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Geometrical interpretation of this property

  1. Jun 23, 2014 #1
    Hi there!

    I have the following property:
    If [itex]x(t)[/itex] is a solution of [itex] \left\{ \begin{array}{l} \dot{x} = f(x) \\ x(t_0) = x_0 \end{array} \right. [/itex] then the function [itex] y(t) = x(t+t_0) [/itex] is a solution of the equation with initial data [itex] y(0) = x_0 [/itex].

    How could it be interpreted geometrically?
  2. jcsd
  3. Jun 23, 2014 #2
    Geometrically, think of [itex]f[/itex] as a vector field. Imagine it as a collection of arrows describing the velocity of a fluid; that is, for each [itex]x[/itex], [itex]f(x)[/itex] is the velocity of the fluid at the point [itex]x[/itex].

    Now, imagine a small pebble moving in this fluid. At each point [itex]x[/itex], the trajectory of the pebble must be tangent to the vector [itex]f(x)[/itex] (why?). Dropping a pebble into the fluid at a specific point [itex]x_0[/itex] represents an initial condition of your ODE. The answer to your question is the following: it doesn't matter what time you drop the pebble at [itex]x_0[/itex]; the pebble's resulting trajectory will always be the same, because the fluid velocity field is unchanging with time.

    The image of the solution of [itex]y[/itex] is just a copy of the image of [itex]x[/itex], but shifted in time -- just like the pebble's motion when dropped at time [itex]t_0[/itex] is identical to its motion when dropped at [itex]0[/itex], but shifted in time.
  4. Jun 23, 2014 #3
    Nice example! So, if the velocity is constant it means that the position [itex] x(t) [/itex] is linear and so [itex] y(t) [/itex] is, and due the way the functions are linked I can say that their graphics are parallels.
    Let me know if I'm wrong.

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