Appendix A2. Picard Iteration
The following first order initial value problem
y'(t) = f ( t , y(t) ) , y(0) = 0
can be reformulated as the an integral equation
.
See Ledder, Appendix A.2.
Assume that f (t, y) and its y partial are continuous on a rectangle containing the point (0,0). Then starting with an initial "guess" for a solution
,
the recursive relation
generates a sequence of functions converging to the solution. These functions are called the Picard iterates generated by g(t).
Maple can generate some of these iterates, use a for..do loop.
The following example is taken from Ledder, see Chapter A.2, Exercise 2.
Example Obtain the first three Picard iterates for the following IVP, generated by the function g(t) = 0.
,
Sketch their graphs and the graph of the exact solution.
Start with the definition of the function f.
> | f := (t,y) -> t + t^2*y; |
Use it to define the diffenential equation.
> | DE := diff(y(t),t) = f(t,y(t)); |
Compute the first three iterates.
> | Y[0] := t -> 0:
for n from 1 to 3 do int(f(s,Y[n-1](s)),s=0..t): Y[n] := unapply(%,t): end do: unassign('n'); |
The formulas for the initial guess and the three iterates are displayed below.
> | for n from 0 to 3 do y[n](t) = Y[n](t) end do; unassign('n'); |
The actual solution formula is obtained next. It is not pretty.
> | soln := dsolve( {DE, y(0)=0} ); |
The plot of the solution, red and thick, and the three iterates, blue, green, black:
> | plot( [rhs(soln),Y[n](t)$n=1..3], t=0..2, y=0..5,
color=[red,blue,green,black], thickness=[2,1$3]); |
The next output reveals that the Picard iterates are the partial sums to the Taylor series for the solution.
> | dsolve( {DE, y(0)=0}, y(t), type=series, order=9); |
Ask the FunctionAdvisor about the Whittaker functions.
> | FunctionAdvisor(Whittaker); |
The 2 functions in the "Whittaker" class are:
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