Maths III - Numerical Methods

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1 Maths III - Numerical Methods Matt Probert matt.probert@york.ac.uk 4 Solving Differential Equations 4.1 Introduction Many physical problems can be expressed as differential s, but solving them is not always easy! Hence it is very useful to know how to solve them numerically. NB An analytical solution is always to be preferred, as it yields much more insight than a single numerical solution. But with a numerical solution, you can do many what if experiments and start to simulate the system, and that too can yield genuinely useful insights. When solving any differential, the boundary conditions are crucial. Whilst analytically we can derive a general solution without any boundary conditions BCs), we can only obtain a particular solution numerically and hence must always know the boundary conditions: Initial value problem: e.g. given y t = 0) find y t) Boundary value problem: e.g. given y x = 0) and y x = L) find y x) In general, the boundary value problems are harder to solve, so we shall focus on initial value problems instead. 4. 1D ODE with initial value BCs The general form of a 1D ordinary differential ODE) is dy = f x, y) 1) dx with given values y x = 0) and you need to find y x). To solve this, we need an iterative solution, y x) y x + h) where h is the step size, and then repeat. Consider the Taylor series: y x + h) = y x) + h dy dx + h d y! dx + h3 d 3 y 3! dx ) so we can use this to solve any 1D ODE. 1

2 4..1 Euler Method The simplest approach to solving a 1D ODE. Just keep first terms of Taylor series: y x + h) y x) + h dy dx + O h ) 3) and we have the dy dx = f x, y) to solve and hence here 4) Fig 1: Euler method in action How effective is this? It is very simple, but awful in practice! Comparison to the Taylor series shows that this method has an error O h ) and hence is called a 1st-order method an n th -order method has an error term O h n+1) ). It is very inefficient, and often unstable unless h is small enough) so you should never use this in practice. However, it is very useful pedagogically as the starting point for many more refined methods Midpoint Method The big problem with the Euler method is that it assumes that the gradient is constant over the step size h. Hence an obvious improvement is to improve the gradient estimate, e.g. [ y x) + y ] x + h) y x + h) y x) + h 5)

3 where y x + h) = f x + h, y x + h)) 6) but y x + h) is the unknown we are trying to find! Hence here NB y x) + h f x, y) is like an Euler trial step to the midpoint of the range x + h / and then correcting the answer to get a better final answer. 7) Fig : Midpoint method in action Introducing a new compact notation: k 1 = k = y x + h) 8) here 3

4 NB This now requires function evaluations per step h but the gain is that this is now a nd-order method as the error is O h 3). This method is known as the midpoint method, a.k.a. the nd-order Runge-Kutta method RK4 Obviously, we can continue to improve the order of the method by generating better approximations. It is often found that the most useful compromise of accuracy and computational cost is known as the 4th order Runge-Kutta method a.k.a. RK4): k 1 = f x, y) h k = f x + h, y + k ) 1 h k 3 = f x + h, y + k ) h k 4 = f x + h, y + k 3 ) h y x + h) 9) here Fig 3: RK4 method in action 4

5 RK4 is the most widely used method for solving differential s. It can sometimes be beaten by specialist method for particular types of but it is a good all rounder and has some useful properties: 1. It is self-starting. Some methods use a history of previous values to extrapolate towards the solution. RK4 does not need any previous points, whereas history -based methods need something else to start them off!. It is not limited to smooth functions. As it does not have a history it can handle functions with discontinuities, etc. 3. It can be used with an adaptive step size. During a calculation you can use a large step size h where the function is smooth and slowly changing, and then reduce h if the function starts to rapidly change, etc. 4.3 Error Can we estimate the accuracy of our answer? Yes! A simple procedure is to repeat the calculation with a successively smaller step size h until there is no change between runs, whereupon you have converged the solution with respect to h. But this might get expensive and/or tedious - so an adaptive step size approach is even better! Adaptive Step Size and RK4 Consider the basic y x + h) y x) O h 5) 10) so if we call the unknown constant of the neglected term α then we have y true x + h) y RK4 x + h) + αh 5 + O h 6) 11) But if we were to use steps of size h / instead we would have y true x + h) y RK4 x + h ) ) 5 h + α + O h 6) 1) as long as α is constant across the step size h. So now if we combine these two s together: y RK4 x + h ) y RK4 x + h) = y true x + h) α ) ) 5 h y true x + h) αh 5) = αh5 + O h 6) 13) we can see how comparing the answer between two runs with different h gives the local error estimate as a guide to the accuracy of the calculation. We 5

6 could use this to produce a 5th order estimate for the answer, but that is not so useful. Much more useful is to use this local error estimate to control the step size even if we do not know the value of α! Define the local error estimate: 14) here then if the target accuracy for a calculation is 0 with corresponding step size h 0 then ) 5 h = 15) 0 h 0 and so the target step size h 0 to achieve the desired accuracy is given by here Hence the procedure is 1. Evaluate y x + h) with a step size of h and also steps of size h / and hence estimate. If > 0 we need to reduce h so repeat with new predicted step size h 0 3. If < 0 then keep the result with step size h and go to new predicted step size h 0 for next step. 4.4 Higher Order Equations Q: What should we do with a nd order ODE? E.g. 16) y + q x) y = r x) 17) A: Rewrite it as a set of coupled 1st order ODEs, e.g. 18) here 6

7 So as long as we are able to solve a 1st order ODE, we can apply this approach to an arbitrarily complex ODE and re-write it as a set of coupled 1st order ODEs and then solve. Example: 1D time-independent Schrödinger can be written as 1 d ψ x) dx + V x) ψ x) = Eψ x) 19) here 0) 4.5 Boundary Value Problems Finally, how can we handle boundary value problems? e.g. standing wave solutions? One simple way to do this is called the shooting method: start from the value at one fixed edge, and guess the value of the unknown system variable e.g. wavelength for standing wave problem), then integrate the towards the other known point, and see if it hits. If it does, then your guess was OK. If not, you should adjust your guess and then start again. Iterate until you have a solution which is consistent with both boundary values, whereupon you have a correct value for the unknown system variable! Fig 4: Shooting method in action 7

8 4.6 Python hints The primary routine in SciPy for solving initial values ODEs is scipy.integrate.odeintf,y0,x,args=)...) where the function f computes the derivatives, the initial conditions y0 may be multi-dimensional vector), the set of points x at which to solve for y and any extra arguments that need to be passed to the function f are given as a tuple. Recommend Textbook: Python for Scientists by John M Stewart, Cambridge University Press 014) Acknowledgments: Figures from 8