- #1

psie

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- 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.

\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.