Is the solution to this 2nd order non-linear ODE actually analytic?

Question

I am working through a project and had to solve this ODE: $$ \ddot{z} - ce^{-z} + d = 0 $$ for $c,d$ being constant, with $c > d > 0$. The project is modelling the height of a blimp over time

After numerically solving this equation for initial conditions $z(0)=0, \dot{z}(0)=0$ (a little Mathematica NDSolve), you get the following graph:

It looks remarkably like a sinusoid. I can play around with the initial conditions or the constants to get shifted and squeezed sinusoids. So, I made an assumption that $z = A\sin\left( \omega t \right) + B$ and plugged it into the differential equation. The exponential term gives quite a bit of difficulty and is ultimately equal to a complex value.

Did I choose a poor ansatz? Is there a better ansatz for me to choose? Is the numerical solution similar but not the same as a sinusoid?

Many Thanks!

Answer 1

$$z'' - ce^{-z} + d = 0$$ Try to reduce the order of the DE. $$z''z' - ce^{-z}z' + dz' = 0$$ Integrate to reduce the order: $$\frac 12(z')^2 +ce^{-z} + dz = K$$

Answer 2

Next after finding the "energy" function add on that $ce^{-z}+dz$ is coercive (definitely growing to infinity at both sides) to conclude that the level surfaces for the function on the left side are bounded in the $(z,z')$ plane. This implies that all non-constant solutions are periodic.

If you substitute $c=d\cdot e^a$ you can find a "small angles" approximation for small oscillations around $a$. The simplified, approximate equation $$ \ddot z+d(z-a)=0\implies z(t)=\ln(c/d)+r\cos(\sqrt{d}t+\phi) $$ results from inserting the exponential series and disregarding quadratic and higher degree terms.

Answer 3

Other answers have pointed to very useful methods, especially if one can notice that a first integral can be formed. Alternatively, you can always analyze these systems in phase space. Let $\dot z = y$, then you have the following coupled system: \begin{align} \dot z & = y,\\ \dot y & = c e^{-z} - d. \end{align} There is a fixed point in $(z,y)$ space at $(\ln \frac{c}{d}, 0)$. Looking at the Jacobian for linear stability, we see that this fixed point is a linear center (implying closed orbits), which explains the periodic behavior. Furthermore, one can notice the system is invariant under the change $t \to -t$ and $y \to -y$, which implies that this fixed point is in fact nonlinear center and hence has infinitely many possible orbits.

Just another way to arrive at this behavior, and is more general if/when the system is not Hamiltonian.