Question

(6.1) de = n1x{1 - -b12%, dy = ray(1 - - byty. (6.2) In Eq. (6.1), is the growth rate of species x, ki is the carrying capacity which limits its growth, and b, is the rate by which the competitor y kills x. Eq. (6.2) has a similar interpretation. The system (6.1)-6.2) has equilibrium points (0,0), (k1,0), (0.k2). (6.3) Note that the equilibrium point (k1,0) means that the second population becomes extinct. Similarly, (0, k2) corresponds to a situation where the first population becomes extinct.

Answer 1

de l = 2x11- 2 L) - biry dy = rzy (1 - / 42 ) - bally equilibrium points are (0,0), lk.,O) € (0,ke) I B, k₂-k, ß2 k, kz 1 ß , ß2-1 ) Pi - Kibi B, B2 -1 [ . - 22,% -bia K2 Jacobian of given system is. ki - big | - bey 220-22] -be/ for 10,0) thos = 10 110,0) 10 RzJ eigenvalues are {, , ? both are positive , so (0,0) es. Unstable Scanned with CamScanner

for ( k,,o) radi tiewe = ( o - ko bi 7 Rz-beki] -R, CO if re- bzki co stable but it ( k o ) is saddle then (ki, o) is re-bek, 70 then point. for (oika) , - bike 0 110,ka) = L-beke & Again two eigenvalues are. - riobike 4 -ų -2, is already negative. if &, -6, Kz co then (0, kel is. stable but it sobiks to then (0, K2) is saddle point Scanned with CamScanner

I g at (+ y ) is. y*) - Iß, K2 - K, BAK, Kz 'B, R- 'BiB-T). - 2 x ?, rix - biya - bint (2*, ) -beya Pc - 2014 - bevey but we have. , - hitte - bigt = 0 re- rett - beret zo s-, * - Eilfi- ß, Kz El Ki ! B, B2 -1 J Beki-K2 IB, B-1 (ME) to | Ki-ß, K2) 7 "B, B, - (Beki -kz Rul B, B.-) Ik - kibi Кz Iki Kibike Ki- hi Kebibz Kibi kabz I -1 hira - Eire &2 | kz-Kikz bz . I K₂ - Kz kibal Rz CS Camócaki K, b, bz & 22 Kok biba hier

rike (ro- Kabi) 7 kikabiber. & 2) bekar, (2 - Kibe) Tkikabibe - rr) bik, R2 (r, - K₂ bi) (kikabiba-2, 6) & R2 (62 - Kibz) (kikabibe - here) T If K, C 8 2 2 4 Kz Ch then. r. -K₂b, yo, & -kibe > and. K. kz bibz-rine co eigenvalues are having negative real parts, so the positive. equilibrium is stable. Scanned with CamScanner