simple domain decomposition, while loop
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Hello,
I need to construct a code that finds solution to a set of equation by domain decomposition. it should be really easy, but I'm just not really an expert in while/for loops, so I need help with this, and I rely appreciate it. I try to explain the problem as best as I can here: so let's say we have 3 equations with 3 unknowns for example a=[2,3,0;4,5,6;0,2,3] and a*x=b with b=[9;2;5] , so I come up with an initial guess for x_2 so we can call it x_0_2, and then MATLAB should run this through a loop/while to do the followings : from the first equation it should solve for x_1_1 which is equal to (9-3*X_0_2)/2 and then solve for x_1_3 by using the third equation as follows x_1_3 = (5 - 2*X_0_2)/3 and then use those two and solve for new x2 which is x_1_2 = (2- 4*x_1_1 - 6*x_1_3)/5 by the second equation, and then give me a matrix with [x_1_1; x_1_2 ; x_1_3] and then it keeps doing this and it will give me [x_2_1; x_2_2 ; x_2_3] until it reaches a limit and I will define that limit in while loop . I think I know how to define that limit, but I need help with writing this process. Thanks
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Roger Stafford
2014-6-7
There is no reason to suppose that such a procedure as you describe will converge properly to a solution. It all depends on the element values in 'a' and 'b'. I hate to see you spend so much time on a method that may not work, when matlab makes available to you a very easy method for solving it using the 'backslash' operator:
x = a\b;
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Roger Stafford
2014-6-7
You don't need indices in this problem unless you wish to store the entire history of convergence. All you meed is an "old set" of three elements and a "new set". At the appropriate point in your loop you copy the new set into the old set and then compute an updated new set from that updated old set using your iteration scheme. When the two sets are sufficiently close, you exit. You should be able to do the rest of this coding yourself.
Don't be surprised if it doesn't converge for other values of the matrix 'a' and vector 'b'.
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