g(x,y)=0$ and $g(x_0,y_0)=0$

$$0 = 0 + (\nabla_{[x,y]} g(x_0,y_0) \cdot\nabla_{[x,y]} \begin{bmatrix}\dot{x}\\ \dot{y}\end{bmatrix}) \begin{bmatrix}{\Delta x}\\ {\Delta y}\end{bmatrix}+\cal{O}(\left|\begin{bmatrix}{\Delta x}^2\\ {\Delta y}^2\end{bmatrix}\right|)$$

and therefore

$$\nabla_{[x,y]} g(x_0,y_0) \cdot \nabla_{[x,y]} \begin{bmatrix}\dot{x}\\ \dot{y}\end{bmatrix}=0+\cal{O}(\left|\begin{bmatrix}{\Delta x}^2\\ {\Delta y}^2\end{bmatrix}\right|)$$

meaning that the Jacobian of the ODE $$\nabla_{[x,y]} \begin{bmatrix}\dot{x}\\ \dot{y}\end{bmatrix}$$ is singular at the manifold $g(x,y)=0$.

Therefore the ODE has a 0 eigenvalue, rendering it marginally stable.

That way, in general, you cannot use a standard ODE algorithm to integrate the resulting ODE because it drifts away from the manifold $g(x,y)=0$.

Appropriate ways to deal this can be to remove $y$ from the system state, ang get $y$ from $g(x,y)=0$ or integrate the ODE and projecting the integrated state into the manifold, after each integration step. Obviously, there are other methods, although less intuitive.