
# 1.2: Ordinary Differential Equations

[ "article:topic" ]

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$

$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

Set

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

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.

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

• Integrated by Justin Marshall.