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}##.