Consider a population that grows according to the logistic u

Consider a population that grows according to the logistic updating function and is harvested at a linear rate h > = 0. The number of individuals of the species satisfies the DTDS (a) Determine all equilibria of the DTDS. (b) Determine conditions on h such that all equilibria are biologically relevant. (c) Determine conditions on h such that the positive equilibrium is stable. (d) Determine conditions on h such that the positive equilibrium is unstable. (e) Determine the value of h. that maximizes the yield and state the resulting maximum yield.

Solution

AT THE OUT SET WE SEE THAT X(T) SHALL BE AN INTEGER > = 0 . X(T+1)=[X(T)][4-H]-[X(T)]^2……………………..1 THIS WILL BE POSITIVE ONLY IF 4-H > X(T) WHAT IS THE INITIAL VALUE OF X(T) AT T= 0 ? SINCE H > = 0 , FOR 4-H > = X(T) WE SHALL HAVE TO RESTRICT X(0) DEPENDING ON H OBVIOUSLY 0 < = X(0) <= 4 ….AND……... 0 < = H <= 4 I ..LET US SEE THE PATTERN …..SETTING H =   0 FIRST X[0] = 0 1 2 3 4 T X(T) X(T) X(T) X(T) X(T) 0 0 1 2 3 4 1 0 3 4 3 0 2 0 3 0 3 0 3 0 3 0 3 0 4 0 3 0 3 0 SO FOR H = 0 , WE GET THE DIFFERENT EQUILIBRIA POSITIONS AS THOSE TERMINATING WITH 0 & 3 I ..LET US SEE THE PATTERN …..SETTING H =   1 FIRST X[0] = 0 1 2 3 4 T X(T) X(T) X(T) X(T) X(T) 0 0 1 2 3 4 1 0 2 2 0 -4 2 0 2 2 0 -28 3 0 2 2 0 -868 4 0 2 2 0 -756028 SO FOR H = 1 , WE GET THE DIFFERENT EQUILIBRIA POSITIONS AS THOSE TERMINATING WITH 0 & 2..AND X(0)=4 IS NOT FEASIBLE III ..LET US SEE THE PATTERN …..SETTING H =   2 FIRST X[0] = 0 1 2 3 4 T X(T) X(T) X(T) X(T) X(T) 0 0 1 2 3 4 1 0 1 0 -3 -8 2 0 1 0 -15 -80 3 0 1 0 -255 -6560 4 0 1 0 -65535 -4.3E+07 SO FOR H = 2 , WE GET THE DIFFERENT EQUILIBRIA POSITIONS AS THOSE TERMINATING WITH 0 & 1 ..X(0)= 3 OR 4 IS NOT FEASIBLE IV ..LET US SEE THE PATTERN …..SETTING H =   3 FIRST X[0] = 0 1 2 3 4 T X(T) X(T) X(T) X(T) X(T) 0 0 1 2 3 4 1 0 0 -2 -6 -12 2 0 0 -6 -42 -156 3 0 0 -42 -1806 -24492 4 0 0 -1806 -3263442 -6E+08 SO FOR H = 3 , WE GET THE DIFFERENT EQUILIBRIA POSITIONS AS THOSE TERMINATING WITH 0 ..AND X(0)= 2 OR 3 OR 4 IS NOT FEASIBLE V ..LET US SEE THE PATTERN …..SETTING H =   5 FIRST X[0] = 0 1 2 3 4 T X(T) X(T) X(T) X(T) X(T) 0 0 1 2 3 4 1 0 -2 -6 -12 -20 2 0 -2 -30 -132 -380 3 0 -2 -870 -17292 -144020 4 0 -2 -756030 -3E+08 -2.1E+10 SO FOR H = 4 , WE GET THE DIFFERENT EQUILIBRIA POSITIONS AS THOSE TERMINATING WITH 0 ..AND X(0)= 1 OR 2 OR 3 OR 4 IS NOT FEASIBLE NOW WE DRAW ALL THE ANSWERS FROM THE ABOVE DATA SET AS BELOW ……. a)…..ALL EQUILIBRIA HAVE BEEN SHOWN ABOVE ……………ANSWER b)… CONDITIONS ON H ARE SHOWN ABOVE ……………ANSWER c)… H CAN BE 0 OR 1 OR 2 FOR POSITIVE EQUILIBRIUM TO BE STABLE d)……H SHALL BE 3 OR 4 FOR POSITIVE EQUILIBRIUM TO BE UNSTABLE e)…..H = 0 MAXIMISES THE YIELD WITH X(T) = 3 AS FEASIBLE IF X(0) = 1