At each time step,the solvers for nonstiff problems allocate vectors of length n,where n is the number of equations in the system. Can I solve ODE systems inwhich there are moreequations than unknowns,or vice versa?No.Problem Size, Memory Use, and Computation SpeedQuestionAnswerHow large a problem can Isolve with the ODE suite?The primary constraints are memory and time. The ODEsolvers handle more general problems y′ = f ( t, y ), or problemsthat involve a mass matrix M(t, y) y′ = f(t, y).
This is because y 2 is much smaller than the other componentsand its major change takes place in a relatively short time. Imposes a much smaller absolute error tolerance on y 2 than on the othercomponents.Note that hb1dae:15-48Initial Value Problems for ODEs and DAEs Otherwise, type hb1dae at the command line. Similarly,although consistent initial conditions are obvious, the example uses an–3inconsistent value y 3(0) = 10 to illustrate computation of consistent initialconditions.To run this example from the MATLAB Help browser, click on the examplename. Thesolver must recognize that the problem is a DAE, not an ODE. The problem has the form of My′ = f ( t, y ) withM =1 0 00 1 00 0 0M is obviously singular, but hb1dae does not inform the solver of this. These differential equations satisfy a linearconservation law that is used to reformulate the problem as the DAE4y′1 = – 0.04 y 1 + 10 y 2 y 347 2y′2 = 0.04y 1 – 10 y 2 y 3 – 3 ⋅ 10 y 20 = y1 + y2 + 圓 – 1Obviously these equations do not have a solution for y ( 0 ) with componentsthat do not sum to 1.
In hb1ode, the problem is solved with initial conditions y 1(0) = 1, y 2(0) = 0 ,y 3(0) = 0 to steady state. The Robertson problem coded in hb1ode is a classic testproblem for codes that solve stiff ODEs.15-4715Differential Equations4y′1 = – 0.04 y 1 + 10 y 2 y 37 24y′2 = 0.04y 1 – 10 y 2 y 3 – 3 ⋅ 10 y 27 2y′3 = 3 ⋅ 10 y 2Note The Robertson problem appears as an example in the prolog toLSODI.