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.


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.

| cite | improve this answer | |
  • $\begingroup$ 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 '14 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 '14 at 18:10
  • $\begingroup$ @Justin: indeed, or in geometric terms the tangent map of the diffeomorphism $\Phi(\cdot,t)$. $\endgroup$ – Jaap Eldering Sep 15 '14 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.