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I searched this and couldn't find my question on here so here it goes:

This is an example from a text I'm reading and I was hoping someone could shed some light on my misunderstanding.

Let $y'=2x-2\sqrt{y_+}$, where $y_+=\mathrm{max}\{y,0\}$ and $y(0)=0$. Then using the Euler-Cauchy Polygon method, we can show that the sequence of functions do not converge. The text goes on to say that indeed, $y_0=0$, $y_1=x^2$ and in general $y_{2n+1}=x^2$ and $y_{2n}=0$. I'm not sure how they got $y_1=x^2$.

Using this method, the line joining the points $(x_0,y_0)=(0,0)$ and $(x_1,y_1)$ is given the line through $(x_0,y_0)$ with slope $f(x_0,y_0)$. Which would give the slope equal to $0$ in this case so $y_1=0$. Maybe I'm not understanding the method correctly though.

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You mixed up two methods:

  1. the Method Of Successive Approximation, mentioned in the title, also known as the Picard iteration
  2. the Euler-Cauchy polygon method, also known as Euler's method.

The computation with $y_{2n+1}=x^2$ uses Picard iteration. It's quite simple: the initial guess $y_0$ is the constant function equal to the initial value, $0$. Plug this function into the right hand side of equation and integrate the result to get $y_1$: $$y_1(x)=\int_0^x \left(2t-\sqrt{y_0(t)^+}\right)\,dt = x^2$$ Then do it again: $$y_2(x)=\int_0^x \left(2t-\sqrt{y_1(t)^+}\right)\,dt = \int_0^x 0\,dt=0 $$ and the process repeats.


The Euler-Cauchy polygon method, as the name suggests, produces polygonal curves (piecewise linear functions), so it could not give you something like $x^2$. Also, it is not a method of iterative approximation: rather, it gives an approximation $y_h$ for every step size $h>0$. The method has its issues with precision, but when $h$ is small enough, it does give a decent approximation in your example.

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  • $\begingroup$ Oh, I see the difference. Thank you! $\endgroup$ – user23793 Sep 3 '14 at 4:04
  • $\begingroup$ I think the reason I confused the two was because right after it explains the Euler method, it gives this example (non Lipschitz continuous). $\endgroup$ – user23793 Sep 3 '14 at 4:08

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