Multistep Methods for IVPs. t 0 < t < T

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1 Multistep Methods for IVPs We are still considering the IVP dy dt = f(t,y) t 0 < t < T y(t 0 ) = y 0 So far we have looked at Euler s method, which was a first order method and Runge Kutta (RK) methods which we could derive with any order desirable (within reason). In RK methods, as well as Euler s method, we use information from the interval [t n, t n+1 ] to predict the solution at t n+1. These are called one-step methods. A natural idea is to use information from previous time intervals also. Methods which use information from previous time intervals are called multistep methods.

2 Of course an obvious problem with this is start up. We only have one starting value Y 0. This is addressed by using a single-step method (such as a RK) to start with in order to generate enough data to start using the desired multistep method. If your multistep method is order r then you should use a starting method which is the same order. As an example, consider the general form of a two-step method, i.e., one which using information from [t n 1, t n ] and [t n, t n+1 ] to predict the value at t n+1. We have [ ] Y n+1 = a 1 Y n +a 2 Y n 1 + t b 0 f(t n+1, Y n+1 )+b 1 f(t n, Y n )+b 2 f(t n 1, Y n 1 ) For Euler s Method [ ] Y n+1 = Y n + t f(t n,y n ) so we have a 1 = 1, a 2 = 0, b 0 = b 2 = 0, b 1 = 1. There is a significant difference between a method where b 0 = 0 and one where it is nonzero.

3 The problem when b 0 0 is that the term f(t n+1, Y n+1 ) is unknown since we do not have Y n+1. In this case the method is called implicit and we will have to iterate to solve it. (More on this later). When b 0 = 0 then all terms on the right hand side are known (once we determine the coefficients) and the method is called explicit because the solution Y n+1 is given explicitly in terms of known quantities. (Recall from calculus the phrase implicit differentiation where we don t have y = f(x) but rather y is given implicitly.) We will look at some common types of explicit and implicit multistep methods. Then we will see how to combine them into a useful type of method called predictor/corrector where we predict with an explicit method and improve or correct this solution with an implicit method. We first look at explicit methods of which the most common group are Adams-Bashford methods.

4 Adams-Bashford (Explicit) Multistep Methods Recall that the general form of a two-step explicit method (b 0 = 0) is [ ] Y n+1 = a 1 Y n + a 2 Y n 1 + t b 1 f(t n, Y n ) + b 2 f(t n 1,Y n 1 ) In the Adams-Bashford methods we take a 1 = 1, a i = 0 for i > 2 so that in general we have methods of the form [ ] Y n+1 = Y n + t b 1 f(t n, Y n ) + b 2 f(t n 1, Y n 1 ) + In order to start a two-step method we need two values. We only have one, Y 0 = y 0. Consequently we must use a one-step method (usually a RK) to obtain Y 1 before we can implement the two-step method. If we have a three-step method then we must perform two steps of a single step method such as a RK method. How do we choose the coefficients for the method? Just as before, we want to choose them so that we can get as high an accuracy as possible. A common way to derive them is to integrate the differential equation and replace the

5 integrand on the right hand side with a polynomial that interpolates f(t, y) at selected points (e.g., t n, t n 1, etc). For example from the fundamental theorem of calculus we have y (t) = f(t, y) = t n+1 t n y dt = t n+1 t n f(t, y)dt = y(t n+1 ) y(t n ) = t n+1 t n f(t, y)dt We will actually not derive a method here but this is the procedure. The two-step Adams Bashforth method is given below where we replace f(t, y) with its linear interpolant on [t n 1, t n ] and then integrate from t n to t n+1

6 Adams-Bashforth 2-step method Given Y 0, Y 1, for n = 1, 2,... Y n+1 = Y n + t 2 Note that here we have the coefficients [ 3f(t n, Y n ) f(t n 1,Y n 1 ) a 1 = 1, a 2 = 0, b 1 = 3 2, b 2 = 1 2 It is derived by replacing the integrand above by the linear interpolating polynomial which interpolates (t n 1,Y n 1 ) and (t n, Y n ) and then integrating. One can show that this is a second order method. The reason it is second order is that when we replace f(t, y) with its linear interpolant we have an error of O( t) but when we integrate this over [t n,t n+1 ] we pick up another factor of t. Since this is a second order method, we should use a second order RK to get Y 1. If we use Euler s method, then it is just O( t) and this could contaminate our solution. ]

7 Higher order Adams-Bashforth Methods In this family, there are also 3-, 4- and higher step methods although the higher step methods are not typically used because the coefficients become large and there are some stability issues. Typically as we add one more step, we pick up another order of accuracy. For a three step method we need to generate Y 1 and Y 2 from a single step method. The three step method given below is third order. Adams-Bashforth 3-step method Given Y 0, Y 1 Y 2, for n = 2, 3,... Y n+1 = Y n + + t [ ] 23f(t n,y n ) 16f(t n 1, Y n 1 ) + 5f(t n 2,Y n 2 ) 12

8 Analogous to the two-step method, we should use a third order RK to generate Y 1, Y 2 to start the method to preserve our accuracy. If we rewrite the Adams-Bashforth formulas in the following way then we can make a table of their coefficients: Y n+1 = Y n + tc p [ b1 f(t n, Y n ) + b 2 f(t n 1, Y n 1 ) + b 3 f(t n 2, Y n 2 ) + the coefficients for schemes up to order 5 are given in the table below.

9 order c p b 1 b 2 b 3 b 4 b An advantage of explicit Adams-Bashforth methods is that they can be implemented such that, after the initial values are chosen, only one function evaluation per step need to be done since we can store the previous function evaluations. Recall in say a third order RK method we have to perform three function evaluations.

10 Implementing an Explicit Adams-Bashforth As we have seen, there are various strategies for implementing methods. We said that one of the advantages to multistep methods is that they require less function evaluations than equivalent order RK methods. To maintain this advantage, we need to store the function values. For example if we have a third order Adams-Bashforth method which uses f evaluated at (t n,y n ), (t n 1, Y n 1 ), and (t n 2,Y n 2 ) then we need to store these values instead of recomputing them. We will store them in a one-dimensional array dimensioned by the number of previous solutions we are using, i.e., the order. In writing this code we will take the approach of writing one function routine where we pass the same variables as in RK plus the number of function values (i.e., the order) of the method. If we enter the coefficients from our table into a one-dimensional array of

11 length, the number of points we are using, then to calculate the solution at Y n+1 all we have to do is take the dot product of the coefficient array with the function values, multiply by t and the appropriate constant and add to Y n. For example for a third order method we will store the coefficients as (23, 16, 5) and dot this into the array holding (f n, f n 1,f n 2 ) where we have used the short hand notation f n = f(t n,y n ). Then Y n+1 = Y n + t 12 [ (23, 16, 5) (f n, f n 1,f n 2 ) This way in the loop to calculate each Y n+1 we can have a statement which works for all order methods. For example, ynew = yold + dt * c p * dot product( coefficients, rhs values) ] You have to be careful that the order of the coefficients match with the order of the function evaluations. Here I have taken the approach of putting f n first (since it is always used) and so I must put b 1 in the first entry of our coefficient array.

12 The other thing that makes multistep methods different from RK methods is that if the order is greater than one (i.e., using more than just f at t n ), then we need to get some starting values using, e.g., RK methods. The way I coded the method was before the loop over the number of steps, I used the case construct to set the coefficients for the problem (both my 1-d array and c p in our table) and get any starting values by calling the appropriate RK method. For example, for a third order method to calculate the solution at Y n+1 we need information from t n, t n 1, t n 2 so we need to calculate two initial initial values Y 1, Y 2 since we are given Y 0. The way we set up the RK I do two calls to the third order RK (in this case) since we may want to write off the information at each time step in our routine. For the third order method we could have the following case (3)! third order multistep AB method rhs values (3) = rhs( t, y init) yold = rk heun( y init, one step, t, t+dt )! get Y 1 t= t + dt

13 rhs values (2) = rhs( t, yold) ynew = rk heun( yold, one step, t, t+dt )! get Y 2 t= t + dt rhs values (1) = rhs (t, ynew ) yold = ynew coefficients = (/ 23.0 prec, prec, five /) c p = one/ 12.0 prec where t is initially set to t init and one step=1. So this initial part of the function is the only section that is dependent on which AB method we are using. What else do we need? To get ready for the next step we always set yold = ynew but we also have to switch the function values that we are storing. For example, after we calculate Y 3 we need to move the function values as follows:

14 rhs values(3) =rhs values(2), rhs values(2)=rhs values(1), rhs values(1) = rhs(t,ynew) This can be achieved by the following loop which is executed in a separate subroutine switch rhs do k = multistep, 2, -1 rhs values(k) = rhs values(k - 1 ) end do rhs values(1) = rhs ynew where multistep is the number of previous function values we are using (and also the order). To actually compute each of the steps after the starting values are computed, we just have do k = multistep, n steps t = t + dt ynew = yold + dt * c p * dot product( coefficients, rhs value rhs ynew = rhs( t, ynew ) call switch rhs values (rhs values, rhs ynew, multistep)

15 yold = ynew end do An easy mistake to make here is to set your loop from k=1,n steps but remember that we have already computed some solutions for the starting values. Try it out. I have included this routine in an updated version of ivp solvers.f90. Download and compile. Run the code for the second order and third order Adams Bashforth method and verify that you get the correct rate of convergence. Add the capability to do a fourth order Adams Bashforth method.

16 Adams-Moulton (Implicit) Multistep Methods Recall that in implicit methods we have the unknown Y n+1 on the right side as well as the left side of the equation. The simplest implicit method is actually Backward Euler which we initially discarded for our IVP. In this case we have Y n+1 = Y n + tf(t n+1,y n+1 ) The reason we discarded it is that the right hand side is a function of our unknown. Although we can t explicitly solve for Y n+1 we could treat it as a nonlinear equation and iterate to approximate the solution. We will see an example below. To derive Adams-Moulton methods one again starts with y (t) = f(t, y) = y(t n+1 ) y(t n ) = t n+1 t n f(t, y)dt except now the interpolating polynomial that replaces the integrand f(t, y)

17 on the right hand side is shifted one point to the right in time to include t n+1 and the solution there. The Adams-Moulton formulas have the general form Y n+1 = Y n + t [ b 0 f(t n+1,y n+1 ) + b 1 f(t n, Y n ) c p where b 0 0. ] +b 2 f(t n 1,Y n 1 ) + b 3 f(t n 2,Y n 2 ) + Note that we have our unknown Y n+1 occurring on both the right and left sides of the equation. For our example where f(t, y) = t + y this occurrence is linear and so it is easy to solve. However, the occurrence of Y n+1 on the right side can be nonlinear. For example if f(t, y) = e y then on the right side we have e Y n+1 yielding a nonlinear equation in Y n+1.

18 The coefficients for schemes up to order 5 are given in the table below. order c p b 0 b 1 b 2 b 3 b For example the first order scheme is Backward Euler Y n+1 = Y n + tf(t n+1,y n+1 ) and the second order scheme is Y n+1 = Y n + t [ ] f(t n+1,y n+1 ) + f(t n, Y n ) 2

19 Notice that for the first and second order schemes we do not need any more starting values than Y 0. However, for higher order schemes we do. How do we solve such equations since the unknowns are on both sides? As an example, lets look at the problem y (t) = t + y 5 y(0) = 2 If we use the second order scheme then Y 1 = Y 0 + t [ ] f(t 1,Y 1 ) +f(t 0,Y 0 ) = 2 + t [ ( t + (Y 1 ) 5) + ( )] 2 2 Y 1 t 2 (Y 1 ) 5 = 2 + t ( t + 4) 2 This is a nonlinear equation for Y 1 which must be solved by an iterative method such as Newton s method. For example, if t = 0.2 and w = Y 1 then our equation is just This is a nonlinear equation w 0.1w 5 = 2 +.1(4.2) = = 2.42 g(w) = 0 for g(w) = w.1w

20 Newton s method for this problem is just given w 0 find w k+1 for k = 0, 1, 2,... from w k+1 = w k g(wk ) g (w k ) = wk+1 = w k wk.1(w k ) (w k ) 4 Recall that for nonlinear methods you need a starting guess. What would you recommend for an initial guess for this nonlinear problem? We will not dwell here on implementing implicit methods because we have another use for implicit methods - in our predictor/corrector schemes. We will see that our implementation of implicit methods in a predictor/corrector setting does not require the solution of a nonlinear equation.

21 Predictor-Corrector Methods A very useful way to solve an IVP is to combine an explicit Adams-Bashforth and an implicit Adams-Moulton of the same order. We will see that this will alleviate the need to iterate to solve the implicit method. How does it work? The first step is to predict a value at time t n+1 with an explicit method. The next step is to improve or correct this solution at time t n+1 with an implicit method. However, on the right hand side of the implicit method we use our predicted solution everywhere there is a Y n+1 which turns our implicit method into an explicit method! As an example, consider a predictor/corrector scheme where we predict with an explicit second order and correct with an implicit second order method. We will add a subscript p on the predicted solution for clarity.

22 We predict the solution at t n+1 with Yp n+1 = Y n + t [ ] 3f(t n,y n ) f(t n 1, Y n 1 ) 2 and correct the solution to get Y n+1 = Y n + t [ ] f(t n+1,yp n+1 ) + f(t n, Y n ) 2 Note that now everything on the right hand side of the corrector scheme is known so our implicit scheme becomes explicit. Note that since we are using multistep methods we still need to get our starting values from single step methods.

23 Stability Some ODEs themselves are not stable. This means that small changes in the initial data produce large changes in the solution. However, we can approximate a stable ODE and our numerical results become unstable! Consider the ODE y (t) = y = y(t) = Ce t As t clearly y 0 no matter what value we choose for C (i.e., the initial condition) so the solution itself is stable. Our numerical solutions may be unstable though if we choose too large of a time step. Below are some results for Euler s method using different uniform time steps, i.e., different t. In each plot we start with different initial conditions. In the first plot we see a time accurate solution but computationally expen-

24 sive delta t In the second plot we see a solution which is not very accurate but computationally inexpensive. 2 delta t In the third plot we see a solution which is not very accurate at all.

25 10 delta t In the fourth plot we see an oscillating solution as we find the final time at T = delta t

26 Clearly we have a numerical instability which is occurring as we take a larger value of t. When you take a numerical analysis class you will see that you can derive stability estimates for various schemes. These typically say that a method is stable if t < ρ for some positive constant ρ. We will not go into this here. Consequently, if you get an oscillating solution or one which blows up in time then you should rerun your calculation with a smaller t. In general, explicit methods have strict stability ranges whereas implicit methods allow much larger time steps. This is true in general for PDEs too.

27 Variable Time Steps In the previous work we have just assumed a uniform time step of length t. Of course if we are using single step or multistep methods, we can easily incorporate a variable time step but how do we choose the time step? Suppose you had an ODE whose solution looks like the plot below. Clearly the solution changes quickly to begin with and then the solution

28 becomes quite boring and changes very little. In order to capture the behavior we need to take a small time step to begin with. But if we use a uniform t then this will make us take too many steps in the part of the solution where nothing much is happening. To avoid this problem, one can use an adaptive time step which just means that we adapt our time step to what is happening in the solution. Of course to implement such a strategy we would need to know when the solution curve is changing a lot. This is usually accomplished by an error estimation. The predictor/corrector setting is the perfect setting to implement this adaptive time step. Suppose we predict our solution Yp n+1 and then we correct it to get Yc n+1. If the difference Yp n+1 Yc n+1 is small then we probably took an appropriate time step. If this difference is really small we can probably increase t and if it is large then we need to reject this step and decrease t. Of course we need to quantify the terms small, etc. but once we do this

29 we could simply add conditionals to test to see when we increase or decrease the time step. We will not do this here but it should be fairly straightforward to implement. This is one of the reasons that predictor/corrector schemes are so popular - it is easy to implement an adaptive time step.

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