MATLAB Mathematics - SERC - Index of

Initial Value Problems for ODEs and DAEs
Evaluating the Solution at Specific Points
The numerical methods implemented in the ODE solvers produce a continuous
solution over the interval of integration [ a,
b]
. You can evaluate the
approximate solution, Sx ( ), at any point in [ a,
b]
using the function deval and
the structure sol returned by the solver. For example, if you solve the problem
described in “Example: Solving an IVP ODE (van der Pol Equation, Nonstiff)”
on page 5-9 by calling ode45 with a single output argument sol,
sol = ode45(@vdp1,[0 20],[2; 0]);
ode45 returns the solution as a structure. You can then evaluate the
approximate solution at points in the vector xint = 1:5 as follows:
xint = 1:5;
Sxint = deval(sol,xint)
Sxint =
1.5081 0.3235 -1.8686 -1.7407 -0.8344
-0.7803 -1.8320 -1.0220 0.6260 1.3095
The deval function is vectorized. For a vector xint, the ith column of Sxint
approximates the solution y( xint(i) ).
Solver for Fully Implicit ODEs
The solver ode15i solves fully implicit differential equations of the form
ftyy′ ( , , ) = 0
using the variable order BDF method. The basic syntax for ode15i is
[t,y] = ode15i(odefun,tspan,y0,yp0,options)
The input arguments are
odefun
tspan
A function that evaluates the left side of the differential equation
of the form ft ( , y,
y′ ) = 0 .
A vector specifying the interval of integration, [t0,tf]. To obtain
solutions at specific times (all increasing or all decreasing), use
tspan = [t0,t1,...,tf].
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