Assume a functional $I[u(x)] = \int\limits_{{x_{\min }}}^{{x_{\max }}} {F(x,u,u')dx} $.

Given I have the Euler-Lagrange (EL) equation $u'(x) = 0$, I need to find the form of $F$ that leads to this EL.

I intuitively tried $F(x,u,u')=u(x)u'(x)$ and this yields \begin{equation} \frac{{\partial F}}{{\partial u}} - \frac{d}{{dx}}\left( {\frac{{\partial F}}{{\partial u'}}} \right) = u'(x) - u'(x) = 0 \end{equation} but I would need $F$ such that \begin{equation} \frac{{\partial F}}{{\partial u}} - \frac{d}{{dx}}\left( {\frac{{\partial F}}{{\partial u'}}} \right) = u'(x). \end{equation} Does such an $F$ that would lead to EL $u'(x) = 0$ exist?

  • $\begingroup$ @MrYouMath I hoped there exists one solution but maybe I could work with a non-unique solution as well. I'm going to use $F$ in a numerical method. The long answer would be that I need to find a variational formulation for $\frac{{\partial u}}{{\partial t}} + r\frac{{\partial u}}{{\partial x}} + \left( {r - \frac{1}{2}{\sigma ^2}} \right)\frac{{{\partial ^2}u}}{{\partial {x^2}}} - ru = 0$ which I can't do without being able to find $F$ for the first order derivative $\frac{{\partial u}}{{\partial x}} = u'(x)$ $\endgroup$ – fragile Sep 29 '17 at 10:00
  • 1
    $\begingroup$ This is related en.wikipedia.org/wiki/Inverse_problem_for_Lagrangian_mechanics $\endgroup$ – MrYouMath Sep 29 '17 at 10:32

I. If $u\in\mathbb{C}$ is complex-valued: Then the Lagrangian $F(x,u,u^{\prime})=i\bar{u}u^{\prime}$ works.

II. If $u\in\mathbb{R}$ is real-valued: Then there is no Lagrangian $F(x,u,u^{\prime},u^{\prime\prime},u^{\prime\prime\prime},\ldots)$.

Sketched proof:

  1. $F(x,u,u^{\prime},u^{\prime\prime},\ldots)$ can not depend on higher derivatives $u^{\prime\prime},u^{\prime\prime\prime},\ldots$, in order for the higher-order EL equation to be independent of them.

  2. The Lagrangian must be an affine function $$F(x,u,u^{\prime})=A(x,u) +u^{\prime}B(x,u) \tag{1}$$ in $u^{\prime}$ in order for the EL equation to be independent of $u^{\prime\prime}$.

  3. It is not hard to see that the EL equation $$ \frac{\partial A}{\partial u}- \frac{\partial B}{\partial x}~=~0 \tag{2}$$ for the Lagrangian (1) does not depend on derivatives. Contradiction. $\Box$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.