Integral curve, flow

We define a specific vector field U. The differential equations for the integral curve of U are

U = {1 + x_1 + x_2, 2 x_1x_2 + Cos[x_1]} ;

E0[ΧIntegral[U]]

{0 == -1 - x_1[] - x_2[] + x_1^′[], 0 == -Cos[x_1[]] - 2 x_1[] x_2[] + x_2^′[]}

Both of the following lines do the same, namely solve an integral curve α:I→^2 numerically. The first curve starts at the initial point α(0)={1,0}. The curve is parametrized on the interval I={-1,1}.

c1 = Χα[U, {1, 0}, {-1, 1}]

c2 = Χ★[2]/.NDSolve[E0[ΧIntegral[U], Χ0[0, {0, 1}]], Χ★[2], {, -1, 1}][[1]]

{InterpolatingFunction[{{-1., 1.}}, <>][], InterpolatingFunction[{{-1., 1.}}, <>][]}

{InterpolatingFunction[{{-1., 1.}}, <>][], InterpolatingFunction[{{-1., 1.}}, <>][]}

We plot the vector field and the integral curves.

PlotShow[PlotLine[#1, {, -1, .5, .02}] &/@{c1, c2}, PlotVectorField[U, {x_1, -π, π}, {x_2, -1, π}, PlotPoints→10, DisplayFunction→Identity]] ;

[Graphics:../HTMLFiles/index_10.gif]

The unit circle gets transported by the flow of U below.

ParametricPlot3D[Ft[Χψ[U, .5 {Cos[t], Sin[t]}, h], h], {t, 0, 2 π}, {h, 0, 1}, PlotPoints→ {30, 10}] ;

[Graphics:../HTMLFiles/index_12.gif]

We compute the differential of the flow d_xψ(t,x)| _ (x = ψ(t, Overscript[x, _])) of a vector field V along the integral curve with initial point Overscript[x, _]∈M. The differential depends on the parameter t∈I. Below, we chose Overscript[x, _]={1,1} and I={-1,1}.

c1 = Χα[V = {3 Sin[x_2], x_1x_2 - x_1^2}, {1, 1}, {-1, 1}] ;

dα = Χdα[V, {1, 1}, {-1, 1}]

dα/.→1//MF

( {{-1.07516, 1.78581}, {-1.47923, -2.56423}} )

Here comes the picture.

PlotShow[PlotLine[c1, {, -.7, .5, .03}], PlotFrame[c1, .3 dα, {-.5, .5, .1}], PlotVectorField[V, {x_1, -1, 2}, {x_2, -1, 2}, PlotPoints→6, DisplayFunction→Identity]] ;

[Graphics:../HTMLFiles/index_23.gif]


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