solve
Description
Examples
Solve ODE Problem at Specified Times
Create an ode
object to integrate the function dydt = @(t,y) (1/2)*t^2
. Specify the initial value of dydt
as 0
.
F = ode(ODEFcn=@(t,y) (1/2)*t^2,InitialValue=0);
Integrate the ode
object by using the solve
method. Specify a vector of times using linspace
to evaluate the solution at specific time points in the interval [0 5]
.
t = linspace(0,5); S = solve(F,t)
S = ODEResults with properties: Time: [0 0.0505 0.1010 0.1515 0.2020 0.2525 0.3030 0.3535 0.4040 0.4545 0.5051 0.5556 0.6061 0.6566 0.7071 0.7576 0.8081 0.8586 0.9091 0.9596 1.0101 1.0606 1.1111 1.1616 1.2121 1.2626 1.3131 1.3636 1.4141 1.4646 1.5152 1.5657 ... ] (1x100 double) Solution: [0 2.1471e-05 1.7177e-04 5.7972e-04 0.0014 0.0027 0.0046 0.0074 0.0110 0.0157 0.0215 0.0286 0.0371 0.0472 0.0589 0.0725 0.0879 0.1055 0.1252 0.1473 0.1718 0.1988 0.2286 0.2612 0.2968 0.3355 0.3774 0.4226 0.4713 0.5237 ... ] (1x100 double)
Plot the results.
plot(S.Time,S.Solution,"-o")
Integrate the ode
object again, this time specifying time points in the interval [-5 5]
. Even though the initial value of dydt
is applied at F.InitialTime=0
by default, the solve
method can evaluate the solution before the initial time.
t2 = linspace(-5,5);
S2 = solve(F,t2);
plot(S2.Time,S2.Solution,"-o")
Solve ODE over Time Interval
Create an ode
object to integrate this system of equations.
Specify initial values of and .
F = ode(ODEFcn=@(t,y) [y(2); -y(1)],InitialValue=[0 1]);
Integrate the ode
object over the interval [0 2*pi]
by using the solve
method. Plot the results.
S = solve(F,0,2*pi);
plot(S.Time,S.Solution,"-o")
Integrate the ode
object again, this time specifying Refine=6
to generate additional solution points in each time step.
S2 = solve(F,0,2*pi,Refine=6);
plot(S2.Time,S2.Solution,"-o")
Input Arguments
F
— ODE problem to solve
ode
object
ODE problem to solve, specified as an ode
object.
t
— Time points to compute solution
real vector
Time points to compute solution, specified as a vector with real elements. The
solver returns the solution evaluated at the given time points. However, the solver does
not step precisely to each point specified in t
. Instead, the solver
uses its own internal steps to compute the solution, and then evaluates the solution at
the requested points in t
. The solutions produced at the specified
points are of the same order of accuracy as the solutions computed at each internal
step.
The time points you specify in t
are not bound by the value of
the InitialTime
property of the ode
object
F
. You can specify time values before or after the value of
InitialTime
.
The solver uses the values in t
to calculate suitable values for
the F.SolverOptions.InitialStep
and
F.SolverOptions.MaxStep
options:
The specified time points give an indication of the scale for the problem, which can affect the value of
InitialStep
used by the solver. Therefore, specifying the initial and final time points of the interval witht0,tf
can lead to different integration results compared to specifying a vector of times.The solver uses the initial and final time points to calculate the maximum step size
MaxStep
. Therefore, changing the initial or final values int
can cause the solver to use a different step sequence, which might change the solution.
Example: S = solve(F,1:10)
solves the ODE problem represented by
F
at the time points 1:10
.
t0,tf
— Interval of integration
real scalars
Interval of integration, specified as two real scalars that indicate the initial
(t0
) and final (tf
) time points. The solver
returns the solution evaluated at each internal integration step within the specified
time interval.
The time points you specify in t0
and tf
are
not bound by the value of the InitialTime
property of the
ode
object F
. You can specify time values before
or after the value of InitialTime
.
The solver uses the initial and final time points to calculate the maximum step size
F.SolverOptions.MaxStep
. Therefore, changing
t0
or tf
can cause the solver to use a different
step sequence, which might change the solution.
Example: S = solve(F,1,10)
solves the ODE problem represented by
F
in the interval [1 10]
.
N
— Solution refinement factor
scalar integer
Solution refinement factor, specified as a scalar integer. The scalar specifies a factor by which the number of output points should increase in each step.
The default value of N
for most solvers is 1
,
but ode45
uses a default value of 4
, while
ode78
and ode89
use a default value of
8
. These solvers use a larger default value to compensate for their
tendency to take large steps.
If the refinement factor is
1
, then the solver returns solutions only at the end of each step.If the refinement factor is greater than
1
, then the solver subdivides each step intoN
smaller intervals and returns solutions at each point.
The extra values produced by the refinement factor are computed by means of continuous extension formulas. The solvers use these specialized formulas to obtain accurate solutions between computed time steps without significant increase in computation time.
Example: S = solve(F,1,10,Refine=5)
solves the ODE problem
represented by F
in the interval [1 10]
and
returns 5
points per time step.
Output Arguments
S
— Computed solution
ODEResults
object
Computed solution, returned as an ODEResults
object.
Version History
Introduced in R2023b
See Also
ode
| solutionFcn
| ODEResults
| odextend
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