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Métodos numericos: ecuaciones diferenciales ordinarias

Métodos numericos: ecuaciones diferenciales ordinarias

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<strong>Métodos</strong>deunpaso<br />

donde O(hn ) es denota una función g(h) para la cual existe una constante positiva k tal que |g(h)| ≤<br />

k|hn |.Entonces<br />

y(t0; t0, y0) =y0,<br />

y 0 (t0; t0, y0) =f(t0, y0) =f1(t0, y0),<br />

y 00 (t0; t0, y0) = d<br />

dt y0 (t0; t0, y0) = d<br />

dt f1(t0, y0)<br />

= ∂<br />

∂t f1(t0, y0)+ ∂<br />

∂y f1(t0, y0)y 0 (t0; t0, y0)<br />

= ∂<br />

∂t f1(t0, y0)+ ∂<br />

∂y f1(t0, y0)f(t0, y0)<br />

= f2(t0, y0),<br />

donde por ∂<br />

∂y f1(t0, y0) denotamos el gradiente de f1(t0, y0).<br />

y 3) (t0; t0, y0) = d<br />

dt y00 (t0; t0, y0) = d<br />

dt f2(t0, y0)<br />

= ∂<br />

∂t f2(t0, y0)+ ∂<br />

∂y f2(t0, y0)y 0 (t0; t0, y0)<br />

= ∂<br />

∂t f2(t0, y0)+ ∂<br />

∂y f2(t0, y0)f(t0, y0)<br />

= f3(t0, y0).<br />

Inductivamente, si y n−1) (t0; t0, y0) =fn−1(t0, y0), entonces<br />

y n) (t0; t0, y0) = d<br />

dt yn−1) (t0; t0, y0) = d<br />

dt fn−1(t0, y0)<br />

= ∂<br />

∂t fn−1(t0, y0)+ ∂<br />

∂y fn−1(t0, y0)y 0 (t0; t0, y0)<br />

= ∂<br />

∂t fn−1(t0, y0)+ ∂<br />

∂y fn−1(t0, y0)f(t0, y0)<br />

= fn(t0, y0).<br />

Así, sustituyendo en la fórmula original<br />

con lo que<br />

y(t1; t0, y0) = y0 + 1<br />

1! f1(t0, y0)h + 1<br />

2! f2(t0, y0)h 2 + ... +<br />

+ 1<br />

n! fn(t0, y0)h n + O(h n ),<br />

y1 = y0 + 1<br />

1! f1(t0, y0)h + 1<br />

2! f2(t0, y0)h 2 + ... + 1<br />

n! fn(t0, y0)h n<br />

es una aproximación de y(t1; t0, y0), esto es<br />

y(t1; t0, y0) ≈ y1 = y0 + 1<br />

1! f1(t0, y0)h + 1<br />

2! f2(t0, y0)h 2 + ... + 1<br />

n! fn(t0, y0)h n . (2.2)<br />

Veamos qué forma particular tiene esta aproximación para diferentes valores de n.<br />

24

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