Ytukyg

I'm working on a problem (self study) to solve the differential equation $$
frac{dx}{dt} = tx^2 + x^3
$$
using successive approximations within a neighborhood interval of $delta < 1/5$. I'm given that the initial condition is $x(0) = 1$.

Toward this end, I've defined the contraction map $$
Phi(f)(t) , triangleq , 1 + int_0^t left{
s (f(s))^2 + (f(s))^3
right} ds.
$$

Let $f_{i+1} = Phi(f_i)$ and $f_0(t) = 1$.

Two things puzzle me:

1. Repeatedly solving the integrals to determine $f_1, f_2, f_3, ldots$, etc. and finding a pattern seems tedious / difficult since the polynomials blow up in degree. Is there an easier route here?




2. How does the given neighborhood radius $delta < 1/5$ figure into the solution? Or is it simply the interval for which the resulting solution is valid?

Contractractivity on a rectangular domain

To apply the fixed-point theorem one needs to fix a domain for the ODE and determine the Lipschitz constant there. In the most simple case the domain is rectangular, $[-δ,δ]times[1-M,1+M]$. The space for the Banach fixed-point theorem is then the set of continuous functions $X={cal C}([-δ,δ]to[1-M,1+M])$.

First one has to guarantee that $Φ$ maps $X$ into itself, that is, $|x(t)-1|<Mimplies |Φ(x)(t)-1|<M$ for $|t|<δ$. Then you can compute the Lipschitz constant $L$ over that rectangle and put as additional condition that the contraction constant for $Φ$ is contracting, that is, $Lδ<1$, possibly by making $δ$ smaller.

The first condition is satisfied if
$$
|Φ(x)(t)-1|leint_0^t |sx(s)^2+x(s)^3|ds<frac{δ^2}2(1+M)^2+δ(1+M)^3overset{!}{le} M.
$$
The last inequality can be transformed as
$$
left(1+fracδ{1+M}right)^2le 1+frac{2M}{(1+M)^4}
iff
δle (1+M)left(sqrt{1+frac{2M}{(1+M)^4}}-1right)
$$
The right side looks like

so that this approach allows to choose $δ=frac18$ with $M=0.5$.

The $x$-Lipschitz constant is the maximum over the $x$-derivative, thus $L=2δ(1+M)+3(1+M)^2=frac38+frac{27}4=7+frac18$ so that actually $Lδ<1$ and the iteration converges on this set.

To reach higher values of $t$ one needs to consider other regions than the rectangular one, or even other integral equations for the iteration, like $1-x(t)^{-1}=int_0^t(s+x(s)),ds$ which will then also need a separate computation of the contraction condition for the fixed-point theorem.

Upper and lower bounds

To get better bounds for the solution, use estimates on the differential equation itself so that the solution of differential inequalities provides the bounds.

For $t>0$ the solution is increasing, so that for $t< δle frac12$ we get $2tle x(t)$ which gives a lower and upper bound for the solution
$$
2tx^2le dot xle 2x^3implies t^2le 1-x(t)^{-1}land 1-x(t)^{-2}le 4t
\~~\
implies frac1{1-t^2}le x(t)le frac1{sqrt{1-4t}}
$$
which shows that the solution will diverge to infinity at some $t^*<1$, but not before $t=frac14<t^*$. Zooming into the numerical solution gives $t^*>0.446$ as close estimate.

For $t<0$ the inverted relations hold as $x(t)>0$ by the uniqueness theorem, as the zero function is also a solution. Thus for $t<0$ $$ frac1{1-t^2}ge x(t)ge frac1{sqrt{1-4t}}$$