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Reading Questions for Boyce and DiPrima, Section 2.8
This is one of the more difficult sections in the text, yet also laden with fruitful ideas. Its content,
in grand scope, is the following:
• For many 1st-order initial value problems
y1 “ f pt, yq, ypt0 q “ y0 ,
(1)
one can know (by Theorem 2.8.1), even without finding it, that a unique solution exists.
• Any solution of the IVP (1) is also ”fixed point” of the function
żt
Gpt, φq :“ y0 `
f ps, φpsqq ds,
t0
in the sense that, if you ”feed” the function φ to G, the output is also φ.
• In those instances in which the IVP (1) is known to have a unique solution (i.e., the conditions
of Theorem 2.8.1 are met), one can start with a simple function φ0 that in no way solves (1),
and generate a sequence of functions (called Picard iterates) φ1 , φ2 , φ3 , . . . iteratively by
setting
żt
φ1 ptq :“ Gpt, φ0 q “ y0 `
f ps, φ0 psqq ds
t0
φ2 ptq :“ Gpt, φ1 q “ y0 `
φ3 ptq :“ Gpt, φ2 q “ y0 `
..
.
φn`1 ptq :“ Gpt, φn q “ y0 `
żt
t0
żt
t0
żt
t0
f ps, φ1 psqq ds
f ps, φ2 psqq ds
f ps, φn psqq ds,
with the result that the limit of this sequence of functions is a fixed point of G (i.e., the solution
of (1).
1. In the text, it is stated that ”it is sufficient to consider only the version of Problem (1) in which
the initial point is at the origin”; i.e., the initial condition is yp0q “ 0. Why should this be so?
2. Suppose, in Problem (1), we take
f pt, yq “ 2t2 p1 ` yq,
with initial condition
yp0q “ 3.
(Note the similarity to Example 1 in the section.) If we start with the function φ0 ptq “ 3, what
is the next Picard iterate φ1 ptq?
3. Identify one item (a concept, a step in an example, a statement, etc.) from this reading
assignment you found difficult or confusing.