- #1

nonequilibrium

- 1,439

- 2

Apparently this has an infinite number of solutions, but I do not understand: can someone exactly pinpoint and explain the fault in my reasoning (that follows)?

**Step 1:**define [tex]DE2 := y(x)' = y^{1/3} \textrm{ with }y \neq 0[/tex].

By integration, we can show that the statement [tex]y(x) = \left( \frac{2}{3} x + c\right)^{3/2} \textrm{ with }y \neq 0[/tex] is

*exactly*equivalent to DE2.

**Step 2:**due to the equivalence of both statements, the solution is unique (otherwise it wouldn't be equivalent). Call this solution S.

**Step 3:**now look at [tex]DE3 := y(x)' = y^{1/3}[/tex].

Here y = 0 is allowed. Say there is a certain solution S' (different from our earlier one (= S)) satisfying DE3. Now because y(x) must be continuous (as it is differentiable, also in x = 0), if S has a different value for y(0) than S' does, then S' also has different values outside of x = 0, but we already stated that S was the unique solution for all except 0. Thus out of the continuity of y(x) follows that S must also be the unique solution for DE3.

**Step 4:**having obtained the unique solution S for all x, we can check y(0) = 0 and put c = 0 and get the particular and unique solution.

Obviously this is in contradiction with the fact that DE has an infinite number of solutions... But why/how/where?

Thank you.