MATHHX A

MATHHX A

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2.6 Anvendelse af differentialligninger

Vi slutter kapitlet af med et simpelt eksempel på anvendelse af differentialligninger. Vi kigger på bestanden af en bestem slags fisk i en sø som funktion af tiden. Hvis der er rigeligt med plads og føde, så vil fiskene kunne formere sig frit. Det kunne f.eks. være at hver fisk i løbet af et år ville blive til \(1{,}2\) fisk (i gennemsnit — man kan ikke have \(0{,}2\) fisk). Sagt på en anden måde kunne det være at der var en tilvækst pr. fisk på \(0{,}2\). Kalder vi antallet af fisk for \(y\), så er tilvæksten af fisk \(y'\) og tilvæksten pr. fisk må så være \(\frac {y'}{y}\). Alt tilvæksten pr. fisk er \(0{,}2\) kan vi altså udtrykke med differentialligningen:

\[\frac {y'}{y} = 0{,}2\]

Hvis vi ganger med \(y\) får vi:

\[y' = 0{,}2\cdot y\]

Sådan en ligning har vi set før og den har ifølge formelsamlingen løsningen;

\[y=c\cdot e^{k\cdot x}\]

Ligningen udtrykker eksponentiel vækst:

(-tikz- diagram)

Figur 2.1: Antal fisk \(y\), som funktion af tiden \(x\), når de for lov at formere sig ubegrænset.

I praksis er der grænser for hvor mange fisk der er plads til i søen. På et eller andet tidspunkt er der ikke nok plads/føde til at bestanden kan vokse. Lad os kalde den størst opnåelige bestand for \(M\) og se på hvordan hvordan bestanden kan udvikle sig? Hvis bestanden er lille, så burde den kunne vokse frit, men kommer den tæt på \(M\) må tilvæksten blive lav. Den mest simple model som passer med det, er en model, hvor tilvæksten pr. fisk er proportional med antallet af ”ledige pladser”. Hvis antallet af fisk er \(y\) og der er plads til \(M\), så må der være \((M-y)\) ledige pladser til nye små fiskebasser. Altså får vi ligningen

\[\frac {y'}{y}=a\cdot (M-y)\]

Vi ganger med \(y\)

\[y'=a\cdot y (M-y)\]

Vi genkender formlen fra tabellen. Det var den som hed logistisk vækst og den har løsningen:

\[y= \frac {M}{1+c\cdot e^{-a\cdot M \cdot x}} \]

Tegner man grafen ser den således ud:

(-tikz- diagram)

Figur 2.2: Antal fisk \(y\), som funktion af tiden \(x\) når der er begrænset plads.

Vi kan se at den i starten ligner eksponentiel vækst og til slut flader ud. Faktisk nærmer den sig linjen \(y = M\).

(-tikz- diagram)

Figur 2.3: i starten ligner grafen eksponentiel vækst og til slut nærmer den sig \(y=M\).
Ekstra

Øvelse 2.6.1

Betragt differentialligningen for logistisk vækst:

\[y'=a\cdot y (M-y)\]

  • a) Argumenter (ud fra ligningen) for at \(y\) må vokse eksponentielt når \(y\) er lille.

  • b) Argumenter (ud fra ligningen) for at \(y\) nærmere sig linjen \(y=M\) når \(y\) er stor.

Løsning 2.6.1

  • a) Når \(y\) er lille vil \(M-y\approx M\). Så ligningen kan skrives som \(y'=a\cdot y \cdot M\). Denne ligning har form som \(y'=k\cdot x\) og udtrykker altså eksponentiel vækst.

  • b) Når \(y\) er stor (dvs tæt på \(M\)) vil \(M-y\approx 0\) og ligningen kan skrives som \(y'=0\), dvs. at når \(y\) er meget tæt på \(M\) er væksten meget tæt på \(0\) og funktionen er tilnærmelsesvis konstant. Altså nærmer funktionen sig linjen \(y=M\).