How can I construct a variation problem whose solution is $f^{(3)}=0$?

149 Views Asked by At

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?

2

There are 2 best solutions below

1
On

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).

2
On
  1. 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 $n\in\mathbb{N}_0$.

  2. 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 $\int \! dx~\lambda(x) f^{(2n+1)}(x)$.