I Finite difference scheme for u'=u^2, u0=1, 0<t<2

[feynman1]

u'(t)=u(t)^2, u(0)=1, 0<t<2
What finite difference scheme can overcome the difficulty when t->1+ and help the solution jump to the negative branch?

 

[anuttarasammyak]

\frac{du}{u^2}=dt
\frac{1}{u}+t=const.=1
Negative branch ?

 

[feynman1]

\frac{du}{u^2}=dt
\frac{1}{u}+t=const.=1
Negative branch ?

when t>1, u<0

 

[anuttarasammyak]

Yes.
u=\frac{1}{1-t}
What's wrong with it ?

 

[feynman1]

u=\frac{1}{1-t}
What's wrong with it ?

Starting from t=0, when t->1+, finite differences fail.

 

[anuttarasammyak]

Yes, but why do you estimate it fails ?

 

[feynman1]

just try numerics yourself and you'll see.

 

[anuttarasammyak]

The solution says
|u|&gt;1
, or
|1/u|&lt;1
if you do not like divergence to $\pm$ infinity. {1/u} (0) = 0.

 

[pasmith]

Starting from t=0, when t->1+, finite differences fail.

The domain of the initial value problem at t = 0 does not extend past t \to 1^{-}.

An antiderivative of 1/(t-1)^2 looks like <br /> f: t \mapsto \begin{cases} 1/(1- t) + c_1 &amp; t &lt; 1 \\<br /> 1/(1 - t) + c_2 &amp; t &gt; 1.\end{cases} The arbitrary constants c_1 and c_2 don't have to be equal, and the constraint that u(0) = 1 only fixes c_1 = 0.

If you want to look at t &gt; 1 then you must start from an initial t_0 &gt; 1 - for example u = -1 when t = 2.