If i have the Lagrange density function $$L(x,\dot{x}) = \frac{1}{2}x^2+ \frac{1}{2}\dot{x}^2,$$

The Euler-Lagrange equation is very simple by

$$ \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = 0$$

In this example, E-L equation is $\ddot{x}-x=0.$

Can i do this in MMA symbol computation? I know i can use and D to take derivative, so i define l=0.5x^2+0.5x'^2 as Lagrange density function, and then Diff[D[l,x']]-D[l,x], the result seems not right, maybe i need define l as a function of t?

I have no experience in symbolic calculations. Can you share some interesting examples? in MMA SE, seems not much symbol computation topic.

And any comments very much appreciate!

ClearAll[x, t]
L = 1/2*x[t]^2 + 1/2*x'[t]^2;
EulerEquations[L, x[t], t]

$$ x(t)-x''(t)=0 $$

can i realize it by ’ and ‘D’?

Sure. You could always apply the definition yourself. The above command just makes it easier.

 D[D[L, x'[t]], t] - D[L, x[t]] == 0

$$ x''(t)-x(t)=0 $$

  • $\begingroup$ Big shot, amazing! Thank you, by the way, can i realize it by and ‘D’? It is more general, suppose E-L have some different definition, and didn’t have EulerEquations. $\endgroup$
    – Ben
    Nov 25 at 12:21

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