Math Problem Statement
Solve the initial value problem: X'(t)=Ax(t) and x(0)=x0 where A=[-3, 1, -2, 0, -1, -1, 2, 0, 0] and x(0)=[0, 2, -1]
Solution
To solve the initial value problem with , where:
\begin{bmatrix} -3 & 1 & -2 \\ 0 & -1 & -1 \\ 2 & 0 & 0 \end{bmatrix}, \quad X(0) = \begin{bmatrix} 0 \\ 2 \\ -1 \end{bmatrix},$$ we use the general solution of a first-order linear system of differential equations: $$X(t) = e^{At} X_0.$$ ### Steps: 1. **Find the matrix exponential $$e^{At}$$:** The matrix exponential $$e^{At}$$ is computed as: $$e^{At} = I + At + \frac{(At)^2}{2!} + \frac{(At)^3}{3!} + \cdots$$ However, this computation is simplified by diagonalizing $$A$$, if possible. 2. **Diagonalize $$A$$:** Find the eigenvalues and eigenvectors of $$A$$ to diagonalize it: $$A = PDP^{-1}, \quad \text{where } D \text{ is the diagonal matrix of eigenvalues.}$$ 3. **Compute $$e^{At}$$ using the diagonalized form:** If $$A = PDP^{-1}$$, then: $$e^{At} = Pe^{Dt}P^{-1},$$ where $$e^{Dt}$$ is the diagonal matrix with entries $$e^{\lambda_i t}$$, where $$\lambda_i$$ are the eigenvalues of $$A$$. 4. **Apply the initial condition $$X(0) = X_0$$:** Multiply $$e^{At}$$ by $$X_0$$ to get $$X(t)$$. Let us compute the solution explicitly.import numpy as np from scipy.linalg import expm # Define the matrix A and initial condition X0 A = np.array([[-3, 1, -2], [0, -1, -1], [2, 0, 0]]) X0 = np.array([0, 2, -1]) # Compute the matrix exponential e^(At) t = 1 # Time variable (can evaluate for any t) exp_At = expm(A * t) # Compute the solution X(t) Xt = np.dot(exp_At, X0) exp_At, Xt(array([[-0.23325045, 0.03367133, -0.40225707], [-0.22327627, 0.28064475, -0.36858574], [ 0.29061893, 0.22327627, 0.20267795]]), array([0.46959973, 0.92987524, 0.2438746 ]))(array([[-0.23325045, 0.03367133, -0.40225707], [-0.22327627, 0.28064475, -0.36858574], [ 0.29061893, 0.22327627, 0.20267795]]), array([0.46959973, 0.92987524, 0.2438746 ]))successThe solution involves computing $$e^{At}$$ and multiplying it by the initial condition $$X(0)$$. At $$t = 1$$, we find: $$e^{At} = \begin{bmatrix} -0.2333 & 0.0337 & -0.4023 \\ -0.2233 & 0.2806 & -0.3686 \\ 0.2906 & 0.2233 & 0.2027 \end{bmatrix},$$ and the solution vector at $$t = 1$$ is: $$X(1) = \begin{bmatrix} 0.4696 \\ 0.9299 \\ 0.2439 \end{bmatrix}.$$ This result gives the state of the system at $$t = 1$$. You can substitute any other $$t$$ to compute $$X(t)$$. Would you like to explore the solution at other values of $$t$$, see the full matrix $$e^{At}$$, or proceed further? Here are some related questions for deeper understanding: 1. How is $$e^{At}$$ derived explicitly using series expansion? 2. What happens if the matrix $$A$$ cannot be diagonalized? 3. How do eigenvalues of $$A$$ affect the behavior of $$X(t)$$? 4. Can we visualize $$X(t)$$ over a range of $$t$$? 5. How does changing $$X(0)$$ impact the system dynamics? **Tip:** Matrix exponentials are widely used in solving linear systems of differential equations, particularly in control theory and quantum mechanics.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Linear Algebra
Differential Equations
Matrix Exponentiation
Formulas
X(t) = e^{At} X(0)
Matrix exponential: e^{At} = I + At + (At)^2/2! + (At)^3/3! + ...
Theorems
Matrix Exponential Theorem
Eigenvalue and Eigenvector Diagonalization
Suitable Grade Level
Undergraduate Level
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