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Solution to system of differential equations

  1. Apr 22, 2006 #1
    In order to find streamlines for vector field, we need to solve this system of differential equations:

    [tex]\frac{dx}{dt} = -x+y+z[/tex]
    [tex]\frac{dy}{dt} = x-y+z[/tex]
    [tex]\frac{dz}{dt} = x+y-z[/tex]
    where
    [tex]x(0) = a[/tex]
    [tex]y(0) = b[/tex]
    [tex]z(0) = c[/tex]

    I have used Mathematica in order to find solutions for these equations and it calculated the following:
    [tex]x = \frac{2a-b-c+ae^{3t}+be^{3t}+ce^{3t}}{3e^{2t}}[/tex]
    [tex]y = \frac{-a+2b-c+ae^{3t}+be^{3t}+ce^{3t}}{3e^{2t}}[/tex]
    [tex]z = \frac{-a-b+2c+ae^{3t}+be^{3t}+ce^{3t}}{3e^{2t}}[/tex]

    it is, of course, right - I have check parametric plot and it is streamline, but I think how it came to that solution? Does anybody can explain how I can solve these equations (step-by-step would be very very helpful :roll: ) without using of Mathematica? I do not understand why there is [tex]3e^{2t}[/tex]?

    Thank you
     
  2. jcsd
  3. Apr 22, 2006 #2
    The simpliest solution can be found using Laplace transform. The system you mentioned can be rewriten in a matrix form as following:

    [tex]\frac{{d\bar x}}{{dt}} = A\bar x[/tex]

    where

    [tex]\bar x = \left( {x,y,z} \right)[/tex]

    Applying Laplace transform to the both sides of the equation yields:

    [tex]s\bar x\left( s \right) - \bar x_0 = A\bar x\left( s \right)[/tex]

    After rearranging and taking out the common terms you get:

    [tex]\bar x\left( s \right) = \left( {sI - A} \right)^{ - 1} \bar x_0[/tex]

    Last, you need to perform the inverse Laplace transform (this is where the [tex]3e^{2t}[/tex] is coming from):

    [tex]\bar x\left( t \right) = L^{ - 1} \left\{ {\bar x\left( s \right)} \right\}[/tex]
     
  4. Apr 22, 2006 #3
    As I understand we derive the matrix from the system, then we calculate det(A-xI) and we get cubic equation and then roots: 1, -2, -2 which are eigenvalues.

    I got an equation: [tex]-x^3-3x^2+4 = 0[/tex] so roots are indeed: 1, -2, -2.

    How to create eigenvectors? I belief that we need so called eigenvector to get linearly independent solutions.
     
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