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Modul 3: Numerisk løsning af differentialligninger - Matematik og ...

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<strong>Modul</strong> 3:<br />

<strong>Numerisk</strong> <strong>løsning</strong> <strong>af</strong> <strong>differentialligninger</strong><br />

<strong>Matematik</strong> <strong>og</strong> modeller 2009<br />

Thomas Vils Pedersen<br />

Institut for Grundvidenskab <strong>og</strong> Miljø<br />

vils@life.ku.dk<br />

Eulers metode<br />

For diff.ligningen y ′ (x) = f (x, y). Startpunkt (x0, y0); steplængde h:<br />

y<br />

y(x2) +<br />

y2 +<br />

y(x1) +<br />

y1 +<br />

y0 +<br />

Hældning f (x0, y0)<br />

<br />

<br />

<br />

<br />

Hældning f (x1, y1)<br />

h h<br />

+ + + x<br />

x0 x1 x2<br />

<br />

Løsningskurven<br />

y = y(x)<br />

x1 = x0 + h, y1 = y0 + f (x0, y0)h,<br />

x2 = x1 + h, y2 = y1 + f (x1, y1)h,<br />

.<br />

.<br />

xn+1 = xn + h, yn+1 = yn + f (xn, yn)h.<br />

.<br />

.<br />

20. maj 2009 — Dias 1/10<br />

Dias 3/10<br />

Oversigt<br />

1 Eulers metode<br />

2 Eulers forbedrede metode<br />

3 Oplæg til Miniprojekt 3<br />

Eksempel<br />

Eulers metode anvendt på<br />

med<br />

y ′ (x) = x + y (= f (x, y))<br />

• begyndelsesbetingelsen y(0.5) = 0.3<br />

• steplængden h = 0.1<br />

[(x−1, y−1), (x−2, y−2) osv. beregnes på tilsvarende vis]<br />

n −3 −2 −1 0 1 2 3 4 5<br />

xn 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0<br />

yn 0.11 0.16 0.22 0.3 0.38 0.48 0.60 0.74 0.90<br />

y(xn) 0.13 0.17 0.23 0.3 0.39 0.50 0.63 0.79 0.97<br />

(Eksakt <strong>løsning</strong> y(x) = 1.8 e −0.5 e x − x − 1.)<br />

Dias 2/10<br />

Dias 4/10


Samme eksempel vha. R<br />

Input:<br />

> # Euler: f(x,y)=x+y; y(0.5)=0.3; h=0.1<br />

> f


Samme eksempel vha. R<br />

> # Forbedret Euler: f(x,y)=x+y; y(0.5)=0.3; h=0.1<br />

> f

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