Problem Sheet 2#
Question 1#
Apply 2nd Order Taylor Method to approximate the solution of the given initial value problems using the indicated number of time steps, N. Compare the approximate solution with the given exact solution, and compare the actual error with the theoretical local and global error:
a. \(y'=t-y, \ \ (0\leq t \leq 4)\) with the initial condition \(y(0)=1\), \(N=4\), with the exact solution \(y(t)=2e^{-t}+t-1.\) The Lipschitz constant is determined on \(D=\{(t,y);0\leq t \leq 4, y\in \rm I\!R \}.\)
b. \(y'=y-t, \ \ (0\leq t \leq 2) \) with the initial condition \(y(0)=2,\) \(N=4\), with the exact solution \(y(t)=e^{t}+t+1\). The Lipschitz constant is determined on \( D=\{(t,y);0\leq t \leq 2, y\in \rm I\!R \}. \)
Question 2#
Apply 3rd Order Taylor Method to approximate the solution of the given initial value problems using the indicated number of time steps. Compare the approximate solution with the given exact solution, and compare the actual error with the theoretical local and global error
a. \(y'=t-y, \ \ (0\leq t \leq 4)\) with the initial condition \(y(0)=1\), \(N=4\), with the exact solution \(y(t)=2e^{-t}+t-1.\) The Lipschitz constant is determined on \(D=\{(t,y);0\leq t \leq 4, y\in \rm I\!R \}.\)
b. \(y'=y-t, \ \ (0\leq t \leq 2) \) with the initial condition \(y(0)=2,\) \(N=4\), with the exact solution \(y(t)=e^{t}+t+1\). The Lipschitz constant is determined on \( D=\{(t,y);0\leq t \leq 2, y\in \rm I\!R \}. \)
Question 3#
Apply the 2nd Order Taylor method using \(N=4\) steps to approximate the solution of initial value problem
Question 4#
Apply 2nd Order Taylor Method to approximate the solution of the given initial value problems using the indicated number of time steps.
a. \(y'=0.2y-0.01y^2, \ \ (2000\leq t \leq 2002)\) with the initial condition \(y(2000)=6,\) and \(h=0.5\).
b. \(y'=0.2y-0.01y^2+\sin(2\pi t), \ \ (2000\leq t \leq 2002)\)with the initial condition \(y(2000)=6,\) and \(h=0.5\).
Question 5#
a. Describe in your own words the statement below and its proof: Let \(y(t)\) denote the unique solution of the Initial Value Problem
(108)#\[\begin{equation} y^{'}=f(t,y), \ \ \ a\leq t \leq b, \ \ \ y(a)=\alpha. \end{equation}\]Explain how that if \(y \in C^{n+1}[a,b]\) the general local error for the Taylor Method is:
(109)#\[\begin{equation}\tau_{i+1}(h)=\frac{h^n}{(n+1)!}f^{n}(\xi_i,y(\xi_i)) \approx O(h^n), \end{equation}\]where \(n\) is the order of the method.
b. Describe in your own words the theorem below and its proof: Assume that the Taylor method satisfies the Lipschitz condition with constant L on \(D=\{(t,y)|a\leq t \leq b, -\infty < y < \infty \}\) and that the local truncation error is of the form:
(110)#\[\begin{equation}\tau_{i+1}(h)=\frac{h^n}{(n+1)!}f^{n}(\xi_i,y(\xi_i)) \approx O(h^n), \end{equation}\]where \(n\) is the order of the method. Let \(y(t)\) denote the unique solution of the Initial Value Problem
(111)#\[\begin{equation} y^{'}=f(t,y), \ \ \ a\leq t \leq b, \ \ \ y(a)=\alpha, \end{equation}\]and \(w_0,w_1,...,w_N\) be the approx generated by the Taylor methods for some positive integer \(N\). Then for \(i=0,1,...,N\)
(112)#\[\begin{equation} |y(t_i)-w_i| \leq \frac{Ch^n}{L}|e^{L(t_i-a)}-1| \end{equation}\]where \(C\) is a constant.
Question 6#
Adapt the Python code for the 2nd Order Taylor method provided to approximate solution of the integrate and fire differential equation:
(113)#\[\begin{equation} \tau_m\frac{dV}{dt} = -(V-E_L) + R_mI(t), \ \ -50\leq t \leq 400, \end{equation}\]where \(E_L = -75\), \(\tau_m = 10\), \(R_m = 10\) and \(I(t)=0.01t\) and the initial condition \(V(-50) = -75\) using a stepsize of \(h=0.5\).