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Thread: differencial equation system

  1. #1
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    differencial equation system

    Now ... I ve made it. Files are smaller.
    I need help with question 2 part (b)
    many thanks
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  2. #2
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    Quote Originally Posted by simfonija View Post
    Now ... I ve made it. Files are smaller.
    I need help with question 2 part (b)
    many thanks
    I'm curious. If all you need is 2 (b) why did you post the first page?

    Anyway, we have a matrix equation:
    $\displaystyle x^{(k)} = Ax^{(k-1)}$
    and an initial population distribution: $\displaystyle x^{(0)} = a_1 x_1 + ... + a_n x_n$
    where $\displaystyle \{ x_n \}$ is the set of eigenvectors of A.

    Let $\displaystyle \{ \lambda _n \}$ be the set of eigenvalues corresponding to the (respective) eigenvector.

    Then
    $\displaystyle x^{(1)} = Ax^{(0)} = A(a_1x_1 + ... + a_n x_n)$

    $\displaystyle x^{(1)} = a_1 \lambda _1 x_1 + ... + a_n \lambda_n x_n$

    Iterating this process gives:
    $\displaystyle x^{(k)} = a_1 ( \lambda _1 )^k x_1 + ... + a_n ( \lambda_n )^k x_n$

    Now, if $\displaystyle \lambda _1$ is the dominant eigenvalue (I think this said the first eigenvalue. It was hard to read.) then we may assume
    $\displaystyle \lambda _1 >> \lambda _k $ for all $\displaystyle k \neq 1$. (Obviously only a positive eigenvalue can dominate the population.) This means that for large k there is effectively only one term in the $\displaystyle x^{(k)}$ that is significant: the first term. So
    $\displaystyle x^{(k)} \approx a_1 ( \lambda_1 )^k x_1$ for large k.

    This means the population is stable in the state represented by eigenvector $\displaystyle x_1$ and has a growth rate represented by $\displaystyle \lambda _1$.

    If there is no dominant eigenvalue then the population will fluctuate in a series of eigenstates with no specific growth rate dominating. However, for large enough k the population should stabilize around a vector that is the sum of eigenvectors with the largest eigenvalues. For example, say that we know that $\displaystyle \lambda _1 = \lambda _2 > \lambda _k$ for $\displaystyle k \neq 1, 2$. Then, given a large enough k we would have:
    $\displaystyle x^{(k)} \approx a_1 ( \lambda _1)^k x_1 + a_2 ( \lambda _2)^k x_2$
    So the population would settle into a growth state vector $\displaystyle a_1 x_1 + a_2 x_2$ with (in this case) a growth of $\displaystyle \lambda _1 = \lambda _2$.

    -Dan
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  3. #3
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    Thank you so much for the detail answer. The first page was mistake, previously I saw there is a part of question 2 (b) on first page. Once again thank you, was great help!
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