Results 1 to 2 of 2

Thread: Linear Algebra/Differential equations

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    20

    Linear Algebra/Differential equations

    How can we show that the matrix differential equation dX/dt=AX+XB has the solution $\displaystyle X(t)=e^{At}X(0)e^{Bt} $ and prove that the solutions of dX/dt=AX-XA keep the same eigenvalues for all time.
    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Joined
    Oct 2007
    Posts
    571
    You can check that that is indeed a solution by differentiating, now suppose we have a solution $\displaystyle X(t)$ we will prove it has to be of that form (uniqueness)

    Remember that the product rule also holds for Matrixes, we have:

    $\displaystyle
    \left( {e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} } \right)^\prime = \left( {e^{ - A \cdot t} \cdot X\left( t \right)} \right)^\prime \cdot e^{ - B \cdot t} + \left( {e^{ - A \cdot t} \cdot X\left( t \right)} \right) \cdot \left( { - B \cdot e^{ - B \cdot t} } \right)
    $

    Hence: $\displaystyle
    \left( {e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} } \right)^\prime = \left[ {\left( { - A \cdot e^{ - A \cdot t} } \right) \cdot X\left( t \right) + e^{ - A \cdot t} \cdot X'\left( t \right)} \right] \cdot e^{ - B \cdot t} +
    $$\displaystyle
    \left( {e^{ - A \cdot t} \cdot X\left( t \right)} \right) \cdot \left( { - B \cdot e^{ - B \cdot t} } \right)
    $

    $\displaystyle
    C
    $ and $\displaystyle
    e^C
    $ commute, for every matrix $\displaystyle C$.

    Thus: $\displaystyle
    \left( {e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} } \right)^\prime = e^{ - A \cdot t} \left[ {\underbrace { - A \cdot X\left( t \right) + X'\left( t \right) - B \cdot X\left( t \right)}_{ = \bold{0}_{n \times n} }} \right]e^{ - B \cdot t} = \bold{0}_{n \times n}
    $

    So it is a constant matrix ( because of the 0 derivative) : $\displaystyle
    e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} = M
    $

    Let $\displaystyle t=0$ we get: $\displaystyle
    e^{ - A \cdot 0} \cdot X\left( 0 \right) \cdot e^{ - B \cdot 0} = {\text{Id}}_{n \times n} \cdot X\left( 0 \right) \cdot {\text{Id}}_{n \times n} = X\left( 0 \right) = M
    $

    $\displaystyle
    e^{ - A \cdot t} \cdot X\left( t \right) \cdot e^{ - B \cdot t} = X\left( 0 \right) \Rightarrow X\left( t \right) = e^{A \cdot t} \cdot X\left( 0 \right) \cdot e^{B \cdot t}
    $ And the uniqueness is proven.


    For the second part of the question, let $\displaystyle B=-A$

    The solutions is of the form: $\displaystyle
    e^{A \cdot t} \cdot X(0) \cdot e^{ - A \cdot t}
    $

    Let's see the characteristic polynomial:

    $\displaystyle
    \det \left( {e^{A \cdot t} \cdot X(0) \cdot e^{ - A \cdot t} - \lambda \cdot {\text{Id}}} \right) = \det \left( {e^{A \cdot t} \cdot X(0) \cdot e^{ - A \cdot t} - \lambda \cdot e^{A \cdot t} \cdot e^{ - A \cdot t} } \right)
    $ $\displaystyle =
    \det \left[ {e^{A \cdot t} \cdot \left( {X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right) \cdot e^{ - A \cdot t} } \right]
    $

    $\displaystyle
    \det \left( {e^{A \cdot t} \cdot X\left( 0 \right) \cdot e^{ - A \cdot t} - \lambda \cdot {\text{Id}}} \right) = \det \left( {e^{A \cdot t} } \right) \cdot \det \left( {X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right) \cdot \det \left( {e^{ - A \cdot t} } \right)
    $ $\displaystyle =
    \det \left( {e^{A \cdot t} } \right) \cdot \det \left( {e^{ - A \cdot t} } \right) \cdot \det \left({X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right) = \det \left( {e^{A \cdot t} \cdot e^{ - A \cdot t} } \right) \cdot \det \left( {X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right)
    $

    $\displaystyle
    = \det \left( {{\text{Id}}} \right) \cdot \det \left({X\left( 0 \right) - \lambda \cdot {\text{Id}}}\right) = \det \left( {X\left( 0 \right) - \lambda \cdot {\text{Id}}} \right)
    $ !!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. TI-89 Differential Equations/Linear Algebra Notes
    Posted in the Calculators Forum
    Replies: 0
    Last Post: Mar 9th 2011, 01:20 PM
  2. Replies: 7
    Last Post: Aug 30th 2009, 10:03 AM
  3. Replies: 1
    Last Post: May 15th 2008, 08:23 PM
  4. Replies: 5
    Last Post: Jul 16th 2007, 04:55 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum