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

MATH221 December 2007

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[hide] 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.

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[show] Hint
Science:Math Exam Resources/Courses/MATH221/December 2007/Question Section 103 10 (a)/Hint 1

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[show] 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 ${\begin{aligned}x_{n}&=2(4)^{n+1}+3(-1)^{n+1}\\y_{n}&=2[4^{n+1}+(-1)^{n}]\end{aligned}}$ .