<msubsup> q i &#x2032; </msubsup> ( 0 ) = 0 , for i = 1 , 2

enrotlavaec

enrotlavaec

Answered question

2022-06-20

q i ( 0 ) = 0, for i = 1 , 2 , 3 , 4.
q 2 ( t ) q 3 ( t ) > q 1 ( t ) q 4 ( t ) for 0 <= t < T. T is finite. At t = T the inequality breaks.
q 4 ( 0 ) q 3 ( 0 ) > q 4 ( 0 ) q 3 ( 0 )
( q 4 ( 0 ) + q 2 ( 0 ) ) ( q 1 ( 0 ) + q 3 ( 0 ) > ( q 4 ( 0 ) + q 2 ( 0 ) ) ( q 1 ( 0 ) + q 3 ( 0 ) )
This also holds but I suspect it is superfluous:
q 2 ( 0 ) q 1 ( 0 ) < q 2 ( 0 ) q 1 ( 0 )
One can also add this without loss of generality:
q 1 ( 0 ) = 1 , q 2 ( 0 ) = 1
Prove or disprove that:
q 4 ( t ) q 3 ( t ) > q 4 ( t ) q 3 ( t )
( q 4 ( t ) + q 2 ( t ) ) ( q 1 ( t ) + q 3 ( t ) > ( q 4 ( t ) + q 2 ( t ) ) ( q 1 ( t ) + q 3 ( t ) )
for 0 <= t < T

Answer & Explanation

Trey Ross

Trey Ross

Beginner2022-06-21Added 30 answers

It can't be true as you have nothing except smoothness to restrict the functions away from zero.
Note that you are trying to prove ( q 1 ( t ) q 2 ( t ) q 1 ( t ) q 2 ( t ) ) > 0 So start them off with q 1 ( t ) = a + b t 2 , q 2 ( t ) = c + d t 2 , q 3 ( t ) = e + f t 2 , q 4 ( t ) = g + h t 2 for t 1. Your second equation can be satisfied if c , e >> a , g
Then for t > 1 add something smooth with positive derivative to q 1 like exp ( a x 2 1 ).
If you pick a large enough, the derivatives will be larger than the function.

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