1.2: Ordinary Differential Equations
- Page ID
- 2130
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$$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
$$\frac{d}{dx}f_{u'}(x,u(x),u'(x))=f_u(x,u(x),u'(x))\]
in \((a,b)\).
Exercise \(\PageIndex{1}\): Proof
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.
Figure 1.2.1.1: 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 \((x_0-\delta,x_0+\delta)\).
Set
$$
\phi(x)
=\left\{\begin{array}{r@{\quad\mbox{if}\quad}l}
\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.
\(\Box\)
Contributors and Attributions
Integrated by Justin Marshall.