1.2: Ecuaciones Diferenciales Ordinarias
- Page ID
- 118056
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$$E (v) =\ int_a^bf (x, v (x), v' (x))\ dx\]
y para dado\(u_a,\ u_b\in{\mathbb R}\)
$$V=\ {v\ en C^2 [a, b]:\ v (a) =u_a,\ v (b) =u_b\},\]
donde\(y\) y\(f\) sea suficientemente regular. Uno de los problemas básicos en el cálculo de la variación es
(P)\(\min_{v\in V}E(v)\).
Ecuación de Euler
Dejar\(u\in V\) ser una solución de (P), entonces
$$\ frac {d} {dx} f_ {u'} (x, u (x), u' (x)) =f_u (x, u (x), u' (x))\]
pulg\((a,b)\).
Ejercicio\(\PageIndex{1}\): Proof
Para fijo\(\phi\in C^2[a,b]\) con\(\phi(a)=\phi(b)=0\) y real\(\epsilon\),\(|\epsilon|<\epsilon_0\), conjunto\(g(\epsilon)=E(u+\epsilon \phi)\). Ya\(g(0)\le g(\epsilon)\) que sigue\(g'(0)=0\). La integración por partes en la fórmula para\(g'(0)\) y el siguiente lema básico en el cálculo de variaciones implica la ecuación de Euler.
Figura 1.2.1.1: Variaciones admisibles
Lema básico en el cálculo de las variaciones. Let\(h\in C(a,b)\) y
$$\ int_a^bh (x)\ phi (x)\ dx=0$$
para todos\(\phi\in C_0^1(a,b)\). Entonces\(h(x)\equiv0\) adelante\((a,b)\).
Comprobante. Asumir\(h(x_0)>0\) por una\(x_0\in (a,b)\), entonces hay\(\delta>0\) tal que\((x_0-\delta,x_0+\delta)\subset(a,b)\) y\(h(x)\ge h(x_0)/2\) en\((x_0-\delta,x_0+\delta)\).
Set
$$
\ phi (x)
=\ left\ {\ begin {array} {r@ {\ quad\ mbox {if}\ quad} l}
\ left (\ delta^2-|x-x_0|^2\ derecha) ^2 & x\ in (x_0-\ delta, x_0+\ delta)\\
0 & x\ in (a, b)\ setmenos [x_0-\ delta, x_0+\ delta]
\ end {array}\ right.
\]
Así\(\phi\in C_0^1(a,b)\) y
$$\ 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,\]
lo que es una contradicción con el supuesto del lema.
\(\Box\)