I am reading Whittaker's Analytical Dynamics. This is chapter 10 Hamiltonian Systems. Paragraph 112 is Variation Equations.

Let ${dx_r\over dt}=X_r(x_1,\dots,x_n,t),\quad (r=1,\dots,n)$ be a system of differential equations. Let $(x_1,\dots,x_n)$ and $(x_1+\delta x_1,\dots,x_n+\delta x_n)$ be the values of the independent variables in the moment $t$, for two neighboring solutions of the system, where $(\delta x_1,\dots,\delta x_n)$ are infinitesimal quantities.

Then $\frac{d}{dt}(x_r+\delta x_r)=X_r(x_1+\delta x_1,\dots,x_n+\delta x_n,t),\quad (r=1,\dots,n)$.

Therefore, $$\frac{d}{dt}\delta x_r = \sum_{q=1}^n\frac{\partial X_r}{\partial x_q}\delta x_q\quad (r=1,\dots,n)$$

Would anyone explain how this last equation follows?

Furthermore let $$\int \sum_r F_r(x_1,\dots,x_n,t)\delta x_r$$ be an integral invariant for the original system. How come is then $${d\over dt}F_r(x_1,\dots,x_n,t)\equiv0$$


1 Answer 1


Subtract the two differential equations:

$$ \begin{align} \frac{d}{dt}\delta x_r &= \frac{d}{dt}(x_r+\delta x_r) - \frac d{dt}(x_r) \\ &= X_r(x_1+\delta x_1,\dots,x_n+\delta x_n,t) - X_r(x_1,\dots,x_n,t). \end{align} $$

This is just the change in $X_r$ when we change $x$ by $\delta x$, which is (from Taylor's theorem/the chain rule) $$\sum_q \frac{\partial X_r}{\partial x_q} \delta x_q.$$

For your second question, remember that the integral of an integral invariant over any domain is constant under the flow; so it must itself be constant along the flow. You can see this by integrating over a tiny domain, where the integrand will be roughly constant.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .