Métodos numericos: ecuaciones diferenciales ordinarias
Métodos numericos: ecuaciones diferenciales ordinarias
Métodos numericos: ecuaciones diferenciales ordinarias
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
<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