91Ó°ÊÓ

Let's find J^2 by multiplying J by itself: \(\mathbf{J}^2 = \mathbf{J}\mathbf{J}\) Use matrix multiplication to get: \(\mathbf{J}^2 = \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} = \begin{pmatrix}\lambda^2 & 2\lambda \\ 0 & \lambda^2 \end{pmatrix}\) Similarly, we can find J^3 and J^4: \(\mathbf{J}^3 = \mathbf{J}\mathbf{J}^2 = \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix}\lambda^2 & 2\lambda \\ 0 & \lambda^2 \end{pmatrix} = \begin{pmatrix}\lambda^3 & 3\lambda^2 \\ 0 & \lambda^3 \end{pmatrix}\) \(\mathbf{J}^4 = \mathbf{J}\mathbf{J}^3 = \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix}\lambda^3 & 3\lambda^2 \\ 0 & \lambda^3 \end{pmatrix} = \begin{pmatrix}\lambda^4 & 4\lambda^3 \\ 0 & \lambda^4 \end{pmatrix}\)

We will use induction to prove that \(\mathbf{J}^n=\left(\begin{array}{cc}{\lambda^{n}} & {n \lambda^{n-1}} \\ {0} & {\lambda^{n}}\end{array}\right)\) - Base Case (n=1): We see that J^1 is already given as \(\mathbf{J} =\left(\begin{array}{cc}{\lambda^1} & {1\lambda^0} \\ {0} & {\lambda^1}\end{array}\right)\) - Inductive Step: Assume that the formula is true for n=k, \(\mathbf{J}^k=\left(\begin{array}{cc}{\lambda^{k}} & {k \lambda^{k-1}} \\ {0} & {\lambda^{k}}\end{array}\right)\) We will show that the formula is true for n=k+1, \(\mathbf{J}^{k+1}=\mathbf{J}\mathbf{J}^k = \begin{pmatrix}\lambda & 1 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix}\lambda^k & k\lambda^{k-1} \\ 0 & \lambda^k \end{pmatrix}\) Using matrix multiplication, we get \(\mathbf{J}^{k+1}=\left(\begin{array}{cc}{\lambda^{k+1}} & {(k+1) \lambda^{k}} \\ {0} & {\lambda^{k+1}}\end{array}\right)\) Thus, the formula holds for n=k+1, and by induction, it holds for all positive integers n.

To calculate the matrix exponential exp(Jt), we first recall the matrix exponential formula: \(e^{\mathbf{A}}=\sum_{n=0}^{\infty}\frac{1}{n!}\mathbf{A}^n\) Applying this to Jt: \(e^{\mathbf{Jt}}=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\mathbf{Jt}\right)^n\) We found the formula for J^n in step 2. So, we can write: \(e^{\mathbf{Jt}}=\sum_{n=0}^{\infty}\frac{1}{n!}\begin{pmatrix}(\lambda t)^n & n(\lambda t)^{n-1}t \\ 0 & (\lambda t)^n\end{pmatrix}\) Now, we sum the series component-wise: \(e^{\mathbf{Jt}}=\begin{pmatrix}\sum_{n=0}^{\infty}\frac{(\lambda t)^n}{n!} & \sum_{n=0}^{\infty}\frac{n(\lambda t)^{n-1}t}{n!} \\ 0 & \sum_{n=0}^{\infty}\frac{(\lambda t)^n}{n!}\end{pmatrix}\) Notice the elements in the diagonal are just the exponential function of λt: \(e^{\mathbf{Jt}}=\begin{pmatrix}e^{\lambda t} & \sum_{n=0}^{\infty}\frac{n(\lambda t)^{n-1}t}{n!} \\ 0 & e^{\lambda t}\end{pmatrix}\) The series in the top-right element can be shown to be te^{\lambda t} (the first term is 0): \(e^{\mathbf{Jt}}=\begin{pmatrix}e^{\lambda t} & te^{\lambda t} \\ 0 & e^{\lambda t}\end{pmatrix}\)

Given \(\mathbf{x'}=\mathbf{Jx}\) and \(\mathbf{x}(0)=\mathbf{x}^0\), we can solve the initial value problem using the matrix exponential: \(\mathbf{x}(t) = e^{\mathbf{Jt}}\mathbf{x}^0\) Substitute the matrix exponential found in step 3: \(\mathbf{x}(t) = \begin{pmatrix}e^{\lambda t} & te^{\lambda t} \\ 0 & e^{\lambda t}\end{pmatrix}\mathbf{x}^0\) Suppose \(\mathbf{x}^0 = \begin{pmatrix}x_1^0 \\ x_2^0\end{pmatrix}\). Then the solution is: \(\mathbf{x}(t) = \begin{pmatrix}e^{\lambda t} & te^{\lambda t} \\ 0 & e^{\lambda t}\end{pmatrix}\begin{pmatrix}x_1^0 \\ x_2^0\end{pmatrix}= \begin{pmatrix}(x_1^0 + tx_2^0)e^{\lambda t} \\ x_2^0e^{\lambda t}\end{pmatrix}\) The solution of the initial value problem is given by: \(\mathbf{x}(t) = \begin{pmatrix}(x_1^0 + tx_2^0)e^{\lambda t} \\ x_2^0e^{\lambda t}\end{pmatrix}\)