Euler-Lagrange equation for function also dependent on $y''$

Question

Given a functional $F[y]=\int_{\alpha}^{\beta}f(x,y(x),y'(x),y''(x))dx$, I need to derive the Euler-Lagrange equation, subject to $y(\alpha),y(\beta),y'(\alpha),y'(\beta)$ taking fixed values.

I understand that a modification of the derivation of the classic Euler-Lagrange equation should lead to this, but I'm not sure how to perform the Taylor expansion with the extra parameter of $f$.

Answer 1

Start by considering the following lemma, which is a variation on the Fundamental Lemma of the Calculus of Variations:

Lemma. Let $a,b\in\mathbb{R}$ be such that $a<b$, let $\mathcal{A}=\{h\in C^2[a,b]:h(a)=h(b)=h'(a)=h'(b)=0\}$, and let $M:[a,b]\to\mathbb{R}$ be a continuous function such that $$\forall h\in\mathcal{A}:\int_a^b M(x)h(x)~\mathrm{d}x=0.$$ Then $M(x)\equiv0$.

The proof is very straightforward, so I'll leave it as an exercise to you. We now use it to find the Euler-Lagrange equation for the problem of minimizing the functional

$$J[y]=\int_a^b F(x,y(x),y'(x),y''(x))~\mathrm{d}x$$

over the set of all $y\in C^2[a,b]$ with $y(a)=\alpha$, $y'(a)=\bar{\alpha}$, $y(b)=\beta$, $y'(b)=\bar{\beta}$.

Suppose $\phi\in C^2[a,b]$ solves the above minimization problem, and consider a weak variation $h=\varepsilon\eta$ with $\varepsilon\in\mathbb{R}$ and $\eta(a)=\eta'(a)=\eta(b)=\eta'(b)=0$. Define

$$\hat{J}(\varepsilon)=J[\phi+\varepsilon\eta].$$

As $\phi$ minimizes $J$, this means that $\hat{J}$ has a minimum at $\varepsilon=0$, and hence $\hat{J}'(0)=0$. Differentiating (I'll leave the details to you) yields that

\begin{align*} 0=\hat{J}'(0)=\int_a^b\biggl(&\eta(x)\partial_2F(x,\phi(x),\phi'(x),\phi''(x))\\&+\eta'(x)\partial_3F(x,\phi(x),\phi'(x),\phi''(x))\\&+\eta''(x)\partial_4F(x,\phi(x),\phi'(x),\phi''(x))\biggr)~\mathrm{d}x. \end{align*}

Doing some integration by parts (I'll leave the details to you here as well) we can rewrite this as

\begin{align*} 0=\int_a^b \eta(x)\biggl(&\partial_2F(x,\phi(x),\phi'(x),\phi''(x))\\&-\frac{\mathrm{d}}{\mathrm{d}x}(\partial_3F(x,\phi(x),\phi'(x),\phi''(x)))\\&+\frac{\mathrm{d}^2}{\mathrm{d}x^2}(\partial_4F(x,\phi(x),\phi'(x),\phi''(x)))\biggr)~\mathrm{d}x. \end{align*}

Now as $\eta$ was chosen arbitarily, we can now apply the above lemma, which yields the Euler-Lagrange equation

$$\partial_2F(x,\phi(x),\phi'(x),\phi''(x))-\frac{\mathrm{d}}{\mathrm{d}x}(\partial_3F(x,\phi(x),\phi'(x),\phi''(x)))+\frac{\mathrm{d}^2}{\mathrm{d}x^2}(\partial_4F(x,\phi(x),\phi'(x),\phi''(x)))=0.$$

A slight end remark that I've been a bit sloppy with what smoothness I require for everything here and there, so just keep that in mind.