# fixpoint iteration to solve $y'(t)=y(t), y(0)=1$

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.

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

## 2 Answers

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 $$(-1,1)$$.

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.