Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

1.2: Ordinary Differential Equations


$$E(v)=\int_a^bf(x,v(x),v'(x))\ dx$$

and for given \(u_a,\ u_b\in{\mathbb R}\)

$$V=\{v\in C^2[a,b]:\ v(a)=u_a,\ v(b)=u_b\},$$

where \(y\) and \(f\) is sufficiently regular. One of the basic problems in the calculus of variation is

(P)                \(\min_{v\in V}E(v)\).

Euler equation. Let \(u\in V\) be a solution of (P), then


in \((a,b)\).

Proof. Exercise. Hints: For fixed \(\phi\in C^2[a,b]\) with \(\phi(a)=\phi(b)=0\) and real \(\epsilon\), \(|\epsilon|<\epsilon_0\), set \(g(\epsilon)=E(u+\epsilon \phi)\). Since \(g(0)\le g(\epsilon)\) it follows \(g'(0)=0\). Integration by parts in the formula for \(g'(0)\) and the following basic lemma in the calculus of variations imply Euler's equation.

Admissible Variations

Figure Admissible Variations

Basic lemma in the calculus of variations. Let \(h\in C(a,b)\) and

$$\int_a^bh(x)\phi(x)\ dx=0$$

for all \(\phi\in C_0^1(a,b)\). Then \(h(x)\equiv0\) on \((a,b)\).

Proof. Assume \(h(x_0)>0\) for an \(x_0\in (a,b)\), then there is a \(\delta>0\) such that \((x_0-\delta,x_0+\delta)\subset(a,b)\) and \(h(x)\ge h(x_0)/2\) on \((x_0-\delta,x_0+\delta)\).

  \left(\delta^2-|x-x_0|^2\right)^2 & x\in(x_0-\delta,x_0+\delta)\\
  0   & x\in (a,b)\setminus[x_0-\delta,x_0+\delta]
  \end{array} \right. .

Thus \(\phi\in C_0^1(a,b)\) and

$$\int_a^b h(x)\phi(x)\ dx\ge \frac{h(x_0)}{2}\int_{x_0-\delta}^{x_0+\delta}\phi(x)\ dx>0,$$

which is a contradiction to the assumption of the lemma.