Show Picard iteration diverges

  • #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.
 
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  • #2
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.
 
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  • #3
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}##.
 

FAQ: Show Picard iteration diverges

What is Picard iteration?

Picard iteration is a method used to find successive approximations to the solutions of differential equations. It is an iterative process that starts with an initial guess and generates a sequence of functions that ideally converge to the actual solution.

Under what conditions does Picard iteration converge?

Picard iteration typically converges if the function involved is Lipschitz continuous with respect to its second argument. This means there exists a constant L such that the absolute difference between the function values is bounded by L times the absolute difference between the arguments. Additionally, the initial guess must be reasonably close to the actual solution.

How can we show that Picard iteration diverges?

To show that Picard iteration diverges, one can demonstrate that the sequence of approximations does not converge to a finite limit. This can be done by showing that the function fails to meet the Lipschitz condition or by providing a counterexample where the sequence of iterates grows without bound or oscillates indefinitely.

Can you provide an example where Picard iteration diverges?

Yes, consider the differential equation dy/dt = y^2 with the initial condition y(0) = 1. The exact solution is y(t) = 1/(1-t), which becomes unbounded as t approaches 1. If Picard iteration is applied to this problem, the iterates will not converge for t ≥ 1, demonstrating divergence.

What are the implications of Picard iteration diverging?

If Picard iteration diverges, it implies that the method is not suitable for solving the given differential equation under the current conditions. This necessitates either modifying the initial guess, altering the method, or using a different approach altogether to find the solution.

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