Can someone tell me exactly where my simple logic breaks down? (infinity \neq 1)

[nonequilibrium]

The simple DE under our attention is DE := y(x)' = y^{1/3} \textrm{ with }y(0) = 0.
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 DE2 := y(x)' = y^{1/3} \textrm{ with }y \neq 0.
By integration, we can show that the statement y(x) = \left( \frac{2}{3} x + c\right)^{3/2} \textrm{ with }y \neq 0 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 DE3 := y(x)' = y^{1/3}.
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.

 

[LCKurtz]

Uniqueness theorems don't apply to the differential equation itself. They apply to the initial value problem. The ivp in question is:

y' = y^1/3, y(0) = 0.

The Picard uniqueness theorem for y' = f(x,y), y(x₀)=y₀ has the hypothesis that f_y be continuous in a neighborhood of (x₀,y₀). This isn't satisfied for the above ivp but it would be if y(0) wasn't 0. And even then Picard only promises a short interval where the solution exists and is unique. You can't mix and match -- different initial values mean different ivp's.

 

[Mute]

S is not a unique solution to DE2, as you didn't specify any initial conditions for DE2, you only specified that y cannot take on the value zero. Until you specify an initial condtion, S cannot be unique.

Edit: Ninja'd.

 

[Hurkyl]

Also, mr vodka has been sloppy about the solution set S, and probably about what he wants DE2 to be.

Many values of c, when plugged into the form S, do not lead to functions of the real line, so they cannot be solutions to DE2.

What it does tell you is that, on any interval where y is nonzero, then y's restriction to that interval must be of the form S, and with a c that actually gives a function on that interval.



I think mr vodka has an idea that can be turned into a correct argument for something, but the omission of these relevant details has prevented him from seeing what his argument is saying.

 

[nonequilibrium]

Thank you for your replies.

It was somewhat sloppy to state "y(x) = \left( \frac{2}{3} x + c\right)^{3/2} \textrm{ with }y \neq 0 is the unique solution". First of all, I mean c to be a real constant, and secondly, with "uniqueness", I mean that whenever an initial condition is given, the resulting y is of that form (remember, I'm excluding y=0 here).

What it does tell you is that, on any interval where y is nonzero, then y's restriction to that interval must be of the form S, and with a c that actually gives a function on that interval.

I agree with this. But since a function of the above form can only have one value (for a certain c) where it turns zero, doesn't the continuity of the differentiable y make that (by taking the limit)
y(0) = \left( \frac{2}{3} 0 + c\right)^{3/2} = c^{3/2}
which would imply that even if y=0 is allowed, the previous expression or y(x) should be the unique solution? Where is the fallacy in this? (Please note that I've had only two lessons of my DE course so far, so I might be missing some crucial subtlety. I understand that the conditions for the uniqueness theorem aren't fulfilled, but that doesn't imply it can't be unique, right?)

NB: What then is another function different from the above that satisfies the original IVP? How did it "slip through the maze"?

EDIT: I only just realized that y(x) = 0 is also an obvious solution... (still don't know how you get an infinite amount of solutions though) and I think I'm realizing the subtleties... I assumed the eventual solution would at least be different from zero somewhere, making it able to rely on my y \neq 0 solution to extrapolate that the solution must also be true for y = 0 due to continuity. But of course y(x) = 0 is never different from zero. The weird thing is that it convinces me that there can only be y(x) = 0 or my earlier solution S, yet I read in my course that there are an infinite number of solutions, meaning I still haven't understood the real problem, as I seem to have excluded other options. What are more solutions?

 

[LCKurtz]

I think I see what is bothering you. Let me use a slightly different example

y' = y^1/2, y(0) = 0.

Of course, y identically 0 solves the DE and the initial condition. Also, solving by separation of variables y = (x+c)²/4 solves the DE, but not the initial condition.

Let's look at y = g(x) = (x-1)²/4. This is a parabola with vertex at x = 1. This is a solution to the DE but not the boundary condition. Now look at the function

f(x) = 0 for x < 1 and f(x) = (x-1)²/4 for x ≥ 1. This piecewise function satisfies both the DE and the initial condition, so it shows that y identically zero isn't a unique solution of the IVP.

Now note that f(2) = 1/4, so f(x) is a solution of the IVP

y' = y^1/2, y(2) = 1/4.

Now the piecewise function f(x) is not the same as the parabola g(x) yet they both solve this last IVP above. Why don't these two solutions to

y' = y^1/2, y(2) = 1/4.

contradict Picard's uniqueness theorem? Do you see the answer to this question?

 

[Hurkyl]

The more I look at it, the more I convince myself that, for any interval where a solution to DE3 is nonzero, y can attain the value zero on only one of its endpoints, which eventually leads to the conclusion there are finitely many solutions.

Why do you think this has infinitely many solutions, by the way? The solution certainly isn't a unique nonzero solution -- you didn't mistakenly infer from that, did you?

 

[nonequilibrium]

Kurtz, that's exactly what I hadn't realized. Wow, that's really interesting! And the funny thing is that it now seems completely logical, whereas my earlier reasoning seemed so logical to me a few minutes ago; a bit scary how whimsical my logical judgement seems to be... And the reason it doesn't contradict the theorem is because the f(x,y) --i.e. the square root-- in y' = f(x,y) is not Lipschitz continuous in zero :)

It's also interesting --from my standpoint at least-- to retrace my original steps to discover any false assumptions that led to my false result. I still agree with Step 1 & 2. I think the essential flaw is in Step 3, where I underestimated the grave importance of y \neq 0. I seem to have implicitly assumed that in any neighbourhood of x = 0, there will be an x value for which y is not zero, making it able to use the continuity of my solution S to claim S was the whole solution. But indeed, there can be no justification for that assumption as Kurtz has showed.

Thanks.

EDIT (after seeing Hurkyl's post): my apologies, I can't really make up a graphical image of what you're implying, but the way I see it --after Kurtz' post at least-- is that for any natural number n, we can have f(x) = 0 for x < n and f(x) = (2/3 x - 2/3 n)^{3/2} for x >= n. This makes for at least an infinite number of solutions. (btw my source of the claim of "infinite amount of solutions" comes from my course notes)

EDIT2:
Earlier post of Kurtz:

This isn't satisfied for the above ivp but it would be if y(0) wasn't 0. And even then Picard only promises a short interval where the solution exists and is unique. You can't mix and match -- different initial values mean different ivp's.

But just to be sure: in this case (as in, in the case of the DE I defined in the first post) if y(0) wasn't 0, we would have a unique solution, correct?

 

[Hurkyl]

Ah right, I forgot to sew the zero solution in as something nonzero vanished. Bad Hurkyl.

 

[LCKurtz]

I think I see what is bothering you. Let me use a slightly different example

y' = y^1/2, y(0) = 0.

Of course, y identically 0 solves the DE and the initial condition. Also, solving by separation of variables y = (x+c)²/4 solves the DE, but not the initial condition.

Let's look at y = g(x) = (x-1)²/4. This is a parabola with vertex at x = 1. This is a solution to the DE but not the boundary condition. Now look at the function

f(x) = 0 for x < 1 and f(x) = (x-1)²/4 for x ≥ 1. This piecewise function satisfies both the DE and the initial condition, so it shows that y identically zero isn't a unique solution of the IVP.

Now note that f(2) = 1/4, so f(x) is a solution of the IVP

y' = y^1/2, y(2) = 1/4.

Now the piecewise function f(x) is not the same as the parabola g(x) yet they both solve this last IVP above. Why don't these two solutions to

y' = y^1/2, y(2) = 1/4.

contradict Picard's uniqueness theorem? Do you see the answer to this question?

Kurtz, that's exactly what I hadn't realized. Wow, that's really interesting! And the funny thing is that it now seems completely logical, whereas my earlier reasoning seemed so logical to me a few minutes ago; a bit scary how whimsical my logical judgement seems to be... And the reason it doesn't contradict the theorem is because the f(x,y) --i.e. the square root-- in y' = f(x,y) is not Lipschitz continuous in zero :)

That isn't quite the reason. This IVP has its boundary condition at x = 2 and f(x,y)=y^1/2 does satisfy the Picard hypotheses at (2, 1/4). Yet we have two solutions to this IVP, f(x) and g(x). So what do you think can be the problem?

 

[nonequilibrium]

Kurtz, good point. I suppose the case is this: if we restrict our interval as to exclude the x for which f(x,y) is not Lipschitz continuous, we necessarily also exclude the point where we can cut a piece out of our function and replace it with f(x) = 0, i.e. in an interval where f(x,y) is Lipschitz continuous, our solution is unique.
In other words: "y' = y^1/2, y(2) = 1/4" has a unique solution for any interval in which no {x}, for which x <= 1, is dense. When we go beyod such an interval, the requirements for the theorem are broken.