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.
Best Answer
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 $y_n > 1+t+t^2+\dots t^n.$