# Derive the Euler–Lagrange equation for a functional a single variable with higher derivatives.

I know how to prove the Euler-Lagrange equation($\frac{\partial f}{\partial y}-\frac{d }{d x} \frac{\partial f}{\partial y_x}$) to minimize the the functional \begin{align} J(y)=\int^{x_2}_{x_1} f(x,y(x),y'(x)) \ dx. \end{align}

My question is how to prove the Euler-Lagrange equation ($\frac{\partial f}{\partial y}-\frac{d }{d x} \frac{\partial f}{\partial y_x}+\frac{d^2}{dx^2} \frac{\partial f}{\partial y_{xx}}$), for the functional
\begin{align} J(y)=\int^{x_2}_{x_1} f(x,y(x),y'(x),y''(x)) \ dx. \end{align}

My Attempt Using $$y(x,\alpha)=y(x,0)+\alpha \eta(x)$$ where $\eta(x)$ is a perturbation, away from $y(x,0)$. The condition for an extrema of $J$ is $\frac{d J(\alpha)}{d \alpha}=0$. $$\frac{d J(\alpha)}{d \alpha}=\frac{d }{d \alpha} \int^{x_2}_{x_1} f(x,y(x),y'(x),y''(x)) dx= \int^{x_2}_{x_1} \frac{d }{d \alpha} f(x,y(x),y'(x),y''(x)) dx$$ We note the total derivative of $\frac{d f}{d \alpha}$ is \begin{align} \frac{d f}{d \alpha}&=\frac{\partial f}{\partial x} \frac{d x}{d \alpha}+\frac{\partial f}{\partial y} \frac{d y}{d \alpha}+\frac{\partial f}{\partial y_x} \frac{d y_x}{d \alpha}+\frac{\partial f}{\partial y_{xx}} \frac{d y_{xx}}{d \alpha} \\ &=\frac{\partial f}{\partial y} \frac{d y}{d \alpha}+\frac{\partial f}{\partial y_x} \frac{d y_x}{d \alpha}+\frac{\partial f}{\partial y_{xx}} \frac{d y_{xx}}{d \alpha} \\ &=\frac{\partial f}{\partial y} \eta(x) + \frac{\partial f}{\partial y_x} \eta'(x)+ \frac{\partial f}{\partial y_{xx}} \eta''(x) \end{align}

so \begin{align} \frac{d f}{d \alpha}&= \int^{x_2}_{x_1} \frac{\partial f}{\partial y} \eta(x) + \frac{\partial f}{\partial y_x} \eta'(x)+ \frac{\partial f}{\partial y_{xx}} \eta''(x) \ dx \end{align} we can use integration by parts and the property $\eta(x_1)=\eta(x_2)=0$ to show that the first two terms are equal to \begin{align} \int^{x_2}_{x_1}[\frac{\partial f}{\partial y}-\frac{d }{d x} \frac{\partial f}{\partial y_x}]\eta(x). \end{align}

For the last part we do integration by parts again using the fact that $\eta'(x)$ vanishes. \begin{align} \int^{x_2}_{x_1} \frac{\partial f}{\partial y_{xx}} \eta''(x) \ dx &= \frac{\partial f}{\partial y_{xx}} \eta'(x)|^{x_2}_{x_1} - \int^{x_2}_{x_1} \frac{d }{dx} \frac{\partial f}{\partial y_{xx}} \eta'(x) \ dx \\ &= 0 - \int^{x_2}_{x_1} \frac{d }{dx} \frac{\partial f}{\partial y_{xx}} \eta'(x) \\ &=-\frac{\partial f}{\partial y_{xx}} \eta(x)|^{x_2}_{x_1} + \int^{x_2}_{x_1} \frac{d^2 }{d x^2} \frac{\partial f}{\partial y_{xx}} \eta(x) \ dx \\ &=\int^{x_2}_{x_1} \frac{d^2 }{d x^2} \frac{\partial f}{\partial y_{xx}} \eta(x) \ dx \\ \end{align}

The final integral is \begin{align} \frac{d f}{d \alpha}&=\int^{x_2}_{x_1}[\frac{\partial f}{\partial y}-\frac{d }{d x} \frac{\partial f}{\partial y_x}+\frac{d^2}{dx^2} \frac{\partial f}{\partial y_{xx}}]\eta(x). \end{align}

with the Fundamental Lemma from the Calculus of Variations, this is enough to prove the answer.

Notes:

• I have seen this post but it does not answer my question.
• I use $y'(x)$ and $y_x$ interchangeably.
• My post should now include full answer.
• Why doesn't the linked post answer your question? It's exactly the same question as far as I can see. Anyway, all you have to do is performing integration by parts 2 times on the final term to get the result. Is the question about how to deal with the boundary term containing $\eta'(x)$ that arises when doing this? – Winther Jan 23 '18 at 16:31
• @Winther Thanks for your comment. I am confused by the introduction of $f_1$,$f_2$. I don't understand the purpose or how this makes the problem any simpler. Also, yes this is a question of how to deal with the $\eta'(x)$ term. – AzJ Jan 23 '18 at 16:39
• You must assume that the $\eta'(x)$ vanish, i.e. only consider variations where both $\eta$ and $\eta'$ are fixed at the boundaries. – Winther Jan 23 '18 at 16:43

When you are considering a functional involving the derivatives of order $n$, you must impose boundary conditions on the derivatives of order $(n-1)$, what implies that the $(n-1)$-th derivatives of the perturbation must be zero at the boundary points. Thus the additional term involving a second derivative of the perturbation can be integrated by parts, because the first derivative of the perturbation is zero at the boundary. Doing one more integration by parts you get the result. See the page 41 of the book Calculus of Variations by Gelfand and Fomin.

• Thank you, a reference as part of your answer is a nice touch. – AzJ Jan 23 '18 at 17:04

Your "attempt" to derive the Euler-Lagrange equation is the first, formally correct, step of the rigorous proof: the second one involves a proper choice of the class of functions to which the "perturbation" $$\eta$$ belongs, and implies also a well defined interpretation of the equation itself.

Precisely, the Euler-Lagrange equation is, for a class of functionals of integral type as $$J$$ is, a condition to be satisfied in order for its first variation $$\delta J(y,\eta) =\lim_{\alpha\to 0}\frac{J(y+\alpha\eta)-J(y)}{\alpha}$$ to vanish, i.e. $$\delta J(y,\eta)=0$$ for all the functions $$y+\alpha\eta$$ which are "near" in a topological sense to $$y$$. This implies that the solutions $$y=y(x)$$ of this equation are stationary points for the functional $$J$$ (a maximum, a minimum or a more complex locus). To guarantee uniqueness or at least restrict the number of solutions, $$y$$ is required to satisfy some conditions, which can be in the form of Dirichlet/Cauchy data prescribed on the boundary of a domain or other, more complex, requirements. These conditions restrict the set of functions where the the solution is to be found: and you want that $$y+\alpha\eta$$ belongs to this set for any "perturbation" $$\eta$$. The easiest way to ensure this is to require that any $$\eta$$ gives a null contribution at the points $$x$$ where $$y$$ already satisfies the required conditions, for example by vanishing up to a given (possibly infinite) order there: examples of this include the vanishing of $$\eta$$ on the boundary of a given domain in the Euclidean space or at the beginning $$x_1$$ and the end $$x_2$$ of a given "time interval".

Due to the particular form of the functional $$J$$, there are two possible choices for the class to which $$\eta$$ should belong in order to fulfill the requirement imposed by the knowledge of $$y$$ at $$x=x_1$$ and $$x=x_2$$: these choices depend on the differentiability properties of the function $$f:[x_1,x_2]\times\mathbb{R}\times\mathbb{R}\to\mathbb{R}$$.

1. $$f$$ is of class $$C^3$$: then, choosing $$\eta\in C^3_0([x_1,x_2])$$ (i.e. vanishing on $$x=x_1$$ and $$x=x_2]$$) and applying the integration by parts formula and the fundamental lemma of the calculus of variations, as you did above, leads to the classical Euler-Lagrange equation,
$$\delta J(y,\eta)=0 \iff \frac{\partial f}{\partial y}\left(x,y^{(i)}\right)-\frac{d }{d x} \frac{\partial f}{\partial y_x}\left(x,y^{(i)}\right)+\frac{d^2}{dx^2} \frac{\partial f}{\partial y_{xx}}\left(x,y^{(i)}\right)=0.$$ where I adopted the notation $$y^{(i)}=\left(y,y^\prime,y^{\prime\prime}\right)$$ and all derivatives shown should be intended in classical sense.

2. $$f$$ is of class $$C^1$$: in this case it is not possible to derive the function $$f$$ a number of times sufficient to apply the integrations by part formula and subsequently the fundamental lemma of the calculus of variations. However, by choosing $$\eta\in C^\infty_0([x_1,x_2])$$, the first variation $$\delta J(y,\eta)$$ can be interpreted as a distribution $$\langle\mathscr{L}(y),\eta\rangle\in\mathscr{D}^\prime,$$ defined as \begin{align} \delta J(y,\eta) &= \langle\mathscr{L}(y),\eta\rangle\\ &= \int^{x_2}_{x_1} \frac{\partial f}{\partial y} \eta(x) + \frac{\partial f}{\partial y_x} \eta'(x)+ \frac{\partial f}{\partial y_{xx}} \eta''(x) dx\\ &= \left\langle\frac{\partial f}{\partial y}-\frac{d }{d x} \frac{\partial f}{\partial y_x}+\frac{d^2}{dx^2} \frac{\partial f}{\partial y_{xx}},\eta\right\rangle\quad \forall \eta\in C^\infty_0([x_1,x_2]) \end{align} Now all derivatives respect to the $$x$$ variable should be interpreted as weak derivatives, and requiring the vanishing of the first variation is requiring the vanishing of the distribution $$\mathscr{L}(y)$$ on the interval $$[x_1,x_2]$$, according for example to the lemma on page 14 of Vladimirov [1].

Some supplementary notes

A more exhaustive treatment (which mainly deals with the multidimensional case) is offered by Giaquinta and Hildebrandt in [2], §2.2-2.3 for the analysis of the first variation of standard variational problems and §5, pp. 59-61 §5 for the analysis of higher order variational problems. Their treatment analyzes also in a refined way the precise differentiability requirements on $$f$$ and the corresponding meaning of the Euler-Lagrange equation.

An addendum: in the recent textbook by Kecs, Teodorescu and Toma [3], the approach sketched in point 2 above is developed, both for the Euler-Lagrange for one-dimensional functionals depending on a function $$y$$ and on its first derivative $$y^\prime$$ ([3], §3.1 pp. 151-156) and for functionals depending also on higher order derivatives $$y^{(j)}$$, $$j=1,\dots,n\geq 1$$ ([3], §3.1 pp. 156-158 and §3.1.1 pp. 158-160).

[1] Vladimirov, V. S. (2002), Methods of the theory of generalized functions, Analytical Methods and Special Functions, 6, London–New York: Taylor & Francis, pp. XII+353, ISBN 0-415-27356-0, MR 2012831, Zbl 1078.46029.

[2] Giaquinta, Mariano; Hildebrandt, Stefan (1996), Calculus of Variations I. The Lagrangian Formalism, Grundlehren der Mathematischen Wissenschaften, 310 (1st ed.), Berlin: Springer–Verlag, pp. xxix+475, ISBN 3-540-50625-X, MR 1368401, Zbl 0853.49001.

[3] Teodorescu, Petre; Kecs, Wilhelm W.; Toma, Antonela (2013), Distribution Theory: With Applications in Engineering and Physics, Weinheim: Wiley-VCH Verlag, pp. XII+394, ISBN 3-527-41083-X, ISBN-13 978-3-527-41083-5, Zbl 1272.46001.