Solution to diff. eqn as a ring

[Alphaboy2001]

Solution to diff. eqn as a ring(killing me - please look at my result!)

Homework Statement

I am presented with the following problem given the direction field

f(u,v) = \left( \begin{array}{ccc} -sin(u) & sin(v) \\ cos(u) & cos(v) \end{array} \right)

as \left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v)

show that

(\{(u,v) \in \mathbb{R}^2|f(u,v) = 0\} = T_{1} \cup T_{2} )


where

T_{1} = \{l \pi, m\pi + \frac{\pi}{2}|l,m \in \mathbb{Z}\}
T_{2} = \{m \pi, l\pi + \frac{\pi}{2}|l,m \in \mathbb{Z}\}

The Attempt at a Solution

Don't I treat T_{1} \cup T_{2} as a ring since any solution to the differential equation \left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v) = 0 must lie on the Ring T_{1} \cup T_{2}?

Thus I must show T_{1} + T_{2} and T_{1} \cdo T_{2} lies
T_{1} \cup T_{2}?

If yes very well.

T_{1} + T_{2} = (\frac{4(l+1)\cdot \pi}{2}, \frac{4(m+1)\cdot \pi}{2})

and

T_{1} \cdot T_{2} = (\frac{(2 \cdot m+1)\cdot \pi}{2}, \frac{(2\cdot l+1)\cdot \pi}{2})

and since l,m \in \mathbb{Z} then 2(l+1) and 2(m+1) and m(2m+1) and l(2l+1) are in T_{1} \cup T_{2} is the Ring which defines in solution for \left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v) which implies that in l,m on the Ring mentioned above will give a solution for the differential equation.

How is that??

Sincerely Yours

Alphaboy

 

[tiny-tim]

Hi Alphaboy! Thanks for the PM

I am presented with the following problem given the direction field

f(u,v) = \left( \begin{array}{ccc} -sin(u) & sin(v) \\ cos(u) & cos(v) \end{array} \right)

as \left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v)

show that

(\{(u,v) \in \mathbb{R}^2|f(u,v) = 0\} = T_{1} \cup T_{2} )


where

T_{1} = \{l \pi, m\pi + \frac{\pi}{2}|l,m \in \mathbb{Z}\}
T_{2} = \{m \pi, l\pi + \frac{\pi}{2}|l,m \in \mathbb{Z}\}

I'm sorry, I'm not following this.

T₁ and T₂, according to your definitions, are the same.

And how can f(u,v) = 0? If the top line is 0, then the bottom line isn't.

And what is the differential equation?

And what is a direction field?

 

[Alphaboy2001]

Hi Alphaboy! Thanks for the PM


I'm sorry, I'm not following this.

T₁ and T₂, according to your definitions, are the same.

And how can f(u,v) = 0? If the top line is 0, then the bottom line isn't.

And what is the differential equation?

And what is a direction field?

Hi Tiny-tim,

you are one of the smartest guys here :D

Anyway if look at my teachers definition of T1 and T2 its their inclusion which gives f(u,v) = 0 for any integer l and m, if and only if l and m are in Z.

and what I am trying to do it to insert T1 + T2 and T1*T2 into f(u,v) and as mentioned if l and m are in Z, then I get f(u,v) = 0.

and it was "vectorfield" and (u', v') is the corresponding differential equation.

I hope this have cleared it up now :)

Sincerely
Alpha