Need help w/ "Given tx' = x, IC x(0) = 0; verify that x(t) = t is sol'n to IVP, and..."

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Cactusguy

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This is from my practice problems for an exam on friday.

Here's what I have so far:

1.
A) Plug in x=t and t=t and x`=1 into the equation to get (t)(1)=(t). Thus x(t)=t is a solution. I'm not sure what maximal interval of existence means. This solution exists everywhere where t doesn't equal zero, because x`=x/t. Is that all I have to say? Or is it something like (0, infinity)?

B) Another solution by inspection is x(t)=0. This is easy to verify graphically, or just logically. No problems there.

C) Wouldn't this invalidate the existence theorem? Because there's no rectangle R you can draw containing (0,0) where x` is continous? Is just saying that enough, or is more needed?

Thanks in advance.
 
An interval of existence for a solution to an IVP with some initial conditions \(x(x_0) = \alpha\) is any neighborhood around \(x_0\) where that solution to the IVP exists. The maximal interval of existence is, therefore, the largest such interval. If you've some reason to restrict \(t\) to always being positive, then yes the maximal interval would be \((0, \infty)\), but if \(t\) can be positive or negative, it would be \((-\infty, 0) \cup (0, \infty)\).

But as far as the "existence/uniqueness theorem" part is concerned, that's difficult for me to say one way or the other because I don't know what, specifically, your class's version of the theorem says. I can guess what it probably says, based on my own knowledge and a bit of research, but I can't say for sure.
 
Well, why not? Is the union of two intervals itself not also an interval? Do I not know what "interval" means? Is that what you're telling me?
 
Cactusguy said:
A) Plug in x=t and t=t and x`=1 into the equation to get (t)(1)=(t). Thus x(t)=t is a solution. I'm not sure what maximal interval of existence means. This solution exists everywhere where t doesn't equal zero, because x`=x/t. Is that all I have to say? Or is it something like (0, infinity)?

B) Another solution by inspection is x(t)=0. This is easy to verify graphically, or just logically. No problems there.

C) Wouldn't this invalidate the existence theorem? Because there's no rectangle R you can draw containing (0,0) where x' is continous? Is just saying that enough, or is more needed?
Click to expand...
A proved theorem should not be able to be invalidated. Rather, what you might be able to do is to show that it doesn't apply to this problem.

Do you see that this is what you have really shown? The theorem (if your theorem is what I expect - you do need to quote it for us) has the existence of that rectangle as a condition -- if the condition is not true, then the theorem does not apply; the solution may then exist but not be unique.

ksdhart2 said:
Is the union of two intervals itself not also an interval?
Click to expand...
No, it is not an interval. The definition of an interval is:
  • In mathematics, a (real) interval is a set of real numbers lying between two numbers, the extremities of the interval. For example, the set of numbers x satisfying 0 ≤ x ≤ 1 is an interval which contains 0, 1 and all numbers in between.
The union of two non-intersecting intervals is not an interval. But that doesn't really matter to the problem, since we are talking about an interval around 0, and there simply is none.

But again, @Cactusguy, we can talk more fully about this once you quote the theorem as you were given it.
 
Dr.Peterson said:
No, it is not an interval. The definition of an interval is:
  • In mathematics, a (real) interval is a set of real numbers lying between two numbers, the extremities of the interval. For example, the set of numbers x satisfying 0 ≤ x ≤ 1 is an interval which contains 0, 1 and all numbers in between.
The union of two non-intersecting intervals is not an interval. But that doesn't really matter to the problem, since we are talking about an interval around 0, and there simply is none.
Click to expand...

Huh. I didn't know that. I thought "interval around [0]" and "neighborhood around [0]" were interchangeable terms. And in the definition of "neighborhood around" that I'm familiar with (although given recent developments, I guess it wouldn't be surprisingly to learn I'm wrong about this too...) the point itself doesn't need to be actually in the neighborhood.

But, yes, I agree that if the interval of existence must include 0, then the "maximal interval of existence" would have to be the empty interval because no such interval can exist.
 
That would seem to be what I was thinking of. Well, I guess at least I'm learning a lot today.
 
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