I'm stuck on generalizing an ODE formula and could use your help!

One way to think about "variation of parameters" is that it bakes the solution $z(t)=e^{At}z_0$ of $z'=Az$ (here $z(t)\in\mathbb{R}^n$, $A\in{\mathbb R}^{n\times n}$) into formulas for nonlinear problems. In particular, to solve $y'=Ay+G[y]$ for some $G:\mathbb{R}^n\rightarrow\mathbb{R}^n$, we can write $$y(t)=e^{At}y_0+\int_0^t e^{A(t-\tau)}G[y(\tau)]\,d\tau.$$ The integral compensates between the closed-form solution of the linear ODE and the solution of the ODE including $G[\cdot]$.

Suppose instead that we wish to solve $y'=F[y]+G[y]$ for two nonlinear functions $F,G:\mathbb{R}^n\rightarrow\mathbb{R}^n.$ Furthermore, let's say we know how to solve $z'=F[z]$ in closed form via some $\Phi(z,t)$ so that $z(t)=\Phi(z_0,t).$ Is there an analogous formula to variation of parameters in this case? E.g. something of the form: $$y(t)=\Phi(y_0,t)+\int_0^t\left[\textrm{something involving $G$}\right]\,d\tau$$

PS: If it helps, we can assume both $F$ and $G$ come from a Hamiltonian dynamics problem. So, $n$ is even and contains both velocity and momentum variables, $\Phi_t$ is inverted by $\Phi_{-t}$, $\Phi_t$ is area-preserving, and so on.


1 Answer 1


Yes, this is called the nonlinear variation of constants formula due to Alekseev: “An estimate for the perturbations of the solutions of ordinary differential equations”, in: Vestnik Moskov. Univ. Ser. I Mat. Meh. 2 (1961), pp. 28–36. I don't think that that article is available in English.

It can also be found in the book by V. Lakshmikantham and S. Leela "Differential and integral inequalities: Theory and applications" Vol. I: Ordinary differential equations.Mathematics in Science and Engineering, Vol. 55-I. New York: Academic Press, 1969, pp. ix+390.

The formula is $$ \tilde{\Phi}(y_0,t) = \Phi(y_0,t) + \int_0^t D\Phi(\tilde{\Phi}(y_0,\tau),t-\tau) G(\tilde{\Phi}(y_0,\tau)) \;d\tau $$ where $\tilde{\Phi}$ denotes the flow of $F+G$.

See also Appendix E in my book for this, a proof, and a few more details.

  • $\begingroup$ Wow, that was fast! Looks like I found a gentleman who literally wrote the book on the matter! I was assuming the formula looked something like that but couldn't get it quite right. $\endgroup$
    – Justin
    Sep 15, 2014 at 18:09
  • $\begingroup$ Incidentally, to make sure I understand, $D\Phi$ here is the Jacobian of $\Phi$ with $t$ fixed, right? [That is, only differentiation in $y$?] $\endgroup$
    – Justin
    Sep 15, 2014 at 18:10
  • $\begingroup$ @Justin: indeed, or in geometric terms the tangent map of the diffeomorphism $\Phi(\cdot,t)$. $\endgroup$ Sep 15, 2014 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.