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How can I show that the Euler method fails to approximate the exact solution $$y(x)=(2x/3)^{3/2}$$ to the IVP $$y'=y^\frac{1}{3}$$ $$y(0)=0$$

Here we have $f(t,y)= y^\frac{1}{3}$, $y_0=0$ and so $f(t_0,y_0)=f(0,0)=0$ and $$y_{n+1}=y_n +h f(t_n,y_n)$$ Thus $$y_1=0 \\ y_2=0 \\ \vdots\\ y_n=0$$

So, I can't understand why it fails. Could you help me?

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Welcome to Math.SE! Please, see here for information about typesetting math on this site. I know it takes a little work, but it makes things much easier to read (and therefore makes it much more likely that you will get help!) – Nick Peterson Jul 20 '13 at 21:55
That initial value problem has more than one solution. The Euler method exactly produces one of the solutions. – Daniel Fischer Jul 20 '13 at 22:02
@ali: was that really the initial condition? If you move it away from the $0$ to be something like $y(0)=0.01$, things are much better. At that point, the derivative of the function DNE. – Amzoti Jul 20 '13 at 22:18

Note that the general solution, $$ y(x) = \left(\frac{2x+C}{3}\right)^{3/2}$$ Can be achieved by integration: $$\int \frac{dy }{y^{1/3}}=\int dx$$ This assumes that $y\neq 0$, which is not necessarily true. You therefore have two solutions at $x=0$, and you can't force the Euler method to "choose" the right one.

The reason you don't have a unique solution at $x=0$ is that $d(y^{1/3})/dx$ does not exist on any open interval containing $x=0$. Choosing an IV point with $x> 0$ will guarantee local uniqueness.

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I am still confused. the approximation will be equal zero no matter how big n I choose, is that right ? If it is right then it means that the approximation fails because it is zero ?? – ali Jul 20 '13 at 22:29
@ali, hope it's clearer now. It's not that the Euler method is failing, it's that it's approximating (correctly) a second solution. – nbubis Jul 20 '13 at 22:33

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