12.864 Inference from Data and Models 18 February 2003 Problem Set No. 1 Due: 26 February 2003 1. You have a dierential equation g2{ (w) 2gw2 n { (w)=0> 0 w 1> subject to the initial conditions, g{ (0){ (0) = 0> =1= gw (a) Solv e it analytically and explicitly, with n =1= (b) You are no w given also a terminal condition { (1) = d= For what values of d is there a solution? (c) Now suppose that both initial and final conditions are missing, but you have instead { (1@2) = 0=521; what are the possible values for the two initial conditions that are consistent with this value (ignore the terminal condition). 2. Now consider the same problem as above, but write the dierential equation in a discrete form having at least rough accuracy. (a) Treating the problem as one of solving a set of simultaneous equations, solve the conventional forward problem with two initial conditions { (0) = 0>{ (w)= w= Comment on its resemblance to the solution to (1a). (b) Let d =1=2> in 1(b). Solve the problem using ordinary least-squares. Solve it insisting, to high accuracy, that { (1) = 1=2= What is the dierence? (c) With { (1@2) = 0=521> find a minimum norm solution for the initial conditions, with the final condition unknown. (Choose a discretization grid so that w =1@2 is a grid poin t. (d) Now suppose the end time is infinitely far into the future and let { (0) = 1= Can you find a value of { (w) so that the solution is bounded for all positive times? Is the result sensitive to the exact value? Hint: One way to proceed is to solve the dierence equation represented by the finite dierence form. 3. Consider the 3 v ectors: e1 =[=9863> =5186> 0=3274]W > e2 =[0=2341> 0=0215> 1=0039]W > e3 =[ 0=9471> 0=3744> 1=1859]W Are these a spanning set (basis)? Explain. How about, e1 =[=9863> =5186> 0=3274]W > e2 =[0=2341> 0=0215> 1=0039]W > e3 =[1=1034> 0=5079> 0=1746]W ? 1Solution: nw + F2h3nw1. F1hF1 + F2 =0 (1) F1 F2 =1 (2) £ ,Solutionis: F1 = 211 >F2 = 2 ¤ 1 h1@2{ =1@2exp (w) 1@2exp( w)= { = 1 h11 h31 =1= 175 2> 21 h31@2 : 0=521 10 2 2 2 2. w = 3. First set is independent. Second set has e3 = e1+1@2e2 = Can find eigenvalues of corresponding matrices, or try expanding e3 = de1 + ee2 (3) and compare to correct value.
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