Euler’s Method — Question 6

Question 6 – Solution

Strategy. A nonlinear change of dependent variable preserves the exact differential equation but generally does not commute with a finite Euler step.

Step 1: Derive and compute the algorithms. Direct Euler gives $Y_{n+1}=Y_n+hY_n^2$, $Y_0=1$. Since $u'=-y'/y^2=-1$, transformed Euler gives $U_{n+1}=U_n-h$, $U_0=1$. Thus $$\begin{array}{c|cc} t&\text{A: direct Euler}&\text{B: }1/U_n\\\hline 1/4&5/4&4/3\\ 1/2&105/64=1.640625&2 \end{array}$$

Step 2: Compare with the exact solution. Separation gives $\boxed{y=1/(1-t)}$, whose maximal interval containing zero is $(-\infty,1)$. Algorithm B is exact at every grid node before $1$, because its transformed equation has the constant slope $-1$: Euler integrates that affine solution exactly. Algorithm A underestimates the two displayed exact values.

Step 3: Compare a single step. From the same positive state, the updates are $$A(y)=y+hy^2,\qquad B(y)=\frac 1{1/y-h}=\frac{y}{1-hy}.$$ Their exact difference is $$\boxed{B(y)-A(y)=\frac{h^2y^3}{1-hy}>0\quad(hy<1)}.$$ For fixed $y$, the discrepancy is of order $h^2$ as $h\to 0$, although both are Euler approximations in their chosen coordinates. The nonlinear reciprocal map changes the finite-step rule.

Step 4: Respect the original blow-up. At $t=1$, algorithm B has $U_4=0$, so its inverse is undefined, matching the exact pole. The direct recurrence uses only addition and multiplication and gives a finite real value after any fixed finite number of exact-arithmetic steps. Those values are numerical iterates, not proof of a classical solution through the pole.

The original solution tends to $+\infty$ as $t\uparrow 1$, so it cannot extend continuously across that time. Beyond $1$, an inverse value from negative $U_n$ would likewise belong to a disconnected formula branch, not a continuation of this IVP.