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)?

  • 2
    $\begingroup$ To me, the last DAE makes sense only if $(\nabla g(x), f(x)) \equiv 0$. If the trajectory stays always on $g(x) = 0$, then $g(x(t)) \equiv 0$ and differentiating this by $t$ gives $\nabla g(x(t)) \cdot \dot{x} = (\nabla g(x), f(x)) \equiv 0$. $\endgroup$ – Evgeny Sep 6 '18 at 22:43
  • $\begingroup$ @Evgeny . That makes sense. if $\{x:g(x)=0\}$ is an invariant manifold for the ODE, then we can analyze the system. $\endgroup$ – Arthur Sep 7 '18 at 15:14

So, we are given that

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


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

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

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


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

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

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


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

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

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


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


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

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 $\{x \mid g(x) = 0 \}$.

  • 1
    $\begingroup$ Thanks for your comprehensive explanations. $\endgroup$ – Arthur Sep 9 '18 at 15:59
  • $\begingroup$ @Arthur: you are most welcome sir. Thanks for a very engaging problem. And thanks for the "acceptance". Cheers! $\endgroup$ – Robert Lewis Sep 9 '18 at 16:02

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