Question

consider the system of differential equations ;

$\fn_jvn \fn_jvn \dot{x}=x(3-2x-y) , \dot{y}=y(2-x-y)$

1) Find the fixed points of the system ,

2) Evaluate the Jacobian Matrix at each fixed point,

3) Classify stability of each fixed point,

4) Sketch the graph of the phase portrait,

Answer 1

The given Autonomous system of differential equations is as under To find the fixed points we put - x (3 - 2 x -y)-0 atter using (iv) an dt atter using Now . (iüi) gives Now , (iv) gives Now , using x-0 from (v) in (vi OW , (vi) , we get so that , two possible fixed points are (vii) Again , using 3- 2x -y- from (v) in (vi) . we get (vii) V111 (ix) 3-2x-y-0 , 2-х-у-0 Now , solving (vii), we get 12 so that , a possible fixed point is , Now , we need to solve (ix) . Now , we get from (ix so that, 3-2x-2-x 1-x Putting x-1 in , say 2-x-y , we get : y=2-1=1 . Thus , another possible fixed point is (xi) Thus , the four possible fixed points of the system are given by V11 , This answers the part (1) of the given question

Now , the Jacobian Matrix J is as under (x (3 -2x -y) (x (3-2x -») after using () . (ii) (x,y) = (xii) (2-x - 2 y) (3- 4 (0) - 0) (2-0-2 (0) (0,02 the Jacobian Matrix at the fixed point Also , (xii) gives J(0,2) = (0,0) 3- 4 (0)-2 (2-0-2 (2) J(0,2) = the Jacobi an Matrix at the fixed point (0,2)

Again , (xii) gives 9-0 2 (0) the Jacobian Matrix at the fixed point Further . (xii) gives 2-12 (1) the Jacobi an Matrix at the fixed point (1,1) This answers the part (2) of the given questi on Now , we will analyze the stability of each fixed point obtained in the part (1 For the Jacobi an Matrix J (0.0) . the trace p is p-the sum of the diagonal elements3 + 2-5 >0 And , q - det [J (0,0)] Now , p>0 , q > 0 the fixed point (0,0) is Unstable and represents Repeller (Source)

For the Jacobi an Matrix J (0.2) the trace p is p= the sum of the diagonal elements -1+(-2)=-1 < 0 And , q=det[J(0,2)] ow. p <0 , q <0 the fixed point (0,2) is Unstable and represents Saddle Point For the Jacobian Matrix JA,0the tra p= the sum of the diagonal elements =-3+-=-<0 And , q=det | JI-,011= Now , p <0,q <0 the fixed point 0s Unstable and represents Saddle Point For the Jacobian Matrix J(1.1) the trace p is p the sum of the diagonal elements2130 And, q=det[J(1,1) 4 2) x(-1)-(-1x(1)-2-1-1>0 Now . p <0. q > → the fixed point (1,1) is Stable and represents Attractor (Sink) This answers the part (3) of the given question Figure for the given problem The following figure shows the Phase Portrait for the given Autonomous System in question Here , the Black colored point is (0,0) and it is Unstable and represents Repeller (Source) The Red colored point is(0,2) and it is Unstable and represents Saddle Point The Blue colored point is: and i is Unstable and represents Saddle Point The Green colored point is and it is Stable and represents Attractor (Sink)

This answers the part (4) of the given question