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This is essentially an extension of this question: Differential Equation and Picard Iteration.

I have the following IVP:

$$ \begin{align} y' &= y^2\\ y(0) &=1, \end{align} $$ which is to be solved on $t \in [0, 1[$ using the Picard iteration scheme

$$ \begin{align} y^{(0)}(t) &\equiv 1\\ y^{(n+1)}(t) &= 1 + \int_{0}^{t} (y^{(n)}(\tau))^2 d\tau. \end{align} $$

Using the identity $(1+t + \cdots + t^n)^2 = \sum_{k=0}^{n}(k+1)t^k + \sum_{k=n+1}^{2n}(2n - k + 1)t^k$ and induction one can show that this results in a sequence of functions $$ y^{(n)}(t) = 1 + t + \cdots + t^n + t^{n+1} p_{n}(t), \quad \mbox{ where }p_{n}(t) \mbox{ is some polynomial in }t. $$

Let's do this:

The base case $y^{(0)}(t) \equiv 1$ obviously satisfies the assertion with $p_{0}(t)\equiv0$. Suppose that $y^{(n)}$ satisfies it too, then

$$ \begin{align} y^{(n+1)}(t) &= 1 + \int_{0}^{t} (y^{(n)}(\tau))^2 d\tau\\ &= 1 + \int_{0}^{t} (1 + \tau + \cdots + \tau^n + \tau^{n+1}p_{n}(\tau))^2 d\tau\\ &= 1 + \int_{0}^{t} (1 + \tau + \cdots + \tau^n)^2 + 2 (1 + \tau + \cdots + \tau^n) \tau^{n+1}p_{n}(\tau) + (\tau^{n+1}p_{n}(\tau))^2 d\tau\\ &= 1 + \int_{0}^{t} (1 + \tau + \cdots + \tau^n)^2 + \tau^{n+1} p_{n}(\tau) (2 (1 + \tau + \cdots + \tau^n) + \tau^{n+1}p_{n}(\tau)) d\tau.\\ \end{align} $$

Using the above mentioned identity the left term in the integral can be evaluated as follows: $$ \begin{align} \int_{0}^{t} (1 + \tau + \cdots + \tau^n)^2d\tau &= \int_{0}^{t}\left(\sum_{k=0}^{n}(k+1)\tau^k + \sum_{k=n+1}^{2n}(2n - k + 1)\tau^k \right)d\tau\\ &= \sum_{k=0}^{n}\int_{0}^{t}(k+1)\tau^k d \tau + \sum_{k=n+1}^{2n}\int_{0}^{t}(2n - k + 1)\tau^k d \tau \\ &= \sum_{k=0}^{n} t^{k+1} + \sum_{k=n+1}^{2n}\frac {2n - k + 1}{k+1}t^{k+1} \\ &= \sum_{k=1}^{n+1} t^{k} + t^{n+2} \sum_{k=0}^{n-1}\frac {n - k}{n+2+k}t^{k}. \\ \end{align} $$

Combining this result with the right part of the integral gives $$ \begin{align} y^{(n+1)}(t) &= \sum_{k=0}^{n+1} t^{k} + t^{n+2} p_{n+1}(t), \end{align} $$ where

$$t^{n+2} p_{n+1}(t) = t^{n+2} \sum_{k=0}^{n-1}\frac {n - k}{n+2+k}t^{k} + \int_{0}^{t} \tau^{n+1} p_{n}(\tau) (2 (1 + \tau + \cdots + \tau^n) + \tau^{n+1}p_{n}(\tau)) d\tau.$$

Now I want to show that on $t \in [0, 1[$ the sequence $(y^{(n)})$ converges point wise to the solution $\frac 1 {1 - t}$:

$$ \begin{align} y^{(n)}(t) - \frac 1 {1-t} &= 1 + t + \cdots + t^n - \frac 1 {1-t} + t^{n+1} p_{n}(t)\\ &= \frac{1-t^{n+1}}{1-t} - \frac 1 {1-t}+ t^{n+1} p_{n}(t)\\ &=t^{n+1}(p_{n}(t) - \frac 1 {1-t}), \end{align} $$ so all left to do is to somehow show that $t^{n+1}p_{n}(t) \rightarrow 0 \;\;(n\rightarrow \infty)$, but here is where I am stuck at the moment as I don't know how to treat the recursive integral part in the expression for $p_n(t)$.

Any help would be greatly appreciated.

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  • $\begingroup$ Just argue that each coefficient of your error polynomial is smaller than $1$ by induction and then it follows because: $$t^{n+1}+\dots +t^{2^{n+1}-1} <\dfrac{t^{n+1}}{1-t}\longrightarrow 0. $$ $\endgroup$
    – dezdichado
    Aug 20, 2022 at 21:45

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Actually, we can simply prove the following bound inductively: $$y_n< 1+t+t^2+\dots + t^{2^n-1}.$$ But this simply follows from some basic integration and the stuff you already written in your work: \begin{align} y_{n+1} < 1 + \int_0^t(1+s+s^2+\dots+s^{2^n-1})^2ds &=\\ 1 + \sum_{k=0}^{2^{n}-1}\int_0^t(k+1)s^kds + \sum_{k=2^n}^{2^{n+1}-2}\int_0^t(2^{n+1}-k-1)s^kds &= \\ 1+t+t^2+\dots + t^{2^n} + \sum_{k=2^n+1}^{2^{n+1}-1}\dfrac{2^{n+1}-k-1}{k+1}t^k &< \\ 1+t+t^2+\dots + t^{2^{n+1}-1} \end{align} This establishes the convergence since you already have $y_n > 1+t+t^2+\dots t^n.$

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