Show Picard iteration diverges

[psie]

Homework Statement

Find the solution of the problem $x'=1+(x-t)^2, x(0)=0$. Then show that there is a number $a>0$ such that the successive approximations diverge when $|t|>a$.

Relevant Equations

The relevant tools are knowing the method of successive approximations, also known as Picard iteration. This is explained on Wikipedia, in the link below.

For an example of a Picard iteration, see here. In this case, we have

\begin{align}
&x_0(t)=x(0)=0,\nonumber\\
&x_1(t)=x_0(t)+\int_0^t \big(1+(x_0(s)-s)\big)^2ds=t+\frac{t^3}{3},\nonumber \\
&x_2(t)=x_0(t)+\int_0^t \big(1+(x_1(s)-s)\big)^2ds=t+\frac{t^7}{3^27},\nonumber\\
&\cdots \nonumber
\end{align}

You can verify by induction that we have, $$x_k(t)=t+\frac{t^{2^{k+1}-1}}{(2^2-1)^{2^{k-1}}(2^3-1)^{2^{k-2}}\cdots(2^{k+1}-1)}.\tag1$$

By inspection, I think the second term goes to $0$ for $|t|\leq 1$. And $x(t)=t$ is indeed a solution to the IVP above. However, what about divergence? Does the numerator grow faster than the denominator for $|t|>1$?

A TA has noted the following. If we denote the second term in $(1)$ by $e_k(t)$, then $$e_k(t)=\frac{(e_{k-1}(t))^2 t}{(2^{k+1}-1)}\geq \frac{(e_{k-1}(t))^2 t}{2^{k+1}}.$$ The TA has suggested that if $e_k(t)$ is to diverge, then it needs to increase by, say, a factor of $2$. So $$\frac{(e_{k-1}(t))^2 t}{2^{k+1}}\geq 2\cdot e_{k-1}(t).$$ However, why would the sequence diverge if it were to increase by a factor of $2$? Moreover I'm unsure how to actually find an $a$ such that $x_k(t)$ diverges for $|t|>a$. Any help is greatly appreciated.

 

[Office_Shredder]

2 is arbitrary. Any number larger than 1 will do, because then the second term grows at least exponentially fast.
If you can find a such that t>a gives youthis exponential growth, then you're done.

 

[psie]

Ok, so the sequence will diverge if $e_{k+1}(t)\geq 2e_k(t)$, because then $$e_{k+1}(t)\geq 2^ke_1(t).\tag2$$ I guess the $t$'s that satisfy this inequality are the ones that satisfy $$\frac{(e_{k-1}(t))^2 t}{2^{k+1}}\geq 2\cdot e_{k-1}(t).\tag3$$ Since $e_{k-1}(t)$ is nonzero for $t\neq 0$, we can divide by $e_{k-1}(t)$ in $(3)$ when $t\neq 0$ ($e_k(t)$ converges at $t=0$ anyway). We can then pick any $k$, so $k=2$ and we get from $(3)$ that, $$\frac{e_{1}(t) t}{2^{3}}=\frac{t^4}{3\cdot 8}\geq 2,$$ which is equivalent to $|t|\geq\sqrt[4]{48}$.