Question

A bifurcation occurs where the number of equilibrium solutions changes as the parameter varies. As 8 increases, the number of equilibrium solutions changes from two to one and eventually there are none. 4. Solve the equation 0 = yd – 2y + Bfor y in terris of B. Describe how the number of solutions depends on B. 5. Sketch the bifurcation diagram, which shows the graph of y versus 8. It is traditional to show stable equilibria with solid lines and unstable equilibria with dotted lines. 6. Sketch the bifurcation diagram for the one parameter family = yu? – ay with parameter a.

Answer 1

4) soli- from the given data we have: y - 2y +B=0 the qoc²tbX+C =0 eq By comparing eq ① with Here a=1, b =-2,c (=p we have to find dis Criminant D=6²- yac :(-2)-4(1)(B) = u-uß D=4(#-B) - when 8=1 then D=0 we get one neal solution when then D is -ve B=1 we gente pain of complex solutions when Bal then D is tve we get reat two real solutions (6) Giver eq is y²-2y+B dt dy ý = we find solublons of ý = 0 y=0 geves to find equilibrium points y²-24 tp So the solution is 2 I Su-uß y = 1 + EST-B 2

to the equilibrium points are Y,= 1 + 5l-B we have we need 92=1-1 M-B to find stability of these points. For that, to plot the flow diagram ( įrsy 4² 24 tp as an upward facing parabola Lowest point (minimum) at y =) and intersecting the y-axis at Y and Y. Note that with its at points we have to consider four cases & case (i) ßco ese, for this case y, is postblue and Yz is negative we see the arrow I y , that Y2 y point towards Y₂ and they point away from Y, ii ye is unstable a stable point and y, is (directl direction of arrows is such that point for plo it points towards reight when ý so and it points towards left when ý ko) an

case (ii) :-ß = case in the V y, 2, Yz =0 we see that again, Yz y 5 .". Iz is stable point and on un stable point for p = 0 Case (90) :- OLBCT is In this case Y and I are positive and destenct again Y2 is ใน 9 y, a stable point ", is an unstable point for ocßel y and case liv):- B =) k. for this case both Y and y are equal to 1 we see that y=1 is y half stable stable point for ß=1 Y,,Y Now or the the for bifurcation diagram we need to plot eque librium points against ß. the equilibrium points are It 92=1-51-8 ५, = / we need Go plot y=1 ES1-ß vs E ß

to note that it is we have a left facing parabola. around Line. y = 1 symmetric branch is The upper unstable (",) and the lower branch is stable (92) ny y=1 > B= yz 6 500:- from the given data we have's IS. dy de ه - 3 y = 0 -> y} – 2y = 0 => y(97-x) =o are ->4-0 con) y=#52 There three equilibrium points 9,= 0, Y2 = 54,93=-55 Wh have only two cases ;- sus caseli) :- = 0 for thes y = y2 =3 = 0 au y the three equilibreum points coalesce and the poent y=0 an unstable point is

Case (*):-620 cu we see that stable point y, is and both Y2 and Y3 unstable points are t 93 > y Yz .:. Bifurcation diagram, 79 ง. Y, →ß Y3 Noth that there are three branches in the bifurcation deagram. Branch y, co Coince des with the ßrases and it is a solid line. ... If you Impres with Answer • please gove me like.