Re: Result to DEQ with WA versus Step-by-Step Yields

*To*: mathgroup at smc.vnet.net*Subject*: [mg132293] Re: Result to DEQ with WA versus Step-by-Step Yields*From*: Itai Seggev <itais at wolfram.com>*Date*: Sat, 1 Feb 2014 23:55:16 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <20140201055416.D6FE16A13@smc.vnet.net>

On Sat, Feb 01, 2014 at 12:54:16AM -0500, amzoti wrote: > When you solve this DEW using WA, you get a result. > > However, when you click step-by-step, the result is different. > > Is this a bug? > > v'' + 10 v' + 125 v = 250 unitstep(t), v(0) = 0, v'(0) = 25 Yes and no, but I think mostly no in this case. The initial answer given by W|A is computed algorithmically, for example by DSolve. In contrast, the step-by-step answer is computed in manner which attempts to simulate how a person might do it. There's no guarantee that these will be produce identical forms for the answer. If they forms are different but the answers agreen, then that's certainly not a bug. Now, let's compare the two answers (v is the step-by-step answer, w the algorithmic answer): In[29]:= FullSimplify[w[t]-v[t]] Out[29]= \[Piecewise] -2+(2 Cos[10 t]+Sin[10 t])/E^(5 t) t<0 0 True So they are the same for t>=0, but not for negative times. Now, looking at the step-by-step solution, it uses the Laplace transform method--a very reasonable choice, because the discontinuity in the forcing function limits the number of methods which will work. (Of methods typically covered in an introductory course, this is probably the only one which will work). However, a Laplace transform is only defined for t>=0. Which likely means that both answers are right for t>=0, but that only w is correct for negative t. Indeed, if put the v back into the original equation we find: In[39]:= (v^\[Prime]\[Prime])[t] + 10 Derivative[1][v][t] + 125 v[t] == 250 UnitStep[t] // Simplify Out[39]= t >= 0 So v only obeys the ODE for t>=0. On the other hand, when is w plugged into the DE you see: In[40]:= (w^\[Prime]\[Prime])[t] + 10 Derivative[1][w][t] + 125 w[t] == 250 UnitStep[t] // Simplify Out[40]= Indeterminate == 0 || t != 0 The Indeterminate == 0 comes from the fact that Mathematica can't compute the second derivative at t == 0 as w only has 1 continuous derivative. (Indeed, since the forcing function is discontinuous you're not actually guaranteed unique solutions, but this is the sort of system which typically does have them.) But for all other times, w is the correct solution to the ODE. So the discrepency comes from choice of method in step-by-step, except under the circumstances I think it's an entirely reasonable choice. If you disagree, you can certainly provide feedback (the plus sign in the bottom right corner of the == output), which will be forwarded to the relevant developers. -- Itai Seggev Mathematica Algorithms R&D 217-398-0700