First entry of the Jacobian of Dynamical Systems with position and velocity

Question

Given a dynamical system of the form $$ \begin{align} &\dot{x}=v\\ &\dot{v}=F(x) \end{align} $$ where $x$ is the position, $v$ is the velocity and $F(x)$ is a function of the position and some parameters, the value of $\frac{\delta \dot{x}}{\delta x}$ is $0$.

However, I do not really understand why it is not $\frac{\delta v}{\delta x}$, which intuitively for me is not always $0$ and would require solving the system first.

Answer 1

Notation such as $\partial \dot x/\partial x$ is quite misleading. It's the right-hand sides of the ODEs that you are taking the partial derivatives of, not the left-hand sides!

Given a planar vector field $$ \begin{pmatrix} f(x,y) \\ g(x,y) \end{pmatrix} , $$ its Jacobian matrix is by definition $$ \begin{pmatrix} f_x(x,y) & f_y(x,y) \\ g_x(x,y) & g_x(x,y) \end{pmatrix} , $$ where the partial derivatives $f_x(x,y) = \frac{\partial f}{\partial x}(x,y)$, etc., by definition are computed by fixing one variable and varying the other. So there is no “coupling” or “dependency” between the variables $x$ and $y$ when computing the partial derivatives. It's simply just the usual partials of the two functions $f$ and $g$.

So if the vector field happens to be of the form $$ \begin{pmatrix} f(x,y) \\ g(x,y) \end{pmatrix} = \begin{pmatrix} y \\ F(x) \end{pmatrix} , $$ then its Jacobian matrix is just $$ \begin{pmatrix} f_x(x,y) & f_y(x,y) \\ g_x(x,y) & g_x(x,y) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ F'(x) & 0 \end{pmatrix} . $$

When you look at the ODE system $(\dot x,\dot y) = \bigl( f(x,y), g(x,y) \bigr)$, then of course you get solutions $(x,y)=(\alpha(t),\beta(t))$ where the functions $\alpha(t)$ and $\beta(t)$ are coupled to each other through the system. But those functions $\alpha(t)$ and $\beta(t)$ are in no way involved when you're computing the partial derivatives of $f(x,y)$ and $g(x,y)$ with respect to the independent variables $x$ and $y$.

(Notice here that I deliberately did not call the solutions “$(x(t),y(t))$”, in order to try to avoid the confusion that may arise from using the same name for the quantity $x$ and for the function $\alpha$ that describes how this quantity depends on some other quantity $t$.)

In your case, when you think of $y$ (or $v$ in your notation) as being the velocity and $x$ the position, so that they are “coupled” to each other in your mind, that coupling comes from the fact that the ODE system has the particular form $(\dot x,\dot y) = \bigl( y, F(x) \bigr)$, so it's really something that has to do with the solutions $(x,y)=(\alpha(t),\beta(t))$ of that particular system, and not with the partial derivatives of the vector field that makes up the system's right-hand side.

The fact that the Jacobian matrix of the vector field may tell us things about the solutions $(\alpha(t),\beta(t))$ of the corresponding ODE system (such as stability of equilibrium points) is a different story, which is not really relevant when talking about how the Jacobian is defined.