Differential Equation - Uniqueness Theroem

cse63146
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Homework Statement



The differential equation that models the volume of a raindrop is \frac{dv}{dt} = kv^{2/3} where k = 3^{2/3}(4 \pi)^{1/3}

A) Why doesn't this equation satisfy the hypothesis of the Uniqueness Theroem?
B) Give a physical interpertation of the fact that solution to this equation with the initial condition v(0) = 0 are not unique. Does this model say anything about the way raindrops begin to form?

Homework Equations





The Attempt at a Solution



A) The equation doesn't satisfy the hypothesis Uniqueness Theroem because when v = 0, the equation's derivative does not exist.

B) At time t = 0, the raindrop does not have volume, but as t increases, it's volume increases as well.

Am I correct for both parts
 
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cse63146 said:

Homework Statement



The differential equation that models the volume of a raindrop is \frac{dy}{dt} = kv^{2/3} where k = 3^{2/3}(4 \pi)^{1/3}
Do you mean "dv/dt", rather than "dy/dt"?

A) Why doesn't this equation satisfy the hypothesis of the Uniqueness Theroem?
B) Give a physical interpertation of the fact that solution to this equation with the initial condition v(0) = 0 are not unique. Does this model say anything about the way raindrops begin to form?

Homework Equations





The Attempt at a Solution



A) The equation doesn't satisfy the hypothesis Uniqueness Theroem because when v = 0, the equation's derivative does not exist.
Strictly speaking an "equation" doesn't have a derivative. What you mean is that the function kv^{2/3} has no derivative at v= 0. That is true and is a reason why the uniqueness theorem does not hold.

B) At time t = 0, the raindrop does not have volume, but as t increases, it's volume increases as well.
How do you conclude that "its volume increases"? Certainly v(t)= 0 for all t satisfies dv/dt= kv^{2/3} as well as v(0)= 0.

Am I correct for both parts
 
so to prove that the volume does increase, I would need to find it's general solution?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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