MATHHX A
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2.7 Anvendelse af differentialligninger
Vi slutter kapitlet af med et simpelt eksempel på anvendelse af differentialligninger. Antag at vi har en afgrænset skov hvor der bor elge (yes de der sjove dyr der ligner en blanding mellem en ko og en hest). Vi vil gerne finde ud af,
hvor mange elge der er i skoven til et bestemt tidspunkt. Hvis der er rigeligt med plads og føde, så vil elgene kunne formere sig frit. Dvs. at hver enkelt elg bliver til f.eks. \(1{,}5\) elg på et år (i gennemsnit - man kan ikke have en
halv elg). Sagt på en anden måde: Tilvæksten er \(\frac {1}{2}\) elg pr. elg. Kalder vi antallet af elge for \(y\), så er tilvæksten \(y'\) og tilvæksten pr. elg må så være \(\frac {y'}{y}\). Vi får altså
differentialligningen
\[\frac {y'}{y} = \frac {1}{2},\]
og hvis vi ganger med \(y\) får vi
\[y' = \frac {1}{2}\cdot y\]
Sådan en ligning har vi set før og den har ifølge formelsamlingen løsningen
\[y=c\cdot e^{k\cdot x}\]
Ligningen udtrykker eksponentiel vækst:
Figur 2.1: Antal elge \(y\), som funktion af tiden \(x\), når de for lov at formere sig ubegrænset.
I praksis er der grænser for hvor mange elge der er plads til i skoven. På et eller andet tidspunkt er der ikke nok plads/føde til at bestanden kan vokse. Kald nu antallet af elge som skoven kan bære for \(M\). Hvordan vil bestanden
udvikle sig? Hvis bestanden er lille, så burde den kunne vokse frit, men kommer tæt på \(M\) må tilvæksten blive lav. Den mest simple model for det vil være hvis tilvæksten pr. elg er proportional med antallet af ”ledige pladser”.
Hvis antallet af elge er \(y\) og der er plads til \(M\), så må der være \((M-y)\) ledige pladser til nye små elgeunger (kalve?, føl?). 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:
Figur 2.2: Antal elge \(y\), 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 \(y = M\).
Figur 2.3: i starten ligner grafen eksponentiel vækst og til slut nærmer den sig \(y=M\) .
Ekstra
Øvelse 2.7.1
Betragt differentialligningen for logistisk vækst:
\[y'=a\cdot y (M-y)\]
Løsning 2.7.1
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 \(y=M\).