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Thread: Matrix Exponentials to solve Differential Equations...

  1. #1
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    Matrix Exponentials to solve Differential Equations...

    Define the exponential of a matrix A to be $\displaystyle e^{A} = I + A + \frac{1}{2!}A^{2}+\frac{1}{3!}A^{3}+...$.

    Let A be the following matrix $\displaystyle \left(\begin{array}{cc}3 & -2 \\ -2 & 3\end{array}\right)$. Show that the eigen values of A are $\displaystyle \lambda_{1} = 1 $ and $\displaystyle \lambda_{2} = 5$ with associated eigenvectors $\displaystyle \left(\begin{array}{c}1\\1\end{array}\right),\left (\begin{array}{c}1\\-1\end{array}\right)$. Hence or otherwise calculate $\displaystyle e^{A}$.

    Use previous to solve the system of differential equations (with respect to t)
    $\displaystyle y'_{1}=3y_{1} + -2y_{2}$
    $\displaystyle y'_{2}=-2y_{1}+3y_{2} $
    with the initial conditions $\displaystyle y_{1}(0)=2$, and $\displaystyle y_{2}(0)=0$, by writing the system as
    $\displaystyle Y'=AY, Y(0)=Y_{0}$
    for the matrix A above, and $\displaystyle Y_{0} = \left(\begin{array}{c}2\\0\end{array}\right)$ and noting that a solution is $\displaystyle Y(t)=e^{tA}Y_{0}$


    I can do the eigenvalues and eigenvectors part, and can calculate $\displaystyle e^{A}$ since $\displaystyle e^{A}=Xe^{D}X^{-1}$,
    but I don't get the differential equations part...

    Any help would be much appreciated...
    Last edited by Unenlightened; Nov 23rd 2009 at 04:44 AM.
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  2. #2
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    Hello,

    As it is said, the system can be written $\displaystyle Y'=AY$, where $\displaystyle Y=\begin{pmatrix} y_1(t) \\ y_2(t) \end{pmatrix}$
    We define $\displaystyle Y'$ as $\displaystyle \begin{pmatrix}y_1'(t) \\ y_2'(t)\end{pmatrix}$

    We indeed have Y'=AY.
    Now, like in a normal DE, a solution to y'=ay is y=y_0 exp(at)

    Here, as it's again said, a solution is $\displaystyle e^{tA}Y_0$

    Now how to calculate $\displaystyle e^{tA}$ ?

    $\displaystyle e^{tA}=\sum_{n=0}^\infty \frac{t^n A^n}{n!}=\sum_{n=0}^\infty \frac{t^nXD^nX^{-1}}{n!}=X \left(\sum_{n=0}^\infty \frac{(tD)^n}{n!} \right) X^{-1}$

    In fact, what is $\displaystyle \sum_{n=0}^\infty \frac{(tD)^n}{n!}$ ?
    $\displaystyle D=\begin{pmatrix}1&0\\0&5\end{pmatrix}$

    So $\displaystyle tD=\begin{pmatrix}t&0\\0&5t\end{pmatrix}$

    And then $\displaystyle \sum_{n=0}^\infty \frac{(tD)^n}{n!}=\begin{pmatrix}\sum_{n=0}^\infty \frac{t^n}{n!} &0 \\0& \sum_{n=0}^\infty \frac{(5t)^n}{n!} \end{pmatrix}=\begin{pmatrix} e^t & 0 \\ 0&e^{5t} \end{pmatrix}$


    Does it look clearer to you ?
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