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Solve the initial value problem $y'(t)=y(t)$, $y(0)=1$ on the interval $[0,1]$ with a fixpoint iteration of the operator $T: Y\to Y, (Ty)(t):=y_0+\int_0^t f(s,y(s))\, ds$. Begin with $y_0(t)=0$ and give the function series $(y_k)$.

The operator $T$ is supposed to be taken from the proof of the theorem of Picard-Lindelöf.

But how do I do the fixpoint iteration here? What is $f(s,y(s))$?

In the proof of Picard-Lindelöf it is $y'(t)=f(t,y(t))$. Since we want to solve $y'(t)=y(t)$ can we set $f(t,y(t))=y(t)$?

So, I set that all together and start the iteration:

We have $y(0)=1$ and $y_0(t)=0$.

$y_1(t)=y(0)+\int_0^t y_0(s)\, ds=1$

$y_2(t)=y(0)+\int_0^t y_1(s)\, ds=t+1$

$y_3(t)=y(0)+\int_0^t y_2(s)\, ds=\frac{1}{2}t^2+t+1$

$y_4(t)=y(0)+\int_0^t y_3(s)\, ds=\frac{1}{6}t^3+\frac12t^2+t+1$

And so on.

We see, that this indeed gives the sum:

$y_n(t)=\sum_{k=0}^n \frac{t^k}{k!}$

Which would give $e^t$ eventually.

Is this done correctly? How comes the interval $[0,1]$ into account here?

Thanks in advance.

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    $\begingroup$ Yes, this is correct. $\endgroup$
    – Ishan Deo
    May 18, 2019 at 19:11
  • $\begingroup$ And what is with the interval $[0,1]$? This does not seem to be important, which is odd. $\endgroup$
    – Cornman
    May 18, 2019 at 19:15

2 Answers 2

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The interval is quite simply a consequence of following the standard proof of Picard-Lindelöf. As the Lipschitz constant is globally $L=1$, one does not need a restriction in the $y$ direction.

In the next step, the Picard iteration is considered on $C([−ϵ,ϵ])$ where it has a Lipschitz constant as a mapping on a function space of $Lϵ=ϵ$, $$ \bigl|P[y_1](t)-P[y_2](t)\bigr|=\left|\int_0^t(y_1(s)-y_2(s))ds\right| \le|t|\,\|y_1-y_2\|\leϵ\,\|y_1-y_2\| $$ demanding that $ϵ<1$ to be a contraction. Thus there is a solution of the ODE on the domain $[−ϵ,ϵ]$.

This sequence of solutions has a limit in the sense of domain extensions of a solution on $(-1,1)$.

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You've correctly interpreted what you're supposed to do. Interestingly, the Picard–Lindelöf theorem simply states that there exists a unique solution to the IVP in some interval $[-\epsilon,\epsilon],$ where $\epsilon>0.$ Given that (and given no other background on what theorems you've got to reference), I can't say for certain what importance (if any) the interval $[0,1]$ might have, since the solution $t\mapsto e^t$ holds everywhere. It could be there to give us a compact domain, which in some cases helps ensure the existence of fixed points of iterations, if I recall correctly.

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