Science:Math Exam Resources/Courses/MATH221/December 2007/Question Section 103 10 (a)

MATH221 December 2007

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Question Section 103 10 (a)
A discrete dynamical system is defined by ${\begin{aligned}x_{n+1}&=x_{n}+3y_{n}\quad \quad x_{0}=5\\y_{n+1}&=2x_{n}+2y_{n}\quad \quad y_{0}=10\end{aligned}}$ Find explicit formulas for $x_{n}$ and $y_{n}$, with the given initial condition.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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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. Let us first write the system in matrix notation: ${\begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}}={\begin{pmatrix}1&3\\2&2\end{pmatrix}}{\begin{pmatrix}x_{n}\\y_{n}\end{pmatrix}}.$ Notice that ${\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}={\begin{pmatrix}1&3\\2&2\end{pmatrix}}{\begin{pmatrix}x_{0}\\y_{0}\end{pmatrix}},$ ${\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}}={\begin{pmatrix}1&3\\2&2\end{pmatrix}}^{2}{\begin{pmatrix}x_{0}\\y_{0}\end{pmatrix}},$ and ${\begin{pmatrix}x_{3}\\y_{3}\end{pmatrix}}={\begin{pmatrix}1&3\\2&2\end{pmatrix}}^{3}{\begin{pmatrix}x_{0}\\y_{0}\end{pmatrix}}...$ Therefore, we hypothesize that for every $n\in \mathbb {N} ,$ ${\begin{pmatrix}x_{n}\\y_{n}\end{pmatrix}}={\begin{pmatrix}1&3\\2&2\end{pmatrix}}^{n}{\begin{pmatrix}x_{0}\\y_{0}\end{pmatrix}}.$ This explicit formula can (and should) be established rigorously by induction (we leave this as an exercise). Now we want to compute ${\begin{pmatrix}1&3\\2&2\end{pmatrix}}^{n}$. Recall that matrix powers are easy to compute via matrix diagonalization. Therefore, we find the eigenvalues and eigenvectors for $A={\begin{pmatrix}1&3\\2&2\end{pmatrix}}$. The eigenvalues are determined from $\det(A-\lambda I)=(1-\lambda )(2-\lambda )-6=(\lambda -4)(\lambda +1)=0$, so the eigenvalues of A are 4 and -1. To find an eigenvector associated with the eigenvalue 4, we solve $(A-4I)x=0$, i.e. the equation ${\begin{pmatrix}-3&3\\2&-2\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}},$ from which it is clear that ${\begin{pmatrix}1\\1\end{pmatrix}}$ is an eigenvector. To find an eigenvector associated with the eigenvalue -1, we solve ${\begin{pmatrix}2&3\\2&3\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}},$ from which we find ${\begin{pmatrix}-3\\2\end{pmatrix}}$ is an eigenvector. Now we form the matrix $P={\begin{pmatrix}1&-3\\1&2\end{pmatrix}}$ whose columns are the eigenvectors of A, and the diagonal matrix $\Lambda ={\begin{pmatrix}4&0\\0&-1\end{pmatrix}}$ with the corresponding eigenvalues. We know that $A=P\Lambda P^{-1}$ and hence $A^{n}=P\Lambda ^{n}P^{-1}=P{\begin{pmatrix}4^{n}&0\\0&(-1)^{n}\end{pmatrix}}P^{-1}.$ The final step is to compute $P^{-1}={\frac {1}{5}}{\begin{pmatrix}2&3\\-1&1\end{pmatrix}}$, and then $A^{n}$: $A^{n}={\frac {1}{5}}{\begin{pmatrix}1&-3\\1&2\end{pmatrix}}{\begin{pmatrix}4^{n}&0\\0&(-1)^{n}\end{pmatrix}}{\begin{pmatrix}2&3\\-1&1\end{pmatrix}}={\frac {4^{n}}{5}}{\begin{pmatrix}2&3\\2&3\end{pmatrix}}+{\frac {(-1)^{n}}{5}}{\begin{pmatrix}3&-3\\-2&2\end{pmatrix}}.$ Check that for n = 1 we recover the matrix A. Finally, the question asks us to find $x_{n}$ and $y_{n}$ explicitly, with the given initial condition. Having computed $A^{n}$, we can now compute ${\begin{pmatrix}x_{n}\\y_{n}\end{pmatrix}}={\begin{pmatrix}1&3\\2&2\end{pmatrix}}^{n}{\begin{pmatrix}x_{0}\\y_{0}\end{pmatrix}}=\left[{\frac {4^{n}}{5}}{\begin{pmatrix}2&3\\2&3\end{pmatrix}}+{\frac {(-1)^{n}}{5}}{\begin{pmatrix}3&-3\\-2&2\end{pmatrix}}\right]{\begin{pmatrix}5\\10\end{pmatrix}}.$ Upon multiplying and simplifying the above equation, we find that ${\begin{aligned}x_{n}&=2(4)^{n+1}+3(-1)^{n+1}\\y_{n}&=2[4^{n+1}+(-1)^{n}]\end{aligned}}$.

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