index_1.html

Integral curve, flow

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

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

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

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

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

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

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

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]

Created by Mathematica (December 22, 2006)