# Deriving time-dependent Euler-Lagrange equation

Given a time dependent function $$b(x,t):\mathbb{R}^n \times [0;+\infty)$$. Derive the Euler-Lagrange equation from the energy function:

$$S = \int_0^T\left(\ddot{x} - b(x,t)\right)^2dt\tag{1}$$ Note that $$𝑥$$ is an $$n$$-dimensional vector.

My solution is the equation: $$x^{(4)} - \left( \frac{\partial b}{\partial x} + \frac{\partial b^T}{\partial x} \right)x'' - \frac{\partial^2 b}{\partial x^2}(x')^2 - 2\frac{\partial b^2}{\partial x \partial t} x' + \frac{\partial b^T b}{\partial x} - \frac{\partial b^2}{\partial t^2} = 0.\tag{2}$$

But I would like to check and see the full reasoning from other people to see if I am on the right track.

• Note that $x$ is an n-dimensional vector. – Huy-Hoàng Nguyễn Jun 8 at 10:31

I'll be a bit more general: Let $$L(t, q, \dot{q},\ddot{q}$$) the lagrange function and let the action be $$S[q]=\int L(t,q,\dot{q},\ddot{q}) \mathrm{d}t$$ And the first variation will be \begin{align} \delta S &=S[q+\delta q]-S[q]\\ &= \int \frac{\partial L}{\partial q_i} \delta q_i+\frac{\partial L}{\partial \dot{q}_i} \delta \dot{q}_i+\frac{\partial L}{\partial \ddot{q}_i}\delta \ddot{q}_i \mathrm{d}t \\ &= \int \frac{\partial L}{\partial q_i}\delta q_i -\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}_i} \delta q_i - \frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \ddot{q}_i} \delta \dot{q}_i \mathrm{d}t\\ &= \int \frac{\partial L}{\partial q_i}\delta q_i -\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}_i} \delta q_i + \frac{\mathrm{d}^2}{\mathrm{d}t^2} \frac{\partial L}{\partial \ddot{q}_i} \delta q_i \mathrm{d}t\\ &= \int \left(\frac{\partial L}{\partial q_i} -\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}_i} + \frac{\mathrm{d}^2}{\mathrm{d}t^2} \frac{\partial L}{\partial \ddot{q}_i}\right) \delta q_i \mathrm{d}t\\ \end{align} Where I assumed that $$\delta q=\delta \dot{q}=0$$ on the boundaries. Which means that $$\frac{\delta S}{\delta q_i}=\frac{\partial L}{\partial q_i} -\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}_i} + \frac{\mathrm{d}^2}{\mathrm{d}t^2} \frac{\partial L}{\partial \ddot{q}_i}=0$$ are the Euler-Lagrange equations.
1. If there are no boundary conditions, it is overkill to use the Euler-Lagrange (EL) equations as it is evident from the very beginning that $$\ddot{x}^i ~=~ b^i(x,t), \qquad i~\in~\{1,\ldots,n\},$$ minimizes OP's functional (1).
2. If OP wants to use EL equations, he must assume adequate boundary conditions ($$x$$ and $$\dot{x}$$ should be fixed at initial and final times). Then EL equations become $$\sum_{i=1}^n \left( \delta^i_j\frac{d^2}{dt^2} - \frac{\partial b^i}{\partial x^j}\right) (\ddot{x}^i - ~b^i)~=~0, \qquad j~\in~\{1,\ldots,n\}.$$
• Thanks, I do forget to mention that due to boundary conditions, we can't always have $x'' = b(x,t)$ – Huy-Hoàng Nguyễn Jun 10 at 13:59