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Euler-Lagrange equation with non-conservative forces

The Euler-Lagrange equation with dissipative forces in the ss coordinate is:

ddt∂L∂s˙−∂L∂s=Qs.\begin{aligned} \dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{s}} - \dfrac{\partial L}{\partial s} &= Q_s. \end{aligned}

The left-hand side of Equation (1) can also be denoted as δSδs\dfrac{\delta S}{\delta s}, the functional derivative of SS with respect to ss. SS is the action and is defined by:

S=∫t0tfL(s,s˙,t)dt\begin{aligned} S=\int_{t_0}^{t_f} L(s, \dot{s}, t) dt \end{aligned}

where t0t_0 is the starting time and tft_f is the finishing time. First, we shall calculate ∂L∂s˙\dfrac{\partial L}{\partial \dot{s}}, the canonical momentum:

∂L∂s˙=ms˙.\begin{aligned} \dfrac{\partial L}{\partial \dot{s}} &= m\dot{s}. \end{aligned}

Hence:

ddt∂L∂s˙=ms¨.\begin{aligned} \dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{s}} &= m\ddot{s}. \end{aligned}

As for the second term in Equation (1), it is:

∂L∂s=mgsin⁡θ(s).\begin{aligned} \dfrac{\partial L}{\partial s} &= mg\sin{\theta(s)}. \end{aligned}

So the functional derivative is therefore:

δSδs=ms¨−mgsin⁡θ(s).\begin{aligned} \dfrac{\delta S}{\delta s} &= m\ddot{s} - mg\sin{\theta(s)}. \end{aligned}

Hence Equation (1) becomes:

ms¨−mgsin⁡θ(s)=−mgCrrcos⁡θ(s)−ρCDAs˙22.\begin{aligned} m\ddot{s} - mg\sin{\theta(s)} &= -mgC_{rr} \cos{\theta(s)} - \dfrac{\rho C_D A\dot{s}^2}{2}. \end{aligned}

Rearranging and factorizing yields:

ms¨=mg(sin⁡θ(s)+Crrcos⁡θ(s))−ρCDAs˙22.\begin{aligned} m\ddot{s} &= mg(\sin{\theta(s)}+C_{rr}\cos{\theta(s)})-\dfrac{\rho C_D A\dot{s}^2}{2}. \end{aligned}

Dividing both sides by mm yields:

s¨=g(sin⁡θ(s)+Crrcos⁡θ(s))−ρCDAs˙22m\begin{aligned} \ddot{s} &= g(\sin{\theta(s)}+C_{rr}\cos{\theta(s)})-\dfrac{\rho C_D A\dot{s}^2}{2m} \end{aligned}

assuming constant gravitational acceleration.

© Brenton Horne. I do not claim to be infallible, so feel free to open issues or pull requests on GitHub if you notice flaws or things that can be improved.
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Last modified: October 23, 2026.