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## Implicit Methods and Differential Equations

Implicit Methods::The Implicit Difference EquationsBy approximating u

_{xx}and u_{t}at t_{j+1}rather than t_{j}as well as using a backwards difference for u_{t}, the equation u_{t}= cu_{xx}is approximated by:Note that every the terms have index j + 1 except one and isolating this term leads to:

u

_{i,j}= −ru_{i−1,j+1}+ (1 + 2r)u_{i,j+1}− ru_{i+1,j+1}for 1 ≤ i ≤ m − 1,Where r = ck/h

^{2}as before.At this time we have ujgiven in terms of u

_{j+1}. This would seem similar to a problem until you consider that the relationship is linear. Utilizing matrix notation we have:u

_{j}= Bu_{j+1}− rb_{j+1,}Where bj+1 represents the boundary situation. Therefore to find uj+1 we need only solve the linear system:

Bu

_{j+1}= u_{j}+ rb_{j+1},Where u

_{j}and b_{j+1}are given andUsing this scheme is called the implicit method since u

_{j+1}is defined implicitly.Since we are solving the most significant quantity is the maximum absolute eigenvalue of B

^{−1}which is 1 divided by the smallest ew of B. The maximum absolute ew’s of B^{−1}as a function of r for different size matrices. Notice that this complete maximum is always less than 1. Therefore errors are always diminished over time and thus this method is always stable. For the same reason it is as well always as accurate as the individual steps.Both this implicit method as well as the explicit method in the previous lecture make O(h

^{2}) error in approximating uxxand O(k) error in approximating ut therefore they have total error O(h^{2}+ k). Therefore although the stability condition permits the implicit method to use arbitrarily large k to maintain accuracy we still need k ∼h^{2}.Maximum complete eigenvalue EW as a function of r for the matrix B

^{−1}from the implicit method for the heat equation calculated for matrices B of sizes m = 2 . . . 10. When EW <1 the method is stable that is, it is always stable.:Crank-Nicholson MethodNow that we encompass two different methods for solving parabolic equation it is natural to ask ‘can we get better by taking an average of the two methods?’ The respond is yes.

We execute a weighted average of the two methods by considering an average of the approximations of u

_{xx}at j and j + 1. This direct to the equations:The implicit process contained in these equations is called the Crank-Nicholson method. Gathering terms acquiesce the equations:

−rλu

_{i−1,j+1 }+ (1 + 2rλ)u_{i,j+1}− rλu_{i+1,j+1}= r(1 − λ)u_{i−1,j}+ (1 − 2r(1 − λ))u_{i,j}+ r(1 − λ)u_{i+1,j.}In matrix notation this is:

Bλu

_{j+1}= Aλu_{j}+ rb_{j+1},Where

and

In this equation u

_{j}and b_{j+1}are known, Aujcan be compute directly as well as then the equation is solved for u_{j+1}.If we choose λ = 1/2 then we are in consequence doing a central difference for ut which has error O(k

^{2}). Our total error is then O(h^{2}+ k^{2}). With a bit of work we are able to show that the method is always stable and so we can use k ∼h without a problem.To get best possible accuracy with a weighted average it is always necessary to use the right weights. For the Crank-Nicholson process with a given r we need to choose:

λ =(r − 1/6)/ 2r

This choice will create the method have truncation error of order O(h

^{4}+ k^{2}) which is really excellent considering that the implicit and explicit methods each have truncation errors of order O(h^{2}+ k). Surprisingly we are able to do even better if we also requireand consequently:

With these options the method has truncation error of order O(h

^{6}) which is absolutely amazing. To appreciate the proposition supposes that we need to solve a problem with 4 significant digits. If we utilize the explicit or implicit method alone then we will need h^{2}≈ k ≈ 10−4. If L = 1 and T ≈ 1 then we require m ≈ 100 and n ≈ 10000. Therefore we would have a total of 1000000 grid points approximately all on the interior. This is plenty.Next presume we solve the same problem using the optimal Crank-Nicholson method. We would need h

^{6}≈ 10^{−4}which would need us to take m ≈ 4.64 therefore we would take m = 5 and have h = 1/5. For k we require k = (√5/10)h^{2}/c. If c = 1 this gives k = √5/250 ≈ 0.0089442 so we would need n ≈ 112 to get T ≈ 1. This offers us a total of 560 interior grid points or a factor of 1785 fewer than the explicit or else implicit method alone.Latest technology based Matlab Programming Online Tutoring AssistanceTutors, at the

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