For a differential equation say $y'=ty^3$ with $y(0)=1$ how would we know if the sequence from Euler's method exists or not?
So I know this particular equation has solution:
$$y=\frac{1}{\sqrt{1-t^2}}$$
So hence has a discontinuity at $t=1$. So does the sequence from Euler's method exist for all $n$? How could I find out? Any hints would be great!
Best Answer
The Euler method for this example reads as $$ y_{n+1}=y_n+ht_ny_n^3. $$ According to the nature of the ODE, consider $y_n^{-2}$. Then the iteration is approximately $$ y_{n+1}^{-2}=y_n^{-2}(1+ht_ny_n^2)^{-2} =y_n^{-2}(1-2ht_ny_n^2+3h^2t_n^2y_n^4+O(h^3)) \approx y_n^{-2}-2ht_n, $$ as long as $hy_n^2$ remains small, $y_n^{-2}\gg h$. As $t_n=nh$ this can be solved as $$ y_n^{-2}\approx y_0^{-2}-n(n-1)h^2=y_0^{-2}-t_n(t_n-h). $$ This means that at $t_n\approx 1$, $n=\lfloor\frac1h\rfloor$, the Euler iteration will reach a value $y_n\approx\frac1{\sqrt h}$, and thus $y_{n+1}\approx\frac2{\sqrt h}$, and from that moment on the second term will dominate, $$ \sqrt{h}y_{n+k+1}\ge(\sqrt{h}y_{n+k})^3\implies y_{n+k}\ge\frac1{\sqrt{h}}2^{3^{k-1}},~~ k\ge 1 $$ Normal 64bit floating point numbers go up to $2^{1024}$, that range gets exceeded before $k=7$.
While in theory the Euler sequence can be indefinitely prolonged, in praxis there is no computer that can do this, the (observable, estimated) universe gets exhausted before $k=200$.