My name is Hanzhang Yin, and I have developed this website as a resource to facilitate the review of key concepts in abstract algebra, and it is not fully completed. Feel free to email me at hanyin@ku.edu if there are any errors or suggestions for improvement.
Problem 1. In this problem, we investigate the numerical method \[ u_{n+1} = \frac{4u_n - u_{n-1}}{3} + \frac{2h}{3} f(t_{n+1}, u_{n+1}) \quad (1) \] for solving the initial value problem \( y'(t) = f(t, y(t)), \, y(t_0) = y_0 \) and \( t \in [t_0, t_0 + T] \). Here \( h = T/N, \, t_n = t_0 + nh \) for \( n = 0, 1, \ldots, N \), with \( u_0 = y_0 \). We assume that \( f(t, y) \) is continuous on \([t_0, t_0 + T] \times \mathbb{R}\) and satisfies the Lipschitz condition with constant \( L > 0 \): \[ |f(t, y) - f(t, z)| \le L |y - z|, \quad \forall t \in [t_0, t_0 + T], \quad \forall y, z \in \mathbb{R}. \]
(a) State whether the method (1) is single-step or multistep, and whether it is implicit or explicit.
(b) How would you compute the value \( u_1 \approx y(t_1) \)?
(c) State and prove whether the method (1) is consistent, zero-stable and/or convergent.
Solution (a). Firstly, we can know that the method (1) is multistep and implicit.
Solution (b).
Solution (c).
Problem 4. Let \( A \in \mathbb{C}^{m \times m} \) be nonsingular and \( 0 \neq b \in \mathbb{C}^m \). Suppose that \( x \) is the solution of \( Ax = b \) and \( \hat{x} \) is an approximation of \( x \). Define \[ r = b - A\hat{x} \quad \text{and} \quad \Delta = \frac{r \hat{x}^*}{\hat{x}^* \hat{x}}, \] where \( \hat{x}^* \) is the conjugate transpose (adjoint) of \( \hat{x} \).
(a) Prove that \( \Delta \) satisfies \( (A + \Delta) \hat{x} = b \).
(b) Prove that \[ \|\Delta\|_2 = \min_{(A+E)\hat{x} = b} \|E\|_2, \] i.e., \( \Delta \) is a matrix with the smallest 2-norm among all the matrices \( E \) that satisfy \( (A + E) \hat{x} = b \).
(c) Prove that \[ (\kappa_2(A))^{-1} \frac{\|r\|_2}{\|b\|_2} \leq \frac{\|x - \hat{x}\|_2}{\|x\|_2} \leq (\kappa_2(A)) \frac{\|r\|_2}{\|b\|_2}, \] where \( \kappa_2(A) = \|A^{-1}\|_2 \|A\|_2 \).
Proof (a). Since \[ \begin{align} \left(A + \Delta\right)\hat{x} &= \left(A + \dfrac{r\hat{x}^*}{\hat{x}^*\hat{x}}\right)\hat{x} = A\hat{x} + \dfrac{r\hat{x}^*\hat{x}}{\hat{x}^*\hat{x}} = A\hat{x} + r. \end{align} \] Given that \(r = b - A\hat{x}\), we have \[ A\hat{x} + r = b. \blacksquare \]