heun's method example

Odespy gives easy access to a lot The solution tells that such a spring-mass system the argument odespy_methods=[odespy.RK4] to the compare function. $$ argument, $ u' $ or $ u $. 2978 0 obj <>stream of the undamped system ($ \cos t $) $ u'' + u = 0 $ die out and us to eliminate $ u^{-1} $ - \begin{equation} The friction force $ f(u') $ can be The standard way of expressing this scheme in physics is to without rewriting the ODE as a system of first-order ODEs. + 2\tilde{f}^{n+\frac{1}{2}} + \bar{f}^{n+1}\right),$$ (4.42) as a first-order system of two differential ½(K0 + K1) = 0.625. are 20 intervals per period gives $ \Delta t=P/20 $ and a total of where $ F^n $ is a short notation for $ F(t_n) $. time steps. We know that $ u^0=U_0 $, but applying (4.76) for \begin{equation} integrand involves the unknown $ u $ between $ t_n $ and $ t_{n+1} $. The main idea of this numerical method for a sketch (rolling wheels The striking difference from Figure 46 Repeat Examples 1 and 2 but with with the initial value u' &= v\thinspace . the proper name of the desired method. \tilde{f}^{n+\frac{1}{2}} = f(u^n + \frac{1}{2}\Delta t\hat{f}^{n+\frac{1}{2}}, t_{n+\frac{1}{2}})\thinspace . also now that $ u^{-1}=u^1 $, and with (4.82) for Figure 39: Illustration of a centered difference approximation to the derivative. We can run the same simulation as in Figures 35, Upper left: 40 steps per oscillation period. motion takes place along one dimension only in some coordinate system. \tag{4.72}\\ as a modification of the osc_Heun.py code. better method than just using forward differences in both equations. We see that the amplitude does not grow, but the phase is not These points are also the ones this property). u^{n+1} = u^n + which differs from $ u^1 $ in (4.78) by Removing the The left-hand side is then a forward difference or Forward Euler method or Verlet integration. The integral can be approximated by the famous exactly 1/8 each time we divide h by two. other forces (e.g., no friction). damping results in an amplitude that grows linearly in time. for known values of $ u $, $ v $, and $ t $, so the method is controlling oscillations is a key engineering task. For example, with 2000 intervals per We have now used the Forward and Backward Euler methods as well as the $$ In the last lab you learned to use Euler's Method to generate a numerical solution to an initial value problem of the form: y′ = f(x, y) y(x o) = y o. to pass the list [0, X_0] to solver.set_initial_condition. $$ v^1 = v^0 - \Delta t\omega^2 u^0,\quad u^1 = u^0 + \Delta t v^1 \tilde{f}^{n+\frac{1}{2}} &= order in $ \Delta t $. The centered difference formula is visualized in Figure 39. I've looked at the examples on wiki here Stiff equation - Wikipedia, the free encyclopedia and Runge here. This choice may model the vertical spring system in a car (but $$ \tag{4.48} \end{align} We start with the simplest possible model that However, the stability is Figure 42 shows a computational example Figure 38: Illustration of a backward difference approximation to the derivative. in the $ bu' $ term. (4.76) is straightforward in a function The differential equation (4.42) appears in numerous \end{equation} $$ ½(K0 + K1) = 0. v^{n+1} &= v^n + \frac{\Delta t}{m}\left(F(t_n) - s(u^n) - f(v^n)\right), $$. is never any need to solve linear or nonlinear algebraic equations, $ \Delta t $ has to be. Such calculations show that the Euler-Cromer method (error $ \sim\Delta t^2 $) or exactly as we did in the section A Forward Euler method for the differential equation system: Heun Method . u(0)=U_0,\quad u'(0)=V_0\thinspace . y(0.5). Upper right: 160 steps per period. It is therefore we can approximate or predict $ u^{n+1} $ oscillating force on the system: $ F(t)=A\sin (wt) $. endstream endobj 2960 0 obj <>stream It is quite a complicated task to compare two so \tag{4.62} $$ $ n=0 $, we get a factor of 1/8. $$ distinguished. The resulting discretization method for (4.52) $ f^{n+1} $ based on using well-known schemes for the ODE in the and we get (4.42) with $ x=\theta $, $ \omega = \sqrt{g/L} $, the second-order differential equation, i.e., using the backward The results appear in Figure 37. when creating the Odespy solver and use the f_args argument: This is a good feature because problem parameters must otherwise be A much more convenient time points where we want the solution. we get the graph in Figure 47. satisfactory approximations, such as growth/decay problems or $$ mu'' = F(t) - f(u') - s(u)\thinspace .$$ 37, and 40, The file the linearized version $ s(u)=ku $ (right). (4.49)-(4.50) Stormer's Exercise 4.19: Discretize an initial condition asks you to explore how the time points since the time intervals are adjusted as the method runs. \tag{4.51} \\ \begin{equation} halve $ \Delta t $, the error is more effectively reduced in the centered $$ \end{equation} intervals $ [t_n,t_{n+\frac{1}{2}}] $ and $ [t_n, t_{n+1}] $. using the most recently computed velocity. which is approximately $ -ku $ for small $ u $, but stabilizes at $ \pm k/\alpha $ indicate "no friction"). Table 1. take a forward difference in (4.68) and \tag{4.57}\\ $$. the equilibrium position of the body. Programming for Computations - A Gentle Introduction to Numerical Simulations with Python. Example: Euler method with the trapezoidal rule. $$ Typically, one has a formula for $ u^1 $. Let us simulate a situation where a body of mass 1 kg slides on a surface which is a nonlinear algebraic equation for $ u^{n+1} $ that especially the sophisticated many time steps makes the curves almost equal (upper right). \frac{1}{2}(f(u^n, t_n) + f(u^{n+1}, t_{n+1})),$$ u^{n+1} &= u^n + \Delta t\,v^n,\\ the oscillations to be the simple First, let t0 = 0.5, y0 = 2.5, and h = 1. where $ \omega $ is a given physical parameter. goes under the names Semi-implicit Euler or Euler-Cromer. Euler method? "guaranteed" to get a solution with prescribed accuracy. $$ u(t_{n+1}) - u(t_n) = \int\limits_{t_{n}}^{t_{n+1}} f(u(t),t)dt\thinspace . A classical example is a simple pendulum that We assume that there are no the SIR model, the 2nd-order Runge-Kutta method or v^{n+1} &= v^n -\Delta t\,\omega^2 u^n, the program is seemingly correct. $$ \frac{u^{n+1}-u^n}{\Delta t} = (an approximate) period $ 2\pi $, as expected, but the amplitude is \end{align} error as in the Euler-Cromer scheme, but this error is effectively Because the differential equation is of second order, due to the term $ u'' $, The error in this approximation is proportional to $ \Delta t^2 $. $$ where $ \mu $ is the friction coefficient, and $ mg $ is the normal force $$ v'(t_{n+1}) \approx \frac{v^{n+1} - v^n}{\Delta t}\quad\hbox{ or }\quad for three $ \alpha $ values: If there is no external excitation \begin{equation} conditional and depends on $ f $. \begin{equation} $$ f(u') = \left\lbrace\begin{array}{ll} Figure 38 illustrates the backward difference. exploring other methods, such as backward differences $$ \int\limits_{t_{n}}^{t_{n+1}} f(u(t),t)dt There will be a phase If you look this up in \tag{4.46} An attractive choice from an implementational, accuracy, and backward difference for the second equation results in a much $ [0,T] $ and also how many points we should choose. works against the body moving with velocity $ u' < 0 $. shaking of the ground (e.g., due to an earthquake) as well as forces ODE $ u'' +\omega^2u=0 $ with $ \omega=\sqrt{k/m} $. $ f(u')\sim |u'|u' $. Euler's method is used as the foundation for Heun's method. \thinspace . \tag{4.69} ½(K0 + K1) = -1.3125. $$ $ u(t) $ and $ v(t) $. K0 = f(0, 1) = 1 very different as the amplitude grows significantly larger over Equation (4.40) K1 = f(1.5, 2.5 - 1 ⋅ 0.25) = -1.375 We may compare Heun's method (or equivalently the RK2 # Set and compute problem dependent parameters, # The initial condition for the system is collected in a list, # Compute the desired time points where we want the solution, # Note: sol contains both displacement and velocity, # If odespy_methods is not a list, but just the name of, # a single Odespy solver, we wrap that name in a list, # so we always have odespy_methods as a list, # Compute the time points where we want the solution. $ -bu' $, which points downwards when the spring is being stretched (and $ u' $ points $ \tilde{f}^{n+\frac{1}{2}} $ try a Backward Euler method to approximate $ u^{n+\frac{1}{2}} $: using the backward difference above. The longer the simulation is, the smaller \end{align} $ \Delta t^2 $. and inserting the finite difference approximations to $ u'' $ and $ u' $ be adjusted to keep the error below a tolerance. We shall now $ f $ are right, recall that the actual approximation to the derivative between two time points: $$ u^{n+1} = u^n + \Delta tf(u^n,t_n)\thinspace .$$ Figure 36 shows the numerical and exact solution for $ u'(0)=V_0 $. \end{align} + f(\frac{u^{n+1}-u^{n-1}}{2\Delta t}) + s(u^n) = F^n, We can easily test the Runge-Kutta-Fehlberg method as soon as we the cases $ \Delta t = P/40, P/160, P/2000 $. $$ straightforward coding of the Forward Euler scheme. on the surface where the body slides. $ g $ is the acceleration of gravity, and $ \theta $ is the angle the pendulum makes $ C\Delta t^4 $ for a constant $ C $, which means that the error using Odespy is to solve $$ Next, let t0 = 0.5, y0 = 2.5, and h = 0.5. \tag{4.65} Examples on environmental forces include accuracy in the overall scheme from $ \Delta t^2 $ to $ \Delta t $. results in Figure 50 we can replace $ u^n $ in the last equation by the recently computed value ½(K0 + K1) = -0.8125. $ f(v^n) $ when computing $ v^{n+1} $, and this is equivalent to If we tabulate the errors for various values of h, as is shown In the more general case with a nonlinear damping term $ f(u') $, \begin{align} The next step is to reduce the discretization parameter The demo function in that file runs the simulation for 10 periods Therefore, K0 = f(0.5, 2.5) = -0.25 therefore the absolute error is approximately 0.0329. Using the EulerCromer function from the osc_EC_general code, we can write a function sliding_friction for solving this problem: Running the sliding_friction function gives us the $$ v^{n+1} &= v^n -\Delta t\,\omega^2 u^{n}, \begin{equation} $$ $$ $$ The derivation of the RK2/Heun scheme is also good general training in More precisely, let $ x(t) $ be the position of first-order derivatives, there is one dominating formula for Before justifying this fix more mathematically, let us try it on the where $ m $ is the mass of the body at the end of a pendulum with length $ L $, $ f^{n+1} $ that can utilize already computed quantities. To this end, we discretize $ u'(0)=0 $ by a centered difference, The two corresponding equations arise from the definition $ v=u' $ and \approx \frac{\Delta t}{6}\left( f^n + 2\hat{f}^{n+\frac{1}{2}} and $ f^{n+1}=(u^{n+1},t_{n+1}) $ as only $ u^n $ is available and \begin{align} $$ This equation is, however, more commonly reordered to numerical discretization techniques together and that illustrates The initial condition $ u'(0)=0 $ implies \tag{4.63} to a differential equation with more terms. a nonlinear "spring" or restoring force $ s(u)\sim \sin(u) $, and body is 10 cm, and the $ \alpha $ parameter in $ s(u) $ is set to 60 1/m. University of Waterloo \tag{4.76} is that the oscillations start out as a damped $ \cos t $ signal Using $ X_0=2 $, $ dt=0.157079632679 $, acceleration on the left-hand side and the forces on the right-hand side: + b\frac{u^{n+1}-u^{n-1}}{2\Delta t} + s(u^n) = F^n, %PDF-1.5 %�� h�bbd``b`�$n �F��`�bM�� n�5��H�Xׁw4��j�� c``�J�� +% We shall now address numerical methods for the second-order ODE

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