# Proving Euler-lagrange for single function of single variable with higher derivatives as several functions of one variable

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.

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.
• 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. Commented Oct 6, 2016 at 11:39
• 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 Commented Oct 6, 2016 at 11:58
• 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 $f_1,f_2$. Commented Oct 6, 2016 at 15:13