Differential Equation and Picard Iteration

Question

Write out $x_n(t)$ for the following differential equation:

$$\frac{dx}{dt} = f(t,x) = x^2, x(0)=1$$

So using the Picard Iteration : $x_n(t) = x_0 + \int_{0}^t f(s,x_{n-1}(s)) ds$,

we have $x_0(t) = 1 \quad \forall t$,

$x_1= 1+ \int_0^t (1)^2 ds = 1+t$

$x_2 = 1+ \int_0^t (1+s)^2 ds = 1+t+t^2+ \frac{t^3}{3}$

$x_3 = 1+ \int_0^t (1+s+s^2+ \frac{s^3}{3})^2 ds = 1+t+t^2+ \frac{t^3}{3}+ \frac{2}{3}t^4+ \frac{1}{3}t^5+ \frac{1}{9}t^6+ \frac{1}{56}t^7$

I am stuck in giving a general $x_n(t)$ since I could not figure out the pattern.

EDIT:

Yes I do know the solution is $\frac{1}{1-t}$, which is why I wonder there should be a recognizable pattern $1+t+t^2+t^3+...$ or something but yet my calculation seems to lead me to nowhere.

Answer 1

I don't see any simple pattern in the coefficients of $t^k$ in polynomial $x_n$ for $k>n$. But there is such a pattern when $k\le n$: the coefficients are all $1$. That is, $$x_n=1+t+t^2+\dots+t^n +O(t^{n+1})\tag1$$ where $O(t^{n+1})$ stands for powers $n+1$ and higher.

It's not hard to prove (1) by induction. The base case is $x_0=1$. If (1) holds for $n$, then squaring (1) yields $$x_n^2=1+2t+3t^2+\dots+(n+1)t^n+O(t^{n+1}) \tag2$$ which integrates to $$x_{n+1}=1+t+t^2+\dots+t^{n+1} +O(t^{n+2})\tag3$$

At least in the sense of convergence of coefficients, (1) is enough to tell you what happens as $n\to\infty$.

Answer 2

Well, the solution is $x(t)=1/(1-t)$, so you should generate the geometric series. Also, note that coefficients can change from iteration to iteration. So, if you have a $t^3/3$ at one step, that doesn't guarantee a $t^3/3$ at the next.

Incidentally, this procedure can be automated very easily with a computer algebra system. Here's how to do so with Mathematica:

F[t_, x_] := x^2;
T[x_] := 1 + Integrate[F[s, x /. t -> s], {s, 0, t}];
NestList[T, 1, 3]

\begin{align} x_0(t)&=1 \\ x_1(t)&=t+1 \\ x_2(t)&=\frac{t^3}{3}+t^2+t+1 \\ x_3(t)&=\frac{t^7}{63}+\frac{t^6}{9}+\frac{t^5}{3}+\frac{2 t^4}{3}+t^3+t^2+t+1 \\ \end{align}

Note that the coefficient of $t^3$ indeed changes from $1/3$ to $1$ as we move from the second to the third iterate. It stabilizes after that, though.