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Determine for the flow of a differential equation

  1. Jan 30, 2014 #1
    In each of the following cases, we define a function
    :
    ##\phi##: ##{\mathbb R} \times {\mathbb R}^3 \rightarrow {\mathbb R}^3 ##
    . Determine in
    each case whether this function could be the flow of a differential equation, and write
    down the differential equation.

    (a) ##\phi_t(\vec{x}) = (8,1,0)##,

    (b) ##\phi_t(\vec{x}) = \vec{x} \ \text{for all } t, ##

    (c) ##\phi_t(\vec{x}) = \vec{x} + (t,t,t).##

    Could anyone helps me to decide how to determine the flow of a differential equation as I have trouble understanding what is the flow of a differential equation, and write down the differential equations?
     
    Last edited: Jan 31, 2014
  2. jcsd
  3. Jan 31, 2014 #2

    haruspex

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    I'll make a stab at fixing up the Latex, but something still seems to be wrong:
    What does ##\phi_t(\vec{x}) = \vec{x} = (8,1,0)## mean? Should it be ##\phi_t(\vec{x}) = \vec{x} + (8,1,0)##?
    As to what the flow of a differential equation means, try a net search. E.g. http://www.math.sjsu.edu/~simic/Fall05/Math134/flows.pdf.
     
  4. Jan 31, 2014 #3
    it should be a) ##\phi_t(\vec{x}) = (8,1,0)## and b) ##\phi_t(\vec{x}) = \vec{x} ##for all t?
     
  5. Jan 31, 2014 #4
    Thank you for the article ... I am still confused by it...Could you show what it means by example?
     
  6. Jan 31, 2014 #5

    haruspex

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    I'll try to explain in words.
    If you have an ODE dX/dt=F(X), where X is an n-dimensional vector and F is an n-dimensional function of it, then you can think of it as a vector field: at each point X in the space there is a vector F(X) pointing from it. You can imagine starting at some point X0 in the space and following the chain of vectors for 'time' t. The point we reach is represented as φt(X0). Thus, φt is a function which takes the whole space and maps each point to where it would be at time t. Or we can write φ(t,X0) = φt(X0), making φ a function :[itex]\Re\times\Re^n\rightarrow\Re^n[/itex]. This is known as the 'flow'.
     
  7. Jan 31, 2014 #6
    i still don't see how dX/dt=F(X) relates to φ(t,X0) = φt? Especially related to setup of my question
     
  8. Jan 31, 2014 #7

    pasmith

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    A flow on [itex]\mathbb{R}^3[/itex] is a differentiable function [itex]f: \mathbb{R} \times \mathbb{R}^3 \to \mathbb{R}^3[/itex] such that the restrictions [itex]f_t : \mathbb{R}^3 \to \mathbb{R}^3 : x \mapsto f(t,x)[/itex] satisfy:
    • [itex]f_0(x) = x[/itex] for all [itex]x \in \mathbb{R}^3[/itex].
    • [itex]f_t(f_s(x)) = f_{t+s}(x) = f_s(f_t(x))[/itex] for all [itex]x \in \mathbb{R}^3[/itex] and all [itex]t \in \mathbb{R}[/itex] and all [itex]s \in \mathbb{R}[/itex].
    Note that these conditions require [itex]f_0 = f_t \circ f_{-t}[/itex] so that [itex]f_{-t} = f_t^{-1}[/itex] and each [itex]f_t[/itex] is invertible.

    To find the differential equation, set [itex]x(t) = \phi_t(x_0)[/itex] for arbitrary constant [itex]x_0[/itex] and differentiate with respect to [itex]t[/itex].
     
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