How can I construct a variation problem whose solution is $f^{(3)}=0$? For example, for the variation problem
$$
\min_{f\in H([a,b])} \int_a^b f''^2(x) \mathrm{d} x
$$
Based on Euler-Lagrange, we can obtain
$$
f^{(4)}=0
$$
Now I need to find a variation problem whose solution is
$$
f^{(3)}=0
$$
However, it seems really hard to construct such a variation problem. Something like $\min \int f' f'' \mathrm{d} x$ and $\min \int f f^{(3)} \mathrm{d} x$ cannot work.
Could anyone please help me with that?
 A: *

*Any even number of differentiations is easy: $f^{(2n)}=0$ is the (higher) Euler-Lagrange equation for the functional $\int \! dx~|f^{(n)}(x)|^2$. Here $n\in\mathbb{N}_0$.


*However OP wants to consider an odd number of differentiations: $f^{(2n+1)}~=~0$.
The remedies are standard:

*

*If $f$ is a complex variable, then use the functional $\int \! dx~\overline{f(x)} f^{(2n+1)}(x)$.


*If a Lagrange multiplier field $\lambda$ is allowed, then use the functional $\int \! dx~\lambda(x) f^{(2n+1)}(x)$.
A: The Euler-Lagrange equation corresponding to the stationary values of the functional
$$
I[f] = \int_a^b \mathcal{L}(x, f, f', \dots, f^{(k)}) \, \text d x
$$
reads
$$
\frac{\partial \mathcal L}{\partial f} - \frac{\text d}{\text d x}\frac{\partial \mathcal L}{\partial f'} + \dots + (-1)^k \frac{\text d^k}{\text d x^k}\frac{\partial \mathcal L}{\partial f^{(k)}} = 0\, .
$$
One notes that the differential equation $f^{(2p)} = 0$ with even-order derivative is deduced from all the Lagrangian functions of the form $$
\mathcal{L} = f^{(k)}f^{(2p-k)}, \qquad k=0, \dots, 2p.
$$
However, it is difficult to perform the same operation for odd-order derivatives, say differential equations of the form $f^{(2p+1)} = 0$. A famous example is the case of the damped harmonic oscillator equation
$$
f'' + 2\zeta\omega f' + \omega^2 f = 0 \, ,
$$
for which one may introduce an $x$-dependent Lagrangian to account for the damping term $2\zeta\omega f'$. Alternatively, one may consider a quasi-variational framework and introduce dissipation functions, or make use of further tricks (see other answers).
