Science:Math Exam Resources/Courses/MATH307/April 2006/Question 04 (d)

MATH307 April 2006

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Question 04 (d)

The following discrete dynamical system describes the yearly migration of wild horse populations among three areas R, G, and B. Let r(t), g(t), and b(t) be the sizes of the horse population in areas R, G, and B respectively at the tth year. ${\begin{aligned}x(t+1)&={\begin{bmatrix}r(t+1)\\g(t+1)\\b(t+1)\end{bmatrix}}\\&={\begin{bmatrix}r(t)/2+g(t)/3+b(t)/3\\r(t)/2+g(t)/3+b(t)/2\\g(t)/3+b(t)/6\end{bmatrix}}\\&={\begin{bmatrix}1/2&1/3&1/3\\1/2&1/3&1/2\\0&1/3&1/6\end{bmatrix}}{\begin{bmatrix}r(t)\\g(t)\\b(t)\end{bmatrix}}\\&=Ax(t)\end{aligned}}$ Where the Markov matrix A describes how the horses move among these areas from one year to the next. The 1st column indicates that each year 1/2 of the horses in area R remain in area R and 1/2 will migrate to area G. The 2nd column shows that horses in area G will be evenly distributed in the three areas one year later. The 3rd column implies that, of the horses in area B, 1/3 will migrate to area R, 1/2 will migrate to area G, and only 1/6 will remain in area B. We assume that no horses are lost and no new horses are added and that initially (i.e. at t = 0), there are a total of 350 horses all located in area B. Thus ${\vec {x}}(0)=[0\ 0\ 350]^{T}$. (d) Find ${\vec {x}}(t)$ in terms of the eigenvalues and eigenvectors of A. Then, calculate $\lim _{t\to \infty }{\vec {x}}(t)$.

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For this problem we want to find $x(t)$. We know that $x(0)=\left[{\begin{array}{c}0\\0\\350\end{array}}\right]$. And $x_{(1)}=Ax_{(0)}$. We can generalize to the form $x_{(n)}=Ax_{(n-1)}$ This can further be generalized to $x_{(n)}=A^{n}x_{(0)}$ From part c we found that $S^{-1}AS=\Lambda$. We can rearrange this to $A=S\Lambda S^{-1}$. So $A^{n}=(S\Lambda S^{-1})^{n}=S\Lambda ^{n}S^{-1}$ because $S$ and $S^{-1}$ will cancel each other out. This means we can easily find $\lim _{n\to \infty }x_{n}=A^{n}x_{0}=S\Lambda ^{n}S^{-1}x_{0}$ Recall that two of our eigenvalues were less than zero so when we take this limit they will go to zero. $\lim _{n\to \infty }x_{n}=\left[{\begin{array}{ccc}{\frac {2}{5}}&-{\frac {1}{2}}&0\\{\frac {3}{7}}&0&-{\frac {1}{2}}\\{\frac {6}{35}}&{\frac {1}{2}}&{\frac {1}{2}}\end{array}}\right]$$\left[{\begin{array}{ccc}1&0&0\\0&0&0\\0&0&0\end{array}}\right]$ $\left[{\begin{array}{ccc}{\frac {2}{5}}&-{\frac {1}{2}}&0\\{\frac {3}{7}}&0&-{\frac {1}{2}}\\{\frac {6}{35}}&{\frac {1}{2}}&{\frac {1}{2}}\end{array}}\right]^{-1}$ $\left[{\begin{array}{c}0\\0\\350\end{array}}\right]$ Once we do this we find the answer $x_{n}=\left[{\begin{array}{c}140\\150\\60\end{array}}\right]$