CONTINUOUS DYNAMICAL SYSTEMS IMath 21b, O. KnillHomework: Section 9.1: 4,8,10,26,32,24*,46* until TuesdayCONTINUOUS DYNAMICAL SYSTMES. A differentialequationddt~x=f(~x) defines a dynamical system. The solu-tions is a curve~x(t) which has thevelocity vectorf(~x(t))for allt. One often writes ˙xinstead ofddtx. So, we have theproblem that we know a formula for the tangent at eachpoint.The aim is to find a curve~x(t) which starts at agiven point~v=~x(0).IN ONE DIMENSION. A system ˙x=g(x, t) is the general differential equation in one dimensions. Examples:•If˙x=g(t), thenx(t) =Rt0g(t)dt. Example: ˙x= sin(t), x(0) = 0 has the solutionx(t) = cos(t)-1.•If˙x=h(x), thendx/h(x) =dtand sot=Rx0dx/h(x) =H(x) so thatx(t) =H-1(t). Example:˙x=1cos(x)withx(0) = 0 givesdxcos(x) =dtand after integration sin(x) =t+Cso thatx(t) = arcsin(t+C).Fromx(0) = 0 we getC=π/2.•If˙x=g(t)/h(x), thenH(x) =Rx0h(x)dx=Rt0g(t)dt=G(t) so thatx(t) =H-1(G(t)).Example:˙x= sin(t)/x2, x(0) = 0 givesdxx2= sin(t)dtand after integrationx3/3 =-cos(t) +Cso thatx(t) =(3C-3 cos(t))1/3. Fromx(0) = 0 we obtainC= 1.Remarks:1) In general, we have no closed form solutions in terms of known functions. The solutionx(t) =Rt0e-t2dtof ˙x=e-t2forexample can not be expressed in terms of functions exp,sin,log,√·etc but it can be solved using Taylor series: becausee-t2= 1-t2+t4/2!-t6/3!+. . .taking coefficient wise the anti-derivatives gives:x(t) =t-t3/3+t4/(32!)-t7/(73!)+. . ..
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