One obtains for a functional for single function of single variable with higher derivatives

$ I[f] = \int_{x_0}^{x_1} \mathcal{L}(x, f, f', f'')~\mathrm{d}x ~;~~ f' := \cfrac{\mathrm{d}f}{\mathrm{d}x}, ~f'' := \cfrac{\mathrm{d}^2f}{\mathrm{d}x^2} $

the Euler-Lagrange equation as $\cfrac{\partial \mathcal{L}}{\partial f} - \cfrac{\mathrm{d}}{\mathrm{d} x}\left(\cfrac{\partial \mathcal{L}}{\partial f'}\right) + \cfrac{\mathrm{d}^2}{\mathrm{d} x^2}\left(\cfrac{\partial \mathcal{L}}{\partial f''}\right) = 0 $

I want to prove the above equation extending Euler-Lagrange for several functions of one variable i.e.

For a given functional of the form :

$ I[f_1,f_2] = \int_{x_0}^{x_1} \mathcal{L}(x, f_1, f_2,f_1', f_2')~\mathrm{d}x ~;~~ f_i' := \cfrac{\mathrm{d}f_i}{\mathrm{d}x}$

the Euler-Lagrange Equation is

$\frac{\partial \mathcal{L}}{\partial f_i} - \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\partial \mathcal{L}}{\partial f_i'}\right) = 0 , i=1,2$

This can be achieved by defining $f_1 = f; f_2 =f'$.

But I'm not able to achieve the same equation as for single function of single variable with higher derivatives.


1 Answer 1


If $f_1$ and $f_2$ are allowed to vary independently, and kept constant at end-point, so $\delta f_1(x_i)=\delta f_2(x_i)=0$, then you obtain the E-L equations stated by using the standard integration by part technique for the linearized variation: $$ \delta I = \int_{x_0}^{x_1} \left( \frac{\partial L}{\partial f_1} \delta f_1 + \frac{\partial L}{\partial f_2} \delta f_2 + \frac{\partial L}{\partial f'_1} \delta f'_1 + \frac{\partial L}{\partial f'_2} \delta f'_2 \right) dx= \int_{x_0}^{x_1} \left((\frac{\partial L}{\partial f_1} - \frac{d}{dx} \frac{\partial L}{\partial f'_1}) \delta f_1 +(\frac{\partial L}{\partial f_2} - \frac{d}{dx} \frac{\partial L}{\partial f'_2}) \delta f_2 \right) dx $$ equating to zero for all variations yields the E-L equations.

If here you set $f_2=f_1'$ and impose $\delta f_2= \delta f'_1$ then you may integrate by part once again to get: $$ \delta I = \int_{x_0}^{x_1} \left(\frac{\partial L}{\partial f_1} - \frac{d}{dx} \frac{\partial L}{\partial f'_1} -\frac{d}{dx} \frac{\partial L}{\partial f_2} + \frac{d^2}{dx^2} \frac{\partial L}{\partial f'_2} \right) \delta f_1 dx $$ hereby obtaining the second order formulation by equating to zero.

  • $\begingroup$ Thanks, but I approached the problem in a certain way i.e. by using a change of variables for $\mathcal{L}(x, f, f', f'') = \mathcal{L}(x, f_1 = f, f_2 = f',f_1' = f', f_2' = f'') $ and then using the chain rule for calculating the partial differentiation in the new space. Why does this approach not lead to the same eqn. $\endgroup$
    – Ricky
    Oct 6, 2016 at 11:39
  • $\begingroup$ Even with your soln, one would like to replace $\frac{\partial }{\partial f_2}$ & $\frac{\partial}{\partial f_2'} $ with the new operators and you would get a different eqn $\endgroup$
    – Ricky
    Oct 6, 2016 at 11:58
  • $\begingroup$ I am not sure about what you precisely get, but a difference might be that my approach imposes $\delta f_1'(x_0)=\delta f_1'(x_1)=0$ which I use during the partial integration. You don't see this if you only substitute after getting the E-L equations for $f_1,f_2$. $\endgroup$
    – H. H. Rugh
    Oct 6, 2016 at 15:13

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