ME 406Distortion of Areas in a 2D FlowWe consider the system of differential equations given by(1)x° = x , y° = -2y ,with solution(2)x(t) = x(0)et, y(t) = y(0)e-2 t .The eigenvalues are 1 and -2, the divergence is -1 and the flow is dissipative. Areas will shrink according to thelaw (3)A(t) = A(0)e-t .In spite of the fact that areas shrink, neighboring solutions can become separated, because any initial separation inx will grow exponentially. To visualize this geometrically, we follow the transformation of a square of initialconditions. The initial vertices of the square are {1,1}, {-1,1}, {-1,-1}, and {1,-1}. The sequence of graphsbelow, which can be animated to form a movie, shows the transformation of this square by the flow.First we define the four vertices of the distorted square.ClearAll@p1, p2, p3, p4Dp1 = 8Exp@tD, Exp@-2tD<;p2 = 8-Exp@tD, Exp@-2tD<;p3 = 8-Exp@tD, -Exp@-2tD<;p4 = 8Exp@tD, -Exp@-2tD<;Now we define parametrically the four sides of the distorted square.side1 = H1 - uL * p1 + u * p2;side2 = H1 - uL * p2 + u * p3;side3 = H1 - uL * p3 + u * p4;side4 = H1 - uL * p4 + u * p1;square = 8side1, side2, side3, side4<;The function squareplot[t], defined below, produces a plot of the distorted square at time t.squareplot@time_D := Module@8plotter, temgraph, i<, temgraph = 8<;Do@plotter = N@square@@iDD ê.tØ timeD;temgraph = Append@temgraph, ParametricPlot@plotter, 8u, 0, 1<,PlotRange Ø 88-5, 5<, 8-5, 5 <<, FrameLabel Ø 8"x", "y"<, Axes ØFalse, PlotLabel Ø SequenceForm@"","t= ",PaddedForm@time, 84, 2<DD, AspectRatio Ø 1, ImageSize Ø 400,Frame Ø True, DisplayFunction Ø IdentityDD, 8i, 1, 4<D;Show@temgraph, DisplayFunction Ø $DisplayFunctionDDThe command below is necessary to turn off a repeated and annoying message from Mathematica. Off@ParametricPlot::ppcomDNow we use a Do loop to produce the graph sequence, which can then be animated as a movie. The printedversion of the file shows only four of the graphs.Do@[email protected] * iD, 8i, 0, 30<D;-4-2 0 24x-4-2024yt = 0.002
View Full Document
Unlocking...